text:
# Time-Speed-Distance: Solution

The solutions to time speed distance: questions are given here. The solutions are detailed with proper explanations of each concept. Also, the best approach is given here that would help to solve the related questions much faster and efficiently.

### TIME SPEED DISTANCE – SOLUTIONS

1) Option (d)

As the Distance covered is same with both the Speeds, the average Speed will be the harmonic mean of the individual Speed.

200/9m/s = (200*3600)/((9*1000) ) = 80 km/h

So, average Speed = (2*a*b)/(a + b) = (2 * 80 * 120)/(80 + 120) = 96 km/h.

2) Option (c)

Distance = Speed × Time

As Distance covered in both the cases is constant, Speed will be inversely proportional to time.

So, if Time by bus: Time by train = 2: 1, Speed of bus: Speed of train = 1: 2

Speed of bus is 40 km/hr, so Speed of train is 80 km/hr.

3) Option (a)

Distance = Speed x Time

In both the cases, Speed is constant so Distance will be proportional to the time.

Vivek takes 20% more than Bharath. So, assume that Bharath takes 100 minutes then Vivek will take 20% more i.e. 20 minutes more, so 120 minutes. So, the ratio of Time taken by Vivek and Bharath = 120/100= 6/5. Distance is proportional to the Time taken, so ratio of Distance of Vivek’s House and Bharath’s House = 6/5, Bharath’s house is at a Distance of 30 km, so Vivek’s house will be at the Distance of (6/5)*30 = 36 km.

4) Option (d)

#### Method 1:

Suppose the normal Time taken is ‘t’ then when he is late by 1 hour, Time taken by him = ‘t – 1’.

Normal Speed = 60/t ,Speed when he wants to reach in t – 1 Time = 60/(t – 1)(60/(t-1))– (60/t) = 5? t2 – t – 12 = 0 ? (t – 4) (t + 3) = 0 ? t = 4 and t = – 3 (-3 is not possible)So, normal Time taken = 4 hours. Hence option (d).

#### Method 2: Reverse Gear Approach

Take a middle answer option, say 5 hours Normal Speed = 60/5 = 12 kmph Speed increased to 12+5=17 kmph. 60/17 = 3.53 hours,

the difference is more than 1 hour. This is not the answer. We can also eliminate 6 hours, as that will result in a higher Time difference.

Answer is either 3 or 4 hours(answer is close to 5) Trying for 4 hours, Normal Speed= 60/4 = 15 km/h

Speed increased to 20 km/h 60/20=3. 4-3 =1. This is the answer.

5) Option (c)

Sajesh increases his Speed from 12 km/hr to 15 km/hr, so increase in Speed = 3 km/hour 1/4. In both the cases, Distance covered remains the same. So, applying constant product rule: His Speed increases by 1/4 so Time will decrease by 1/5.

Here Time is decreasing by 1 hour and 1 hour is 1/5 of the actual Time taken. So, actual Time taken = 5 hours. So, Distance between Office and House = 5 * 12 = 60 kms. Hence option (c). Alternatively, you can also check the answer options.

(a) Distance = 40 kms, actual Speed = 12. Actual Time taken = 40/12. New Time taken = 40/15.

(40/12)– (40/15)≠ 1

(b) Distance = 50 km, actual Speed = 12. Actual Time taken = 50/12. New Time taken = 50/15. (50/12)– (50/15)≠1

(c) Distance = 60 kms, actual Speed = 12. Actual Time taken = 60/12= 5 hrs. New Time taken = 60/15 = 4 hrs.

Difference = 1 hour.

6) Option (b)

Let x,y and z be the Distances uphill, on flat ground and downhill respectively. Then

7) Option (a)

Let AC be 3 km more than DE. Then

Thus, if the Distance is decreased by 20 km, the Time taken will be the same.

8) Option (d)

Their Speeds are not given. So the ratio of the Time taken cannot be determined.

9) Option (d)

There is no traveling Distance because they can travel anywhere in the path. So the ratio of their Speeds cannot be determined i.e., for example, if player P2 will take the path N-J-F-B, or N-d-a-B etc. can never be determined.

10)Option (c)

If you observe the options carefully, you can decide which answer option to start with. Since the fraction 5/6 is involved. Start with an option which is a multiple of 6; option (b) and option (c) Now one has 2 Distances 20 km and 22 km A’s original Speed = 6kmph. In 2 hours he will travel 12 km B’s original Speed= 4 kmph. In 2 hours he will travel 8 km 12+8=20 and not 22. Here itself, you can mark the answer as option (c) Verifying A’s new Speed= 5 kmph B’s new Speed = 5 kmph After 1 hour A travels 5 km, B travels 0 After 2 hours A travels 10 km (total), B travels 5. Total =15 km. They thus meet after 2.5 hours.

11) Option (a)

It is given that the platoon and the last person moved with uniform Speed. Also, they both moved for the identical amount of Time. Hence, the ratio of the Distance they covered – while person moving forward and backward – are equal. Let’s assume that when the last person reached the first person, the platoon moved X meters forward. Thus, while moving forward the last person moved (50+X) meters whereas the platoon moved X meters. Similarly, while moving back the last person moved [50-(50-X)] = X meters whereas the platoon moved (50-X) meters. Now, as the ratios are equal

(50+X)/X = X/(50-X) (50+X)*(50-X) = X*X Solving, X=35.355 meters Thus, total Distance covered by the last person = (50+X) + X = 2*X + 50 = 2*(35.355) + 50 = 120.71 meters =120 m (approximately).

12) Option (b)

Using the concept of ratios of Speed (Speed directly proportional to Distance), we can easily arrive at the totalDistance. The images below give a picture of the Distances when Amar and Akbar finish the race

Length of the race track = 400 +x The ratio of Akbar and Anthony’s Speed can be arrived at as (240+x)/x = (400+x)/(100+x)

Now go from answer options, answer will be option (b) 800 m Verifying, x = 400 640/400 = 800/500. Thus, answer is option (b).

13) Option (c)

The ratio of Speeds of A and B= 3:1, This means that when A makes 1.5 rounds, B makes 0.5 round. For B to complete the race of 15 rounds, it would take him 15/0.5 x 5 =150 minutes.

14) Option (c)

Solve the question using assumption If B’s Speed is 3, then A’s Speed is 3+(3/3) = 4 If A travels 12 km, B travels 6 km Ratio of Time taken = A:B = 3:2. Option (c).

15) Option (b)

In 2/3 of an hour A travels 10 km in 40 min B travels 8 km in 40 min After one hour, A would have still traveled only 10 km and B would have traveled 20 km, their relative Speed = 15-12= 3 km/hr A gains this 10 km in 10/3 =3.33 hours.

16) Option (b)

the total Distance traveled before they come to a halt can be calculated as 40(1+1/2+1/4+1/8…)/6 = 40((1/(1-1/2))/6)=13.33 km They cover a Distance of 750 m for the first time they meet, and subsequently cover a Distance of 1500 m each Time they meet, hence if they meet 8 Times, then the Distance covered will be 1500 * 8 + 750 = 12750, which is lesser than 13,333.33. If they meet 10 Times then the Distance would be 14250>13333.33. Hence they meet 9 Times.

17) Option (d)

Let T be Time Katrina takes to make 25 steps. Then Katrina takes 3T to make 75, and Priyanka takes 2T to make 150. Suppose the escalator has N steps visible and moves n steps in Time T. Then Priyanka covers N + 2n = 150, N – 3n = 75. Hence N = 120, n = 15 => Choice (d) is the right answer.

18) Option (c)

Let the escalator moves x steps when A walks down 90 steps. Total number of steps on a stationary escalator = x + 90. When A takes 90 steps, B should have taken 60 steps and the escalator x steps. So when B takes 80 steps, the escalator should have taken 4/3*x steps. So, 4/3*x + 80 = x + 90 = Total number of steps in the escalator when it is stationary. So x = 30. Hence, total number of steps = 120. By the Time they meet, together they will 120 steps in the ratio 3:2. i.e 72 and 48 steps. So, the difference is 24. Option (d).

19) Option (b)

Time taken by Ship 1 = 3x/40 hr Time taken by Ship 2 = 2x/60 hr 3x/40–2x/60= 1

10x=240 x=24 Distances traveled are 48 & 72= 120.

20) Option (d)

Consider that Ishan’s journey will end when the group reaches the end of the walkway (as long as he catches up with the crowd before the walkway ends). When he steps on the walkway, the crowd is 180 metres from the end. The walkway travels this Distance in (180/3) = 60 seconds, and Ishan’s average rate of movement is 300/60 = 5 metres per second.

21) Option (d)

Let x be the number of hours Sana walks at 10kmph. x will be the no. of hours Peter and Gavin traveled by car at 50kmph. Let x1 be the number of hours that Gavin walks till Peter reaches Sana. Let x2 be the number of hours when Peter and Sana travel in the car till the destination. The total Distance traversed by each of them can be evaluated as follows

Peter = 50x-50*1+50*2=200——————(1)

Sana= 10x+10*1+50*2=200—————— (2)

Gavin = 50x+10*1+10*2=200—————–(3)

Solving we get t1= 2 hours, t=3 hours, t2=3 hours. Total = 8 hours.

22) Option (c)

This problem can be solved by noticing that the Speeds are in a ratio of 2:1. The Distances covered will, therefore, be in a ratio of 2:1. When they meet for the 8th Time, they would have covered the entire Distance of the track 8 Times = 8 (2pr) .Anu would have covered 8 Times the Distance and Varsha 4 Times the Distance (Total= 12, ratio 2:1) Difference = 4(2pr) = 4x 2 x (22/7)x(140/2)= 1760.

23) Option (d)

1400 m track is divided into 7 parts of 200 m each. The Speeds of the two bikes are in a ratio of 3:4, each time they meet, they will cover a Distance in the ratio of 3:4 They will thus meet for the first Time at 3/7th the Distance which will happen at point 4 is 600 m. The second time they will meet at the starting point. 3rd Time at point 4… and so on.so the 15th Time, they meet at 600 m which is 800 m from A.

24) Option (c)

Cost incurred (maximum ) a Speed x capacity x cost/km Required ratio= (7x* 5x* 3x)/( 4x* 2x* 2x)=105/16.

25) option (a)

As the Distance covered is constant in both the cases, the Time taken will be inversely proportional to the Speed. In the problem, Speed of Amit and Aman is in ratio 60: 90 or 2:3. So the ratio of the Time taken by Amit to that taken by Aman will be in the ratio 3:2. So if Amit takes 9 hrs, Aman will take 6 hrs.

26) Option (b)

This is an instance of Case 1 as illustrated above. When Ram meets Shyam for the third Time, they together would have covered a Distance of 5d, i.e 5x30m = 150 m. Ratio of Speed of Ram and Shyam = 2:1, so the total Distance traveled by them will also be in the ratio 2:1 as the Time taken is constant. So the Distance traveled by Ram will be 2/3 x150= 100 m.

27) Option (d)

A takes 1 hour after meeting and B takes 4 hours after meeting. Hence Time traveled before meeting = v1.4 = 2 hours Time taken by A= 2+1 = 3 hours

Study tuned with Byju’s for a better understanding of topics and proper preparation for the CAT exam, At Byju’s candidates, are also provided with the CAT Syllabus and other free Study materials to learn more effectively.


text:
COMP 1002 Winter 2019
Homework Assignment # 5
Due: March 21, by 10pm
(Type it up and upload on D2L)

1. Induction examples\ Prove the following by math. induction. The format of the proof should be as follows: 1) state what statement you are proving (it should have a variable on which you do induction as a parameter). 2) state the base case(s) and prove them. 3) state the induction hypothesis 4) state and prove the induction step.

1. Show that for all $n \geq 2$, $5^n +9 \leq 6^n$.

Solution:

2. Show that $\forall n \geq 0$,   the number $7^{n+2}+8^{2n+1}$ is divisible by 57.

Solution:

3. Consider a sequence defined as follows: $s_0 = 1$, $s_1 = 2$ and for $n \geq 2$, $s_n = 1 + \max{s_{\lfloor {n/2}\rfloor}, s_{\lceil{n/2}\rceil}}$ (Recall that $\lfloor x \rfloor$ is the floor of $x$, that is, largest integer $y \leq x$, and $\lceil x \rceil$ is the ceiling of $x$, smallest integer $y \geq x$.)

Prove by strong induction that $\forall n > 0, s_{n} \geq s_{n-1}$. Hint: compare $s_{\lfloor (n+1)/2 \rfloor }$, $s_{\lceil (n+1)/2 \rceil}$, $s_{\lfloor n/2 \rfloor }$ and $s_{\lceil n/2 \rceil}$. Which of them are the same and when? What is the relationship among ones that are not the same?

Solution:

2. Structural induction

1. Give a recursive definition of a set $S$ of all binary that have the same number of 0s and 1s.

Solution:

2. Now show, using structural induction on your recursive definition, that every string in your set has even number of symbols.

Solution:

format: plain


text:
# Have something like figure but with its own counter and listof

Given the following MWE:

\documentclass[english]{scrartcl}

\usepackage{syntax}

\begin{document}

\begin{figure}
\begin{grammar}
<example> ::= $[ \lit{def} <identifier> \lit{=} <expr>$]
\end{grammar}
\caption{an example grammar}
\label{grm:example}
\end{figure}

\listoffigures

\end{document}

This draws me a nice (partial) definition of the non-terminal “example” in a fictive language and I can see it in the list of figures at the end. But in the real document there are also other figures, which should be in another listofwhatever.

So how can I define such environments by myself with the corresponding counters and lists?

edit

If it matters: I am using xelatex 3.1415926-2.5-0.9999.3-2013060708 (TeX Live 2013).

-
Have a look at the float package. –  Paul Gessler Jun 20 at 21:11

This functionality is offered by a number of packages: float, newfloat, and trivfloat for new floating objects.

Here's an example using newfloat (refer to the package documentation for the customization options it offers):

\documentclass[english]{scrartcl}
\usepackage{syntax}
\usepackage{newfloat}

\DeclareFloatingEnvironment{example}

\begin{document}

\begin{example}
\begin{grammar}
<example> ::= $[ \lit{def} <identifier> \lit{=} <expr>$]
\end{grammar}
\caption{an example grammar}
\label{grm:example}
\end{example}

\listofexamples

\end{document}

-


text:
# Do permanent magnets have an electric current surrounding them?

Permanent magnets seems to have different properties to electromagnets, such that electromagnets can be used for induction and energy transfer if a conductor is placed within their changing magnetic field.

So I understand that an electromagnet will have a changing magnetic field, which in turn generates a changing electrical current in a conductor placed within this magnetic field.

My question is, do permamant magnets generate this same electric current if a conductor is placed within their magnetic field?

I.e. if I had a strong permanent magnet, would I be able to generate a current in a coil if it was brought into the permananet magnet's magnetic field?

Now, onto the explanation:-

According to Faraday's Law, you will get a current in a conductor when the amount of magnetic flux linked with the conductor changes.

Note that it is immaterial whether the source of the magnetic field is a permanent magnet or an electromagnet. All that needs to happen for you to detect a current in the conductor is for the flux to change.

In case of the electromagnet, when you change the strength of the field; the flux linked with the coil changes and current is induced.

In case of the permanent magnet, your field is fixed; you can't change it. But, if there is relative motion between the permanent magnet and the coil; then the flux linked with the coil will change and if that happens,current will definitely be induced.

According to Faraday's law of induction, $$\mathcal{E} = -N {{d\Phi} \over dt}$$ you will need a change in the magnetic flux $\Phi$ in order to get an EMF or an electric field.

So if you just put your coil in the magnetic field of the permanent magnet, you will not measure a current. There will only be a current, if you move the coil around so that the flux changes.


text:
# Take advantage of external RTC to avoid 32.768KHz duplicity?

Hello. I decided to use an external solution for RTC. Sometimes the RTC chip offers an buffered 32.768KHz output. Can I use this output to supply the 32.768KHz to the nRF52832 and gain a litle power saving? Thanks in advance.

edit retag close delete

Sort by » oldest newest most voted

Hi,

In general, it’s no problem using an external 32.768 kHz clock source. But, if the external 32.768 kHz clock is only sometimes available, you would need to switch between the RC oscillator clock and the external clock.

Unfortunately, you cannot change the SoftDevice LF clock source on the fly, since this would disrupt the protocol timing. You would therefore have to disable the SoftDevice, and enable it again with the new clock source. If you are in an active connection, you will be disconnected, and therefore have to re-connect every time you change the clock source. In most applications this would not be acceptable. The potential power saving is also very small, with the RC oscillator only using about 0.6 µA.

more

Hello Sigurd. The external RTC chip would be powered by the same power supply than the nRF, its seems not a problem to lose oscillation except at startup than can be 1s. I think the power saving can be considerable because the external RTC only needs about 200nA. Can the long startup be a problem? For the connection, I guess input is pin2 P0.00/XL1. Is correct? Thank you.

( 2017-01-13 10:04:25 +0100 )editconvert to answer

P0.00/XL1 is correct. What do you mean by "its seems not a problem to lose oscillation except at startup than can be 1s ?" ? It takes 1s to start up the bufferd 32K clock ? Does the RTC chip provide a stable 32K clock after startup?

( 2017-01-13 15:13:45 +0100 )editconvert to answer

Sorry. I mean that the clock would be always available because the RTC chip would be powered by the same power supply than the nRF, but there is a time at the RTC start where the clock is not ready. From NXP datasheet: "...The oscillator is considered to be stopped during the time between power-on and stable crystal resonance. This time can be in the range of 200 ms to 2 s depending on crystal type, temperature, and supply voltage..."

( 2017-01-13 17:36:38 +0100 )editconvert to answer

In that case it should not be a problem. The SoftDevice will wait for the clock to startup. Remember to specify the correct clock accuracy, e.g. NRF_CLOCK_LF_XTAL_ACCURACY_20_PPM.

( 2017-01-16 12:46:46 +0100 )editconvert to answer

Hello Sigurd. It's OK. I will check the overall power comsumption with this solution. Thankyou.

more

[hide preview]

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text:
# Newton's Method (as applied to Auto Financing)

## Homework Statement

A car dealer sells a new car for $18,000. He also offers to sell the same car for monthly payments of$375.00 for five years. What monthly rate is this dealer charging?

## Homework Equations

A = [R(1 - (1 + i))^-60] / i

where A = the present value, R = the monthly payment, i = the interest rate, n = the number of monthly payments

Replacing i by x, show that 48x(1 + x)^60 - (1 + x)^60 + 1 = 0.

## The Attempt at a Solution

I thought that, "Replacing i by x, show that 48x(1 + x)^60 - (1 + x)^60 -1 = 0.", seemed to suggest that they wanted me to solve for i (or x), but in hindsight I don't think that's what I need to do. I don't know where to start now.  Or I suppose rather that I'm just supposed to set it equal to 0 somehow. In which case I guess I don't know what they're doing to (1 + x)^60.

Last edited:

Ray Vickson
Homework Helper
Dearly Missed

## Homework Statement

A car dealer sells a new car for $18,000. He also offers to sell the same car for monthly payments of$375.00 for five years. What monthly rate is this dealer charging?

## Homework Equations

A = [R(1 - (1 + i))^-60] / i

where A = the present value, R = the monthly payment, i = the interest rate, n = the number of monthly payments

Replacing i by x, show that 48x(1 + x)^60 - (1 + x)^60 + 1 = 0.

## The Attempt at a Solution

I thought that, "Replacing i by x, show that 48x(1 + x)^60 - (1 + x)^60 -1 = 0.", seemed to suggest that they wanted me to solve for i (or x), but in hindsight I don't think that's what I need to do. I don't know where to start now.  Or I suppose rather that I'm just supposed to set it equal to 0 somehow. In which case I guess I don't know what they're doing to (1 + x)^60.

You are using an incorrect formula. If you have 60 monthly payments of R, the payment in month 2 is discounted by 1/(1+r), that in month 3 is discounted by 1/(1+r)^2, ... and the final payment in month 60 is discounted by 1/(1+r)^59. Here, r is the monthly interest rate. So, equating the net present value to the car's price P we have
$$P = R \left( 1 + \frac{1}{1+r} + \frac{1}{(1+r)^2} + \cdots + \frac{1}{(1+r)^{59}} \right)\\ = R \left( \frac{1+r}{r} - \frac{1}{r(1+r)^{59}} \right).$$
So, we need to solve
$$18000 = 375 \left( \frac{1+r}{r} - \frac{1}{r(1+r)^{59}} \right).$$

RGV


text:
# Substitution Reactions

This page lists the learning units from the topic "Substitution Reactions" in a reasonable order with regard to contents.

## Substitution Reactions

### $SN2$ - Second-order Nucleophilic Substitution 60 min.

This learning unit deals with second-order nucleophilic substitutions ($SN2$). The mechanism, applications, and parameters that influence the $SN2$ reaction are discussed.

### $SN1$ - First-order Nucleophilic Substitution 45 min.

This learning unit deals with first-order nucleophilic substitutions ($SN1$). The mechanism, applications, and parameters that influence the $SN1$ reaction are discussed.

### SN/E Competition - $SN2$/$E2$ Competition 40 min.

First, this learning unit describes the mechanistic reasons for the competition of elimination and nucleophilic substitution. Then, the different parameters that influence the $SN2$/$E2$ competition are discussed. After working through the learning unit, one should be able to predict, depending on given parameters, the predominant mechanism of a reaction or to choose the parameters of a reaction, so that a mechanism desired actually mainly occurs, respectively.

### SN/E Competition - $SN1$/$E1$ Competition 30 min.

First, this learning unit describes the mechanistic reasons for the competition of elimination and nucleophilic substitution. Then, the different parameters that influence the $SN1$/$E1$ competition are discussed. After working through the learning unit, one should be able to predict, depending on given parameters, the predominant mechanism of a reaction or to choose the parameters of a reaction, so that a mechanism desired actually mainly occurs, respectively.

### SN/E Competition - Role of the Temperature 20 min.

First, this learning unit describes the mechanistic reasons for the competition of elimination and nucleophilic substitution. Then, the influence of the reaction temperature on the $SN2$/$E2$ competition as well as the $SN1$/$E1$ competition is discussed.

## Substitution Reactions - collected basic and advanced subjects in larger chapters

### SN/E Competition (overall) 90 min.

First, this learning unit describes the mechanistic reasons for the competition of elimination and nucleophilic substitution. Then, the different parameters that influence the $SN2$/$E2$ competition as well as the $SN1$/$E1$ competition are discussed. After working through the learning unit, one should be able to predict, depending on given parameters, the predominant mechanism of a reaction or to choose the parameters of a reaction, so that a mechanism desired actually mainly occurs, respectively.

### Answers to the Exercises "SN/E Competition" 30 min.

This learning unit just contains the answers to the exercises "SN/E Competition". Please, do not read this learning unit, but try to answer the exercises of the learning unit "SN/E Competition (overall)".


text:
# Set/Examples

Let $P$ denote the set of living people.
 $\displaystyle \text {The person reading this web page}$ $\in$ $\displaystyle P$ $\displaystyle \text {Julius Caesar}$ $\notin$ $\displaystyle P$ $\displaystyle -4$ $\notin$ $\displaystyle P$


text:
Browse Questions
Home  >>  AIMS

# Factors of $\;(x^{2} + \large\frac{x}{6}- \large\frac{1}{6})\;$ are

$(a)\;\large\frac{1}{6} (2x+1)(3x+1)\qquad(b)\;\large\frac{1}{6} (2x+1)(3x-1)\qquad(c)\;\large\frac{1}{6} (2x-1)(3x-1)\qquad(d)\;\large\frac{1}{6} (2x-1)(3x+1)$

Answer : $\;\large\frac{1}{6} (2x+1)(3x-1)$
Factors of $\;(x^{2} + \large\frac{x}{6}- \large\frac{1}{6})\;$ are $\;\large\frac{1}{6} (2x+1)(3x-1)$


text:
# Choosing font for bold small caps (or any other particular family+series+shape combination)

What is the best way (most proper and general solution) to get rid of the warning

LaTeX Font Warning: Some font shapes were not available, defaults substituted.

by explicitly setting which font should be used for particular family+series+shape combination in edited document?

Bold small-caps are not widely available font-wise. E.g. they are available in CM-Super (which the default font for T1 encoding), but not in overall better Latin Modern.

Let's say I use Latin Modern as the default font (\usepackage{lmodern}). What should be done to make LaTeX fallback to CM-Super in case of bold small caps? (And only in the case of bold small caps, medium small caps should remain typed using Latin Modern.)

-
As always a minimal working example (MWE) as reference would be very much appreciated. People could then use it to implement, test and compare their solutions. –  Martin Scharrer Jul 4 '11 at 0:40
@Martin: I've avoided giving MWE on purpose. I prefer to make this a more general question, but I still gave a specific example w/ bold small caps and lmodern+cm-super, because sometimes people just prefer to think about concrete case. –  przemoc Jul 4 '11 at 0:48

AFAIK you just need to declare that font shape using \DeclareFontShape. You can just declare the font shape to use the font shape of the standard font. This should lead to the same result as before just without the warning.

Using lmodern with T1 font encoding you get the following warning with \scshape\bfseries:
LaTeX Font Warning: Font shape T1/lmr/bx/sc' undefined
(Font)              using T1/lmr/bx/n' instead on input line 19.

LaTeX Font Warning: Some font shapes were not available, defaults substituted.

So you need to declare the font shape T1/lmr/bx/sc. As Ulrike pointed out you can substitute another font using the following code:

\documentclass{article}
\usepackage[T1]{fontenc}

\DeclareFontShape{T1}{lmr}{bx}{sc} { <-> ssub * cmr/bx/sc }{}

\begin{document}

{\normalfont normal font}
{\scshape small caps}
{\ttfamily tt family}
{\bfseries bold font}
{\scshape\bfseries bold small caps}% works and uses cmr font instead of lmr

\end{document}

-
Effort appreciated. It is good as it works for my exemplary case, but at the same time I cannot say I like it, because it requires checking and copying some stuff from .fd file, which I believe should be avoidable, as we're ultimately reusing family+series+shape already defined out there. –  przemoc Jul 4 '11 at 1:14
Mmm, Adobe Acrobat Reader crashes with the above MWE when I try to display Properties->Fonts! pdffonts actually displays the fonts nicely, so does evince. –  Martin Scharrer Jul 4 '11 at 1:15
Works in okular too, where I see shiny new SFXC1200. :-) –  przemoc Jul 4 '11 at 1:22
@przemoc: There's nothing else to do in this case; the font you desire is not available in the Latin Modern family. If their developers don't add themselves such a substitution, it's impossible for LaTeX to define it by itself: no default substitution can be allowed to change the family. –  egreg Jul 4 '11 at 8:10
You can use the sub (or ssub) function to declare the substituation: \usepackage{lmodern} \normalfont %to load T1lmr.fd \DeclareFontShape{T1}{lmr}{bx}{sc} { <-> ssub * cmr/bx/sc }{} –  Ulrike Fischer Jul 4 '11 at 9:06


text:
Indico search will be reestablished in the next version upgrade of the software: https://getindico.io/roadmap/

This search is only for public events. Restricted events are not available.

IMPORTANT! Indico has been upgraded. Please let us know as soon as possible if you find any issues and email indico-support@fnal.gov

36th Annual International Symposium on Lattice Field Theory

22-28 July 2018
Kellogg Hotel and Conference Center
EST timezone

Axial U(1) symmetry and Dirac spectra in high-temperature phase of $N_f=2$ lattice QCD

Jul 24, 2018, 2:00 PM
20m
Centennial (Kellogg Hotel and Conference Center)

Centennial

Kellogg Hotel and Conference Center

219 S Harrison Rd, East Lansing, MI 48824
Nonzero Temperature and Density

Speaker

Dr Kei Suzuki (KEK)

Description

We investigate the axial $U(1)$ symmetry in the phase above the critical temperature in $N_f=2$ lattice QCD, where the ensembles are generated with Mobius domain-wall fermions, and the overlap/domain-wall reweighting is applied. We show the $U(1)_A$ susceptibility extracted from the spectra of the overlap Dirac eigenmodes and discuss its temperature, quark-mass, and spatial volume dependence. The behavior of $U(1)_A$ susceptibility is compared with that of topological susceptibility and their relation is discussed.

Primary author

Dr Kei Suzuki (KEK)

Co-authors

Dr Guido Cossu (The University of Edinburgh) Prof. Hidenori Fukaya (Osaka University) Dr Shoji Hashimoto (KEK) Prof. Sinya Aoki (YITP) Yasumichi Aoki (KEK)

 Slides


text:
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text:
# Snowfall in HTML

I am a beginner and I have made Snowfall in HTML for my mom. I'm pretty sure it will not look that awesome to any developer out there, but hey, that's why I've posted it.

I'd like a general review of this. I'm especially concerned about the quality and enhancements of this HTML.

Happy Winter!


text:
# Chaos HackTheBox Walkthrough

Today we’re going to solve another boot2root challenge called “Chaos“. It’s available at HackTheBox for penetration testing practice. This laboratory is of a medium level, but with adequate basic knowledge to break the laboratories and if we pay attention to all the details we find during the examination it will not be complicated. The credit for making this lab goes to felamos. Let’s get started and learn how to break it down successfully.

Level: Medium

Since these labs are available on the HackTheBox website.

### Penetration Testing Methodology

Reconnaissance

• Nmap

Enumerate

• Information relevant in SSL certificate
• Dirsearch
• WPScan
• Gobuster
• Nikto

Exploiting

• Information disclosure in drafts mail
• Use of encryption and weak password
• Read and command execute in LaTeX software

Privilege Escalation

• Cracking to passwords stored in Firefox
• Capture the flag

### Reconnaissance

We will use the following commando to perform a scan to all ports with scripts and versions.

nmap -A -p- 10.129.98.125

We access web service on port 80 and the server we are not allowed the connection with direct IP.

We access other web service on port 10000, here we enumerate a Webmin panel. We tested typical passwords (admin:admin, admin:1234, etc…) but not work.

### Enumeration

We ever have to review an SSL certificate, here we enumerate an email corporate and domain name.

We insert name domains “chaos” and “chaos.htb” in the “/etc/hosts” file. We do this because it is very likely that they use virtual hosting on the machine.

We revise the web site in “chaos.htb“, but not found the information relevant. So we use the dirsearch tool with IP address. Is very import to perform scans in both targets (IP and Host), is possible that virtual hosting uses and we obtains different results.

We enumerate a WordPress directory.

We found a post published but we need the password unlocked.

We use the WPScan tool and we enumerate a username.

We use the username as a password and “Voilà!”, we have credentials for webmail… But? Where is webmail?

We use gobuster tool with a subdomains wordlist and we found a subdomain “webmail.chaos.htb“.

Correct! We found webmail!

We use the credentials found in WordPress and we reading the mail which is in the “Drafts folder“.

Here we found a Python script and a message encrypted, also view a message from Ayush user telling us that the password is “sahay

We download both files and we read the content, we have a file cipher and a python script not finished.

We found the function “encrypt” in google and we found the complete python script.

We copy libraries and function “decrypt“, we add a line with “sahay” password and filename for decrypt.

We execute the python script, we read the “t” file, we found a string in “base64” coding, we decoding the file and we found a URL.

### Exploiting

We visit the website, we found a form for creating PDF files. This leads us to believe that there may also be directories containing PDF files with confidential information.

So we use nikto tool and we enumerate the directory interesting.

We enumerate files with information disclosure, software name and version.

We use Burp, we intercept request and we testing read “/etc/passwd” file as proof of concept.

But not work, it this server blacklist use with words forbidden.

We use the following code for protection bypass and read “/etc/passwd” file.

\newread\file
\openin\file=/etc/passwd
\text{\line}
\closein\file

Yeah! We now read a PDF file on web browser

So, we now command execute to obtain a reverse shell.

We have problems with the path system, so we use route absolute for binary call and we read the “user.txt” file.

### Privilege Escalation (root)

We found a “.mozilla” folder hidden, this is an uncommon folder.

We compress the “firefox” folder and transfer it to our computer with netcat.

Unzip the folder, download the “firefox_decrypt” tool and use it. We will be asked for the “master password” we will use the one found in WordPress.

This will return us some credentials for the “Webmin” service, but we can reuse the password with the system user “root” and read the flag.

Author: David Utón is Penetration Tester and security auditor for Web applications, perimeter networks, internal and industrial corporate infrastructures, and wireless networks. Contacted on LinkedIn  and Twitter


text:
Home/Class 12/Maths/

Find the equation of the line joining (1, 2) and (3, 6) using determinant.
Speed
00:00
03:53

## QuestionMathsClass 12

Find the equation of the line joining (1, 2) and (3, 6) using determinant.

Let $$p\left( {x,y} \right)$$ be any point on the line joining the points (1, 2) and (3, 6).
Then, Area of triangle that could be formed by these points is zero.
$$\therefore$$ Area of triangle = $$\frac{1}{2}\left| {\begin{array}{*{20}{c}} {{x_1}}&{{y_1}}&1 \\ {{x_2}}&{{y_2}}&1 \\ {{x_3}}&{{y_3}}&1 \end{array}} \right| = 0$$
$$\Rightarrow$$  $$\frac{1}{2}\left| {\begin{array}{*{20}{c}} x&y&1 \\ 1&2&1 \\ 3&6&1 \end{array}} \right| = 0$$
$$\Rightarrow \frac{1}{2}\left[ {x\left( {2 - 6} \right) - y\left( {1 - 3} \right) + 1\left( {6 - 6} \right)} \right] = 0$$
$$\Rightarrow - 4x + 2y = 0$$
$$\Rightarrow - 2x + y = 0$$
$$\Rightarrow y = 2x$$ which is required line.


text:
# Remarks on WS1S Complexity

## Complexity of One Algorithm

The construction in Using Automata to Decide WS1S determinizes automaton whenever it needs to perform negation. Moreover, existential quantifier forces the automaton to be non-deterministic. Therefore, with every alternation between and we obtain an exponential blowup. For formula with n alternations we have complexity with a stack of exponentials of height . Is there a better algorithm?

## Lower Bound

The following paper shows that, in the worst case such behavior cannot be avoided, because of such high expressive power of MSOL over strings.


text:
# Combinations of characteristic functions: $\alpha\phi_1+(1-\alpha)\phi_2$

Suppose we are given two characteristic functions: $\phi_1,\phi_2$ and I want to take a weighted average of them as below:

$\alpha\phi_1+(1-\alpha)\phi_2$ for any $\alpha\in [0,1]$

Can it be proven that the result is also a characteristic function? If so, I am guessing this result could extend to any number of combinations $\alpha_i$ as long as $\sum_i\alpha_i=1$

Secondly, if $\phi$ is again a characteristic function, then $\mathfrak{R}e\phi(t)=\frac12(\phi(t)+\phi(-t))$ is also a characteristic function. I don't even know how to begin attempting this proof as I am not sure what the $\mathfrak{R}$ represents.

Lastly, regarding the symmetry of characteristic functions,

$\phi$ is symmetric about zero iff it is real-valued iff the corresponding distribution is symmetric about zero.

Once again, my lack of familiarity with the complex plane leaves me in the dark here. Why can a complex-valued function not be symmetric about zero?

-
The symbol $\Re$ means "real part". Thus, if $z = x + iy$, $\Re z = x$. Similarly, $\Im z = y$ is the imaginary part. –  Calvin McPhail-Snyder Aug 4 '12 at 22:03
thanks...makes much more sense now! –  Justin Aug 4 '12 at 22:15

To prove these are characteristic functions, let us use random variables. This yields simpler, and more intuitive, proofs.

In the first case, assume that $\phi_1(t)=\mathrm E(\mathrm e^{itX_1})$ and $\phi_2(t)=\mathrm E(\mathrm e^{itX_2})$ for some random variables $X_1$ and $X_2$ defined on the same probability space and introduce a Bernoulli random variable $A$ such that $\mathrm P(A=1)=\alpha$ and $\mathrm P(A=0)=1-\alpha$, independent of $X_1$ and $X_2$. Then:

The function $\alpha\phi_1+(1-\alpha)\phi_2$ is the characteristic function of the random variable $AX_1+(1-A)X_2$.

The extension to more than two random variables is direct. Assume that $\phi_k(t)=\mathrm E(\mathrm e^{itX_k})$ for every $k$, for some random variables $X_k$ defined on the same probability space and introduce an integer valued random variable $A$ such that $\mathrm P(A=k)=\alpha_k$ for every $k$, independent of $(X_k)_k$. Then:

The function $\sum\limits_k\alpha_k\phi_k$ is the characteristic function of the random variable $\sum\limits_kX_k\mathbf 1_{A=k}$.

In the second case, assume that $\phi(t)=\mathrm E(\mathrm e^{itX})$ for some random variable $X$ and introduce a Bernoulli random variable $A$ such that $\mathrm P(A=1)=\frac12$ and $\mathrm P(A=0)=\frac12$, independent of $X$. Then:

The function $t\mapsto\frac12(\phi(t)+\phi(-t))$ is the characteristic function of the random variable $AX+(1-A)(-X)=(2A-1)X$.

-

By Bochner's theorem, a function $\phi : \mathbb{R} \to \mathbb{C}$ is the characteristic function of a probability measure if and only if

1. $\phi$ is positive definite,
2. $\phi(0) = 1$, and
3. $\phi$ is continuous at the origin.

Since these properties are conserved under convex combination, your second statement is true whenever $\alpha_i$ are non-negative.

The symbol $\Re(z)$ means the real part of a complex number $z$. If $\phi$ is a characteristic function, then $\phi(-t) = \bar{\phi}(t)$, thus we have

$$\Re \phi(t) = \frac{\phi(t)+\bar{\phi}(t)}{2} = \frac{\phi(t)+\phi(-t)}{2}.$$

Now let $\phi$ be the characteristic function of a probability measure $\mu$. Then clearly the mapping $t \mapsto \phi(-t)$ is the characteristic function of the measure $\tilde{\mu} : E \mapsto \mu(-E)$. Thus in view of the first answer, $\Re \phi(t)$ is also a characteristic function.

If your symmetry about zero means $\phi(-t) = \phi(t)$, then the first assertion follows by our second answer. Now since $\phi(t) = \phi(-t)$, the corresponding measure must coincide, that is, we must have $\mu(E) = \tilde{\mu}(E) = \mu(-E)$ for any Borel measurable $E \subset \mathbb{R}$. Thus $\mu$ is symmetric.

-


text:
Relativistic effect on two-body reaction inducing atomic displacement - Archive ouverte HAL Access content directly
Journal Articles Journal of Nuclear Materials Year : 2019

## Relativistic effect on two-body reaction inducing atomic displacement

(1, 2) , (1)
1
2
Shengli Chen
• Function : Correspondent author
• PersonId : 1069305

Connectez-vous pour contacter l'auteur
David Bernard
• Function : Author
• PersonId : 1062794

#### Abstract

The effect of the relativistic treatment of two-body reaction kinematics on atomic recoil energy and subsequent primary damage parameters, such as damage energy and number of atomic displacements, has been studied. If the emitted particle has a mass larger than the incident particle, the ratios of relativistic quantities in comparison to the classic mechanical ones can be null and infinite due to the null recoil energies. However, the relativistic corrections on recoil energies are limited in [-0.6%, 0.5%] and [-6%, 5%] for 20 MeV and 200 MeV neutron discrete (n, $\alpha$) reactions of $^{56}$Fe. For (n,n') and (n,p) reactions of any nucleus, the relativistic corrections are about 1% on recoil energy for a 20 MeV incident neutron, while the corrections can be more than 30% (about 10% on average) for 200 MeV incident energy. The relativistic effect is less than 5% on ($\alpha$,n) reaction inducing recoil energy of any nucleus (and within 0.6% on atomic displacement damage for $^{56}$Fe) for secondary energy lower than the incident energy of 200 MeV. On the other hand, about 10 keV and 1500 keV broader Primary Knock-on Atom (PKA) spectra are respectively found for 20 MeV and 200 MeV neutron-induced (n,n'), (n,p), and (n, $\alpha$) reactions of $^{56}$Fe. Therefore, the relativistic treatment of two-body reactions should be applied for computing PKA spectra and subsequent radiation damage for high energy neutrons.

### Dates and versions

cea-02534706 , version 1 (30-04-2020)

### Identifiers

• HAL Id : cea-02534706 , version 1
• DOI :

### Cite

Shengli Chen, David Bernard. Relativistic effect on two-body reaction inducing atomic displacement. Journal of Nuclear Materials, 2019, 522, pp.236-245. ⟨10.1016/j.jnucmat.2019.05.020⟩. ⟨cea-02534706⟩

### Export

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21 View


text:
# thraxil.org:

## My Own Javascript Y Combinator

by anders pearson Mon 15 Sep 2008 19:09:58

I came across a link to the slides for Xavier Leroy's course on Functional programming languages this weekend and have been slowly making my way through them.

It's just slides from the lectures so it can be a bit of a struggle to follow them without, you know, the actual lectures. On the other hand, having to struggle a bit and figure out what's going on from just the slides has been kind of good for forcing me to really pay attention and not gloss over the parts that aren't immediately obvious.

Judging by the slides, the lectures are fantastic. They're incredibly ambitious in scope. In the first lecture, it starts by introducing the Lambda Calculus (well, that part is fast enough that I think it's assumed that the reader is already basically familiar), shows reduction strategies, implements a lambda calculus interpreter in a couple lines of Caml, makes it efficient, adds lexical scoping and closures, and keeps it all tied back to the Lambda Calculus. Further lectures go on to describe abstract machines, compile the enriched Lambda Calculus for the abstract machine, do tail call elimination and lazy evaluation, proves the correctness of the abstract machine and compiler, gets into exception returning vs state passing vs continuation passing implementations, monads and monad transformations (including the concurrency monad transformer), and then gets into some even deeper optimizations. It basically connects the theoretical, mathematical underpinnings of programming all the way to the tricks that are used in compiler optimization. Highly recommended if that sounds interesting to you.

Anyway, in the course of making my way through that material, I decided that I needed to refresh my understanding of the Lambda Calculus and combinators since I haven't really done much of that since undergrad. I've been writing a lot of JavaScript lately for work and, while I still don't really like the syntax that much, JavaScript does have proper closures so it's suitable for lambda calculus experimentation.

So I decided to implement the Y Combinator in JavaScript as an exercise. I remember it being covered in class long ago, and, while I could see why it was interesting, I don't think it ever quite clicked for me exactly how it worked. There's a good page explaining the how and why of the Y Combinator here. The example code is all in Scheme, so I just went through the page and translated each example to JavaScript, making sure I understood what was going on at each step. I kind of like this approach to learning CS. Just reading a text with code samples, it's all too easy to accept that the samples do what it says they do and not really spend the extra effort to understand exactly what's going on. Translating it into a different programming languages forces you to really grasp every detail.

So, this is my JavaScript Y Combinator. There are many like it, but this one is mine:

function Y (X) {
return (function(procedure) {
return X(function(arg) {
return procedure(procedure)(arg);
});
})(function(procedure) {
return X(function(arg) {
return procedure(procedure)(arg);
});
});
}

It's not as pretty as the scheme version:

(define Y
(lambda (X)
((lambda (procedure)
(X (lambda (arg) ((procedure procedure) arg))))
(lambda (procedure)
(X (lambda (arg) ((procedure procedure) arg)))))))

And neither is as pretty as the original untyped Lambda Calculus:

Y = λf·(λx·f (x x)) (λx·f (x x))

But it works.


text:
# Free abelian groups clarification

I'm having troubles understanding the following definition of a free abelian group: $$\mathbb{Z}^{(A)} := \{f: A\rightarrow \mathbb{Z} \ |\ f(a)\ne 0\ \ \text{for finitely many elements}\ a \in A\}$$

I don't get how to write an element of this set as a formal sum. On the internet I could find such a representation: $$f = \sum_{a \in A}{n_a\phi_a}$$ Where $\phi_a(x) = \begin{cases} \text{1,} &\quad\text{if x}\ = \text{a} \\ \text{0,} &\quad\text{if x}\ \ne \text{a}\\ \end{cases}$

The problem is, I don't udnerstand what $n_a$ is here, many resources don't mention what this is. For example, on Wikipedia it's writtten as a function, but this doesn't feel natural to me. Is $n_a\phi_a$ a function composition then? An integer value seems to fit here more.

I am totally fine with less general definition of a free group $\mathbb{Z}^{\oplus n} = \underbrace{\mathbb{Z}\oplus\ .\ .\ .\ \oplus\mathbb{Z}}_\text{n-times}$. With an element of $\mathbb{Z}^{\oplus n}$ being a tuple $(m_1, ..., m_n)$ that can be written in a unique way as a sum: $$\sum_{i = 1}^n{m_i\phi(i)}$$ With $\phi(i) := \underset{\text{i-th place}}{(0,\ .\ .\ .,\ 0,\ 1,\ 0,\ .\ .\ .\ ,\ 0)} \in \mathbb{Z}^{\oplus n}$ and $m_i$ being some integer coefficient.

I understand that $\mathbb{Z}^{(A)}\cong \mathbb{Z}^{\oplus n}$ if, for example, if $A =\{1,\ .\ .\ .\ , n\}$, then $(m_1, ..., m_n)$ may be identified with a function $A\rightarrow \mathbb{Z}$ taking $i$ to $m_i$. But I don't understand how to put this information together and define an element of $\mathbb{Z}^{(A)}$ via sum.

Moreover, should group homomorphism $\psi:\mathbb{Z}^{(A)} \rightarrow G$ ($G$ is an arbitrary abelian group) be defined as $$\psi(\sum_{a \in A}{n_a\phi_a}):= \sum_{a \in A}{n_af(a)}$$ where $f:A\rightarrow G$ is a set function (a forgetful functor, in fact). Again, only integer value $n_a$ feels ok for me.

Another point of confusion is the fact that only finite sums are used, $\mathbb{Z}^{(A)}$ is defined to have function with only finitely many non-zero values. Why is that? I don't have an intuitive understanding in what way infinities could 'break' free groups as I was only given a definition but no explanation why.

$\mathbb{Z}^{(A)}$ is not the definition of a free abelian group, but rather an explicit construction.
A basis for $\mathbb{Z}^{(A)}$ is given by $\{\phi_a : a \in A\}$. The representation of elements in that basis is explicit: $$f = \sum_{a \in A}{f(a)\phi_a}$$ Since $f$ is zero at almost every point of $A$, this is a finite sum. So, the answer to your first question is that $n_a$ is just an integer: $n_a=f(a)$.
About your last point, the simplest example is the Baer–Specker group $\mathbb Z ^{\mathbb N}$ of all integer sequences. This is the same as the set of all functions $\mathbb N \to \mathbb Z$ with componentwise addition. Baer proved that the Baer–Specker group is not free abelian, a nontrivial theorem.*


text:
# Analog-to-digital converter

An analog-digital converter is an electronic device, component or part of a component for converting analog input signals into a digital data stream , which can then be further processed or stored. Other names and abbreviations are ADC , analogue-to-digital converter or A / D converter , English ADC ( a nalogue-to- d igital c onverter ) or shortly A / D.

A variety of repositioning methods are in use. The counterpart is the digital-to-analog converter (DAU).

Analog-digital converters are elementary components of almost all devices in modern communication and entertainment electronics, such as B. cell phones , digital cameras , or camcorders . They are also used to record measured values in research and industrial production facilities, in machines and everyday technical objects such as motor vehicles or household appliances.

4-channel multiplex stereo analog-digital converter, used on a Sound Blaster X-Fi sound card for a PC

## Working method

An ADC converts a time and value-continuous input signal (analog signal) into a time-discrete and value-discrete sequence of digitally represented values. Due to a finite number of possible output values, quantization always takes place . The result of an AD conversion can be imagined in a signal-time diagram in a sequence of points with stepped horizontal and vertical distances. The main parameters of an ADC are its bit depth and its maximum sampling rate . The conversion time is usually much shorter than the reciprocal of the sampling rate.

The bit depth of an AD converter already limits the maximum possible accuracy with which the input signal can be converted. The usable accuracy is lower due to other sources of error in the ADC. In addition to the fastest possible methods, there are also slow (integrating) methods for suppressing interference.

### Time discretization (sampling)

Sampling of an analog signal using a sample-and-hold circuit
Spectrum of an on band-limited signal (blue) and which by sampling at the frequency resulting image frequencies (Green) -, without overlapping, that is at the correct sampling${\ displaystyle f _ {\ mathrm {max}}}$${\ displaystyle f _ {\ mathrm {s}}}$

The minimum required sampling frequency for loss-free discretization results from the bandwidth of the input signal.

In order to be able to completely reconstruct the signal later, the sampling frequency must be greater than twice the maximum possible frequency in the input signal (see Nyquist frequency ). Otherwise, there is a sub-sampling and leads in the reconstructed signal to the input signal is not existing frequencies. Therefore the input signal must be band-limited . Either it is this on its own or it is made into such a signal by low-pass filtering .

Sometimes, however, the signal to be sampled is so high-frequency that this condition cannot be technically implemented. However, if the input signal is periodic, a reconstruction can still be made possible by multiple sampling with a time offset without violating the sampling theorem, since intermediate points are determined when the signal is passed through several times and a larger number of interpolation points is created, which in the end results in an increase in the Sampling rate.

In many conversion processes, the input signal must not change during signal conversion. Then switch to the actual AD converter, a sample and hold circuit (sample and hold) before which the signal value ( English sample ) analog so caches that it remains constant during quantization. This is especially true of the step-by-step and bit-wise converters, which require longer conversion times. If a converter requires this sample-and-hold circuit, it is usually included when it is implemented as an integrated circuit .

In many applications, the input signal should always be sampled at exactly the same time intervals. However, due to random variations in the distances, an effect occurs that is known as jitter . It falsifies the original signal during the later reconstruction, since this is again equidistant - that is, with the same time intervals.

The conversion time must not be confused with the latency of a converter, i.e. H. the time that elapses after the acquisition until an AD converter has passed on the date. This time can be much longer than the conversion time, which can be particularly disruptive in control technology. It is caused by pipelining of the converter, post-processing of the data and the serial data transmission .

### Quantization

Digital signal (red dots) after sampling and quantization of an analog signal (gray line)

The quantization of the previously time-discrete signal represents the actual transition from an analog signal to a digital signal. Due to the finite bit depth of the converter, there is only a certain number of code words and their associated input voltage. The signal is quantized. The deviation between the true input voltage and the quantized input voltage is called the quantization deviation . The more bits or code words that are available, the smaller this unavoidable deviation. In an ideal AD converter, each additional bit reduces this noise by 6.02 dB. With real AD converters, the effective number of bits (ENOB) can be used to estimate what a further bit would bring in the converter under consideration (for example, a further bit in a 12-bit converter with an ENOB of 11 bits would be approx , 15 bit or 0.9 dB).

The ratio of the maximum possible undistorted input voltage and the noise with a signalless input is called the dynamic range . Converters that deliver a constant code word in the absence of an input signal have an infinitely high dynamic range. It makes more sense to specify the signal-to-noise ratio (or the SINAD , signal to noise and distortion ratio, ratio of the signal to the sum of noise and distortion).

### Reference value

Since the analog signal fed to the ADC is converted into a variable digital value, it must be evaluated with a specified value or signal (input signal range or measuring range ). In general, a fixed reference value (e.g. an internally generated reference voltage) is used. The analog input signal is mapped digitally, the reference defines the permissible peak value of the input signal. ${\ displaystyle U _ {\ mathrm {r}}}$

### Quantization characteristic

In analog-digital converters, there is always a non-linear relationship between the input and output variables . However, if the digital value changes at constant intervals with increasing input voltage or if the characteristic curve approaches a straight line with extremely fine gradations, one still speaks of a linear analog-digital converter. There are

 unipolar versions, for example in the dual system 000… 000 for 0 100… 000 for / 2    ${\ displaystyle U _ {\ mathrm {r}}}$ 111… 111 for - 1 LSB ${\ displaystyle U _ {\ mathrm {r}}}$ bipolar versions, for example in the dual system with offset 000… 000 for / 2    ${\ displaystyle -U _ {\ mathrm {r}}}$ 100… 000 for 0 111… 111 for / 2 - 1 LSB ${\ displaystyle U _ {\ mathrm {r}}}$

other codings, for example two's complement , BCD code, can also be used.

### Interfaces

In addition to the sample-and-hold circuit already mentioned , additional circuits are required for the interface to the analog world, so that these are often integrated on a chip together with the actual converter. These can be, for example, buffer or amplifier circuits, possibly with switchable gain ( Programmable Gain Amplifier (PGA)) and inputs for differential signal transmission . There are also variants without a real subtraction amplifier at the input; Instead, the two lines of the differential signal are processed one after the other and only then is the difference formed (so-called pseudo differential inputs ).

Digital data is made available at the output. Classically, each bit of the output variable appears on its own connection pin; the size is output in parallel  - not to be confused with the parallel conversion . If the size is to be displayed on a display , integrated seven-segment encoders are also used, or the size is output as a BCD code in a multiplex process . The disadvantage of the parallel output, especially when further processing by microprocessors or controllers , is the large number of connection pins required. Therefore serial connections are often implemented, for example with the protocols I²C , SPI or I²S . With appropriate data rates, LVDS or JESD204B technology is used , for example .

## Deviations

In addition to the unavoidable quantization error, real AD converters have the following errors:

### Zero point errors, gain errors and non-linearity errors

Deviations from the proportional relationship
a) additive, b) multiplicative, c) non-linear

The following errors are defined as deviations in the characteristic curves between the real and ideal converter (see figure):

The gain error is often given as a fraction of the current value, the zero point error together with the quantization error and the non-linearity error as fractions of the final value or as a multiple of an LSB.

a) if one step is unequal,
b) if the width of a higher step is incorrect

Individual steps can be of different widths.

With a continuously increasing input variable, depending on the implementation method, it can happen that a value of the output variable is skipped, especially when there is a carry over several binary digits, for example from 0111 1111 to 1000 0000. This is called “missing codes”.

### Temporal and aperture errors

When every non-constant input signal is converted, fluctuations in the converter clock Δ t ( clock jitter ) result in an error proportional to the change in the input signal over time. In the case of a sinusoidal signal of frequency f and amplitude A, it is . Any jitter creates additional noise - there is no threshold below which there is no deterioration in the signal-to-noise ratio. Many current converters (especially delta-sigma converters) have internal clock processing. The reason for this is that many converters require a higher internal clock rate or, in the case of delta-sigma converters, that jitter there directly (i.e. even with a constant input signal) causes conversion errors. ${\ displaystyle E_ {ap} \ leq | x '(t) \ Delta t | \ leq 2A \ pi f_ {0} \ Delta t}$

## Implementation process

There are a large number of methods that can be used to convert analog to digital signals. The following are the main principles. In all examples, the electrical voltage is used as the input variable.

The blocks themselves control the internal process of a conversion. For the cooperation with a computer, an ADC can be provided with a start input for the request for a new conversion, with a “busy” output for the notification of the ongoing conversion and with bus-compatible data outputs for reading out the resulting digital value.

### Integrating converter (counting method)

There are two processes involved in these procedures:

• an analog process influenced by the measured variable, in which a capacitor is continuously charged and discharged again alternately,
• a digital process that measures times or pulse densities depending on the charging process, which is counted.

The follow-up converter also counts. This is operated without a capacitor as a feedback converter and is explained further below.

#### Single-slope converter (sawtooth / single ramp method)

Functional principle of an ADC using the sawtooth method

In the sawtooth method, the output voltage of a sawtooth generator is compared with the ground potential (0 V) and the ADC input voltage via two comparators K 1 and K 2 . During the period in which the sawtooth voltage passes through the range between 0 V and the voltage , the pulses of a quartz oscillator are counted. Due to the constant slope of the sawtooth voltage, the elapsed time and thus the counter reading when reaching is proportional to the level of the ADC input voltage. At the end of the counting process, the counting result is transferred to a register and is available as a digital signal. The counter is then reset and a new translation process begins. ${\ displaystyle U_ {r}}$${\ displaystyle U_ {e}}$${\ displaystyle U_ {e}}$${\ displaystyle U_ {r} = U_ {e}}$

The conversion time for this ADC depends on the input voltage. Rapidly changing signals cannot be captured with this type of converter. Converters using the sawtooth method are imprecise, as the sawtooth generator works with the help of a temperature and age-dependent integration capacitor. Due to their relatively low circuit complexity, they are used for simple tasks, for example in game consoles, in order to digitize the position of a potentiometer that is moved by a joystick or a steering wheel.

#### Dual and Quadslope converter (multi-ramp method)

Dual and quad slope converters essentially consist of an integrator and several counters and electronic switches . The integrator works with an external, high-quality capacitor that is charged and discharged in two or more cycles. With the two-ramp procedure (dual slope), the integrator input is first connected to the unknown ADC input voltage, and charging takes place over a fixed time interval. For the subsequent discharge, the integrator is connected to a known reference voltage of opposite polarity. Details are explained under digital metrology . The required discharge time to reach zero voltage at the integrator output is determined by a counter; the meter reading directly represents the input voltage if dimensioned appropriately. With this method, the size of the capacity is reduced from the result. In order to correct the zero point error of the ADC, a further charge / discharge cycle is carried out with the integrator input short-circuited with the four-ramp method. The reference voltage is the determining factor for the accuracy; this means, for example, that thermally induced fluctuations must be avoided.

Such converters based on the multi-ramp method are slow, do not require a sample-and-hold circuit and offer high resolution as well as good differential linearity and good suppression of interference signals such as noise or network coupling. The typical area of application are display measuring devices ( digital multimeters ) which hardly need a conversion time of less than 500 ms and can eliminate superimposed 50 Hz disturbances of the mains frequency with a suitable integration time.

#### Charge balance converter

In the charge balancing process, the capacitor of an integrator is charged by an electric current proportional to the input variable and discharged by short current surges in the opposite direction so that no charge builds up on average. The greater the charging current, the more frequently it is discharged. The frequency is proportional to the input variable; the number of discharges in a fixed time is counted and provides the digital value. In terms of its behavior, the procedure is similar to the dual slope procedure. Other analog input stages that contain a voltage-frequency converter with sufficient high-quality accuracy also lead to a digital value via a frequency count.

### Feedback converter (serial process)

Analog-digital conversion using the DAU setting

These work with a DAU that provides a comparison value . This is approximated to the analog input signal using a suitable strategy . The digital value set at the end on the DAU is the result of the ADC. Since the process requires a period of time in which the input signal must not change, a sample is taken from it by means of a sample and hold circuit (S / H) and recorded during the implementation. ${\ displaystyle U_ {v}}$${\ displaystyle U_ {e}}$

#### Follow-up relocator

Here a counter is used as a data memory. Depending on the sign of , the system counts up or down by one step and then compares it again - counts and compares it again until the difference is smaller than the smallest adjustable step. These converters simply “follow” the signal, whereby the conversion time depends on the distance between the current input signal and the signal during the last conversion. ${\ displaystyle U_ {e} -U_ {v}}$

#### Successive approximation

These work with a DAU that builds up a comparison value every time. The input signal is limited by nesting intervals . Simple successive approximation converts one bit per step. A conversion that is orders of magnitude more precise and faster can be achieved in that the conversion is carried out redundantly by converting with a smaller step size than corresponds to one bit. ${\ displaystyle U_ {v}}$

##### Weighing method
Time curve for the weighing procedure with four bits at = 6.5 V and a smallest adjustable voltage step = 1 V. Result = 0110
B ⋅ = 6 V.${\ displaystyle U_ {m}}$${\ displaystyle U_ {q}}$${\ displaystyle U_ {q}}$

One possible approximation method is the weighing method . First, all bits are set to zero in a data memory ( successive approximation register , SAR). Beginning with the most significant bit ( Most Significant Bit , MSB) to the least significant bit will be down to least significant bit (LSB) determined one after the other all the bits of the digital value.

The control unit sets the bit in progress to one on a trial basis; the digital-to-analog converter generates the comparison voltage corresponding to the current digital value. The comparator compares this with the input voltage and causes the control unit to reset the bit in progress to zero again if the comparison voltage is higher than the input voltage. Otherwise the bit is at least necessary and remains set. After setting the least significant bit is smaller than the smallest adjustable step. ${\ displaystyle U_ {e} = U_ {m}}$${\ displaystyle U_ {m} -U_ {v}}$

The input signal must not change during the conversion , since otherwise the lower-order bits would be obtained on the basis of the higher-order bits that were determined but no longer valid. Therefore a sample-and-hold circuit (S / H) is connected upstream of the input. The ADC needs one clock cycle conversion time for each bit of accuracy. Such converters achieve resolutions of 16 bits at a conversion rate of 1 MHz. ${\ displaystyle U_ {m}}$

##### Redundant converters

Redundant analog-to-digital converters similar to the weighing process assume that there is no exact halving of the still open interval around the target value, but that this interval is only limited by a portion thereof. To do this, they have a digital-to-analog converter, the elements of which are not graded according to the dual system, i.e. always by a factor of 2, but by a smaller factor. On the one hand, you accept that more elements are required to cover the same range of values, but on the other hand, you allow the converter to work an order of magnitude faster and achieve an accuracy that is several orders of magnitude higher: The faster function comes from the fact that the comparator In each step there is no need to wait until its amplifier has settled up to a multiple of the target accuracy (always on the order of magnitude as many settling time constants as the converter is supposed to convert bits), but rather make a decision after the short 50 percent settling time which is then faulty in a fairly large area within the remaining interval. However, this is more than compensated for by the redundantly designed converter elements. The total conversion time of such a converter is in the order of magnitude a power of ten below that of its simple model. Due to the redundant implementation process, such a converter has a much lower level of inherent noise than its purely dual counterpart.

In addition, such an ADC can calibrate itself up to an accuracy that is only limited by the noise. By letting the self-calibration run much more slowly than the implementation in the practical application, the influence of noise in this process can be reduced by an order of magnitude. The resulting characteristic of such a converter is absolutely linear apart from a noise-like deviation of a few multiples of the smallest element used for self-calibration. By placing two converters of this type next to each other on the same chip and one always in the calibration mode, converters of this type can be made almost resistant to manufacturing tolerances, changes in temperature and operating voltage. The achievable resolution is exclusively noise-limited.

#### Delta-Sigma method

Values in different levels of the delta-sigma converter

The delta-sigma method, also known as a 1-bit converter, is based on delta-sigma modulation . In the simplest form (first-order modulator), the input signal comes to the integrator via an analog subtracter and causes a signal at its output, which is weighted with one or zero by a comparator. A flip-flop generates a discrete-time binary signal with which a 1-bit digital-to-analog converter supplies a positive or negative electrical voltage, which pulls the integrator back to zero via the subtractor ( control loop ). A downstream digital filter converts the serial and high-frequency bit stream into data with a low refresh rate, but a large bit width (16 or 24 bits) and a high signal-to-noise ratio (94 to 115 dB). In practice, delta-sigma converters are constructed as third or fourth order systems, that is to say by means of several serially arranged differential and integrator stages. This allows better noise shaping and thus a higher gain in resolution with the same oversampling.

One advantage of the delta-sigma converter is that the dynamics can be interchanged within certain limits due to the bandwidth. Due to the continuous sampling at the input, no sample and hold circuit is required. In addition, low demands are made on the analog anti-aliasing filter.

The advantages come at the price of the disadvantage of the comparatively high latency period, which is mainly due to the digital filter stages. Delta-Sigma converters are therefore used where continuous signal curves and only moderate bandwidths are required, such as in the audio sector. Practically all audio devices in the field of entertainment electronics, such as DAT recorders , use these converters.

It is also increasingly used for data converters in communication technology and measurement technology due to falling prices. However, due to the high oversampling required in this case, there are limits to the method at higher frequencies. The limit is around 2.5 MHz.

### Parallel converter

Two-bit parallel converter with code conversion; a comparator detects overflow, three comparators generate the sum code , the AND gates enclose it in a 1-of-n code in order, from which the OR gate to the desired binary code generate

#### Single-stage parallel converters (flash converters)

While the successive approximation carries out several comparisons with just one comparator, the direct method or Flash implementation can do with just one comparison. In the case of flash converters, however, a separately implemented comparator is required for each possible output value (except for the largest). For example, an 8-bit flash converter therefore requires 2 8 −1 = 255 comparators.

The analog input signal is simultaneously compared in the flash converter by all comparators with the comparison values (generated by a linear voltage divider). This is followed by a code conversion of the 2 n −1 comparator signals into an n  bit wide binary code (with n : resolution in bits). The result is immediately available after the throughput delays (switching time of the comparators and delay in the decoding logic). As a result, the flash converters are very fast, but generally also involve high power losses and acquisition costs (especially with the high resolutions).

Characteristic curve of a two-bit parallel converter.
H stands for HIGH = positively overdriven, L for LOW = negatively overdriven.

Regardless of the resolution, the code conversion only requires one column with AND gates and one column with OR gates (see picture). It converts the result of the comparators into a binary number. It works with a very short transit delay that is the same for all binary digits. Three comparators are required for the four possible values of a two-bit converter. The fourth only has the function of signaling that the measuring range has been exceeded and supporting code conversion.

#### Multi-stage parallel converters (pipeline converters)

Pipeline converters are multi-stage analog-to-digital converters with several independent stages that are constructed in a pipeline architecture . Their stages usually consist of flash converters over a few bits.

A rough quantization is carried out in each pipeline stage, this value is converted into an analog signal with a DAC and subtracted from the buffered input signal. The residual value is increasingly fed to the next stage. The advantage lies in the greatly reduced number of comparators, e.g. B. 30 for a two-stage eight-bit converter. Furthermore, a higher resolution can be achieved. The multistage increases the latency but does not significantly reduce the sampling rate. The pipeline converters have replaced the real parallel converters except for extremely time-critical applications. These multi-stage converters achieve data rates of 250 MSPS (Mega Samples Per Second) with a resolution of 12 bits (MAX1215, AD9480) or a resolution of 16 bits with 200 MSPS (ADS5485).

The values of the quantization levels are added, taking their weighting into account. Most of the time, a correction ROM also contains calibration data which are used to correct errors that arise in the individual digitization stages. In some versions, these correction data are also generated in response to an external signal and stored in a RAM .

Pipeline converters are typically used in all digital oscilloscopes and when digitizing video signals. As an example, the MAX109 enables a sampling rate of 2.2 GHz with a resolution of 8 bits. In the meantime, however, there are even faster (4 GSPS) and more precise converters (16 bit @ 1 GSPS). In today's digital oscilloscopes with possible sampling rates of 240 GSPS, demultiplexers are also connected upstream.

### Hybrid converter

A hybrid converter is not an independent converter, but rather it combines two or more conversion methods, for example based on a SAR structure, with the original comparator being replaced by a flash converter. As a result, several bits can be determined simultaneously in each approximation step.

## Market situation

There are essentially four processes on the market:

• Pipeline converters are used for high sampling rates. Speeds 40 MSPS to 5 GSPS. Usual dynamics 8 bit, 12 bit (up to 4 GSPS) or 16 bit (up to 1 GSPS).
• If high accuracy with moderate sampling rates is important and latency is not an issue, sigma-delta converters are used. Speeds from a few PLCs to 2.5 MSPS. Bit depths 16 bits to 24 bits.
• If the latency is significant or interferes with the comparatively steep low-pass behavior, successively approximating converters are used. Speeds 0.1 MSPS to 10 MSPS.
• In simple display measuring devices such as multimeters, counting methods that dampen slow interference, such as dual-slope converters, are used.

With this procedure you can cover almost all practical requirements and with moderate requirements (e.g. 12 bit, 125 KSPS, 4 channels) these converters can be obtained inexpensively (approx. 1 €).

## Important parameters

• Sampling (sample rate) - indication of the frequency of transposition.
• Resolution (Resolution) - width of the steps (even number of stages, or number of points), which are used to display the output signal.
• Zero point error - The converter characteristic (without taking the graduation into account) is shifted. The digital value differs from the correct value by a constant amount.
• Sensitivity error , gain error - The converter characteristic (without taking the gradation into account) is twisted (slope error). The digital value differs from the correct value by a constant percentage of the input variable.
• Integral non-linearity - the error due to the fact that a converter characteristic curve, which is taken as a linear basis (without taking the gradation into account), is not straight.
• Differential non-linearity - deviation of the width of the conversion stages from one another
• Quantization curve - graphical representation of the relationship between the digital output values and the analog input values, e.g. B. following a linear or logarithmic course.
• Quantization error - Due to the limited resolution, the output signal deviates from the functional (continuous) curve.
• Information gap (missing code) - If a continuous increase in the input signal does not result in consecutively numbered values of the output code, but rather a value is skipped; possible with a differential non-linearity of more than 1 LSB.
• Latency - propagation delay from the acquisition of the input signal to the provision of the associated output signal.
• Signal-to-noise ratio in dB
• Dynamic range in dB
• Dynamic parameters
• Intermodulation interference in dB
• Operating current - the faster the converter electronics have to work, the higher its supply current (self-heating).

## literature

• Rudy J. van de Plassche: CMOS integrated analog-to-digital and digital-to-analog converters. 2nd edition. Kluwer Academic, Boston 2003, ISBN 1-4020-7500-6 (English)
• Ulrich Tietze, Christoph Schenk: Semiconductor circuit technology. 12th edition. Springer, Heidelberg 2002, ISBN 3-540-42849-6
• Rüdiger Klein, the new workbook electronics. Bercker, Kevelaer, ISBN 978-3-645-65094-6


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• Characterization of Hamamatsu 64-channel TSV SiPMs(1804.00897)

April 3, 2018 physics.ins-det, astro-ph.IM
The Hamamatsu UV-light enhanced 64-channel SiPM array of the newest generation (S13361-3050AS-08) has been examined for the purpose of being used for the Silicon Elementary Cell Add-on (SiECA) of the EUSO-SPB balloon experiment. Characterization measurements have been performed with the newly installed Single Photon Calibration Stand at KIT (SPOCK). The results of the characterization measurements including the breakdown voltage, the gain, the PDE, the dark-count rate and the crosstalk probability of all 64 SiPM channels are presented. Additional measurements of the SiPM sensitivity to photons with wavelengths lower than 400nm show an improved PDE for small wavelengths compared to the SiPM array S12642-0808PA-50, which was also investigated for comparison. The response dynamics have been investigated for low numbers of incoming photons. Temperature dependent measurements of the gain, the PDE, the dark-count rate and the crosstalk probability have been performed for one channel of the SiPM array.
• We measure the energy emitted by extensive air showers in the form of radio emission in the frequency range from 30 to 80 MHz. Exploiting the accurate energy scale of the Pierre Auger Observatory, we obtain a radiation energy of 15.8 \pm 0.7 (stat) \pm 6.7 (sys) MeV for cosmic rays with an energy of 1 EeV arriving perpendicularly to a geomagnetic field of 0.24 G, scaling quadratically with the cosmic-ray energy. A comparison with predictions from state-of-the-art first-principle calculations shows agreement with our measurement. The radiation energy provides direct access to the calorimetric energy in the electromagnetic cascade of extensive air showers. Comparison with our result thus allows the direct calibration of any cosmic-ray radio detector against the well-established energy scale of the Pierre Auger Observatory.
• Neutrinos in the cosmic ray flux with energies near 1 EeV and above are detectable with the Surface Detector array of the Pierre Auger Observatory. We report here on searches through Auger data from 1 January 2004 until 20 June 2013. No neutrino candidates were found, yielding a limit to the diffuse flux of ultra-high energy neutrinos that challenges the Waxman-Bahcall bound predictions. Neutrino identification is attempted using the broad time-structure of the signals expected in the SD stations, and is efficiently done for neutrinos of all flavors interacting in the atmosphere at large zenith angles, as well as for "Earth-skimming" neutrino interactions in the case of tau neutrinos. In this paper the searches for downward-going neutrinos in the zenith angle bins $60^\circ-75^\circ$ and $75^\circ-90^\circ$ as well as for upward-going neutrinos, are combined to give a single limit. The $90\%$ C.L. single-flavor limit to the diffuse flux of ultra-high energy neutrinos with an $E^{-2}$ spectrum in the energy range $1.0 \times 10^{17}$ eV - $2.5 \times 10^{19}$ eV is $E_\nu^2 dN_\nu/dE_\nu < 6.4 \times 10^{-9}~ {\rm GeV~ cm^{-2}~ s^{-1}~ sr^{-1}}$.
• A measurement of the cosmic-ray spectrum for energies exceeding $4{\times}10^{18}$ eV is presented, which is based on the analysis of showers with zenith angles greater than $60^{\circ}$ detected with the Pierre Auger Observatory between 1 January 2004 and 31 December 2013. The measured spectrum confirms a flux suppression at the highest energies. Above $5.3{\times}10^{18}$ eV, the "ankle", the flux can be described by a power law $E^{-\gamma}$ with index $\gamma=2.70 \pm 0.02 \,\text{(stat)} \pm 0.1\,\text{(sys)}$ followed by a smooth suppression region. For the energy ($E_\text{s}$) at which the spectral flux has fallen to one-half of its extrapolated value in the absence of suppression, we find $E_\text{s}=(5.12\pm0.25\,\text{(stat)}^{+1.0}_{-1.2}\,\text{(sys)}){\times}10^{19}$ eV.


text:
The School has a very sizable teaching load, and each semester it hires 30-45 undergraduates to TA for its courses. This is a paying job, and can frequently be a very rewarding experience.

Graduate students are not eligible for these positions.

### Fall 2017 Undergraduate Teaching and Lecture Assistant Positions

UTAs are not hired by course.  Because class availabilty varies term by term, we expect teaching assistants to be comfortable teaching various courses.  ULAs will assist with differential or multivariable calculus, or linear algebra.  The minimum requirements at the time of application are:

• Completion of or concurrently enrolled in Multivariable Calculus/Calculus III (Math 2551 or 2550 or 2401 or 2605) or Differential Equations (Math 2552 or 2403) for UTA positions
• Completion of Integral Calculus or Calculus II (Math 1552 or 1502/1512) or Linear Algebra (Math 1553 or 1554 or 1564 or 1522 or Math 2406) for ULA positions; additionally, all 1000-level mathematics courses (1551, 1552, 1553) should be completed by the end of the spring term
• A high overall GPA with grades of all A's and B's (preferably A's) in mathematics courses
• Excellent people and communication skills
• Fall 2017 will be the beginning of at least your second year as a college student (no first-year students are eligible for UTA or ULA positions)

Most spring UTAs/ULAs are selected from the fall UTA/ULA team, but there are sometimes a few positions for new UTAs/ULAs during the spring term.

Applications are now closed for the 2017-2018 academic year.  Applications for the Fall 2018 term will open in December, 2017.

Please do not search for old links as the application materials change each year.

An undergraduate teaching assistant (UTA) has many responsibilities.

1. Teaching two 50-minute recitation sessions each week.
2. Grading homework, quiz/exam papers, and projects (including proctoring and grading during the final exams week) for the lead instructor.
3. Holding office hours (at least one hour per week).
4. Attending organizational meetings with the lead instructor.
5. Attending the CETL/Mathematics TA Orientation (tentatively scheduled on Sunday afternoon, August 20).
6. Enrolling in CETL 2000 MAT, TA Development Seminar (11:05 am -11:55 am on Tuesdays or Thursdays during the semester).

An undergraduate lecture assistant (ULA) has slightly different responsibilities.

1. Attending lecture and assisting the instructor during lecture once per week.
2. Grading homework, quiz/exam papers, and projects (including proctoring and grading during the final exams week) for the lead instructor.
3. Holding office hours (two to three hours per week).
4. Holding regularly scheduled review sessions.
5. Attending organizational meetings with the lead instructor.
6. Attending the CETL/Mathematics TA Orientation (tentatively scheduled on Sunday afternoon, August 20).
7. Enrolling in CETL 2001 (11:05-11:55 on Tuesdays or Thursdays during the semester).

The stipend for the 2017-2018 academic year will be approximately \$1,490 per semester for UTAs teaching 4-credit hour courses and \$1,345 per semester for UTAs and ULAs teaching 2-credit hour courses.

Upon a satisfactory completion of the UTA duties, a UTA will receive one credit hour (P/F) of CETL 2000 MAT and a ULA will receive one credit hour (P/F) of CETL 2001.

Go to the UTA/ULA Application page

Please note: ULA applicants will fill out the same application materials as the UTA applicants.  Please see the problem set instructions for the differences in the application process.


text:
# Lesson 11

Probabilities in Games

## 11.1: Rock, Paper, Scissors (20 minutes)

### Optional activity

The mathematical purpose of this activity is for students to collect data, look for patterns in data, and to use probability to decide if events are independent.

### Launch

Arrange students in groups of two. Invite two students to demonstrate the game “Rock, Paper, Scissors.”

Students should be encouraged to think of their own hypotheses to study, but here are some examples of things for students who struggle to think of something.

• One of the options (for example, rock) tends to result in a win for me.
• The person who won last tends to win again more often.
• Playing the same option two games in a row tends to win more often on the second time.

### Student Facing

There is a classic game called “Rock, Paper, Scissors.” Two people play by counting to 3 together, then making a hand gesture to resemble paper (hand flat, palm down), rock (fist), or scissors (two fingers extended).

• When paper and rock are shown, paper wins the round.
• When paper and scissors are shown, scissors wins the round.
• When rock and scissors are shown, rock wins the round.
• When both players show the same thing, the round is a tie.

Find a partner and play the game with them 10 times in a row. Record the number of times you have played the game, the name of your opponent, what each person shows in each round, and who is the winner. Find another partner and play another 10 times in a row.

Is the event “win the round” dependent on another event? Explain your reasoning.

Choose an event that you think might influence the probability of winning, then analyze the data using probability to determine whether the event you chose to study is independent of winning. Provide evidence to support your claim.

### Activity Synthesis

The purpose of this discussion is for students to communicate how they used mathematics to justify their findings.

• “Which events did you choose to compare?” (I chose to compare the event of me displaying paper and winning.)
• “How did you determine whether or not the two events are independent?” (I computed $$P(\text{win | played rock})$$ and compared that to $$P(\text{win})$$. Since the probabilities were not close to the same, I think the events might be dependent.)
• “What was challenging or interesting about this activity?” (I thought it was interesting because we collected our own data and actually used probability to answer questions.)
Speaking, Representing: MLR8 Discussion Supports. Use this routine to support whole-class discussion. After each student shares, provide the class with the following sentence frames to help them respond: "I agree because . . .” or "I disagree because . . . .” If necessary, revoice student ideas to demonstrate mathematical language use by restating a statement as a question in order to clarify, apply appropriate language, and involve more students.
Design Principle(s): Support sense-making

## 11.2: Guess Which Card (20 minutes)

### Optional activity

The mathematical purpose of this activity is for students to apply conditional probability in the context of a problem. It is important that students use the ruler to draw the X on each card so that slight differences in the way the Xs are drawn are not recognizable.

### Launch

Arrange students in groups of 2. Give 3 cards and one ruler to each group.

Demonstrate the game by playing one round with the students.

Tell students, "The first card will be blank on both sides, the second card will have an X on one side, and the third card will have an X on both sides. It is important to use the ruler when drawing the Xs on the cards since you should not be able to guess the card based on small differences in how the X is drawn."

Place all three cards in the bag. Remove one and hold it against the wall so that only one side can be seen by the students. Ask them which card they think this card is based on what they are seeing. Is it the blank card, the card with one X on one side, or the card with an X on both sides?

Engagement: Provide Access by Recruiting Interest. Begin with a small-group or whole-class demonstration of how to play the game. Check for understanding by inviting students to rephrase directions in their own words.
Supports accessibility for: Memory; Conceptual processing

### Student Facing

Your teacher will give you 3 index cards.

• Leave one card blank.
• Use a ruler to draw lines connecting opposite corners to make an X on one side of the second card.
• Use a ruler to draw lines connecting opposite corners to make an X on both sides of the third card.
• Put all three cards in the bag.

One partner will remove a card from the bag and place it on the desk immediately so that only one side of the card can be seen. The goal is to guess correctly which card is on the desk: the blank card, the card with an X only on one side, or the card with an X on both sides.

1. Noah is playing the game and is looking at a card that shows an X. He says, “I have a fifty-fifty chance of correctly guessing which card it is.” Do you agree with Noah? Explain your reasoning.
2. Play the game many times with your partner, taking turns for who takes a card out of the bag. Record whether the side that shows has an X or is blank, which card you guess, and which card it actually is when you check. Continue to play until your teacher tells you to move on.
3. Use the results from your games to estimate $$P(\text{the card had an X on both sides | the side showing had an X on it})$$. Explain or show your reasoning.
4. Since all of the outcomes are known, find the actual probability $$P(\text{the card had an X on both sides | the side showing had an X on it})$$. Explain or show your reasoning.

### Anticipated Misconceptions

Some students may have difficulty figuring out the answer to question 4. Prompt students to think about the total number of sides rather than the number of cards. Emphasize that the number of sides are the sample space, not the number of cards.

### Activity Synthesis

The purpose of this discussion is for students to communicate their reasoning with each other.

Poll the class to collect information on about what was shown and which card it turned out to be.

both sides blank one X two Xs
blank side shown
X shown

Students should notice that two of the positions (top right and bottom left) must contain zeros.

Here are some questions for discussion.

• “How did you figure out the answer to question 4?” (I figured out that it was $$\frac{2}{3}$$ since $$P(\text{X on both sides | saw an X}) = \frac{P(\text{X on both sides and saw an X})}{P(\text{saw an X})}$$. We know $$P(\text{the card had an X on both sides and the side showing had an X on it}) = \frac{2}{6}$$ (there are 6 sides that could be showing and 2 of them are Xs from the card with an X on both sides) and $$P(\text{the side showing had an X on it}) = \frac{3}{6}$$ (there are 6 sides that could be showing and 3 of them have an X). Then we can find $$P(\text{the card had an X on both sides | the side showing had an X on it}) = \frac{ \left( \frac{2}{6} \right)}{\left( \frac{3}{6} \right)}$$ which is $$\frac{2}{3}$$.
• “Do you think that playing the game 100 times in question 2 would have helped you arrive at the answer to question 4? Explain your reasoning.” (I think it might have made me feel more confident in my answer to question 4. Since my group only played the game 12 times, it really did not give me much insight. I wonder what would have happened if we played it 100 times. Maybe we could combine the data from the whole class to get a better ideas of what would happen in the long run.)
Speaking: MLR8 Discussion Supports. Use this routine to support whole-class discussion. For each observation that is shared, ask students to restate what they heard using precise mathematical language. Consider providing students time to restate what they hear to a partner before selecting one or two students to share with the class. Ask the original speaker if their peer was accurately able to restate their thinking. Call students' attention to any words or phrases that helped to clarify the original statement. This provides more students with an opportunity to produce language as they interpret the reasoning of others.
Design Principle(s): Support sense-making

## Lesson Synthesis

### Lesson Synthesis

Here are some questions for discussion.

• “What did you find important, challenging, or otherwise notable about today’s lesson?” (It was really interesting to collect data to answer a question. I really felt like I was able to better understand the concept of probability.)
• “What mathematical work from this unit did you use in today’s activity?” (I used probability, conditional probability, the addition rule, and relative frequencies.)
• “How might the mathematical tools that you have learned in this unit help you outside of mathematics class?” (I think that when I see data that compares data using percentages in the news that I will really have to stop and think about the different factors that might account for any differences. I wonder why I never hear about independent and dependent events when I look at data in the news. I think it would be really interesting if the graphs or tables used by the news had to explain them using the concept of probability.)


text:
Rules for Solving the Problems based on Time and Work

1) If a person can complete a work in n days, then in 1 day he does nth of the work. We have

$A\ Person's \ 1\ Day\ Work=$ $\frac{1}{Total\ number\ of\ days\ to\ complete\ the\ work}$

2) If a person completes of a work in 1 day, then he requires n number of days to complete the work

$Total\ no. \ of\ days\ to\ complete\ the\ work=$ $\frac{1}{A \ Person's \ 1\ Day\ Work}$

3) If part of the work has been done and the remaining part of the work has to be done by a person A, then

$No. \ of\ days \ to\ complete\ the\ work=$ $\frac{Remaining\ Part\ of\ the\ Work}{A's\ 1\ Day\ Work}$

4) If $A\ is\ m$ times as good ·a workman as $B$, then

• i) Ratio of work done by $A \ and\ B$ in the same time $= m : 1$
• ii) Ratio of time taken by $A \ and\ B$ to finish the work $=1: m$

5) If some persons work together to do a piece of work, then the money paid to them is divided among them in the ratio of their one day’s (or one hour’s) work.