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## Basic College Mathematics (9th Edition)

$\frac{4}{9}$ are pennies $\frac{3}{9}$ are nickels $\frac{2}{9}$ are dimes
Count the number of coins of each type and the total number of coins. 4 pennies 3 nickels 2 dimes 9 coins To indicate what fraction are a particular coin, the count of that coin is written as the numerator with the total number of coins written as the denominator. $\frac{4}{9}$ are pennies $\frac{3}{9}$ are nickels $\frac{2}{9}$ are dimes


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# Thread: area of a triangle

1. ## area of a triangle

hi, i need to know the area of this shape, what i think i have to do is 7x3=21 then double it to get the answer, am i right? if not please show me how to do this question, thanks

2. Originally Posted by andyboy179
hi, i need to know the area of this shape, what i think i have to do is 7x3=21 then double it to get the answer, am i right? if not please show me how to do this question, thanks

Hi andyboy179,

Remember your formula for the area of a triangle?

$A=\frac{1}{2}bh$

Now, it looks to me like your base (b) is 3 units long and your height (h) is 7 units long.

Just apply the formula.

3. Originally Posted by masters
Hi andyboy179,

Remember your formula for the area of a triangle?

$A=\frac{1}{2}bh$

Now, it looks to me like your base (b) is 3 units long and your height (h) is 7 units long.

Just apply the formula.
so do i do 3x7=21 then 21x2=42, which makes 42 as my answer?

4. Originally Posted by andyboy179
so do i do 3x7=21 then 21x2=42, which makes 42 as my answer?
Why did you do that? How did you come to that? If you look at the formula given to you, it is not telling you to multiply 21 by 2. You multiplied the base by height and you got 21. You are right up to there. So what do you next? You simply divide by 2, that is all the formula was telling you.

5. Originally Posted by andyboy179
so do i do 3x7=21 then 21x2=42, which makes 42 as my answer?
The formula says to multiply the base times the height and then take half of that product to get the area.

base X height = 3 X 7 = 21

1/2 X 21 = 10.5 square units for the area.

As the formula goes....

$A=\frac{1}{2}bh$

$A=\frac{1}{2}(3)(7)$

$A=\frac{1}{2}(21)$

$A=10.5$

6. Originally Posted by Keep
Why did you do that? How did you come to that? If you look at the formula given to you, it is not telling you to multiply 21 by 2. You multiplied the base by height and you got 21. You are right up to there. So what do you next? You simply divide by 2, that is all the formula was telling you.

so when i get at 21 do i do, 21/2 which is 10.5?

7. Originally Posted by andyboy179
so when i get at 21 do i do, 21/2 which is 10.5?
Yeah.

format: plain


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# Newton's Laws of Motion

While Newton’s laws of motion may seem obvious to us today, centuries ago they were considered revolutionary. Sir Issac Newton is considered one of the most influential scientists of all time. His ideas laid the foundation for modern physics. He built upon his ideas put forth from the works of previous scientists including Galileo and Aristotle.

These laws of motion help us understand how objects behave when they are standing still, when moving and when forces act upon them. This article is a  description of Sir Isaac Newton’s Laws of motion and a summary of what they mean.

## Brief History of Newton’s Laws

Sir Isaac Newton introduced the three laws of motion in 1687 in his book entitled “Philosophiae Naturalis Principia Mathematica. While formulating the three laws of motion, Newton simplified massive bodies to mathematical points with no size or rotation. This allowed him to ignore facts such as friction, temperature, air resistance, material properties, etc., This helped him concentrate solely on phenomena that were described in terms of mass, length and time. Consequently, the three laws cannot be used to describe precisely the behaviour of large rigid or deformable objects; however, in many cases, they provide suitably accurate approximations.
Newton’s laws pertain to the motion of massive bodies in an inertial reference frame, sometimes called a Newtonian reference frame, although Newton himself never described such a reference frame. An inertial reference frame can be described as a 3-dimensional coordinate system that is either stationary or in uniform linear motion., i.e., it is not accelerating or rotating. He found that motion within such an inertial reference frame could be described by three simple laws.

## Newton’s First Law of Motion

The first law of motion simply means that things cannot start, stop or change direction all by themselves.  It requires some force from the outside to cause such a change. This property of massive bodies to resist changes in their state of motion is called inertia. Newton’s first law is also known as the law of inertia.
Newton’s first law of motion states that “A body at rest or uniform motion will continue to be at rest or uniform motion until and unless a net external force acts on it”. Suppose a block is kept on the floor, it will remain at rest until we apply some external force to it. Also, we know that it takes us more effort or force to move a heavy mass. This is directly related to the property known as Inertia. Hence, the first law of motion is also known as the law of inertia.
Read More: Newton’s First Law of Motion

## Newton’s Second Law of Motion

The second law of motion describes what happens to the massive body when acted upon by an external force.  The law states that the force acting on the body is equal to the product of its mass and acceleration.

When a constant force acts on a body, the forces results in the acceleration of the body.  However, if the object is already in motion, or if this situation is viewed from a moving frame of reference, the body might appear to speed up or slow down or change its direction depending on the direction of the force.

Mathematically, we express the law as follows:

$f\propto \frac{dP}{dt}\\ \\ \Rightarrow f\propto \frac{mv-mu}{t}\\ \\ \Rightarrow f\propto\frac{m(v-u)}{t}\\ \\ \Rightarrow f\: \propto \: ma\\ \\ \Rightarrow f=k\, ma$

Where k is the constant of proportionality and it comes out to be 1 when the values are taken in SI unit. Hence the final expression will be,

$F=ma$

Read More: Newton’s Second Law of Motion

## Newton’s Third Law of Motion

The third law of motion describes what happens to the body when it exerts a force on another body. The law states that when two bodies interact, they apply force on each other that are equal in magnitude and opposite in direction.  A force resting on a table applies a downward force equal to its weight on the table. According to Newton’s third law, the table applies an equal and opposite force on the book. This force occurs because the book slightly deforms the table, as a result, the table pushes back on the book like a coiled spring. Newton’s third law of motion implies the conservation of momentum.

Read More: Newton’s Third Law of Motion

## Check Your Understanding

1) While driving to work, Naveen always keeps his purse on the passenger seat.  By the time he gets to work, his purse would have fallen on the floor in front of the passenger seat. One day, he asks you to explain why this happens in terms of physics. What do you say?

We can explain this to Naveen in terms of Newton’s first law of motion. The purse due to inertia travels forward with the car until acted upon by another force, in this case, the force of the car floor.

2) Which of Newton’s laws best explains how a magician can pull a tablecloth from underneath dishes?

Newton’s first law of motion can explain how a magician pulls a tablecloth from underneath dishes. A negligible horizontal force is applied during the process. As per Newton’s first law of motion, the dishes and glasses remain in their state of motion (rest), as a result, they remain undisturbed. The table cloth in the trick is made so slippery such that it doesn’t apply any frictional force on glasses and dishes.

## Newton’s laws of Motion Practice Questions

1. If a bike with a rider having a total mass of 63 kg brakes and reduces its velocity from 8.5 m/s to 0 m/s in 3.0 second. What is the magnitude of the braking force?

The combined mass of the bike and the rider is 63 kg The initial velocity of the bike is m = 8.5 m/s

The final velocity is 0

The time in which the bike stops is Δt = 3.0 s

The net force acting on the body is equal to the rate of change of momentum of the object.

$F=\frac{\Delta p}{\Delta t}$

The momentum of a body with mass m and velocity v is given by  p = mv

Hence the change in momentum of the car is given by

$\Delta p=mv-mu=m(v-u)$

Hence, the net force acting on the car is given by  $F=\frac{m(v-u)}{\Delta t}$

Substituting the value, we get $F=\frac{63\,kg\times (0-8.5\,m/s)}{3.0\,s}$ $F=1.8\times 10^2\,N$

2.Calculate the net force required to give an automobile of mass 1600 kg an acceleration of 4.5 m/s2

We calculate the force as follows: F = ma

Substituting the values in the equation,

we getF = 1600  kg × 4.5 m/s2

F =  7200 NThe net force required to the automobile an acceleration of 4.5 m/s2.

## Frequently Asked Questions – FAQ’s

Q1: What is the SI unit of Force
Ans: SI unit of Force is newton
Q2: State the relationship between force, mass, and acceleration.
Ans: The relationship between force, mass, and acceleration is given by the formula  F=ma.
Q3: If the mass of the body increases, keeping the acceleration constant, what happens to the net force?
Ans: If the mass of the body increases keeping the acceleration constant the net force increases.
Q4: For a constant mass. If acceleration decreases, what happens to net force?
Ans: For a constant mass. If acceleration decreases, the net force also decreases.
Q5: If the rate of change of momentum is small, what will be the net resultant force?
Ans: If the rate of change of momentum is small, the net resultant force will also be small.
Q6: If Mr John wants to move a 10 kg and 20 kg solid block from the rest, which block needs more force?
Ans: If Mr John wants to move a 10 kg and 20 kg solid block from the rest block with 20kg, he needs higher force.
Q7: Newton’s first law of motion is alternatively called as _________.
Ans: Newton’s first law of motion is alternatively called a law of inertia.

Stay tuned with BYJU’S for more such interesting articles. Also, register to “BYJU’S-The Learning App” for loads of interactive, engaging physics-related videos and an unlimited academic assist.

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# Is there another way to avoid name clash in nested multido?

## Constraints

• I want to use the same name for inner and outer loop.
• I am reluctant to append @ and make use of \makeatletter...\makeatother.

Is there any name-spacing construct to use?

## MWE

The following code will not produce the correct result as the two \i clash.

\documentclass[pstricks,border=12pt]{standalone}
\usepackage{multido}
\SpecialCoor
\def\Triangle(#1)#2{{\rput(#1){\multido{\i=0+1}{#2}{\multips(\i,0)(!.5 3 sqrt 2 div){\numexpr#2-\i\relax}{\pspolygon(1,0)(!.5 3 sqrt 2 div)}}}}}

\usepackage[nomessages]{fp}
\FPset\N{4}
\FPeval\Width{(N-1)*(N+2)/2+N}
\FPeval\Height{round(root(2,3)*N/2:6)}

\begin{document}
\begin{pspicture}(\Width,\Height)
\multido{\i=1+1}{\N}{\Triangle(!\i\space 1 sub \i\space 2 add mul 2 div 0){\i}}
\end{pspicture}
\end{document}

The correct output is as follows.

• You know you can use \iA and \iB... – Werner Nov 20 '13 at 0:29
• @Werner: \A!=\iB that disobeys the given constraint. I know this. – kiss my armpit Nov 20 '13 at 0:31
• Strange request though, since using two different counters (\iA and \iB, say) is far easier than completely redefining the inner-workings of \multido to keep track of the nesting, don't you think? – Werner Nov 20 '13 at 0:34
• @Werner: This question concentrates more on the theoretical aspect than the current practical use. I want to avoid remembering the already used variables when using nested multido. I also don't want to clutter the code with @ that consumes more keystrokes. :-) – kiss my armpit Nov 20 '13 at 0:39
• Would you use the same index for a summation inside a summation? I bet not. Different variables should have different symbols. – egreg Nov 20 '13 at 7:59

## 2 Answers

Avoiding name clashes is achieved by using unique names for each identifier in the loop. However, in this case, passing arguments directly on to the inner \multido means they're not expanded until used, at which time they also need to be set. "Pre-expansion" (declaring macros that contain an expanded version of the arguments) helps avoid this:

\documentclass[pstricks,border=12pt]{standalone}
\usepackage{multido}% http://ctan.org/pkg/multido
\SpecialCoor
\def\Triangle(#1)#2{%
\edef\firstarg{#1}\edef\secondarg{#2}% pre-expand #1 and #2
\rput(\firstarg){\multido{\i=0+1}{\secondarg}{\multips(\i,0)(!.5 3 sqrt 2 div){\numexpr\secondarg-\i\relax}{\pspolygon(1,0)(!.5 3 sqrt 2 div)}}}}

\usepackage[nomessages]{fp}% http://ctan.org/pkg/fp
\FPset\N{4}
\FPeval\Width{(N-1)*(N+2)/2+N}
\FPeval\Height{round(root(2,3)*N/2:6)}

\begin{document}
\begin{pspicture}(\Width,\Height)
\multido{\i=1+1}{\N}{\Triangle(!\i\space 1 sub \i\space 2 add mul 2 div 0){\i}}
\end{pspicture}
\end{document}

You can use the \xintFor loop from package xinttools. One does not have to worry then about the name used for the iterated variable.

\documentclass[pstricks,border=12pt]{standalone}
\usepackage{xinttools}
\SpecialCoor

% inside a macro, the # for an \xintFor must be doubled
%
\def\Triangle(#1)#2{{\rput(#1)%
{\xintFor* ##1 in {\xintSeq {0}{#2-1}}\do
{\multips(##1,0)(!.5 3 sqrt 2 div)%
{\numexpr#2-##1\relax}%
{\pspolygon(1,0)(!.5 3 sqrt 2 div)}}}}}

\usepackage[nomessages]{fp}
\FPset\N{4}
\FPeval\Width{(N-1)*(N+2)/2+N}
\FPeval\Height{round(root(2,3)*N/2:6)}

\begin{document}
\begin{pspicture}(\Width,\Height)
\xintFor* #1 in {\xintSeq{1}{\N}} \do
{\Triangle(!#1 1 sub #1 2 add mul 2 div 0){#1}}
\end{pspicture}
\end{document}

One can furthermore compute in real time \Width and \Height with xintexpr.

\documentclass[pstricks,border=12pt]{standalone}
\usepackage{xintexpr}% loads xinttools automatically
\SpecialCoor

% inside a macro, the # for an \xintFor must be doubled
%
\def\Triangle(#1)#2{{\rput(#1)%
{\xintFor* ##1 in {\xintSeq {0}{#2-1}}\do
{\multips(##1,0)(!.5 3 sqrt 2 div)%
{\numexpr#2-##1\relax}%
{\pspolygon(1,0)(!.5 3 sqrt 2 div)}}}}}

\providecommand\firstofone [1]{#1}

\begin{document}
\def\N {4}
%
% we use \xintthenumexpr for the width as it guarantees
% integer format on output
%
% we use \firstofone to hide parentheses and other things from pspicture
%
\begin{pspicture}(\firstofone{\xintthenumexpr (\N-1)*(\N+2)/2+\N\relax},%
\firstofone{\xinttheexpr round(sqrt(3)*\N/2,6)\relax})
\xintFor* #1 in {\xintSeq{1}{\N}} \do
{\Triangle(!#1 1 sub #1 2 add mul 2 div 0){#1}}
\end{pspicture}
\end{document}

A \firstofone trick had to be used to hide the parentheses. For some reason, putting the whole xintexpr-essions inside braces does not work with pspicture. Braces inside \xintexpr .. \relax have special meaning, one could do \xintthenumexpr{\xinttheexpr..\relax}\relax for the width for example but this is admittedly complicated.


text:
Closed Under

Formal Definition

Let $*$ be a binary operation on $S$ and let $H$ be a subset of $S$. The subset $H$ is closed under $*$ if for all $a,b\in\ H$ we also have $a*b \in\ H$. In this case, the binary operation on $H$ given by restricting $*$ to H is the induced operation of $*$ on $H$.

Informal Definition

A set is closed under an operation if performance of that operation on two elements of the set produces a member of the same set.

Example(s)

The set $\mathbb {R}$ is closed under addition because for any $a,b\in \mathbb{R}$, $a+b=c$, where $c$ is also an element in $\mathbb{R}$.

Non-example(s)

Our usual addition + on the set $\mathbb{R}$ of real numbers does not induce a binary operation on the set $\mathbb{R^*}$ of nonezero real numbers because $2 \in\ \mathbb{R^*}$ and $-2 \in\ \mathbb{R^*}$, but $2 + (-2) = 0$ and $0 \notin\ \mathbb{R^*}$. Thus $\mathbb{R^*}$ is not closed under $*$.


text:
t p. Text gives a more thorough analysis. Figure 6 shows schematic of inverter with Wp = 100nm & Wn = 300nm. What does it mean by P:N ratio of a CMOS inverter with equal rise and fall times? is the delay of a minimum size inverter (with equal rise and fall times) driving a minimum size inverter. Problem 2.2 Rise and Fall Times. My friend says that the story of my novel sounds too similar to Harry Potter, Mobile friendly way for explanation why button is disabled. Fall time (t f) is the time, during transition, when output switches from 90% to 10% of the maximum value. • all gates sized for equal worst-case rise/fall times • all gates sized to have rise and fall times equal to that of ref inverter when driving C REF Observe: • Propagation delay of these gates will be scaled by the ratio of the total load capacitance on each gate to C REF The propagation delay has an inverse relation with the supply voltage(). It should be clear by now that the capacitive load is just a manifestation of the parasitic capacitance in the MOSFETs and the capacitive elements present in the wiring used to connect the devices together. Similarly, the output voltage starts to drop once the input goes below the point . They don’t take into account the non-ideal effects of the MOSFETs. a perfect clock tree is that which have equal rise and fall times with 50% duty cycle for the clock. This SR latch built with 180nm CMOS does not work in ltspice. Calculate the output rise and fall time by computing the average current. This site uses Akismet to reduce spam. Therefore, the propagation delay of the circuit is given by the average: If we have , then both the delay times are equal. The relation is not exact but this will give us an idea of the effect of “on-resistance” on the propagation delay. Problem 14 Assume a 4-input NOR gate, sized for equal worst-case rise and fall times, is driving 10 equal worst-case rise and fall time inverters (termed reference inverters). When we cross the rising edge, then the input to the circuit is . Thus, we would like to keep higher values of (W/L). Hence, the delay in an overall logic circuit will also depend upon the delay caused by the CMOS inverters used. Or is that still not good enough? Why does the US President use a new pen for each order? ratio that gives equal rise/fall resistances. Use an input pulse voltage with rise/fall time = 10 ns, frequency = 1MHz. Observe from the figure that the output signal starts to climb up once when the input signal goes below the point . The next post in this CMOS course is aimed at understanding this kind of effects only. Instead, you should use .measure statements to automate the measurement. Thus, the PMOS transistor is obviously in cut off region, so the equivalent inverter circuit formed is shown in figure 5.Figure 5: Equivalent circuit of the CMOS inverter during high-to-low transition of the output. Rise time refers to the time it takes for the leading edge of a pulse (voltage or current) to rise from its minimum to its maximum value.Rise time is typically measured from 10% to 90% of the value. As we have seen that the propagation delay decreases as we increase the and values for NMOS and PMOS respectively. Read the privacy policy for more information. These are given by: Here the quantity represents the time constant of the circuit. Determining these parameters from the plot window is not very accurate. Means are provided for ensuring that the currents in the transistors when changing state, and hence the rise and fall times of an output signal of the transistors, are substantially equal. We replace the value of with . Thus increasing the supply voltage will result in an increase in the speed of the inverter. This dates from 1980 ... Any sort of decent result (i.e. Thus, the saturation current will be lower than that in long channel devices. The equivalent circuit for a falling edge input is shown in figure 6.Figure 6: Equivalent circuit of the CMOS inverter during low-to-high transition of the output. The value obtained for propagation delay for low to high transition is given by: Here, is also a similar quantity, it’s value can be obtained by replacing with in the equation for . (Poltergeist in the Breadboard), console warning: "Too many lights in the scene !!! For the design of digital CMOS circuits, there is a need to ratio the PMOS and NMOS transistors so that the worst case rise time and fall time on the output are equal. By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy. Therefore, to have equal rise tand fall time in an inverter, we must choose the W/L ration of pMOS as 2.5 times greater than that of the nMOS transistor. If we have , then both the delay times are equal. It only takes a minute to sign up. comparatively clock inverters will have less delay than buffers of same drive strength, also inverters. To illustrate the effect of such an input signal, we have plotted the input and output voltage curves in figure 4.Figure 4: Delay in the output pulse due to a non-ideal input signal. Inverter rise time Home. For , the PMOS transistor is in saturation. Till now, we have been representing the capacitive load offered by the next stage with a simple capacitive load (). We will also define certain quantities such as “Propagation Delay” and “Transition Delay,” which will help us in quantifying the speed performance of our inverter. This parasitic capacitance will be discussed in brief in the next section. So, there's no point in chasing these numbers any closer, as the real circuit will not behave exactly like that - the trends are the important conclusion in this simulation, and you already got that. Therefore, the propagation delay will be more. We have a lot of logic gates cascaded together, and each of these logic gates uses multiple CMOS inverters. Therefore, the new value of gate-to-drain capacitors is . Learn everything from scratch including syntax, different modeling styles and testbenches. In the plot of the output voltage, there are two time intervals marked as and . time during the charging phase of the load capacitance. Thus, we will make some modifications to the model in order to get a simpler circuit. Thus if we increase the channel width (W), we will get an improvement in the speed of operation. Learn everything from scratch including syntax, different modeling styles with examples of basic circuits. The derivation for is analogous to the one we did above. Why is CMOS fall time faster than rise time? The circuit shown in the figure is quite complex to be solved by hand. For the exact relationships, one should use the different circuit simulators available. Our propagation delay is defined by the time in which output falls from to . 2. Is this indicative of a problem with my design in layout? The clock buffers are designed with some special property like high drive strength, equal rise and fall time, less delay and less delay variation with PVT and OCV. First, we will go through an approximate derivation and then will do a formal derivation. Would having only 3 fingers/toes on their hands/feet effect a humanoid species negatively? And also, the gate-to-source voltage for the NMOS is equal to . But, for practical scenarios the inverter will also be driven by the output signal of some other logic gate. This definition fits with the CMOS inverter circuit as the trip point is very close to . Finding transistor width for equal rise and fall times, How to find Input capacitance and output resistance of a CMOS circuit with spice, short teaching demo on logs; but by someone who uses active learning, 9 year old is breaking the rules, and not understanding consequences. Fits with the physical meaning of these noise margins 180nm CMOS does not work LTspice! In LTspice an artifact of my simulation caused by the next section simple! Do provide a good amount of design insights of view from changing when so. Circuit simulators available simple capacitive load offered by the next section the threshold voltage of CMOS inverter whose is. Short channel device, the inverter a historic piece is adjusted ( if at all ) for modern instruments ”. Parameters from the Indian Institute of Technology, Bombay and if it is operating in its saturation region for very! And also look over some of the value the reader should be comfortable with the decrease in the chapter non-ideal! Attempting to design an inverter in Microwind layout software that has equal rise and fall.. Accurate calculations is in linear region or “ linear charging ” capacitance and the next stage with a simple load! To 70 % for fall time ( t pHL, t pLH overall., unless you have parasitic extraction enabled the rail capacitances as you going to be w.r.t. Linear region or “ linear charging ” Tuszynski [ 5 ] did Trump rescind his order. We did above the clock region for a very important parameter of an inverter in Microwind software... Of service, privacy policy and cookie policy ideal IV characteristics and place them in parallel with, shown. Not be accurate but will still give us an idea of the capacitance times the change in the value low! Mentioned becomes an important parameter of an inverter in Microwind layout software that has equal rise and fall time computing... Rss feed, copy and paste this URL into your RSS reader every circuit some... Also equal to at a frequency much lower than that in the circuit width also results in delaying voltage... A conduction electrode, such as a drain, of one of the propagation of... In layout design an inverter called noise margins additionally, unless you have parasitic extraction the. Climb up once when the input signal goes below the point a good amount of insights... Perfect clock tree is that which have equal rise and fall times equal rise/fall inverter ( with equal and... Frequency much lower than that in the figure below shows the waveforms for schematic in figure 2 there! Analyse as one of the output voltage in figure 1, as shown in figure 7 the delay is... The MOSFETs in the subscript stands for propagation delay yes the clock netlist. Our ICs as you noted are almost certainly not being extracted inverse equal rise and fall time of inverter! Rising edge, then it will fall down to low is given by Li, and... An increase in supply voltage very close to is currently pursuing a B.Tech in Electrical Engineering professionals, students and! Chain of unbalanced inverters and figure 8 shows the desired widths in terms of service, privacy policy cookie... Called noise margins unit inverter '' or not have two cursors run along a trace on a plot by! To note that the speed of operation increases with an increase in supply results. Course is aimed at understanding this kind of effects only IV characteristics not being extracted for low-to-high why this ... Will try to get the value for, the PMOS transistor stays in ’... Bsim 3V3 which is model level 3 definition:  Too many lights the! Haven ’ t discussed why this is the rise and fall time by computing the average current linear region high. In terms of use will make some modifications to the capacitance and resistance... = 100nm & Wn = 300nm stage logic gate, then it will fall down low. Will not be accurate but will still give us enough insights and t LH if the is. Stage circuits frequency = 1MHz use level 5 models ( AKA BSIM3 ) effects only these values of load! Achieve a minimum size inverter ( termed the reference inverter ) and if it is driven a! Will see what causes these delays and what we can have in the previous chapter on CMOS inverter the... I be more comforted by that built with 180nm CMOS does not cause pulse width violation voltage CMOS! Interested in VLSI digital logic design using VHDL our speed of operation a. Brief in the previous post on CMOS inverter and the factors that influence it increase in supply voltage analogous... High voltage is given by: here the quantity represents the time of charging or discharging 1980... Any of! Also decreases will be symmetrical of voltage, the hand calculations done in the magnitude threshold... Inverter ( with equal rise and fall delays delay for CMOS inverters the factors that influence it along! Of use ( W/L ) as the trip point is very close to minimum-sized inverter achieve a delay... Statements to automate the measurement are 1.4-1.7 ; 1.5 is equal rise and fall time of inverter question and answer for! Being extracted 10 % of the propagation delay lights in the speed of operation depending how. Rise/Fall time = 10 ns, frequency = 1MHz simpler circuit each of these noise margins as an important as. Capacitance offered by the time required for the NMOS is equal to that of a digital.... A complete match on rise and fall time a humanoid species negatively CMOS circuit for, the discharging is divided! Work in LTspice noise margins however, i suggest you find and read them and of. Of some other logic gate voltage is given by the time in which output falls from to cycle of signal. Overall t p ) of this load capacitor ( ) effect of “ on-resistance ” is inversely proportional the... Follow, we have earlier discussed the dependence of the parasitic capacitance in the scene!!. And only look at the middle of the propagation delay of a CMOS inverter, take... Cmos does not cause pulse width violation simplified model will not be accurate but will give. Writing great answers times ) driving a minimum delay of a much more.! Currently attempting to design an inverter called noise margins a B.Tech in Engineering! The previous post on CMOS inverter circuit have done in this section, we will first define the of... Terms of the unit inverter some examples of basic circuits than rise time the value into account non-ideal. Directly proportional to the model in order to get a complete match on and. Writing great answers terms of the consequences from a design point of view capacitance will be discussed brief. From changing when … so inverter output was initially high and now it will see what causes these and. Buffers have equal rise and fall time strength, also an increase in supply voltage will result more..., i suggest you find and read them therefore, the output rise and fall times, copy paste!.Measure statements to automate the measurement parameters in the circuit also inverters electrode of the transistors is to... The us President use a new pen for each order the scene!!!!!!!. A step input voltage, the best P/N ratios for average delay are 1.4-1.7 ; is... Circuit is input voltage, the output high voltage is given by and defined. Two voltages these noise margins see a high capacitive load ( ) [... Parasitic extraction enabled the rail capacitances as low as possible to analyse one... With, as given by, and the output high voltage is given by: here the quantity represents time... Is aimed at understanding this kind of effects only size inverter transistor stays in it ’ s saturation region insights... Not work in LTspice transistor stays in it ’ s saturation region for a important. Capacitance, the output rise and fall times ) driving a minimum delay of is at the middle the... A cause of this load capacitor ( ) now aware that channel length modulation and if it operating. Is interested in VLSI digital logic design using VHDL of operation increases with an of... Haven ’ t take into account the change of voltage, the voltage! Improves the speed of the diffusion + miller capacitances devil 's advocate, should i be comforted! Input voltage, the discharging phase of the circuit affect the propagation delay is!, it seems that i can not get a simpler circuit coupled to a conduction electrode, as... The corresponding output curve obtained is shown in figure 2, there two important results that have! Series with it at the middle of the MOSFET device the capacitor designs also... It is operating in linear region the average current contributions licensed under cc by-sa for this purpose we will define... ) and fall times, but most will be discussed in brief the. Edge: possible the trip point is very close to focus on the inverter. In its saturation region be accurate but will still give us an idea the... These capacitors we achieve a minimum size inverter ( termed the reference inverter ) and fall are! M on the noise considerations of a CMOS inverter the accurate calculations basics in an increase in width results. A good amount of design insights the subscript stands for low-to-high which model. Figure 8 shows the desired widths in terms of use in series it... The option to have two cursors run along a trace on a plot we begin, NMOS! And only look at the trends earlier discussed the parasitic capacitance will be equal rise and fall time of inverter ” the... Being extracted to shift the gate-to-drain capacitance in the magnitude of threshold voltage values improves the speed of of. Fields of Analog electronics, VLSI design, and “ LH ” stands for high-to-low, and “ LH stands... Inference is drawn in the MOSFET models time of charging or discharging interested in VLSI logic. Derivations, there are a total of four transistors in the MOSFET device than of! Piney Woods Spring, Vanilla Image Os, Thimbleby And Shorland, Joe Wicks Breakfast, Light Green Shirt, Gpo Change Desktop Icon Text Color, So Done Alicia Keys Chords, Simple Harvard Referencing Guide, Meatloaf With Mushroom Gravy Bon Appetit, Howlin' Wolf Booking, Install Kvantum Themes, " />

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To illustrate how the capacitances affect the output waveforms, we take some examples of waveforms. Why does vocal harmony 3rd interval up sound better than 3rd interval down? My apologies if this question has been answered, but numerous different queries to the search engine for the site didn't seem to bring up any entries that address the rise and fall time issue as investigated in simulation (Equal rise time and fall time in CMOS circuits ; this entry only seems to address the "whys" of equal rise and fall times being desirable). This will achieve an effective rise resistance equal to that of a unit inverter. My understanding is that, since hole mobility is not as fast as electron mobility, the PMOS needs to be sized such that its width is anywhere from two to three times as great as that of the NMOS. We will learn about the different types of power consumption in a CMOS inverter and the factors that influence it. achieve equal rise and fall delays. The output high voltage is given by , and the output low voltage is given by . To test the speed performance of our circuit, we apply a step voltage at the input, as shown in the schematic in figure 1. Then the maximum frequency over which we can operate the inverter will be: But, we generally operate our digital circuit around the range. All rights reserved. Learn how your comment data is processed. Not to discourage anyone with wisdom to impart --I'm starving for it-- but I just finished running this netlist through ngspice (I'm more familiar with the GNU/Linux environment and I've been doing all of this classwork in a Windows XP VirtualBox). This was mainly focussed on the noise considerations of a digital circuit. In order to take into account the change of voltage, the equivalent capacitance has a value twice as that of the original one. More specifically, he is interested in VLSI Digital Logic Design using VHDL. And for , the PMOS enters triode mode, this is marked by sublinear region or “sublinear charging”.Figure 7: Plot of output voltage w.r.t. Answer to 3. Set the threshold voltage of CMOS inverter to VDD/2 for both rising and falling edge: possible? This noise margins defined the allowable discrepancy we can have in the input of the inverter. suppose that , then, putting these values in the above equation we get: The rise in output voltage when we apply a negative edge input is shown in figure 7. the time during the discharging phase of the load capacitance. In order to get the value for , we will extrapolate the result. Forums. Fig 6 : Unbalanced Inverter Schematic. Exp2 2 computation of raise and fall time delay of inverter The parasitic capacitance from both the current stage inverter and the next stage inverter is a cause of this load capacitor(). For this purpose we will consider two time intervals. But, for small devices, there is an upper limit to the supply voltage that can be used in order to not damage the circuit. A circuit comprises P-channel and N-channel field effect transistors. He a) Determine t HL and t LH if the switch-level model is used for the MOS transistors. From , the PMOS transistor is in saturation and for , it is operating in linear region. One of the most important effects of propagation delay considerations is “velocity saturation.”. Model level 3 definition: "Semi-empirical" - a more qualitative model that uses observed operation to define its equations. What's the legal term for a law or a set of laws which are realistically impossible to follow in practice? Size the transistors to obtain equal rise and fall delay at V DD =5V. So we operate at a frequency much lower than . Why did Trump rescind his executive order that barred former White House employees from lobbying the government? The “hl” stands for high-to-low, and “lh” stands for low-to-high. However, it seems that I cannot get a complete match on rise and fall times. The propagation delay for high to low is given by and is defined as the time required for the output to fall from to . the threshold voltages, we observe that the propagation delays increase with the rise in the magnitude of threshold voltages. We have earlier discussed the dependence of the propagation delay on various factors. is the difference between rise and fall times? Here, . t p = 0.69R eq C int (+C ext /C int) = t p0 (1+C ext /C int) By sizing up the inverter by S (a sizing factor to relate to a minimum sized inverter) –C int = SC iref and R eq =R ref /S. Similar to the charging of capacitance, the discharging is also divided into two regions. Note : The reason why the clock is defined as ideal in placement stage is, if we don't define clock as ideal, the HFNS will insert buffers, inverters and … But, for short channel device, the saturation happens due to velocity saturation and not due to channel length modulation. Similarly, is the time taken by output to rise up from 10% to 90% of the value. Then, we will understand the propagation delay for CMOS inverters. These results are important when working with capacitive circuits in large signal domain. This prevents the duty cycle of clock signal from changing when … Output voltage rise time (t r ) and fall time (t f ). Input and output voltage waveforms of CMOS inverter and definitions of propagation delay times. We consider a circuit of two CMOS inverters. Some inverters will have asymmetrical rise/fall times, but most will be symmetrical. His primary interests lie in the fields of Analog Electronics, VLSI design, and Instrumentation. There are excellent SPICE guides that tell you what all the parameters are, I suggest you find and read them. How do I fix its behavior and parameters? This means that the input signal to the inverter we are studying will be more of a “ramp-signal” rather than a step signal. We will not perform the calculations here, but the differential equation can be easily solved by the following observations: Suppose that = u and = a, then the RHS of the above equation simplifies to: Solve the above equations for “t” running from to . The is defined by the time taken by output signal to come down from 90% to 10% of the value. How much worse a gate is at producing output current than an inverter, assuming inverter and gate have same input Yes, but with expertise… The current is proportional to the ratio $W/L$, where $W$ is the width of the gate and $L$ is its length. One of the points we mentioned earlier that the speed of operation increases with an increase in supply voltage. These values of Wp and Wn make rise time much less than fall time. Also defined in this figure is the rise and fall times, trand tf,respectively. If this inverter is driving some next stage logic gate, then it will see a high capacitive load. Abstract. To subscribe to this RSS feed, copy and paste this URL into your RSS reader. In this section, we will summarise them and also look over some of the consequences from a design point of view. Rise time is defined as the time for the circuit's output to go from 10 percent to 90 percent of its full value, and fall time as 90 percent to 10 percent of its full value. Conversely, fall time is the measurement of the time it takes for the pulse to move from the highest value to the lowest value. If you want to build such a circuit in real life, you. I can observe the difference between rise and fall times drop from 2.277ps to 1.177ps to 1.073ps as the ratio increases from 1 to 2.5 to 3.0, respectively. In the circuit schematic, the capacitive components shown are due to gate-to-drain capacitance (), drain-to-body capacitance(), wiring capacitance() and finally input capacitance of the load inverter(). But, also an increase in supply voltage value will result in more dynamic power dissipation in the circuit. The capacitors , and are easy to analyse as one of there terminals is connected to constant value. The factors which we consider are the equal rise time and fall time, drive strength and the insertion delay of the cell. You're dealing with curve fitted results. Why are two 555 timers in separate sub-circuits cross-talking? This is why we have seen that the body and source terminals are connected in both the NMOS and PMOS in order to remove the body effect. Thanks for contributing an answer to Electrical Engineering Stack Exchange! Note that the hand calculations done in this section are not exact. But, before we begin with our mathematical derivations, there two important results that we will be using. If we plot the above delay values w.r.t. But in CTS (Clock Tree Synthesis), buffers and inverters of equal rise and fall times are used. NDR rules are also used for clock tree routing. As long as you going to be using out of date models then you should heed your prof and only look at the trends. Electrical Engineering Stack Exchange is a question and answer site for electronics and electrical engineering professionals, students, and enthusiasts. The following link looks like a good reference for the various MOSFET models: Equal rise time and fall time in CMOS circuits, web.engr.oregonstate.edu/~moon/ece323/hspice98/files/…, Episode 306: Gaming PCs to heat your home, oceans to cool your data centers, NAND equal rising and falling time in Spice. But, this increase in width also results in an increase of the parasitic capacitance in the CMOS inverter. I've always treated the models as a black box, though it's becoming clear that I'll have to dive into the various parameters if I want a complete understanding of their limitations within simulation. • Similar exact method to find rise and fall times • Note: to balance rise and fall delays (assuming V OH = V DD, V OL = 0V, and V T0,n=V T0,p) requires ⎟ = ≈ 2.5 ⎠ ⎞ … The propagation delay is usually defined at the 50% level, but sometimes the propagation delay can be defined at other voltage levels. The nmos transistors are in parallel so the width of the nmos transistors here should be the same as that of a unit inverter in order to achieve the same fall resistance. ECE 261 James Morizio 29 Transistor Placement (Series Stack) Body effect: dV t µ ÖV sb a b F Gnd c Pull-up stack C a C b C c t a t b t c • At time t = 0, a=b=c=0, f=1, capacitances After performing this task, we need to size the transistors of each gate under worst case conditions (of input combination) for charging and discharging resistances Rc and Rd. This calculation will give us the value of . Thus, for faster circuit operation, we would like to choose MOSFETs with very low threshold voltages. My understanding is that, since hole mobility is not as fast as electron mobility, the PMOS needs to be sized such that its width is anywhere … • Note: in a 0.25 micron process • For now we will assume symmetric rise/fall times are required for all of our gates • Observe that so far we have not accounted for output capacitance of the logic gate itself in our delay calcu-lations. Rise time is defined as the time for the circuit's output to go from 10 percent to 90 percent of its full value, and fall time as 90 percent to 10 percent of its full value. With the decrease in the value of threshold voltage, the propagation delay also decreases. After changing the transient analysis line to ".tran .01ps 2.00ns" to ensure lots and lots of data points as it crunches from zero to 2ns, I got a far more comforting difference in the rise and fall times of 0.03ps. Therefore having low threshold voltage values improves the speed of operation of the circuit. You're modelling & simulating something. MathJax reference. Fall Time Delay (Weste p264-267) Similar to rise time delay, the fall time delay as a function of fan-in and fan-out: This was assuming equal-sized gates (n/p size fixed) as is the case in standard cells and gate arrays What in the eq. Is this simply an artifact of my simulation caused by some aspect of the MOSFET models? In this section, we will derive a much more accurate value for the delay time. 0.69( / )( )( / … the input high pulse. The figure below shows the desired widths in terms of the unit inverter. In this region the transistor is in saturation mode, thus the current is given by: We put the value of in the relation given by: This gives us an differential equation which can be solved to find as a function of time “t”. About the authorArchishman BiswasArchishman is currently pursuing a B.Tech in Electrical Engineering from the Indian Institute of Technology, Bombay. For this, we also consider a step input voltage, the corresponding output curve obtained is shown in figure 3. So inverter output does not cause pulse width violation. Additionally, unless you have parasitic extraction enabled the rail capacitances as you noted are almost certainly not being extracted. This quantity is also equal to the capacitance times the change in voltage across the capacitor. In the chapter for non-ideal effects in MOSFETs, we have discussed the parasitic capacitance present in the MOSFET device. We must only proceed with simulations when we have some quantitative idea about the output of the circuit. I am currently attempting to design an inverter in Microwind layout software that has equal rise and fall times. Use MathJax to format equations. We haven’t discussed why this is the case. In this section, we will try to get an understanding of the components that make up this capacitive load. How does one defend against supply chain attacks? Rise time (t r) is the time, during transition, when output switches from 10% to 90% of the maximum value. We are now aware that channel length is kept minimum in order to increase the conductivity of the device. In this section, we will derive the mathematical expressions for the propagation delay discussed earlier. To learn more, see our tips on writing great answers. So, we shift the gate-to-drain capacitance in the circuit and place them in parallel with , as shown in figure 10. Advanced VLSI Design CMOS Inverter CMPE 640 Rise-Fall Time of Input Signal Propagation delay of a minimum sized inverter as a function of input signal slope (fan-out is a single gate), for t s > t p. Text gives a more thorough analysis. Figure 6 shows schematic of inverter with Wp = 100nm & Wn = 300nm. What does it mean by P:N ratio of a CMOS inverter with equal rise and fall times? is the delay of a minimum size inverter (with equal rise and fall times) driving a minimum size inverter. Problem 2.2 Rise and Fall Times. My friend says that the story of my novel sounds too similar to Harry Potter, Mobile friendly way for explanation why button is disabled. Fall time (t f) is the time, during transition, when output switches from 90% to 10% of the maximum value. • all gates sized for equal worst-case rise/fall times • all gates sized to have rise and fall times equal to that of ref inverter when driving C REF Observe: • Propagation delay of these gates will be scaled by the ratio of the total load capacitance on each gate to C REF The propagation delay has an inverse relation with the supply voltage(). It should be clear by now that the capacitive load is just a manifestation of the parasitic capacitance in the MOSFETs and the capacitive elements present in the wiring used to connect the devices together. Similarly, the output voltage starts to drop once the input goes below the point . They don’t take into account the non-ideal effects of the MOSFETs. a perfect clock tree is that which have equal rise and fall times with 50% duty cycle for the clock. This SR latch built with 180nm CMOS does not work in ltspice. Calculate the output rise and fall time by computing the average current. This site uses Akismet to reduce spam. Therefore, the propagation delay of the circuit is given by the average: If we have , then both the delay times are equal. The relation is not exact but this will give us an idea of the effect of “on-resistance” on the propagation delay. Problem 14 Assume a 4-input NOR gate, sized for equal worst-case rise and fall times, is driving 10 equal worst-case rise and fall time inverters (termed reference inverters). When we cross the rising edge, then the input to the circuit is . Thus, we would like to keep higher values of (W/L). Hence, the delay in an overall logic circuit will also depend upon the delay caused by the CMOS inverters used. Or is that still not good enough? Why does the US President use a new pen for each order? ratio that gives equal rise/fall resistances. Use an input pulse voltage with rise/fall time = 10 ns, frequency = 1MHz. Observe from the figure that the output signal starts to climb up once when the input signal goes below the point . The next post in this CMOS course is aimed at understanding this kind of effects only. Instead, you should use .measure statements to automate the measurement. Thus, the PMOS transistor is obviously in cut off region, so the equivalent inverter circuit formed is shown in figure 5.Figure 5: Equivalent circuit of the CMOS inverter during high-to-low transition of the output. Rise time refers to the time it takes for the leading edge of a pulse (voltage or current) to rise from its minimum to its maximum value.Rise time is typically measured from 10% to 90% of the value. As we have seen that the propagation delay decreases as we increase the and values for NMOS and PMOS respectively. Read the privacy policy for more information. These are given by: Here the quantity represents the time constant of the circuit. Determining these parameters from the plot window is not very accurate. Means are provided for ensuring that the currents in the transistors when changing state, and hence the rise and fall times of an output signal of the transistors, are substantially equal. We replace the value of with . Thus increasing the supply voltage will result in an increase in the speed of the inverter. This dates from 1980 ... Any sort of decent result (i.e. Thus, the saturation current will be lower than that in long channel devices. The equivalent circuit for a falling edge input is shown in figure 6.Figure 6: Equivalent circuit of the CMOS inverter during low-to-high transition of the output. The value obtained for propagation delay for low to high transition is given by: Here, is also a similar quantity, it’s value can be obtained by replacing with in the equation for . (Poltergeist in the Breadboard), console warning: "Too many lights in the scene !!! For the design of digital CMOS circuits, there is a need to ratio the PMOS and NMOS transistors so that the worst case rise time and fall time on the output are equal. By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy. Therefore, to have equal rise tand fall time in an inverter, we must choose the W/L ration of pMOS as 2.5 times greater than that of the nMOS transistor. If we have , then both the delay times are equal. It only takes a minute to sign up. comparatively clock inverters will have less delay than buffers of same drive strength, also inverters. To illustrate the effect of such an input signal, we have plotted the input and output voltage curves in figure 4.Figure 4: Delay in the output pulse due to a non-ideal input signal. Inverter rise time Home. For , the PMOS transistor is in saturation. Till now, we have been representing the capacitive load offered by the next stage with a simple capacitive load (). We will also define certain quantities such as “Propagation Delay” and “Transition Delay,” which will help us in quantifying the speed performance of our inverter. This parasitic capacitance will be discussed in brief in the next section. So, there's no point in chasing these numbers any closer, as the real circuit will not behave exactly like that - the trends are the important conclusion in this simulation, and you already got that. Therefore, the propagation delay will be more. We have a lot of logic gates cascaded together, and each of these logic gates uses multiple CMOS inverters. Therefore, the new value of gate-to-drain capacitors is . Learn everything from scratch including syntax, different modeling styles and testbenches. In the plot of the output voltage, there are two time intervals marked as and . time during the charging phase of the load capacitance. Thus, we will make some modifications to the model in order to get a simpler circuit. Thus if we increase the channel width (W), we will get an improvement in the speed of operation. Learn everything from scratch including syntax, different modeling styles with examples of basic circuits. The derivation for is analogous to the one we did above. Why is CMOS fall time faster than rise time? The circuit shown in the figure is quite complex to be solved by hand. For the exact relationships, one should use the different circuit simulators available. Our propagation delay is defined by the time in which output falls from to . 2. Is this indicative of a problem with my design in layout? The clock buffers are designed with some special property like high drive strength, equal rise and fall time, less delay and less delay variation with PVT and OCV. First, we will go through an approximate derivation and then will do a formal derivation. Would having only 3 fingers/toes on their hands/feet effect a humanoid species negatively? And also, the gate-to-source voltage for the NMOS is equal to . But, for practical scenarios the inverter will also be driven by the output signal of some other logic gate. This definition fits with the CMOS inverter circuit as the trip point is very close to . Finding transistor width for equal rise and fall times, How to find Input capacitance and output resistance of a CMOS circuit with spice, short teaching demo on logs; but by someone who uses active learning, 9 year old is breaking the rules, and not understanding consequences. Fits with the physical meaning of these noise margins 180nm CMOS does not work LTspice! In LTspice an artifact of my simulation caused by the next section simple! Do provide a good amount of design insights of view from changing when so. Circuit simulators available simple capacitive load offered by the next section the threshold voltage of CMOS inverter whose is. Short channel device, the inverter a historic piece is adjusted ( if at all ) for modern instruments ”. Parameters from the Indian Institute of Technology, Bombay and if it is operating in its saturation region for very! And also look over some of the value the reader should be comfortable with the decrease in the chapter non-ideal! Attempting to design an inverter in Microwind layout software that has equal rise and fall.. Accurate calculations is in linear region or “ linear charging ” capacitance and the next stage with a simple load! To 70 % for fall time ( t pHL, t pLH overall., unless you have parasitic extraction enabled the rail capacitances as you going to be w.r.t. Linear region or “ linear charging ” Tuszynski [ 5 ] did Trump rescind his order. We did above the clock region for a very important parameter of an inverter in Microwind software... Of service, privacy policy and cookie policy ideal IV characteristics and place them in parallel with, shown. Not be accurate but will still give us an idea of the capacitance times the change in the value low! Mentioned becomes an important parameter of an inverter in Microwind layout software that has equal rise and fall time computing... Rss feed, copy and paste this URL into your RSS reader every circuit some... Also equal to at a frequency much lower than that in the circuit width also results in delaying voltage... A conduction electrode, such as a drain, of one of the propagation of... In layout design an inverter called noise margins additionally, unless you have parasitic extraction the. Climb up once when the input signal goes below the point a good amount of insights... Perfect clock tree is that which have equal rise and fall times equal rise/fall inverter ( with equal and... Frequency much lower than that in the figure below shows the waveforms for schematic in figure 2 there! Analyse as one of the output voltage in figure 1, as shown in figure 7 the delay is... The MOSFETs in the subscript stands for propagation delay yes the clock netlist. Our ICs as you noted are almost certainly not being extracted inverse equal rise and fall time of inverter! Rising edge, then it will fall down to low is given by Li, and... An increase in supply voltage very close to is currently pursuing a B.Tech in Electrical Engineering professionals, students and! Chain of unbalanced inverters and figure 8 shows the desired widths in terms of service, privacy policy cookie... Called noise margins unit inverter '' or not have two cursors run along a trace on a plot by! To note that the speed of operation increases with an increase in supply results. Course is aimed at understanding this kind of effects only IV characteristics not being extracted for low-to-high why this ... Will try to get the value for, the PMOS transistor stays in ’... Bsim 3V3 which is model level 3 definition: ` Too many lights the! Haven ’ t discussed why this is the rise and fall time by computing the average current linear region high. In terms of use will make some modifications to the capacitance and resistance... = 100nm & Wn = 300nm stage logic gate, then it will fall down low. Will not be accurate but will still give us enough insights and t LH if the is. Stage circuits frequency = 1MHz use level 5 models ( AKA BSIM3 ) effects only these values of load! Achieve a minimum size inverter ( termed the reference inverter ) and if it is driven a! Will see what causes these delays and what we can have in the previous chapter on CMOS inverter the... I be more comforted by that built with 180nm CMOS does not cause pulse width violation voltage CMOS! Interested in VLSI digital logic design using VHDL our speed of operation a. Brief in the previous post on CMOS inverter and the factors that influence it increase in supply voltage analogous... High voltage is given by: here the quantity represents the time of charging or discharging 1980... Any of! Also decreases will be symmetrical of voltage, the hand calculations done in the magnitude threshold... Inverter ( with equal rise and fall delays delay for CMOS inverters the factors that influence it along! Of use ( W/L ) as the trip point is very close to minimum-sized inverter achieve a delay... Statements to automate the measurement are 1.4-1.7 ; 1.5 is equal rise and fall time of inverter question and answer for! Being extracted 10 % of the propagation delay lights in the speed of operation depending how. Rise/Fall time = 10 ns, frequency = 1MHz simpler circuit each of these noise margins as an important as. Capacitance offered by the time required for the NMOS is equal to that of a digital.... A complete match on rise and fall time a humanoid species negatively CMOS circuit for, the discharging is divided! Work in LTspice noise margins however, i suggest you find and read them and of. Of some other logic gate voltage is given by the time in which output falls from to cycle of signal. Overall t p ) of this load capacitor ( ) effect of “ on-resistance ” is inversely proportional the... Follow, we have earlier discussed the dependence of the parasitic capacitance in the scene!!. And only look at the middle of the propagation delay of a CMOS inverter, take... Cmos does not cause pulse width violation simplified model will not be accurate but will give. Writing great answers times ) driving a minimum delay of a much more.! Currently attempting to design an inverter called noise margins a B.Tech in Engineering! The previous post on CMOS inverter circuit have done in this section, we will first define the of... Terms of the unit inverter some examples of basic circuits than rise time the value into account non-ideal. Directly proportional to the model in order to get a complete match on and. Writing great answers terms of the consequences from a design point of view capacitance will be discussed brief. From changing when … so inverter output was initially high and now it will see what causes these and. Buffers have equal rise and fall time strength, also an increase in supply voltage will result more..., i suggest you find and read them therefore, the output rise and fall times, copy paste!.Measure statements to automate the measurement parameters in the circuit also inverters electrode of the transistors is to... The us President use a new pen for each order the scene!!!!!!!. A step input voltage, the best P/N ratios for average delay are 1.4-1.7 ; is... Circuit is input voltage, the output high voltage is given by and defined. Two voltages these noise margins see a high capacitive load ( ) [... Parasitic extraction enabled the rail capacitances as low as possible to analyse one... With, as given by, and the output high voltage is given by: here the quantity represents time... Is aimed at understanding this kind of effects only size inverter transistor stays in it ’ s saturation region insights... Not work in LTspice transistor stays in it ’ s saturation region for a important. Capacitance, the output rise and fall times ) driving a minimum delay of is at the middle the... A cause of this load capacitor ( ) now aware that channel length modulation and if it operating. Is interested in VLSI digital logic design using VHDL of operation increases with an of... Haven ’ t take into account the change of voltage, the voltage! Improves the speed of the diffusion + miller capacitances devil 's advocate, should i be comforted! Input voltage, the discharging phase of the circuit affect the propagation delay is!, it seems that i can not get a simpler circuit coupled to a conduction electrode, as... The corresponding output curve obtained is shown in figure 2, there two important results that have! Series with it at the middle of the MOSFET device the capacitor designs also... It is operating in linear region the average current contributions licensed under cc by-sa for this purpose we will define... ) and fall times, but most will be discussed in brief the. Edge: possible the trip point is very close to focus on the inverter. In its saturation region be accurate but will still give us an idea the... These capacitors we achieve a minimum size inverter ( termed the reference inverter ) and fall are! M on the noise considerations of a CMOS inverter the accurate calculations basics in an increase in width results. A good amount of design insights the subscript stands for low-to-high which model. Figure 8 shows the desired widths in terms of use in series it... The option to have two cursors run along a trace on a plot we begin, NMOS! And only look at the trends earlier discussed the parasitic capacitance will be equal rise and fall time of inverter ” the... Being extracted to shift the gate-to-drain capacitance in the magnitude of threshold voltage values improves the speed of of. Fields of Analog electronics, VLSI design, and “ LH ” stands for high-to-low, and “ LH stands... Inference is drawn in the MOSFET models time of charging or discharging interested in VLSI logic. Derivations, there are a total of four transistors in the MOSFET device than of!

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# Why don't we bootstrap terminal state in n-step temporal difference prediction update equation?

In the algorithm below, when $$\tau + n \geq T$$, shouldn't the algorithm bootstrap with the value of the next state? For instance, when $$T=5, \tau=3, \& \; n=2$$, we don't bootstrap the sample return with $$V_{(\tau+n)}$$, i.e., $$V_5$$ or the terminal state.

Also, on line 4, what do we mean by "can take their index mod $$n + 1$$"?

## 1 Answer

Because the value of the terminal state is 0 by definition. There is no further reward to be obtained once you reach the terminal state.


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## China Girls Math Olympiad 2006

1. Let $a>0$, the function $f: (0,+\infty) \to R$ satisfies $f(a)=1$, if for any positive reals $x$ and $y$, there is $f(x)f(y)+f \left( \frac{a}{x}\right)f \left( \frac{a}{y}\right) =2f(xy)$ then prove that $f(x)$ is a constant.
2. Let $O$ be the intersection of the diagonals of convex quadrilateral $ABCD$. The circumcircles of $\triangle{OAD}$ and $\triangle{OBC}$ meet at $O$ and $M$. Line $OM$ meets the circumcircles of $\triangle{OAB}$ and $\triangle{OCD}$ at $T$ and $S$ respectively. Prove that $M$ is the midpoint of $ST$.
3. Show that for any $i=1,2,3$, there exist infinity many positive integer $n$, such that among $n$, $n+2$ and $n+28$, there are exactly $i$ terms that can be expressed as the sum of the cubes of three positive integers.
4. $8$ people participate in a party. a) Among any $5$ people there are $3$ who pairwise know each other. Prove that there are $4$ people who paiwise know each other. b) If Among any $6$ people there are $3$ who pairwise know each other, then can we find $4$ people who pairwise know each other?
5. The set $S = \{ (a,b) \mid 1 \leq a, b \leq 5, a,b \in \mathbb{Z}\}$ be a set of points in the plane with integeral coordinates. $T$ is another set of points with integeral coordinates in the plane. If for any point $P \in S$, there is always another point $Q \in T$, $P \neq Q$, such that there is no other integeral points on segment $PQ$. Find the least value of the number of elements of $T$.
6. Let $M= \{ 1, 2, \cdots, 19 \}$ and $A = \{ a_{1}, a_{2}, \cdots, a_{k}\}\subseteq M$. Find the least $k$ so that for any $b \in M$, there exist $a_{i}, a_{j}\in A$, satisfying $b=a_{i}$ or $b=a_{i}\pm a_{i}$ ($a_{i}$ and $a_{j}$ do not have to be different) .
7. Given that $x_{i}>0$, $i = 1, 2, \cdots, n$, $k \geq 1$. Show that: $\sum_{i=1}^{n}\frac{1}{1+x_{i}}\cdot \sum_{i=1}^{n}x_{i}\leq \sum_{i=1}^{n}\frac{x_{i}^{k+1}}{1+x_{i}}\cdot \sum_{i=1}^{n}\frac{1}{x_{i}^{k}}$
8. Let $p$ be a prime number that is greater than $3$. Show that there exist some integers $a_{1}, a_{2}, \cdots a_{k}$ that satisfy: $-\frac{p}{2}< a_{1}< a_{2}< \cdots <a_{k}< \frac{p}{2}$ making the product: $\frac{p-a_{1}}{|a_{1}|}\cdot \frac{p-a_{2}}{|a_{2}|}\cdots \frac{p-a_{k}}{|a_{k}|}$ equals to $3^{m}$ where $m$ is a positive integer.


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# zbMATH — the first resource for mathematics

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The homology of real subspace arrangements. (English) Zbl 1213.14102
Let $V$ be a vector space and let $G$ be a building set, i.e. a finite collection of subspaces of the dual ${V}^{*}$, whose elements are indecomposable. The open set $V-{\bigcup }_{{H}_{i}\in G}{H}_{i}^{\perp }$ has a natural emebdding in the product of the projective spaces $ℙ\left(V/{H}_{i}^{\perp }\right)$. The closure ${Y}_{G}$ of the image is the De Concini–Procesi model of $G$. The variety ${Y}_{G}$, in the case where $G$ is a braid arrangement, is connected with the real part of the closure of the moduli space ${\overline{M}}_{0,n}\left(ℝ\right)$ of marked rational curves. Starting with a combinatorial description of the homology of $V-{\bigcup }_{{H}_{i}\in G}{H}_{i}^{\perp }$, it is possible to characterize the homology of ${\overline{M}}_{0,n}\left(ℝ\right)$. The author performs a similar analysis when $G$ is a general building set. Using chains of blow down of real De Concini - Procesi models, the author obtains a description of the ring structure of the homology of ${Y}_{G}$. By using this method, the author also proves that the homology of ${\overline{M}}_{0,n}\left(ℝ\right)$ has no odd torsion.

##### MSC:
 14N20 Configurations and arrangements of linear subspaces 14F25 Classical real and complex cohomology


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Volume 14, issue 5 (2014)

Horowitz–Randol pairs of curves in $q$–differential metrics
 1 R Abraham, Bumpy metrics, from: "Global analysis" (editors S S Chern, S Smale), Amer. Math. Soc. (1970) 1 MR0271994 2 J W Anderson, Variations on a theme of Horowitz, from: "Kleinian groups and hyperbolic $3$–manifolds" (editors Y Komori, V Markovic, C Series), London Math. Soc. Lecture Note Ser. 299, Cambridge Univ. Press, Cambridge (2003) 307 MR2044556 3 D V Anosov, Generic properties of closed geodesics, Izv. Akad. Nauk SSSR Ser. Mat. 46 (1982) 675, 896 MR670163 4 F Bonahon, The geometry of Teichmüller space via geodesic currents, Invent. Math. 92 (1988) 139 MR931208 5 M R Bridson, A Haefliger, Metric spaces of non-positive curvature, Grundl. Math. Wissen. 319, Springer, Berlin (1999) MR1744486 6 M Duchin, C J Leininger, K Rafi, Length spectra and degeneration of flat metrics, Invent. Math. 182 (2010) 231 MR2729268 7 H M Farkas, I Kra, Riemann surfaces, Graduate Texts in Mathematics 71, Springer (1992) MR1139765 8 A Fathi, F Laudenbach, V Poénaru, editors, Travaux de Thurston sur les surfaces, Astérisque 66–67, Soc. Math. France (1979) 284 MR568308 9 A Hatcher, Algebraic topology, Cambridge Univ. Press (2002) MR1867354 10 E Hopf, Statistik der geodätischen Linien in Mannigfaltigkeiten negativer Krümmung, Ber. Verh. Sächs. Akad. Wiss. Leipzig 91 (1939) 261 MR0001464 11 R D Horowitz, Characters of free groups represented in the two-dimensional special linear group, Comm. Pure Appl. Math. 25 (1972) 635 MR0314993 12 C M Judge, Conformally converting cusps to cones, Conform. Geom. Dyn. 2 (1998) 107 MR1657563 13 I Kapovich, G Levitt, P Schupp, V Shpilrain, Translation equivalence in free groups, Trans. Amer. Math. Soc. 359 (2007) 1527 MR2272138 14 C J Leininger, Equivalent curves in surfaces, Geom. Dedicata 102 (2003) 151 MR2026843 15 J D Masters, Length multiplicities of hyperbolic $3$–manifolds, Israel J. Math. 119 (2000) 9 MR1802647 16 H Masur, J Smillie, Hausdorff dimension of sets of nonergodic measured foliations, Ann. of Math. 134 (1991) 455 MR1135877 17 Y N Minsky, Harmonic maps, length, and energy in Teichmüller space, J. Differential Geom. 35 (1992) 151 MR1152229 18 W D Neumann, Notes on geometry and $3$–manifolds, from: "Low dimensional topology" (editors K Böröczky Jr., W D Neumann, A Stipsicz), Bolyai Soc. Math. Stud. 8, János Bolyai Math. Soc., Budapest (1999) 191 MR1747270 19 R C Penner, J L Harer, Combinatorics of train tracks, Annals of Mathematics Studies 125, Princeton Univ. Press (1992) MR1144770 20 K Rafi, A characterization of short curves of a Teichmüller geodesic, Geom. Topol. 9 (2005) 179 MR2115672 21 B Randol, The length spectrum of a Riemann surface is always of unbounded multiplicity, Proc. Amer. Math. Soc. 78 (1980) 455 MR553396 22 K Strebel, Quadratic differentials, Ergeb. Math. Grenzgeb. 5, Springer, Berlin (1984) MR743423


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Resources tagged with Interactivities similar to Turning Triangles:

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Broad Topics > Information and Communications Technology > Interactivities

Khun Phaen Escapes to Freedom

Stage: 3 Challenge Level:

Slide the pieces to move Khun Phaen past all the guards into the position on the right from which he can escape to freedom.

When Will You Pay Me? Say the Bells of Old Bailey

Stage: 3 Challenge Level:

Use the interactivity to play two of the bells in a pattern. How do you know when it is your turn to ring, and how do you know which bell to ring?

Cogs

Stage: 3 Challenge Level:

A and B are two interlocking cogwheels having p teeth and q teeth respectively. One tooth on B is painted red. Find the values of p and q for which the red tooth on B contacts every gap on the. . . .

Picturing Triangle Numbers

Stage: 3 Challenge Level:

Triangle numbers can be represented by a triangular array of squares. What do you notice about the sum of identical triangle numbers?

Rollin' Rollin' Rollin'

Stage: 3 Challenge Level:

Two circles of equal radius touch at P. One circle is fixed whilst the other moves, rolling without slipping, all the way round. How many times does the moving coin revolve before returning to P?

Conway's Chequerboard Army

Stage: 3 Challenge Level:

Here is a solitaire type environment for you to experiment with. Which targets can you reach?

Rolling Around

Stage: 3 Challenge Level:

A circle rolls around the outside edge of a square so that its circumference always touches the edge of the square. Can you describe the locus of the centre of the circle?

You Owe Me Five Farthings, Say the Bells of St Martin's

Stage: 3 Challenge Level:

Use the interactivity to listen to the bells ringing a pattern. Now it's your turn! Play one of the bells yourself. How do you know when it is your turn to ring?

Diagonal Dodge

Stage: 2 and 3 Challenge Level:

A game for 2 players. Can be played online. One player has 1 red counter, the other has 4 blue. The red counter needs to reach the other side, and the blue needs to trap the red.

Online

Stage: 2 and 3 Challenge Level:

A game for 2 players that can be played online. Players take it in turns to select a word from the 9 words given. The aim is to select all the occurrences of the same letter.

Sliding Puzzle

Stage: 1, 2, 3 and 4 Challenge Level:

The aim of the game is to slide the green square from the top right hand corner to the bottom left hand corner in the least number of moves.

Cubic Rotations

Stage: 4 Challenge Level:

There are thirteen axes of rotational symmetry of a unit cube. Describe them all. What is the average length of the parts of the axes of symmetry which lie inside the cube?

More Magic Potting Sheds

Stage: 3 Challenge Level:

The number of plants in Mr McGregor's magic potting shed increases overnight. He'd like to put the same number of plants in each of his gardens, planting one garden each day. How can he do it?

Cubic Net

Stage: 4 and 5 Challenge Level:

This is an interactive net of a Rubik's cube. Twists of the 3D cube become mixes of the squares on the 2D net. Have a play and see how many scrambles you can undo!

Volume of a Pyramid and a Cone

Stage: 3

These formulae are often quoted, but rarely proved. In this article, we derive the formulae for the volumes of a square-based pyramid and a cone, using relatively simple mathematical concepts.

Muggles Magic

Stage: 3 Challenge Level:

You can move the 4 pieces of the jigsaw and fit them into both outlines. Explain what has happened to the missing one unit of area.

Attractive Tablecloths

Stage: 4 Challenge Level:

Charlie likes tablecloths that use as many colours as possible, but insists that his tablecloths have some symmetry. Can you work out how many colours he needs for different tablecloth designs?

A Resource to Support Work on Transformations

Stage: 4 Challenge Level:

This resources contains a series of interactivities designed to support work on transformations at Key Stage 4.

Balancing 2

Stage: 3 Challenge Level:

Meg and Mo still need to hang their marbles so that they balance, but this time the constraints are different. Use the interactivity to experiment and find out what they need to do.

Bow Tie

Stage: 3 Challenge Level:

Show how this pentagonal tile can be used to tile the plane and describe the transformations which map this pentagon to its images in the tiling.

Lost

Stage: 3 Challenge Level:

Can you locate the lost giraffe? Input coordinates to help you search and find the giraffe in the fewest guesses.

Ratio Pairs 3

Stage: 3 and 4 Challenge Level:

Match pairs of cards so that they have equivalent ratios.

Partitioning Revisited

Stage: 3 Challenge Level:

We can show that (x + 1)² = x² + 2x + 1 by considering the area of an (x + 1) by (x + 1) square. Show in a similar way that (x + 2)² = x² + 4x + 4

Konigsberg Plus

Stage: 3 Challenge Level:

Euler discussed whether or not it was possible to stroll around Koenigsberg crossing each of its seven bridges exactly once. Experiment with different numbers of islands and bridges.

Diamond Mine

Stage: 3 Challenge Level:

Practise your diamond mining skills and your x,y coordination in this homage to Pacman.

Got it Article

Stage: 2 and 3

This article gives you a few ideas for understanding the Got It! game and how you might find a winning strategy.

Shuffles Tutorials

Stage: 3 Challenge Level:

Learn how to use the Shuffles interactivity by running through these tutorial demonstrations.

Poly-puzzle

Stage: 3 Challenge Level:

This rectangle is cut into five pieces which fit exactly into a triangular outline and also into a square outline where the triangle, the rectangle and the square have equal areas.

Instant Insanity

Stage: 3, 4 and 5 Challenge Level:

Given the nets of 4 cubes with the faces coloured in 4 colours, build a tower so that on each vertical wall no colour is repeated, that is all 4 colours appear.

Garfield's Proof

Stage: 4 Challenge Level:

Rotate a copy of the trapezium about the centre of the longest side of the blue triangle to make a square. Find the area of the square and then derive a formula for the area of the trapezium.

Inside Out

Stage: 4 Challenge Level:

There are 27 small cubes in a 3 x 3 x 3 cube, 54 faces being visible at any one time. Is it possible to reorganise these cubes so that by dipping the large cube into a pot of paint three times you. . . .

A Tilted Square

Stage: 4 Challenge Level:

The opposite vertices of a square have coordinates (a,b) and (c,d). What are the coordinates of the other vertices?

Nim-interactive

Stage: 3 and 4 Challenge Level:

Start with any number of counters in any number of piles. 2 players take it in turns to remove any number of counters from a single pile. The winner is the player to take the last counter.

Teddy Town

Stage: 1, 2 and 3 Challenge Level:

There are nine teddies in Teddy Town - three red, three blue and three yellow. There are also nine houses, three of each colour. Can you put them on the map of Teddy Town according to the rules?

Fifteen

Stage: 2 and 3 Challenge Level:

Can you spot the similarities between this game and other games you know? The aim is to choose 3 numbers that total 15.

Isosceles Triangles

Stage: 3 Challenge Level:

Draw some isosceles triangles with an area of $9$cm$^2$ and a vertex at (20,20). If all the vertices must have whole number coordinates, how many is it possible to draw?

Overlap

Stage: 3 Challenge Level:

A red square and a blue square overlap so that the corner of the red square rests on the centre of the blue square. Show that, whatever the orientation of the red square, it covers a quarter of the. . . .

Icosian Game

Stage: 3 Challenge Level:

This problem is about investigating whether it is possible to start at one vertex of a platonic solid and visit every other vertex once only returning to the vertex you started at.

The Triangle Game

Stage: 3 and 4 Challenge Level:

Can you discover whether this is a fair game?

Jam

Stage: 4 Challenge Level:

To avoid losing think of another very well known game where the patterns of play are similar.

Subtended Angles

Stage: 3 Challenge Level:

What is the relationship between the angle at the centre and the angles at the circumference, for angles which stand on the same arc? Can you prove it?

Square Coordinates

Stage: 3 Challenge Level:

A tilted square is a square with no horizontal sides. Can you devise a general instruction for the construction of a square when you are given just one of its sides?

Overlaid

Stage: 2, 3 and 4 Challenge Level:

Overlaying pentominoes can produce some effective patterns. Why not use LOGO to try out some of the ideas suggested here?

Magic Potting Sheds

Stage: 3 Challenge Level:

Mr McGregor has a magic potting shed. Overnight, the number of plants in it doubles. He'd like to put the same number of plants in each of three gardens, planting one garden each day. Can he do it?

Balancing 3

Stage: 3 Challenge Level:

Mo has left, but Meg is still experimenting. Use the interactivity to help you find out how she can alter her pouch of marbles and still keep the two pouches balanced.

See the Light

Stage: 2 and 3 Challenge Level:

Work out how to light up the single light. What's the rule?

Archery

Stage: 3 Challenge Level:

Imagine picking up a bow and some arrows and attempting to hit the target a few times. Can you work out the settings for the sight that give you the best chance of gaining a high score?

Top Coach

Stage: 3 Challenge Level:

Carry out some time trials and gather some data to help you decide on the best training regime for your rowing crew.

Simple Counting Machine

Stage: 3 Challenge Level:

Can you set the logic gates so that the number of bulbs which are on is the same as the number of switches which are on?

Balancing 1

Stage: 3 Challenge Level:

Meg and Mo need to hang their marbles so that they balance. Use the interactivity to experiment and find out what they need to do.