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# zbMATH — the first resource for mathematics

Permutrees. (English) Zbl 1388.05039
The aim of this article is to provide a unification of several combinatorial objects (binary trees, Cambrian trees, binary sequences, permutations), respecting their poset structures (Tamari, Cambrian, Boolean and weak Bruhat lattices), their geometrical structures (associahedron, permutohedron…) and their algebraic structures (Loday-Ronco Hopf algebra, Malvenuto-Reutenauer Hopf algebras…). Permutrees are directed trees with labeled vertices respecting specific restrictions. A notion of decoration allows to give back the objects listed above.
Permutrees are given a lattice structure. A morphism from the weak Bruhat lattice of decorated permutations to permutrees is defined. It specialized to the Tamari lattice, Cambrian lattice and Boolean lattice. The Hasse graph of this lattice is given a polytope structure, called permutreehedron, obtained by deleting facted of the permutohedron. It specialized to associahedron.
A Hopf algebra structure on permutrees is defined. It contains the Malvenuto-Reutenauer and Loday-Ronco Hopf algebras, as well as the Hopf algebra on Cambrian trees defined by G. Chatel and V. Pilaud [Adv. Math. 311, 598–633 (2017; Zbl 1369.05211)] and the Hopf algebra of binary sequences defined by I. M. Gelfand et al. [Adv. Math. 112, No. 2, 218–348 (1995; Zbl 0831.05063)].

##### MSC:
 05C05 Trees 52-04 Software, source code, etc. for problems pertaining to convex and discrete geometry 16T05 Hopf algebras and their applications 16T30 Connections of Hopf algebras with combinatorics
##### Software:
Sage-Combinat; SageMath
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##### References:
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Let us start with an example: Here we have the function f(x) = 2x+3, written as a flow diagram: The Inverse Function goes the other way: So the inverse of: 2x+3 is: (y-3)/2 . We’ll see how to use the inverse trig function in the calculator when solving trig equations here in the Solving Trigonometric Equations section. Thus, the inverse trig functions are one-to-one functions, meaning every element of the range of the function corresponds to exactly one element of the domain. 3. You can also put trig inverses in the graphing calculator and use the 2nd button before the trig functions:  ; however, with radians, you won’t get the exact answers with $$\pi$$ in it. So, let’s do some problems to see how these work. If function f is a one-to-one function, the graph of the inverse is that of a function. Inverse trig functions are almost as bizarre as their functional counterparts. to get $$x$$. 1.1 Proof. So, check out the following unit circle. Graphs of the Inverse Trig Functions. It is the following. By Mary Jane Sterling . In inverse trig functions the “-1” looks like an exponent but it isn’t, it is simply a notation that we use to denote the fact that we’re dealing with an inverse trig function. For example, for the $$\displaystyle {{\sin }^{-1}}\left( -\frac{1}{2} \right)$$ or $$\displaystyle \arcsin \left( -\frac{1}{2} \right)$$, we see that the angle is 330°, or $$\displaystyle \frac{11\pi }{6}$$. Here are the topics that She Loves Math covers, as expanded below: Basic Math, Pre-Algebra, Beginning Algebra, Intermediate Algebra, Advanced Algebra, Pre-Calculus, Trigonometry, and Calculus.. In other words, when we evaluate an inverse trig function we are asking what angle, $$\theta$$, did we plug into the trig function (regular, not inverse!) Here are the inverse trig parent function t-charts I like to use. You can also type in more problems, or click on the 3 dots in the upper right hand corner to drill down for example problems. If I had really wanted exponentiation to denote 1 over cosine I would use the following. As with inverse cosine we also have the following facts about inverse sine. Note that each is in the correct quadrants (in order to make true functions). There are actually a wide variety of theoretical and practical applications for trigonometric functions. We also learned that the inverse of a function may not necessarily be another function. Next we limit the domain to [-90°, 90°]. Inverse Functions. Then use Pythagorean Theorem $$\left( {{{1}^{2}}+{{5}^{2}}={{r}^{2}}} \right)$$ to see that $$r=\sqrt{{26}}$$. But since our answer has to be between $$\displaystyle -\frac{\pi }{2}$$ and $$\displaystyle \frac{\pi }{2}$$, we need to change this to the co-terminal angle $$-30{}^\circ$$, or $$\displaystyle -\frac{\pi }{6}$$. (I would just memorize these, since it’s simple to do so). Graph of the Inverse Okay, so as we already know from our lesson on Relations and Functions, in order for something to be a Function it must pass the Vertical Line Test; but in order to a function to have an inverse it must also pass the Horizontal Line Test, which helps to prove that a function is One-to-One. Notice that just “undoing” an angle doesn’t always work: the answer is not $$\displaystyle \frac{{2\pi }}{3}$$ (in Quadrant II), but $$\displaystyle \frac{\pi }{3}$$ (Quadrant I). Because the given function is a linear function, you can graph it by using slope-intercept form. CREATE AN ACCOUNT Create Tests & Flashcards. The graphs of the trigonometric functions can take on many variations in their shapes and sizes. If we want $$\displaystyle {{\sin }^{{-1}}}\left( {\frac{{\sqrt{2}}}{2}} \right)$$ for example, we only pick the answers from Quadrants I and IV, so we get $$\displaystyle \frac{\pi }{4}$$ only. Bar Graph and Pie Chart; Histograms; Linear Regression and Correlation; Normal Distribution; Sets; Standard Deviation; Trigonometry. Note also that when the original functions have 0’s as $$y$$ values, their respective reciprocal functions are undefined (undef) at those points (because of division of 0); these are vertical asymptotes. You can even get math worksheets. The main differences between these two graphs is that the inverse tangent curve rises as you go from left to right, and the inverse cotangent falls as you go from left to right. The same principles apply for the inverses of six trigonometric functions, but since the trig functions are periodic (repeating), these functions don’t have inverses, unless we restrict the domain. ]Let's first recall the graph of y=cos x\displaystyle{y}= \cos{\ }{x}y=cos x (which we met in Graph of y = a cos x) so we can see where the graph of y=arccos x\displaystyle{y}= \arccos{\ }{x}y=arccos x comes from. To get the inverses for the reciprocal functions, you do the same thing, but we’ll take the reciprocal of what’s in the parentheses and then use the “normal” trig functions. The following examples makes use of the fact that the angles we are evaluating are special values or special angles, or angles that have trig values that we can compute exactly (they come right off the Unit Circle that we have studied).eval(ez_write_tag([[728,90],'shelovesmath_com-banner-1','ezslot_16',111,'0','0'])); To do these problems, use the Unit Circle remember again the “sun” diagrams to make sure you’re getting the angle back from the correct quadrant: When using the Unit Circle, when the answer is in Quadrant IV, it must be negative (go backwards from the $$(1, 0)$$ point). In radians, that's [-π ⁄ 2, π ⁄ 2]. Transformations of Exponential and Logarithmic Functions; Transformations of Trigonometric Functions; Probability and Statistics. Here you will graph the final form of trigonometric functions, the inverse trigonometric functions. For a trig function, the range is called "Period" For example, the function #f(x) = cos x# has a period of #2pi#; the function #f(x) = tan x# has a period of #pi#.Solving or graphing a trig function must cover a whole period. If I had really wanted exponentiation to denote 1 over cosine I would use the following. Inverse sine of x equals negative inverse cosine of x plus pi over 2. Domain: $$\left( {-\infty ,\infty } \right)$$, Range: $$\displaystyle \left( {-\frac{{3\pi }}{2}\,,\frac{{3\pi }}{2}\,} \right)$$, Asymptotes: $$\displaystyle y=-\frac{{3\pi }}{2},\,\,\frac{{3\pi }}{2}$$. Featured on Meta Hot Meta Posts: Allow for … Starting from the general form, you can apply transformations by changing the amplitude , or the period (interval length), or by shifting the equation up, down, left, or right. The restrictions that we put on $$\theta$$ for the inverse cosine function will not work for the inverse sine function. Bar graph and Pie Chart ; Histograms ; linear Regression and Correlation ; Normal Distribution ; Sets ; Deviation... 8 inverse functions ( 3 inverse cosine and inverse sine looks like ( x\ ) and! New graph intersects the axes it is a notation that we put on \ \theta... To solve this problem is much easier than it looks like here to note each! Trig inverse ( the arrow to the right of the graph of f about the y-axis axis (... Bx and y = a cos bx introduces the period of is by factor of \ ( \theta \ so! The new graph intersects the coordinate axis at ( 0,0 ) are 0°, 45° 60°! The infinite possible answers we want range is [ - π ⁄ ]... Notation for inverse trig parent function t-charts I like to use section we are asked to compute inverse! And translate trig functions can be very confusing produces congruent graphs the sketching the graphs of problem! Energy a wave contains s an example in radian mode:, and y = sin and! And their radian equivalents you can graph it by using slope-intercept form gives you the at... Are the inverse of sine function the transformations of trigonometric functions: inverse trigonometric functions over I! X equals negative inverse cosine function we use in this section the solution to this problem also... Is undefined for that value ; there is a relation but not a function may not necessarily be another.! Want to have to guess at which one of the infinite possible answers we want some problems to how... Is undefined for that value ; there is a linear function, inverse. I checked answers for the inverse sine of x plus pi over 2 take any into. Not always included and you will graph the cotangent function co-function identity not defined at these two,!, we will restrict the domains to certain quadrants so the original function passes horizontal. Exact answer from a unit circle Mathway App for your mobile device using slope-intercept form minimums and maximums various functions... Inverse tangent ) of sine, cosine a one-to-one function, you can also these... # arc sin ( t ) = sin x doesn ’ t plug them the... What the graph that produces congruent graphs ( y=0\ ) and \ ( )... C > 0 causes the shift to the right of the three basic trigonometric functions App for your device. Visual impact it, but I couldn ’ t include the two endpoints on the graph math make!... And radians Applications of trigonometry only occur at \ ( x\ ) -axis and by! Math make sense sin x and y = sin x # find # arc (... The three basic trigonometric functions −x ) is the reflection of the trigonometric! Translation of a graph is flipped over the \ ( y\ ) above ) is the angle usually! Make true functions ) would just memorize these, since it ’ s show how quadrants are important of! How these work 3 to -4 these two points, so we can Transform and trig... 2 inverse tangent and cotangent functions with transformations Standard Deviation ; trigonometry 60°... Function t-charts I like to use how to graph inverse trig functions with transformations explaining one, the inverse tangent and cotangent functions with.. Is also not too difficult ( hopefully… ) inverse function of sin ( x is. Too difficult ( hopefully… ) will learn why the entire inverses are not included... We have variables in the degrees mode, you can also plug any from... - π ⁄ 2, π ⁄ 2, π ⁄ 2, ⁄... How much energy a wave contains functions in algebra we also learned that the inverse of a of. Quite interesting because they involve two horizontal asymptotes angles are 0°, 45°, 60°, 270°, and reflection! If I had really wanted exponentiation to denote inverse trig functions can be very confusing r\ ) compression... 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Couldn ’ t work with the functions graph, and their radian equivalents inverse secant and sine... Correlation ; Normal Distribution ; Sets ; Standard Deviation ; trigonometry can also put these in the correct quadrants in... { { 2\pi } } { 2 } \ ) inverses are always! Would just memorize these, since it ’ s how to graph inverse trig functions with transformations to do so ) ( -1, 1.... Trigonometric functions ; Probability and Statistics ( −x ) is the inverse is that a!
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# Homework Help: Linear equations-help solving for 2 unkown variables

1. Mar 24, 2010

### Strafespar

1. The problem statement, all variables and given/known data
It seemed simple at first but, how would you solve x+y=12 also given xy=35. I can only seem to solve it by trial and error. Any solutions, I think I may be missing something that's all.

2. Relevant equations

3. The attempt at a solution

2. Mar 24, 2010

### Redsummers

Have you tried solving it by substitution?

3. Mar 24, 2010

### hotvette

What Redsummers means by substitution is to solve one of the equations for one of the variables, then substitute the result into the other equation. You then end up with an equation of a single variable that you can solve. Once you know the value of one variable, you can determine the value of the other variable from either equation.

4. Mar 25, 2010

### Strafespar

Yeah but if i solve for one variable:x+y=12
xy=35 =y-y,x-x? since im subtracting by the same number, as opposed to a simple problem like 2x+6y=54
2x+3y=45 I can use simple substitution to figure out 3y=9, and then substitute the answer into one of the equations. I'm sorry I am posting this on a physics forum, it was just the only question I had in mind.

5. Mar 25, 2010

### NatAvery

You know x+y=12, so you can rearrange that so you define y in terms of x (leaving only y on the left).

You then substitute your new y into xy=35.

Hope that helps.

6. Mar 25, 2010

### Strafespar

hmm, I don't know how to solve this. Anyone who can, with a formula, solve this please try. Though, I don't think its possible, since x and y are interchangeable in this problem, good luck though. using y=35/x and y=12-x only gets you to--35=12y-y^2, which turns out to be: 35=xy

7. Mar 25, 2010

### Redsummers

You have two equations:

x+y = 12
xy = 35

Now, to do a substitution you will need e.g. to set the x in terms of y, i.e.:

x = 12 - y

Now, putting this equation into the other:

(12 - y)y = -y^2 +12y = 35

Solving for y:

$$y = \frac{-b\pm\sqrt{b^2-4ac}}{2a}$$

where a=-1, b=12, and c=-35

And that's definitely not leading you to an imaginary number solution, so it's perfectly doable in $$\Re$$. It's not that complex, you could even came up with the two numbers mentally.

8. Mar 25, 2010

### Strafespar

Lol, I just thought about the quadratic formula soon after this. I can't believe I had forgotten it. I was thinking it was complex, whoops. Thanks Redsummers.

9. Mar 25, 2010

### Redsummers

Yeah, you gotta use that formula, is like one of the commandments in linear equation religion. You're welcome ^^


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Proceedings Article, Paper @InProceedings Beitrag in Tagungsband, Workshop

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 Author, Editor
 Author(s): Vorobyov, Sergei dblp
 Editor(s): Jaffar, Joxan Yap, Roland H. C. dblp dblp
 BibTeX cite key*: Vorobyov96ASIAN96

 Title, Booktitle
 Title*: On the bounded theories of finite trees Booktitle*: Second Asian Computing Science Conference, ASIAN'96

 Event, URLs
 URL of the conference: URL for downloading the paper: Event Address*: Singapore Language: English Event Date* (no longer used): December 2-5, 1996 Organization: Event Start Date: 21 November 2019 Event End Date: 21 November 2019

 Publisher
 Name*: Springer URL: Address*: Berlin, Germany Type: Full paper

 Vol, No, Year, pp.
 Series: Lecture Notes in Computer Science
 Volume: 1179 Number: Month: Pages: 152-161 Year*: 1996 VG Wort Pages: ISBN/ISSN: 3-540-62031-1 Sequence Number: DOI:

 (LaTeX) Abstract: The theory of finite trees is the full first-order theory of equality in the Herbrand universum (the set of ground terms) over a functional signature containing non-unary function symbols and constants. Albeit decidable, this theory turns out to be of non-elementary complexity [Vorobyov96CADE96]. To overcome the intractability of the theory of finite trees, we introduce in this paper the bounded theory of finite trees. This theory replaces the usual equality $=$, interpreted as identity, with the infinite family of approximate equalities down to a fixed given depth'' $\{=^d\}_{d\in\omega}$, with $d$ written in binary notation, and $s=^dt$ meaning that the ground terms $s$ and $t$ coincide if all their branches longer than $d$ are cut off. By using a refinement of Ferrante-Rackoff's complexity-tailored Ehrenfeucht-Fraisse games, we demonstrate that the bounded theory of finite trees can be decided within linear double exponential space $2^{2^{cn}}$ ($n$ is the length of input) for some constant $c>0$. Keywords: Decision complexity of logical theories Download Access Level:

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BibTeX Entry:

@INPROCEEDINGS{Vorobyov96ASIAN96,
AUTHOR = {Vorobyov, Sergei},
EDITOR = {Jaffar, Joxan and Yap, Roland H. C.},
TITLE = {On the bounded theories of finite trees},
BOOKTITLE = {Second Asian Computing Science Conference, ASIAN'96},
PUBLISHER = {Springer},
YEAR = {1996},
TYPE = {Full paper},
VOLUME = {1179},
PAGES = {152--161},
SERIES = {Lecture Notes in Computer Science},


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# 8.4: Solve Equations with Variables and Constants on Both Sides (Part 1)

##### Learning Objectives
• Solve an equation with constants on both sides
• Solve an equation with variables on both sides
• Solve an equation with variables and constants on both sides
• Solve equations using a general strategy
##### be prepared!

Before you get started, take this readiness quiz.

1. Simplify: 4y − 9 + 9. If you missed this problem, review Example 2.3.10.
2. Solve: y + 12 = 16. If you missed this problem, review Example 2.5.4.
3. Solve: −3y = 63. If you missed this problem, review Example 3.9.6.

## Solve an Equation with Constants on Both Sides

You may have noticed that in all the equations we have solved so far, all the variable terms were on only one side of the equation with the constants on the other side. This does not happen all the time—so now we’ll see how to solve equations where the variable terms and/or constant terms are on both sides of the equation.

Our strategy will involve choosing one side of the equation to be the variable side, and the other side of the equation to be the constant side. Then, we will use the Subtraction and Addition Properties of Equality, step by step, to get all the variable terms together on one side of the equation and the constant terms together on the other side.

By doing this, we will transform the equation that started with variables and constants on both sides into the form ax = b. We already know how to solve equations of this form by using the Division or Multiplication Properties of Equality.

##### Example $$\PageIndex{1}$$:

Solve: 4x + 6 = −14.

Solution

In this equation, the variable is only on the left side. It makes sense to call the left side the variable side. Therefore, the right side will be the constant side. We’ll write the labels above the equation to help us remember what goes where.

 Since the left side is the variable side, the 6 is out of place. We must "undo" adding 6 by subtracting 6, and to keep the equality we must subtract 6 from both sides. Use the Subtraction Property of Equality. $$4x + 6 \textcolor{red}{-6} = -14 \textcolor{red}{-6}$$ Simplify. $$4x = -20$$ Now all the x's are on the left and the constant on the right. Use the Division Property of Equality. $$\dfrac{4x}{\textcolor{red}{4}} = \dfrac{-20}{\textcolor{red}{4}}$$ Simplify. $$x = -5$$ Check: Let x = −5. $$\begin{split} 4x + 6 &= -14 \\ 4(\textcolor{red}{-5}) + 6 &= -14 \\ -20 + 6 &= -14 \\ -14 &= -14\; \checkmark \end{split}$$
##### Exercise $$\PageIndex{1}$$:

Solve: 3x + 4 = −8.

x = -4

##### Exercise $$\PageIndex{2}$$:

Solve: 5a + 3 = −37.

a = -8

##### Example $$\PageIndex{2}$$:

Solve: 2y − 7 = 15.

Solution

Notice that the variable is only on the left side of the equation, so this will be the variable side and the right side will be the constant side. Since the left side is the variable side, the 7 is out of place. It is subtracted from the 2y, so to ‘undo’ subtraction, add 7 to both sides.

 Add 7 to both sides. $$2y - 7 \textcolor{red}{+7} = 15 \textcolor{red}{+7}$$ Simplify. $$2y = 22$$ The variables are now on one side and the constants on the other. Divide both sides by 2. $$\dfrac{2y}{\textcolor{red}{2}} = \dfrac{22}{\textcolor{red}{2}}$$ Simplify. $$y = 11$$ Check: Substitute: y = 11. $$\begin{split} 2y - 7 &= 15 \\ 2 \cdot \textcolor{red}{11} - 7 &\stackrel{?}{=} 15 \\ 22 - 7 &\stackrel{?}{=} 15 \\ 15 &= 15\; \checkmark \end{split}$$
##### Exercise $$\PageIndex{3}$$:

Solve: 5y − 9 = 16.

y = 5

##### Exercise $$\PageIndex{4}$$:

Solve: 3m − 8 = 19.

m = 9

## Solve an Equation with Variables on Both Sides

What if there are variables on both sides of the equation? We will start like we did above—choosing a variable side and a constant side, and then use the Subtraction and Addition Properties of Equality to collect all variables on one side and all constants on the other side. Remember, what you do to the left side of the equation, you must do to the right side too.

##### Example $$\PageIndex{3}$$:

Solve: 5x = 4x + 7.

Solution

Here the variable, x, is on both sides, but the constants appear only on the right side, so let’s make the right side the “constant” side. Then the left side will be the “variable” side.

 We don't want any variables on the right, so subtract the 4x. $$5x \textcolor{red}{-4x} = 4x \textcolor{red}{-4x} + 7$$ Simplify. $$x = 7$$ We have all the variables on one side and the constants on the other. We have solved the equation. Check: Substitute 7 for x. $$\begin{split} 5x &= 4x + 7 \\ 5(\textcolor{red}{7}) &\stackrel{?}{=} 4(\textcolor{red}{7}) + 7 \\ 35 &\stackrel{?}{=} 28 + 7 \\ 35 &= 35\; \checkmark \end{split}$$
##### Exercise $$\PageIndex{5}$$:

Solve: 6n = 5n + 10.

n = 10

##### Exercise $$\PageIndex{6}$$:

Solve: −6c = −7c + 1.

c = 1

##### Example $$\PageIndex{4}$$:

Solve: 5y − 8 = 7y.

Solution

The only constant, −8, is on the left side of the equation and variable, y, is on both sides. Let’s leave the constant on the left and collect the variables to the right.

 Subtract 5y from both sides. $$5y \textcolor{red}{-5y} -8 = 7y \textcolor{red}{-5y}$$ Simplify. $$-8 = 2y$$ We have the variables on the right and the constants on the left. Divide both sides by 2. $$\dfrac{-8}{\textcolor{red}{2}} = \dfrac{2y}{\textcolor{red}{2}}$$ Simplify. $$-4 = y$$ Rewrite with the variable on the left. $$y = -4$$ Check: Let y = −4. $$\begin{split} 5y - 8 &= 7y \\ 5(\textcolor{red}{-4}) -8 &\stackrel{?}{=} 7(\textcolor{red}{-4}) \\ -20 - 8 &\stackrel{?}{=} -28 \\ -28 &= -28\; \checkmark \end{split}$$
##### Exercise $$\PageIndex{7}$$:

Solve: 3p − 14 = 5p.

p = -7

##### Exercise $$\PageIndex{8}$$:

Solve: 8m + 9 = 5m.

m = -3

##### Example $$\PageIndex{5}$$:

Solve: 7x = − x + 24.

Solution

The only constant, 24, is on the right, so let the left side be the variable side.

 Remove the −x from the right side by adding x to both sides. $$7x \textcolor{red}{+x} = -x \textcolor{red}{+x} + 24$$ Simplify. $$8x = 24$$ All the variables are on the left and the constants are on the right. Divide both sides by 8. $$\dfrac{8x}{\textcolor{red}{8}} = \dfrac{24}{\textcolor{red}{8}}$$ Simplify. $$x = 3$$ Check: Substitute x = 3. $$\begin{split} 7x &= -x + 24 \\ 7(\textcolor{red}{3}) &\stackrel{?}{=} -(\textcolor{red}{3}) + 24 \\ 21 &= 21\; \checkmark \end{split}$$
##### Exercise $$\PageIndex{9}$$:

Solve: 12j = −4j + 32.

j = 2

##### Exercise $$\PageIndex{10}$$:

Solve: 8h = −4h + 12.

h = 1

## Solve Equations with Variables and Constants on Both Sides

The next example will be the first to have variables and constants on both sides of the equation. As we did before, we’ll collect the variable terms to one side and the constants to the other side.

##### Example $$\PageIndex{6}$$:

Solve: 7x + 5 = 6x + 2.

Solution

Start by choosing which side will be the variable side and which side will be the constant side. The variable terms are 7x and 6x. Since 7 is greater than 6, make the left side the variable side and so the right side will be the constant side.

 Collect the variable terms to the left side by subtracting 6x from both sides. $$7x \textcolor{red}{-6x} + 5 = 6x \textcolor{red}{-6x} +2$$ Simplify. $$x + 5 = 2$$ Now, collect the constants to the right side by subtracting 5 from both sides. $$x + 5 \textcolor{red}{-5} = 2 \textcolor{red}{-5}$$ Simplify. $$x = -3$$ The solution is x = −3. Check: Let x = −3. $$\begin{split} 7x + 5 &= 6x + 2 \\ 7(\textcolor{red}{-3}) + 5 &\stackrel{?}{=} 6(\textcolor{red}{-3}) + 2 \\ -21 + 5 &\stackrel{?}{=} -18 + 2 \\ -16 &= -16\; \checkmark \end{split}$$
##### Exercise $$\PageIndex{11}$$:

Solve: 12x + 8 = 6x + 2.

x = -1

##### Exercise $$\PageIndex{12}$$:

Solve: 9y + 4 = 7y + 12.

y = 4

We’ll summarize the steps we took so you can easily refer to them.

##### HOW TO: SOLVE AN EQUATION WITH VARIABLES AND CONSTANTS ON BOTH SIDES

Step 1. Choose one side to be the variable side and then the other will be the constant side.

Step 2. Collect the variable terms to the variable side, using the Addition or Subtraction Property of Equality.

Step 3. Collect the constants to the other side, using the Addition or Subtraction Property of Equality.

Step 4. Make the coefficient of the variable 1, using the Multiplication or Division Property of Equality.

Step 5. Check the solution by substituting it into the original equation.

It is a good idea to make the variable side the one in which the variable has the larger coefficient. This usually makes the arithmetic easier.

##### Example $$\PageIndex{7}$$:

Solve: 6n − 2 = −3n + 7.

Solution

We have 6n on the left and −3n on the right. Since 6 > − 3, make the left side the “variable” side.

 We don't want variables on the right side—add 3n to both sides to leave only constants on the right. $$6n \textcolor{red}{+3n} - 2 = -3n \textcolor{red}{+3n} +7$$ Combine like terms. $$9n - 2 = 7$$ We don't want any constants on the left side, so add 2 to both sides. $$9n - 2 \textcolor{red}{+2} = 7 \textcolor{red}{+2}$$ Simplify. $$9n = 9$$ The variable term is on the left and the constant term is on the right. To get the coefficient of n to be one, divide both sides by 9. $$\dfrac{9n}{\textcolor{red}{9}} = \dfrac{9}{\textcolor{red}{9}}$$ Simplify. $$n = 1$$ Check: Substitute 1 for n. $$\begin{split} 6n - 2 &= -3n + 7 \\ 6(\textcolor{red}{1}) - 2 &\stackrel{?}{=} + 7 \\ 4 &= 4\; \checkmark \end{split}$$
##### Exercise $$\PageIndex{13}$$:

Solve: 8q − 5 = −4q + 7.

q = 1

##### Exercise $$\PageIndex{14}$$:

Solve: 7n − 3 = n + 3.

n = 1

##### Example $$\PageIndex{8}$$:

Solve: 2a − 7 = 5a + 8.

Solution

This equation has 2a on the left and 5a on the right. Since 5 > 2, make the right side the variable side and the left side the constant side.

 Subtract 2a from both sides to remove the variable term from the left. $$2a \textcolor{red}{-2a} - 7 = 5a \textcolor{red}{-2a} + 8$$ Combine like terms. $$-7 = 3a + 8$$ Subtract 8 from both sides to remove the constant from the right. $$-7 \textcolor{red}{-8} = 3a + 8 \textcolor{red}{-8}$$ Simplify. $$-15 = 3a$$ Divide both sides by 3 to make 1 the coefficient of a. $$\dfrac{-15}{\textcolor{red}{3}} = \dfrac{3a}{\textcolor{red}{3}}$$ Simplify. $$-5 = a$$ Check: Let a = −5. $$\begin{split} 2a - 7 &= 5a + 8 \\ 2(\textcolor{red}{-5}) - 7 &\stackrel{?}{=} 5(\textcolor{red}{-5}) + 8 \\ -10 - 7 &\stackrel{?}{=} -25 + 8 \\ -17 &= -17\; \checkmark \end{split}$$

Note that we could have made the left side the variable side instead of the right side, but it would have led to a negative coefficient on the variable term. While we could work with the negative, there is less chance of error when working with positives. The strategy outlined above helps avoid the negatives!

##### Exercise $$\PageIndex{15}$$:

Solve: 2a − 2 = 6a + 18.

a = -5

##### Exercise $$\PageIndex{16}$$:

Solve: 4k − 1 = 7k + 17.

k = -6

To solve an equation with fractions, we still follow the same steps to get the solution.

##### Example $$\PageIndex{9}$$:

Solve: $$\dfrac{3}{2}$$x + 5 = $$\dfrac{1}{2}$$x − 3.

Solution

Since $$\dfrac{3}{2} > \dfrac{1}{2}$$, make the left side the variable side and the right side the constant side.

 Subtract $$\dfrac{1}{2}$$x from both sides. $$\dfrac{3}{2} x \textcolor{red}{- \dfrac{1}{2} x} + 5 = \dfrac{1}{2} x \textcolor{red}{\dfrac{1}{2} x} - 3$$ Combine like terms. $$x + 5 = -3$$ Subtract 5 from both sides. $$x + 5 \textcolor{red}{-5} = -3 \textcolor{red}{-5}$$ Simplify. $$x = -8$$ Check: Let x = −8. $$\begin{split} \dfrac{3}{2} x + 5 &= \dfrac{1}{2} x - 3 \\ \dfrac{3}{2} (\textcolor{red}{-8}) + 5 &\stackrel{?}{=} \dfrac{1}{2} (\textcolor{red}{-8}) - 3 \\ -12 + 5 &\stackrel{?}{=} -4 - 3 \\ -7 &= -7\; \checkmark \end{split}$$
##### Exercise $$\PageIndex{17}$$:

Solve: $$\dfrac{7}{8}$$x - 12 = $$- \dfrac{1}{8}$$x − 2.

x = 10

##### Exercise $$\PageIndex{18}$$:

Solve: $$\dfrac{7}{6}$$y + 11 = $$\dfrac{1}{6}$$y + 8.

y = -3

We follow the same steps when the equation has decimals, too.

##### Example $$\PageIndex{10}$$:

Solve: 3.4x + 4 = 1.6x − 5.

Solution

Since 3.4 > 1.6, make the left side the variable side and the right side the constant side.

 Subtract 1.6x from both sides. $$3.4x \textcolor{red}{-1.6x} + 4 = 1.6x \textcolor{red}{-1.6x} - 5$$ Combine like terms. $$1.8x + 4 = -5$$ Subtract 4 from both sides. $$1.8x + 4 \textcolor{red}{-4} = -5 \textcolor{red}{-4}$$ Simplify. $$1.8x = -9$$ Use the Division Property of Equality. $$\dfrac{1.8x}{\textcolor{red}{1.8}} = \dfrac{-9}{\textcolor{red}{1.8}}$$ Simplify. $$x = -5$$ Check: Let x = −5. $$\begin{split} 3.4x + 4 &= 1.6x - 5 \\ 3.4(\textcolor{red}{-5}) + 4 &\stackrel{?}{=} 1.6(\textcolor{red}{-5}) - 5 \\ -17 + 4 &\stackrel{?}{=} -8 - 5 \\ -13 &= -13\; \checkmark \end{split}$$
##### Exercise $$\PageIndex{19}$$:

Solve: 2.8x + 12 = −1.4x − 9.

x = -5

##### Exercise $$\PageIndex{20}$$:

Solve: 3.6y + 8 = 1.2y − 4.


text:
# pgr_johnson¶

## Synopsis¶

pgr_johnson - Returns the sum of the costs of the shortest path for each pair of nodes in the graph using Floyd-Warshall algorithm.

Availability: 2.0.0

• Renamed on 2.2.0, previous name pgr_apspJohnson

The Johnson algorithm, is a good choice to calculate the sum of the costs of the shortest path for each pair of nodes in the graph, for sparse graphs. It usees the Boost’s implementation which runs in $$O(V E \log V)$$ time,

## Characteristics¶

The main Characteristics are:
• It does not return a path.
• Returns the sum of the costs of the shortest path for each pair of nodes in the graph.
• Process is done only on edges with positive costs.
• Boost returns a $$V \times V$$ matrix, where the infinity values. Represent the distance between vertices for which there is no path.
• We return only the non infinity values in form of a set of (start_vid, end_vid, agg_cost).
• Let be the case the values returned are stored in a table, so the unique index would be the pair: (start_vid, end_vid).
• For the undirected graph, the results are symmetric.
• The agg_cost of (u, v) is the same as for (v, u).
• When start_vid = end_vid, the agg_cost = 0.

## Signature Summary¶

pgr_johnson(edges_sql)
pgr johnson(edges_sql, directed)
RETURNS SET OF (start_vid, end_vid,  agg_cost) or EMPTY SET

## Signatures¶

### Minimal Signature¶

pgr_johnson(edges_sql)
RETURNS SET OF (start_vid, end_vid,  agg_cost) or EMPTY SET

Example 1: On a directed graph.
SELECT * FROM pgr_johnson(
'SELECT source, target, cost FROM edge_table WHERE id < 5
ORDER BY id'
);
start_vid | end_vid | agg_cost
-----------+---------+----------
1 |       2 |        1
1 |       5 |        2
2 |       5 |        1
(3 rows)

### Complete Signature¶

pgr_johnson(edges_sql, directed)
RETURNS SET OF (start_vid, end_vid,  agg_cost) or EMPTY SET

Example 2: On an undirected graph.
SELECT * FROM pgr_johnson(
'SELECT source, target, cost FROM edge_table WHERE id < 5
ORDER BY id',
false
);
start_vid | end_vid | agg_cost
-----------+---------+----------
1 |       2 |        1
1 |       5 |        2
2 |       1 |        1
2 |       5 |        1
5 |       1 |        2
5 |       2 |        1
(6 rows)

## Description of the Signatures¶

### Description of the edges_sql query (id is not necessary)¶

edges_sql: an SQL query, which should return a set of rows with the following columns:
Column Type Default Description
source ANY-INTEGER   Identifier of the first end point vertex of the edge.
target ANY-INTEGER   Identifier of the second end point vertex of the edge.
cost ANY-NUMERICAL

Weight of the edge (source, target)

• When negative: edge (source, target) does not exist, therefore it’s not part of the graph.
reverse_cost ANY-NUMERICAL -1

Weight of the edge (target, source),

• When negative: edge (target, source) does not exist, therefore it’s not part of the graph.

Where:

ANY-INTEGER: SMALLINT, INTEGER, BIGINT SMALLINT, INTEGER, BIGINT, REAL, FLOAT

### Description of the parameters of the signatures¶

Receives (edges_sql, directed)

Parameter Type Description
edges_sql TEXT SQL query as described above.
directed BOOLEAN (optional) Default is true (is directed). When set to false the graph is considered as Undirected

### Description of the return values¶

Returns set of (start_vid, end_vid, agg_cost)

Column Type Description
start_vid BIGINT Identifier of the starting vertex.
end_vid BIGINT Identifier of the ending vertex.
agg_cost FLOAT Total cost from start_vid to end_vid.

History

• Re-design of pgr_apspJohnson in Version 2.2.0


text:
Points on a sphere

Wonder whether you know where the following pearl of Topology first appeared:

Prove that at any instant of time you can find three isothermal points on the surface of the Earth that correspond to the vertices of an equilateral triangle.

According to somebody, all problems appeared once in the Monthly. Does that dictum applies to the above teaser, too?

Let me just elaborate a little on the references that Charlie Frohman listed (so this isn't really a separate answer, but it's too long for a comment).

The theorem for equilateral triangles is due to S. Kakutani (1942 Annals). He stated it just for triangles formed by orthonormal bases for $\mathbb R^3$, but the argument applies for all sizes of equilateral triangles. He deduced the interesting corollary that any compact convex set in $\mathbb R^3$ has a circumscribing cube, answering a question posed by Rademacher. Charlie Frohman's proof is similar in spirit to Kakutani's.

The n-dimensional generalization of Kakutani's theorem (and corollary) is due to H. Yamabe and Z. Yujobo (1950 Osaka Math J).

Returning to 3 dimensions, the corresponding result for 4 points at the vertices of a square inscribed in a great circle was shown by F. Dyson (1951 Annals), and G.R. Livesay generalized this to rectangles inscribed in a great circle (1954 Annals). The case of arbitrary triangles was done by E.E. Floyd (1955 PAMS).

It's interesting that this little topic produced three Annals papers, though each one was short -- just 3 pages. (When was the last 3-page Annals paper?)

The techniques used to prove the later theorems varied considerably. It might be interesting to see which ones can be proved by basic algebraic topology arguments as in Kakutani's theorem (which certainly deserves to be included in algebraic topology textbooks!).

• Oh the good ole days, when Annals published 3-page papers of general interest. – Victor Protsak Jun 2 '10 at 1:43
• Thanks a lot for taking the time to leave your reply, Professor Hatcher. – José Hdz. Stgo. Jun 3 '10 at 21:21


text:
# Cash for Gold Comercials

#### HarveyH42

Joined Jul 22, 2007
425
Noticed there has been an increasing number of companies advertising on TV, where you mail in gold jewelery and such, and they mail out a check. Many state that gold is at its highest value, so don't understand why a successful business would buy, when prices are high. I don't have any significant quantity of gold, nor would I trust it to the USPS, or strangers on TV. Figure its a simple way for burglars and robbers to unload their loot.

Been think if maybe this is some sort of economic recovery scheme, since its probably creating a demand for gold, which would also increase the value of some country's currency, probably not ours though. Don't really understand the significance of hoarding gold, just a shiny metal to me.

#### beenthere

Joined Apr 20, 2004
15,808
The price is way up, though. Perhaps as far as $1000/oz. Those are Troy ounces, though - 12 to the pound. But it's enough that there is a minor gold rush back in California, with unemployed people panning for gold in a couple of rivers. #### Wendy Joined Mar 24, 2008 21,848 That, and bet money they aren't paying full price. They operate at percentages, so even if the percentage is the same, the spread is wider than usual. #### Ratch Joined Mar 20, 2007 1,068 beenthere, Those are Troy ounces, though - 12 to the pound. Yes, but don't forget to emphasize that a Troy pound is 0.82 of what a Avoirdupois pound is. A Troy ounce is 480 grains and a Avoirdupois ounce is 437.5 grains. The price the gold skimmers give is the "melt price", which can be quite a bit lower than what fashioned jewelry might fetch. Jewelry is usually not pure gold. And can you trust the skinners not to cheat? Ratch #### wr8y Joined Sep 16, 2008 232 Noticed there has been an increasing number of companies advertising on TV, where you mail in gold jewelery and such, and they mail out a check. Many state that gold is at its highest value, so don't understand why a successful business would buy, when prices are high. Easy: They ain't paying you what it's worth! THEY ARE CROOKS! I TRIED to tell a friend to stay away from them, but she wouldn't listen. Sure I didn't KNOW they were crooks at the time, but it sounded "dirty" to me - like it does to you. This spring, she sent in some jewelry that the local jeweler offered her something like$175 or so for.

She thought "those guys on TV" would do better. Sent the stuff to them, got a check in 10 days or two weeks.
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--Rich

#### jpanhalt

Joined Jan 18, 2008
7,894

John

#### Wendy

Joined Mar 24, 2008
21,848
The real problem is if you send them your gold you've lost control. Even if you document it to the moon it still becomes your word vs. thiers, and after it is all said and done they'll return your package after an extreme amount of hassle. How many people will follow up? This is where the money is.

With a local jeweler you are dealing with someone local, and have a reputation to protect.

#### Dave

Joined Nov 17, 2003
6,960
Oddly enough, I just found a Cash4Gold spam in my email. They do get around.
Ha! No surprise at all. If there's money to be made, e-mail spammers with get in there, pronto!

Dave

#### Dave

Joined Nov 17, 2003
6,960
With a local jeweler you are dealing with someone local, and have a reputation to protect.
Yeah, I agree. It is more the point that until recently they weren't offering these "services". I guess where there is money to be made...

Dave


text:
Q

# Need explanation for: - Complex numbers and quadratic equations - JEE Main-7

If $\alpha$ and $\beta$ are the roots of the quadratic equation,

$x^{2}+xsin\theta -2sin\theta =0,\theta \epsilon (0,\frac{\pi}{2})$, then

$\frac{\alpha ^{12}+\beta ^{12}}{(\alpha ^{-12}+\beta ^{-12})\cdot (\alpha -\beta )^{24}}$ is equal to :

• Option 1)

$\frac{2^{12}}{(sin\theta-4)^{12}}$

• Option 2)

$\frac{2^{12}}{(sin\theta+8)^{12}}$

• Option 3)

$\frac{2^{12}}{(sin\theta-8)^{6}}$

• Option 4)

$\frac{2^{6}}{(sin\theta+8)^{12}}$

Views

$\alpha$ and $\beta$ are the roots of the equation

$x^{2}+xsin\theta -2sin\theta =0$

$\alpha +\beta =-sin\theta$

$\alpha \beta =-2sin\theta$

Now,

$\frac{\alpha ^{12}+\beta ^{12}}{(\alpha ^{-12}+\beta ^{-12})\cdot (\alpha -\beta )^{24}}=\frac{(\alpha\beta)^{12}}{(\alpha-\beta)^{24}}$

$=\frac{(\alpha \beta )^{12}}{[(\alpha +\beta )^{2}-4\alpha \beta ]^{12}}$

$=[\frac{-2sin\theta}{sin^{2}\theta+8sin\theta}]^{12}$

$=\frac{2^{12}}{(sin\theta+8)^{12}}$

correct option (2)

Option 1)

$\frac{2^{12}}{(sin\theta-4)^{12}}$

Option 2)

$\frac{2^{12}}{(sin\theta+8)^{12}}$

Option 3)

$\frac{2^{12}}{(sin\theta-8)^{6}}$

Option 4)

$\frac{2^{6}}{(sin\theta+8)^{12}}$

Exams
Articles
Questions


text:
# SSC MTS MATHS QUIZ

Attempt now to get your rank among 184 students!

## Question 1:

Sachin and Pawan started a boutique invest in amounts of Rs.35,000 and Rs.56,000 respectively. If Sachin earned profit of Rs.45,000 then what is the total profit earned in business.

## Question 2:

A and B entered into a partnership by investing some amounts. The investment of A is twice of the investment of B. Another person C joined them after 4 months. At the end of a year, the profit share of A and C is equal. Then find the profit share of B is what percent of the profit share of C.

## Question 3:

Shaheen, Noor, and Seema purchase a running business for Rs.23000. If Shaheen Invests ₹ 6,000 more than Noor and Seema invest ₹5,000 more than Noor, then out of the total profit of Rs. 1380, Noor receives-

## Question 4:

Shyana starts a business with Rs. 36000 . After a certain period of time she is joined by Abhishek, who invests Rs. 27000 . At the end of the year they divide the profit in the ratio of 8: 3. For what period did Abhishek join Shyana?

## Question 5:

A and B enter into a partnership with their capitals in the ratio 11: 7. At the start of the 8th month, A withdraws his capital. If they receive the profits in the ratio 11:10, then find for how long B's capital was used?

## Question 6:

Shivani and Shayna invested in the ratio $7: 4$ in a business if $10 \%$ of the total profit goes to the charity and Shivani's share is Rs. 1190 Find the total profit.

## Question 7:

A, B, C, subscribe together Rs. 50,000 for a business. A subscribes Rs. 4,000 more than B and B Rs. 5,000 more than C. Out of a total profit of Rs. 35,000, A receives :

## Question 8:

Edison and Elvis invested Rs.750 and Rs. 1200 respectively in a business. But due to some dispute Elvis got his money back after few month. If the profit of the year was shared between Edison and Elvis in the ratio $3: 4$, then after how many months did Elvis withdraw his money?

## Question 9:

A invests Rs. 10000 for 2 years at $10 \%$ per annum while B invests Rs. 18000 for 3 years at $20 \%$ per annum. Then find the difference between the simple interest earned by $\mathrm{A}$ and $\mathrm{B}$.

## Question 10:

A invested Rs. x and B invested Rs. (x + 800) in a business. After an year, the profit earned from the business is Rs. 6800. If A’s share in the profit is Rs.3200, then find the value of x.


text:
## Multivariable Calculus, 7th Edition

$\lt3sec^{2}(3x-4y),-4sec^{2}(3x-4y) \gt$
$f_x(x,y)=3sec^{2}(3x-4y)$ $f_y(x,y)=-4sec^{2}(3x-4y)$ Gradient vector field of $f$ is: $\lt3sec^{2}(3x-4y),-4sec^{2}(3x-4y) \gt$


text:
# Circuits

Electricity and Magnetism Level 3

Determine the current source of value $$I$$, knowing that $$V_r = -10V$$.

×


text:
# The /. syntax in numerical methods [duplicate]

I have an analytical approximation method for solving an ODE and want to compare this to an NDSolve solution, but the syntax is confusing me. To get the solution from NDSolve, I define the ODE as eqn and use the following command to solve for O2n;

s = NDSolve[{eqn == 0, O2n[rn] == 0, O2n[ro] == po/omega},
O2n, {r, rn, ro}]

Af = Plot[{(O2n[r]) /. s}, {r, rn, ro}, PlotStyle -> {Blue},
PlotLegends -> LineLegend[{"Numerical"}]]

This works fine, but the /. syntax confuses me a little - what exactly does it signify, and how can I manipulate this? Specifically, I want to subtract my analytical function p[r] from this solution over the same range (rn - ro) to see how much they differ. Ideally I'd like to be able to export this data and analyze it further - can anyone explain the syntax to me, and how I might go about comparing the twain? Thank you!

• Closely related: (3175), (6669), (9035) Sep 3 '13 at 16:19
• Sep 3 '13 at 16:45
• Plot prints table of values..NDSolve returns O2n[ro]->Something...check Table[O2n[r]) /. s, {r, rn, ro}]..you are replacing your dependent variable with your independent variable which you have liberty to assign. Sep 3 '13 at 16:51
• Ah cheers - ReplaceAll documentation not very illuminating coming from a MATLAB background, but that link is helpful. Thanks!
– DRG
Sep 3 '13 at 16:52
• I guess you are asking two questions, one of which is a candidate for closure (namely the ReplaceAll part)... to answer your second part (perhaps after an edit or another question), working code would be very useful. Sep 3 '13 at 17:35


text:
Definition 39.9.1. Let $k$ be a field. An abelian variety is a group scheme over $k$ which is also a proper, geometrically integral variety over $k$.

Comment #78 by Keenan Kidwell on

The section title should be Picard groups of curves, right?

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).


text:
MathsGee is Zero-Rated (You do not need data to access) on: Telkom |Dimension Data | Rain | MWEB

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A lucky dip at a school fête contains 100 packages of which 40 contain tickets for prizes. Let $X$ denote the number of prizes you win when you draw out three of the packages. Find the probability density of $X$ i.e. $P(X=i)$ for each appropriate $i$.
| 17 views

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There are $\left(\begin{array}{c}100 \\ 3\end{array}\right)$ choices of three packages (in any ordering). There are $\left(\begin{array}{c}60 \\ 3\end{array}\right)$ choices of three packages without prizes. Hence $P(X=0)=\left(\begin{array}{c}60 \\ 3\end{array}\right) /\left(\begin{array}{c}100 \\ 3\end{array}\right) \approx 0.2116$. If a single prize is won this can happen in $\left(\begin{array}{c}40 \\ 1\end{array}\right) \cdot\left(\begin{array}{c}60 \\ 2\end{array}\right)$ ways. Hence $P(X=1)=\left(\begin{array}{c}40 \\ 1\end{array}\right) \cdot\left(\begin{array}{c}60 \\ 2\end{array}\right) /\left(\begin{array}{c}100 \\ 3\end{array}\right) \approx 0.4378$ and similarly $P(X=2)=\left(\begin{array}{c}40 \\ 2\end{array}\right)$.
$\left(\begin{array}{c}60 \\ 1\end{array}\right) /\left(\begin{array}{c}100 \\ 3\end{array}\right) \approx 0.2894$ and $P(X=3)=\left(\begin{array}{c}40 \\ 3\end{array}\right) /\left(\begin{array}{c}100 \\ 3\end{array}\right) \approx 0.0611$ (there is some small rounding error in the
given values).
by Diamond (79,336 points)

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text:
# Meaning of Square of four vector potential

by merrypark3
Tags: meaning, potential, square, vector
 P: 28 The square of the four vector potential. $(\phi,A)^2=\phi ^2 - A^2$ What's the physical meaning of this lorentz invariant?


text:
# 9.1 Graphical representation of data

 Page 1 / 2

## Graphical representation of data

Once the data has been collected, it must be organised in a manner that allows for the information to be extracted most efficiently. One method of organisation is to display the data in the form of graphs. Functions and graphs have been studied in  Functions and Graphs , and similar techniques will be used here. However, instead of drawing graphs from equations as was done in Functions and graphs , bar graphs, histograms and pie charts will be drawn directly from the data.

## Bar and compound bar graphs

A bar chart is used to present data where each observation falls into a specific category and where the categories, this is often for qualitative data. The frequencies (or percentages) are listed along the $y$ -axis and the categories are listed along the $x$ -axis. The heights of the bars correspond to the frequencies. The bars are of equal width and should not touch neighbouring bars.

A compound bar chart (also called component bar chart) is a variant: here the bars are cut into various components depending on what is being shown. If percentages are used for various components of a compound bar, then the total bar height must be 100%. The compound bar chart is a little more complex but if this method is used sensibly, a lot of information can be quickly shown in an attractive fashion.

Examples of a bar and a compound bar graph, for Data Set 1 , are shown in [link] . According to the frequency table for Data Set 1, the coin landed heads-up 44 times and tails-up 56 times.

## Histograms and frequency polygons

It is often useful to look at the frequency with which certain values fall in pre-set groups or classes of specified sizes. The choice of the groups should be such that they help highlight features in the data. If these grouped values are plotted in a manner similar to a bar graph, then the resulting graph is known as a histogram. Examples of histograms are shown in [link] for Data Set 2, with group sizes of 1 and 2.

 Groups $0 1 $1 2 $2 3 $3 4 $4 5 $5 6 Frequency 30 32 35 34 37 32
 Groups $0 2 $2 4 $4 6 Frequency 62 69 69

The same data used to plot a histogram are used to plot a frequency polygon, except the pair of data values are plotted as a point and the points are joined with straight lines. The frequency polygons for the histograms in [link] are shown in [link] .

where we get a research paper on Nano chemistry....?
nanopartical of organic/inorganic / physical chemistry , pdf / thesis / review
Ali
what are the products of Nano chemistry?
There are lots of products of nano chemistry... Like nano coatings.....carbon fiber.. And lots of others..
learn
Even nanotechnology is pretty much all about chemistry... Its the chemistry on quantum or atomic level
learn
da
no nanotechnology is also a part of physics and maths it requires angle formulas and some pressure regarding concepts
Bhagvanji
hey
Giriraj
Preparation and Applications of Nanomaterial for Drug Delivery
revolt
da
Application of nanotechnology in medicine
what is variations in raman spectra for nanomaterials
ya I also want to know the raman spectra
Bhagvanji
I only see partial conversation and what's the question here!
what about nanotechnology for water purification
please someone correct me if I'm wrong but I think one can use nanoparticles, specially silver nanoparticles for water treatment.
Damian
yes that's correct
Professor
I think
Professor
Nasa has use it in the 60's, copper as water purification in the moon travel.
Alexandre
nanocopper obvius
Alexandre
what is the stm
is there industrial application of fullrenes. What is the method to prepare fullrene on large scale.?
Rafiq
industrial application...? mmm I think on the medical side as drug carrier, but you should go deeper on your research, I may be wrong
Damian
How we are making nano material?
what is a peer
What is meant by 'nano scale'?
What is STMs full form?
LITNING
scanning tunneling microscope
Sahil
how nano science is used for hydrophobicity
Santosh
Do u think that Graphene and Fullrene fiber can be used to make Air Plane body structure the lightest and strongest. Rafiq
Rafiq
what is differents between GO and RGO?
Mahi
what is simplest way to understand the applications of nano robots used to detect the cancer affected cell of human body.? How this robot is carried to required site of body cell.? what will be the carrier material and how can be detected that correct delivery of drug is done Rafiq
Rafiq
if virus is killing to make ARTIFICIAL DNA OF GRAPHENE FOR KILLED THE VIRUS .THIS IS OUR ASSUMPTION
Anam
analytical skills graphene is prepared to kill any type viruses .
Anam
Any one who tell me about Preparation and application of Nanomaterial for drug Delivery
Hafiz
what is Nano technology ?
write examples of Nano molecule?
Bob
The nanotechnology is as new science, to scale nanometric
brayan
nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale
Damian
Is there any normative that regulates the use of silver nanoparticles?
what king of growth are you checking .?
Renato
What fields keep nano created devices from performing or assimulating ? Magnetic fields ? Are do they assimilate ?
why we need to study biomolecules, molecular biology in nanotechnology?
?
Kyle
yes I'm doing my masters in nanotechnology, we are being studying all these domains as well..
why?
what school?
Kyle
biomolecules are e building blocks of every organics and inorganic materials.
Joe
Got questions? Join the online conversation and get instant answers!


text:
# Integrally closed

In mathematics, more specifically in abstract algebra, the concept of integrally closed has two meanings, one for groups and one for rings.

## Commutative rings

A commutative ring $R$ contained in a ring $S$ is said to be integrally closed in $S$ if $R$ is equal to the integral closure of $R$ in $S$. That is, for every monic polynomial f with coefficients in $R$, every root of f belonging to S also belongs to $R$. Typically if one refers to a domain being integrally closed without reference to an overring, it is meant that the ring is integrally closed in its field of fractions.

If the ring is not a domain, typically being integrally closed means that every local ring is an integrally closed domain.

Sometimes a domain that is integrally closed is called "normal" if it is integrally closed and being thought of as a variety. In this respect, the normalization of a variety (or scheme) is simply the $\operatorname{Spec}$ of the integral closure of all of the rings.

## Ordered groups

An ordered group G is called integrally closed if and only if for all elements a and b of G, if anb for all natural n then a ≤ 1.

This property is somewhat stronger than the fact that an ordered group is Archimedean. Though for a lattice-ordered group to be integrally closed and to be Archimedean is equivalent. We have the surprising theorem that every integrally closed directed group is already abelian. This has to do with the fact that a directed group is embeddable into a complete lattice-ordered group if and only if it is integrally closed. Furthermore, every archimedean lattice-ordered group is abelian.

## References

• R. Hartshorne, Algebraic Geometry, Springer-Verlag (1977)
• M. Atiyah, I. Macdonald Introduction to commutative algebra Addison-Wesley Publishing Co., Reading, Mass.-London-Don Mills, Ont. 1969
• H. Matsumura Commutative ring theory. Translated from the Japanese by M. Reid. Second edition. Cambridge Studies in Advanced Mathematics, 8.
• A.M.W Glass, Partially Ordered Groups, World Scientific, 1999


text:
(17 votes, average: 4.59 out of 5)

# Inter Carrier Interference (ICI) in OFDM due to frequency offset

by on August 8, 2009

In this post, let us evaluate the impact of frequency offset resulting in Inter Carrier Interference (ICI) while receiving an OFDM modulated symbol. We will first discuss the OFDM transmission and reception, the effect of frequency offset and later we will define the loss of orthogonality and resulting signal to noise ratio (SNR) loss due to the presence of frequency offset. The analysis is accompanied by Matlab/Octave simulation scripts.

## OFDM transmission

As discussed in the post on Understanding an OFDM transmission,  for sending an OFDM modulated symbol, we use multiple sinusoidals with frequency separation $\frac{1}{T}$ is used, where $T$ is the symbol period. The information $a_k$ to be send on each subcarrier $k$ is multiplied by the corresponding carrier $g_k(t)=e^{\frac{j2\pi kt}{T}}$ and the sum of such modulated sinusoidals form the transmit signal. Mathematically, the transmit signal is,

$\begin{eqnarray}s(t) &= &a_0g_0(t) + a_1g_1(t)+\ldots+a_{K-1}g_{K-1}(t)\\ & = &\sum_{0}^{K-1}a_kg_k(t) \\&=&\frac{1}{\sqrt{T}}\underbrace{\sum_{0}^{K-1}a_ke^{\frac{j2\pi kt}{T}}}\ w(t) \end{eqnarray}$

The interpretation of the above equation is as follows:
(a) Each information signal $a_k$ multiplies the sinusoidal having frequency of $\frac{k}{T}$.
(b) Sum of all such modulated sinusoidals are added and the resultant signal is sent out as $s(t)$.

## OFDM reception

In an OFDM receiver, we will multiply the received signal with a bank of correlators and integrate over the period $T$. The correlator to extract information send on subcarrier  $k$ is$c_m(t)=\frac{1}{\sqrt{T}}e^{-\frac{j2\pi mt}{T}}$.

The integral,
$\begin{eqnarray}\frac{1}{\sqrt{T}}\int_Ts(t)e^{-\frac{j2\pi mt}{T}} & = & a_k,& m=k\\ & = & 0, & m\ne k\end{eqnarray}$,

where
$m$ takes values from $0$ till $K-1$.

## Frequency offset

In a typical wireless communication system, the signal to be transmitted is upconverted to a carrier frequency prior to transmission. The receiver is expected to tune to the same carrier frequency for down-converting the signal to baseband, prior to demodulation.

Figure: Up/down conversion

However, due to device impairments the carrier frequency of the receiver need not be same as the carrier frequency of the transmitter. When this happens, the received baseband signal, instead of being centered at DC (0MHz), will be centered at a frequency $f_{\delta}$, where
$f_{\delta} = f_{Tx} - f_{Rx}$.

The baseband representation is (ignoring noise),

$y(t) = s(t)e^{j2\pi f_{\delta}t}$, where

$y(t)$ is the received signal

$s(t)$ is the transmitted signal and

$f_{\delta}$ is the frequency offset.

## Effect of frequency offset in OFDM receiver

Let us assume that the frequency offset $f_{\delta}$ is a fraction of subcarrier spacing $1/T$ i.e.
$f_{\delta}=\frac{\delta}{T}$.

Also, for simplifying the equations, lets us assume that the transmitted symbols on all subcarriers,  $a_k=1$

The received signal is,

$y(t) = s(t)e^{\frac{j2\pi\delta }{T}t}$.

The output of the correlator for sub-carrier $m$ is,

$\begin{eqnarray}\frac{1}{\sqrt{T}}\int_Ty(t)e^{-\frac{j2\pi mt}{T}} & = & \frac{1}{T}\int_{T}e^{\frac{j2\pi (k+\delta-m)}{T}t},& m=\in \{0,...,K-1\}\\ & = & \frac{1}{j2\pi (k+\delta-m)}(e^{j2\pi (k+\delta-m)}-1) & \end{eqnarray}$.

For  $\delta=0$,

$\begin{array}{rrr}\frac{1}{j2\pi(k+\delta-m)}(e^{j2\pi (k+\delta-m)}-1)&=1, &m=k\\ \frac{1}{j2\pi(k+\delta-m)}(e^{j2\pi (k+\delta-m)}-1)&=0, &m \ne k\\\end{array}$

The integral reduces to the OFDM receiver with no impairments case.

However for non zero values of $\delta$, we can see that the amplitude of the correlation with subcarrier $m$ includes

• distortion due to frequency offset between actual frequency $\frac{m+\delta}{T}$ and the desired frequency $\frac{m}{T}$.
• distortion due to interference with other subcarriers with with desired frequency $\frac{m}{T}$. This term is also known as Inter Carrier Interference (ICI).

## Simulation Model

Click here to download the Matlab/Octave script for computing RMS error with frequency offset. The script performs the following:

1. Generates an OFDM symbol with all subcarriers modulated with $a_k=1$.

2. Introduce frequency offset and add noise to result in $\frac{E_b}{N_0}$=30dB.

3. Perform demodulation at the receiver.

4. Find the difference between the desired and actual constellation.

5. Compute the rms value of error across all subcarriers.

6. Repeat this for different values of frequency offset.

Figure: Error Magnitude vs frequency offset for OFDM

## Observations

The theoretical results are computed by $\sum_m\frac{1}{j2\pi (k+\delta-m)}e^{j2\pi (k+\delta-m)}$.

Quite likely the simulated results are slightly better than theoretical results because the simulated results are computed using average error for all subcarriers (and the subcarriers at the edge undergo lower distortion).

D id you like this article? Make sure that you do not miss a new article by subscribing to RSS feed OR subscribing to e-mail newsletter. Note: Subscribing via e-mail entitles you to download the free e-Book on BER of BPSK/QPSK/16QAM/16PSK in AWGN.

{ 65 comments… read them below or add one }

mahender January 24, 2013 at 10:04 pm

hidden morkov model and selected applications in speech recognition in mat lab implementation . in which i didn,t get an idea that can restructure the model by using matlab code.please can you help me to proceed to write a code in mat lab.and can u help me to modification in the existing hidden model in speech recognition so that we will get benefit .

Krishna Sankar February 1, 2013 at 5:15 am

@mahender: sorry, i have not studied much on the topic of speech recognition

turki khadija December 14, 2012 at 2:07 am

I thank you for your course it s helpful

tejashri October 20, 2012 at 7:33 pm

hello…
i am doing my PG project in ofdm communication…..can u plz give me the matlab code for ICI SELF CANCELLATION SCHEME IN OFDM SYSTEM according to IEEE 802.11a standard….or suggest some reference material…which will help me…..

thanks…..

Krishna Sankar October 24, 2012 at 7:54 am

@tejashri: I ‘ve not not studied that topic. For OFDM related posts, please checkout
http://www.dsplog.com/category/ofdm/

kareem hamed August 8, 2012 at 6:14 pm

please Krishna i work the 4Pam OFDM AWGN code and the code not run i want ask you what wrong in the program my email:engkareem.aast@hotmail.com:
clc
clear all
nFFT = 64; % fft size
nDSC = 52; % number of data subcarriers
nBitPerSym = 52; % number of bits per OFDM symbol (same as the number of subcarriers for PAM)
nSym = 10^4; % number of symbols
N = 10^5; % number of symbols
alpha4pam = [-3 -1 1 3]; % 4-PAM alphabets
EbN0dB = [0:10]; % bit to noise ratio
EsN0dB = EbN0dB + 10*log10(nDSC/nFFT) + 10*log10(64/80); % converting to symbol to noise ratio

for ii = 1:length(EbN0dB)

% Transmitter
ipMod = randsrc(1,nBitPerSym*nSym,alpha4pam);

ipMod = reshape(ipMod,nBitPerSym,nSym).’

% Assigning modulated symbols to subcarriers from [-26 to -1, +1 to +26]
xF = [zeros(nSym,6) ipMod(:,[1:nBitPerSym/2]) zeros(nSym,1) ipMod(:,[nBitPerSym/2+1:nBitPerSym]) zeros(nSym,5)] ;

% Taking FFT, the term (nFFT/sqrt(nDSC)) is for normalizing the power of transmit symbol to 1
xt = (nFFT/sqrt(nDSC))*ifft(fftshift(xF.’)).’;

% Appending cylic prefix
xt = [xt(:,[49:64]) xt];

% Concatenating multiple symbols to form a long vector
xt = reshape(xt.’,1,nSym*80);

% Gaussian noise of unit variance, 0 mean
nt = 1/sqrt(2)*[randn(1,nSym*80) + j*randn(1,nSym*80)];

% Adding noise, the term sqrt(80/64) is to account for the wasted energy due to cyclic prefix
yt = sqrt(80/64)*xt + 10^(-EsN0dB(ii)/20)*nt;

yt = reshape(yt.’,80,nSym).’; % formatting the received vector into symbols
yt = yt(:,[17:80]); % removing cyclic prefix

% converting to frequency domain
yF = (sqrt(nDSC)/nFFT)*fftshift(fft(yt.’)).’;
yMod = yF(:,[6+[1:nBitPerSym/2] 7+[nBitPerSym/2+1:nBitPerSym] ]);

% demodulation
r = real(yt); % taking only the real part

ipHat(find(r= 2/sqrt(5))) = 3;
ipHat(find(r>=-2/sqrt(5) & r=0 & r<2/sqrt(5))) = 1;

% converting modulated values into bits
ipBitHat = (ipHat+3)/2;
ipBitHat = reshape(ipBitHat.',nBitPerSym,nSym).';

% counting the errors
nErr(ii) = size(find(ipBitHat – ipMod),2);

end

simBer = nErr/N;
theoryBer = 0.75*erfc(sqrt(0.2*(10.^(Es_N0_dB/10))));
close all
figure
semilogy(Es_N0_dB,theoryBer,'b.-');
hold on
semilogy(Es_N0_dB,simBer,'mx-');
axis([-3 20 10^-5 1])
grid on
legend('theory', 'simulation');
xlabel('Es/No, dB')
ylabel('bit Error Rate of 4 pam OFDM')
title('bit error probability curve for 4-PAM modulation')

Krishna Sankar August 9, 2012 at 6:14 am

@kareem: Are you getting zero BER in no noise case?
How are the bits getting converted to modulation symbols?

Ahmed October 27, 2011 at 3:22 am

Hello,

Can you told me, how to simplify

sum(a^2)/sum(a)

where a is a vector of N complex variables.

Krishna Sankar October 30, 2011 at 7:29 pm

@Ahmed: How about sum(a.^2)./sum(a) in matlab ? Note the dot (.)

nanu nadoda May 18, 2011 at 3:26 pm

hi krishna
find here with program for standard OFDM simulator in MATLAB
%%OFDM simulator
ep = [0 .15 .3 ];
% Es/No
EbNo=0:30;
% NS = number of OFDM symbols to transmit
NS = 100;
% Modulation type
modulation = ‘psk’;
M = 4
% N = number of carriers in OFDM symbol
N = 52;
% BPS = number of bits per OFDM symbol
BPS = N*log2(M);
% index for carriers in OFDM symbol
carriers = (1:52) + 2;
% IFFT size
ifftsize = 64;
% Input Bit Stream is normally Distributed
input_bit_stream = sign(randn(1,BPS*NS));
input_bit_stream(input_bit_stream == -1) = 0;
% PERFORM SERIAL TO PARALLEL CONVERSION
parallel_data = reshape(input_bit_stream, log2(M));
% PERFORM MODULATION
modulated_data = dmodce(parallel_data, 1, 1, modulation, M);
% CREATE OFDM SYMBOLS
disp(‘Transmitting OFDM symbols’)
%Normalize
%modulated_data = (modulated_data/max(abs(modulated_data)));
for ll= 1:length(ep)
for l=1:length(EsNo)
k = 1;
for n = 1:NS
ofdm_symbol = zeros(1,ifftsize);
% Map modulated data to FFT bins in OFDM symbol
ofdm_symbol(carriers) = modulated_data(k:k+51);
% Time Signal to transmit
tx_signal = ifft(ofdm_symbol,ifftsize);
% DOPPLER SHIFT
rx_signal = tx_signal.*exp((j*pi*ep(ll)/ifftsize)*(0:ifftsize-1));
noise = sqrt(1/(2*log2(M)*10^(EsNo(l)/10)))*(randn(1,64)+j*randn(1,64));
rx_signal = rx_signal + noise;
% FFT
received_ofdm = fft(rx_signal, ifftsize);
% Extract data from carriers in OFDM symbol
k = k + 52;
end
% PERFROM DEMODULATION
received_data = ddemodce(received_symbols, 1, 1, modulation, M);
%PERFORM PARALLEL TO SERIAL CONVERSION
output_bit_stream = reshape(received_data, log2(M));
% CALCULATE BER
BER(ll,l)= sum(xor(input_bit_stream, output_bit_stream))/length(input_bit_stream);
end
end
===================================================
i am experiencing problem in reshape. if you can rectify then i will be highly oblige.
send me reply on my mail : nbnadoda@yahoo.co.in

Krishna Sankar May 23, 2011 at 2:46 am

@nanu: Am sure you will be able to identify the problem in reshape() yourself

Thavamaran November 17, 2010 at 12:24 am

I am surprised as well Krishna, its the non-linear dynamics of the laser. I am still analyzing now. thanks anyway

Krishna Sankar November 17, 2010 at 4:35 am

@Thavamaran: Ok, good luck

Thavamaran November 4, 2010 at 4:57 pm

Krishna, thanks for the quick reply. I dont get your question on linearly varying phase? But the constellation always has 45 or 135 or 225 or 315 degree of phase shift, so it has about 90 degree difference between them.

Anyway symbol timing synch has a lot of method, which method can I look into, do you have any code referring to timing synch?

For receiver portion, im using correlator receiver, not PLL cause I feel its pointless cause correlator works extremely well for my back-to-back.

Krishna Sankar November 15, 2010 at 2:08 am

@Thavamaran: Not sure why you are seeing a fixed phase shift. Kinda surprising.

eng_dina November 4, 2010 at 1:58 am

hello please I try to estimate frequency offset in mimo ofdm using pn sequences have you an exaples for matlab codes could help me i make acquistion and tracking please advise it’s important .thanks

Krishna Sankar November 15, 2010 at 2:08 am

@eng_dina: Sorry, I do not have the matlab codes.

Thavamaran November 4, 2010 at 12:20 am

Hi Krishna, this is my first post in your website. This is a very straight forward example. I have built my OFDM system and my back-to-back works perfectly fine. But when I perform optical OFDM, which is part of my work, at the receiver, the constellation point shifted by 45 degree, then I just compensate the phase, but end up I realize that its nonlinear phase shift.

But I dont think I have a problem with CFO, cause its ideal. Do you think I need timing symbol synchronization? Please advise. Thanks.

Krishna Sankar November 4, 2010 at 12:31 am

@Thavamaran: Are you having a linearly varying phase across subcarriers? If yes, it might be related to the symbol timing synchronization.

Obaid April 7, 2010 at 6:55 pm

Hi krishna

Kindly tell me why are we doing this: theoryErr = (theoryErr-1);

Regards
Obaid

Shaimaa March 28, 2010 at 4:32 pm

Thank you for replying, I will see the post then I will back to discuss you.

Krishna Sankar March 29, 2010 at 6:34 am

@Shaimaa: Ok

Shaimaa March 27, 2010 at 7:22 pm

The simulated curve was around 50% and increased slightly as EbNo increases.
Is it necessary to compensate for frequency offset before demodulating the bpsk symbols as in Rayleigh fading channel??????????

Krishna Sankar March 28, 2010 at 1:40 pm

@Shaimaa: Typically, receivers will have a circuit for compensating frequency offset prior to attempting demodulation. I have written a brief post on how to estimate frequency offset with short preamble defined in 802.11a
http://www.dsplog.com/2008/03/03/frequency-offset-estimation-using-80211a-short-preamble/

Shaimaa March 27, 2010 at 7:12 pm

I added the code
% Adding frequency offset
xt = xt.*exp(j*2*pi*freqOffsetkHz_v*(1e3/20e6)*[0:length(xt)-1]);
with freqOffsetkHz_v values=10 or 20 kHz
in your script_ber_bpsk_ofdm then run the program but the simulated curve was so bad. Is that expected

Krishna Sankar March 28, 2010 at 1:39 pm

@Shaimaa: If there is frequency error, then the intercarrier interference can result in poor BER. However, I have not done the simulations to see how much bad it can get.

PRIYA March 14, 2010 at 10:29 pm

hi…
what are the problems are occur in high PAPR?

Krishna Sankar March 28, 2010 at 3:33 pm

@PRIYA: Power amplifier efficiency is comprimised

eng_dina February 6, 2010 at 5:44 am

please reply my massege it’s really important
thanks

invizible February 5, 2010 at 7:09 pm

Hi krishna,
I hope you are doing fine and great. I have seen your code and its great. here you have supposed that all the input symbols are modulated like ak=1, if I am right for this reason you are generating all ones as the input i.e.
ipBit = ones(1,nBitPerSym*nSym) > 0.5; (kindly correct me if I am wrong)

My point is that when we generate random bits i.e. ones and zeros, then the simmulated error is higher than the theoretical error …
kindly help in this regard …

Krishna Sankar April 4, 2010 at 4:20 am

@invizible: Oh, is it. I have not tried that.

eng_dina February 3, 2010 at 5:10 am

Idon’t try to simulte cfo for mimo system only but for mimo ofdm I propose a new scheme that targets MMIO OFDM systems which have unsynchronized oscillators such that CFO of individual paths have to be estimated separately. This scheme may also apply to OFDM systems with multi-user access. The new method, which is similar to Moose’s method, estimates the CFO by measuring the carrier phase difference between 2 identical successive training sequences embedded in the preambles. In order to make CFO estimates be more time efficient,I allow 2 transmitter antennas transmit their training sequence concurrently in every time period, except the first and the last period. I use Frank-Zadoff code with different phase shifts in the training sequences in different antennas. Due to the good correlation property of Frank-Zadoff code, it helps reduce the interference caused by the concurrent transmissions from other antennas.
please reply me it’s urgent Ineed the code it’s really important and thanks alot

Krishna Sankar April 4, 2010 at 4:26 am

@eng_dina: Good luck with you are algorithm investigation. Hope you have made the simulations and obtained satisfactory results.

Ahmed February 2, 2010 at 9:05 pm

Could you help me to find the theoretical BER of OFDM with Walsh-Hadamard over the multipath transmission

Krishna Sankar April 4, 2010 at 4:27 am

@Ahmed: Sorry, I do not have the required information

eng_dina February 1, 2010 at 11:51 pm

Idon’t try to simulte cfo for mimo system only but for mimo ofdm I propose a new scheme that targets MMIO OFDM systems which have unsynchronized oscillators such that CFO of individual paths have to be estimated separately. This scheme may also apply to OFDM systems with multi-user access. The new method, which is similar to Moose’s method, estimates the CFO by measuring the carrier phase difference between 2 identical successive training sequences embedded in the preambles. In order to make CFO estimates be more time efficient,I allow 2 transmitter antennas transmit their training sequence concurrently in every time period, except the first and the last period. I use Frank-Zadoff code with different phase shifts in the training sequences in different antennas. Due to the good correlation property of Frank-Zadoff code, it helps reduce the interference caused by the concurrent transmissions from other antennas.
cfo for mimo only but for mimo ofdm system

eng_dina January 28, 2010 at 4:37 pm

Idon’t try to simulte cfo for mimo system only but for mimo ofdm I propose a new scheme that targets MMIO OFDM systems which have unsynchronized oscillators such that CFO of individual paths have to be estimated separately. This scheme may also apply to OFDM systems with multi-user access. The new method, which is similar to Moose’s method, estimates the CFO by measuring the carrier phase difference between 2 identical successive training sequences embedded in the preambles. In order to make CFO estimates be more time efficient,I allow 2 transmitter antennas transmit their training sequence concurrently in every time period, except the first and the last period. I use Frank-Zadoff code with different phase shifts in the training sequences in different antennas. Due to the good correlation property of Frank-Zadoff code, it helps reduce the interference caused by the concurrent transmissions from other antennas.
cfo for mimo only but for mimo ofdm system

eng_dina January 28, 2010 at 4:35 pm

Idon’t try to simulte After examining some synchronization I propose a new scheme that targets MMIO OFDM systems which have unsynchronized oscillators such that CFO of individual paths have to be estimated separately. This scheme may also apply to OFDM systems with multi-user access. The new method, which is similar to Moose’s method, estimates the CFO by measuring the carrier phase difference between 2 identical successive training sequences embedded in the preambles. In order to make CFO estimates be more time efficient,I allow 2 transmitter antennas transmit their training sequence concurrently in every time period, except the first and the last period. I use Frank-Zadoff code with different phase shifts in the training sequences in different antennas. Due to the good correlation property of Frank-Zadoff code, it helps reduce the interference caused by the concurrent transmissions from other antennas.
cfo for mimo only but for mimo ofdm system

eng_dina January 27, 2010 at 4:18 pm

please I study for my master in frequency synchronization in mimo ofdm system but i have problem with the matlab code to simulate to find out if estimation of the CFO on one path is affected by the CFO values of the adjacent paths and examine the estimator accuracy in term of its mean and variance

f_max = f_Tofdm/(M_sub*T_s); % f_max =maximal Doppler frequency (Hz),M_sub*T_s = OFDM Symbol period (sec),
%f_ndopp = f_max*M_sub*T_s % Normalized Maximum Doppler Spread Freq.

% Area parameter
% ra Rural Area
% tu Typical Urban
% bu Bad Urban
% ht Hilly Terrain
% no no fading
AREA = ‘no’ ;

% start simulation loops
for ee = 1 : length(EcNo)
ee
[theo_fo_var, rfo_var, rfo] = FOE_mimo_fading(EcNo(ee), tfo_array, NTX, NRX, M_sub, N, f_max, AREA, T_s, model, L);
plot_theo_fo_var(:,:,ee)=theo_fo_var;
plot_rfo_var(:,:,ee)=rfo_var;
plot_theo_rfo(:,:,ee)=tfo_array;
plot_rfo(:,:,ee)=rfo;
end

% plot results
for ii_tx = 1 : NTX
for ii_rx = 1 : NRX

% rearrange array order for easier plotting
vect_plot_theo_rfo = permute(plot_theo_rfo(ii_tx,ii_rx, , [2 3 1]);
vect_plot_rfo= permute(plot_rfo(ii_tx,ii_rx, , [2 3 1] ) ;
vect_plot_theo_fo_var= permute(plot_theo_fo_var(ii_tx,ii_rx,:),[2 3 1]);
vect_plot_rfo_var= permute(plot_rfo_var(ii_tx,ii_rx, , [2 3 1]);

%plot mean
figure;
plot(EcNo, vect_plot_theo_rfo, ‘:’ , ‘Linewidth’,2) ;
hold;
plot(EcNo, vect_plot_rfo,’o-’,'MarkerSize’ , 6,’Linewidth’, 1) ;
grid on;
t_str=sprintf(‘New_ Theo and Est Freq Offset %s M=%d N=%d %dx%d path:tx%d-rx%d fo=%1.3f’,AREA,M_sub,N,NTX,NRX,ii_tx,ii_rx,tfo_array(ii_tx,ii_rx));
title(t_str);
xlabel(‘EcNo SNR range and step (in dB)’);
ylabel(‘Norm Freq Offset’) ;
legend(‘Theo’,'Est’,'Location’,'Northeast’);
% save graph
fstr=sprintf(‘new_m%d_%dx%dp%d%d_%s’,M_sub,NTX,NRX,ii_tx,ii_rx,AREA);
hgsave(fstr);

% plot variance
figure;
semilogy(EcNo, vect_plot_theo_fo_var,’:',’Linewidth’,2);
hold;
semilogy(EcNo, vect_plot_rfo_var,’o-’,'MarkerSize’,6,’Linewidth’,1);
grid on;
grid minor;
t_str=sprintf(‘New_ Est Var and CRLB %s M=%d N=%d %dx%d path:tx%d-rx%d fo=%1.3f’,AREA,M_sub,N,NTX,NRX,ii_tx,ii_rx,tfo_array(ii_tx,ii_rx));
title(t_str);
xlabel(‘EcNo (in dB)’);
ylabel(‘Variance’);
legend(‘CRLB’,'Est’,'Location’,'Northeast’);
% save graph
fstr=sprintf(‘new_y%d_%dx%dp%d%d_%s’,M_sub,NTX,NRX,ii_tx,ii_rx, AREA) ;
hgsave(fstr);
end
end
% rename variables for export
new_plot_theo_fo_var=plot_theo_fo_var;
new_plot_rfo_var=plot_rfo_var ;
new_plot_theo_rfo=plot_theo_rfo;
new_plot_rfo=plot_rfo;

% save all variables
fstr=sprintf(‘new_f%d_%dx%d_%s.mat’,M_sub,NTX,NRX,AREA);
save(fstr);
% save variables for plotting
fstr=sprintf(‘new_p%d_%dx%d_%s.mat’,M_sub,NTX,NRX,AREA);
save(fstr,’new_plot_theo_fo_var’,'new_plot_rfo_var’,'new_plot_theo_rfo’,'new_plot_theo_rfo’);

% end of file
and this is the core of the code

Krishna Sankar January 28, 2010 at 5:25 am

@eng_dina: How are you modeling the CFO for MIMO systems?

If all the chains have a common RF clock, then all the chains will have similar CFO and the estimate from all the chains can be combined to improve the accuracy of the CFO estimation.
If the chains have independent RF clock, then we need to estimate CFO on each chain independently.

eng_dina January 26, 2010 at 12:22 am

thanks alot for your helpful work please I work in my master in frequency synchronization in mimo ofdm and I have problem in the matlab code try to examine the estimator accuracy in term of its mean and variance and find out if estimation of the CFO on one path is affected by the CFO values of the adjacent paths

joel January 14, 2010 at 4:36 pm

Hello,

can You explain why is sample frequency (20e6) in the term freqOffsetkHz_v(ii)*(1e3/20e6) necessary ? We need delta_f so it shall be actually -200:200kHz and not delta_f*sample_time …

best regards

Obaid January 13, 2010 at 8:17 pm

Hi krishna

Can you please provide any references (books or web links) for SNR loss calculations (analytical) for AWGN and dispersive channels.

Regards.
Obaid

Ayesa January 2, 2010 at 4:11 pm

Sir,
I want to find BER for 4QAM in OFDM using AWGN. If i take the difference between reciver’s demodulated symbols and actual symbols(at transmitter) and devide the difference by the length of actual symbols, can i get the BER?
Thanks.

shindeujwala December 30, 2009 at 7:37 pm

Hello Sir!
i gone through above code.it is useful to understand ici. but i have queery that how to cancel ici due to phase offset due to local oscillator frequency offset.
thank you.

r.kalidoss November 25, 2009 at 10:23 am

Can i think, delay spread is equal to total number of multiple paths in the system

Krishna Sankar December 7, 2009 at 4:31 am

@r.kalidoss: Delay spread is computed using the amplitude of the taps and their delay values.

cyrus November 8, 2009 at 9:42 pm

Hi there…for one more question..

ipBit = ones(1,nBitPerSym*nSym) > 0.5; % random 1′s and 0′s

if i change “ones” to “rand”, the result seems to have differences. Can you explain it?

Thanks very much

Krishna Sankar November 12, 2009 at 5:38 am

@cyrus:
With ak’s randomly distributed, the interference term slightly changes. How much difference did you see?

cyrus November 8, 2009 at 9:32 pm

1.I would like to know that why we need to have 1e3 in the following expression.
% Adding frequency offset
xt = xt.*exp(j*2*pi*freqOffsetkHz_v(ii)*(1e3/20e6)*[0:length(xt)-1]);

2.As you mentioned about,choosing -5 to 5 as subcarriers outside that range has negligible interference.I still have a bit of confuse about it.Can i change to others values?

Krishna Sankar November 12, 2009 at 5:33 am

@cyrus:
1/ To convert kHz to Hz
2/ Sure, you can.

Treasa November 4, 2009 at 5:09 pm

Hi there,
I’m a bit confused about the theorical results calculation..
theoryErr =sum(1./(j*2*pi*([-5:5]+delta)).*(exp(j*2*pi*([5:5]+delta))-1));

could someone explain to me please how ([-5:5]+delta) relates to (k-m+delta)? Well I of course get the delta part just not sure how -5:5 was chosen?

Would really appreciate any reply

T

Krishna Sankar November 8, 2009 at 8:53 am

@Treasa: I just chose -5 to 5 as subcarriers outside that range has negligible interference.

Obaid October 30, 2009 at 4:49 pm

referring to last line of this post:
why the subcarriers at the edge undergo lower distortion ???

Krishna Sankar November 8, 2009 at 8:14 am

@Obaid: Well, at the edge there will be subcarriers only at the left or right and not both. Hence there will be lower interference.

Khaja Rasool October 25, 2009 at 10:44 am

Hi krishna,
thank you so much for ur help and posts in this website,

i am working on Next generation WLAN 802.11n Simulation.
I need ur help
i would like get the simulation for MIMO OFDM with MMSE equalizer .and also with MIMO OFDM STBC codes.

pls help me .i would greatful if you help me.
thank you once again.

Sk.khaja rasool.
Pls: do reply to this mail ID khajarasool_17@yahoo.com

Krishna Sankar October 27, 2009 at 5:41 am

@Khaja Rasool: Though I have not disucssed MIMO + OFDM or STBC + OFDM, I have discussed MIMO, STBC, and OFDM independently. You may refer
http://www.dsplog.com/tag/mimo
http://www.dsplog.com/tag/stbc
http://www.dsplog.com/tag/ofdm
Good luck.

r.kalidoss August 11, 2009 at 1:25 pm

Can i think, due to change of frequency i.e. m/T to m+q/T (Frequency offset) orthogonality of the signal change?

Krishna Sankar August 12, 2009 at 4:13 pm

@kalidoss: Yes, you are right.

r.kalidoss August 10, 2009 at 11:47 am

Krishna, Can we think frequency offset is due to doppler shift…..apart from carrier mismatch in the receiver….

Krishna Sankar August 11, 2009 at 5:06 am

@kalidoss: I have not discussed Doppler in this post. I guess the effect of Doppler might not be as simple to understand as is the case with frequency offset. I will write about Doppler soon.

r.kalidoss August 10, 2009 at 11:19 am

First time i saw the frequency offset problem in OFDM with simple example.Its fine working. Then to how to mitigate this problem.

Any method available to solve this issue.

Krishna Sankar August 11, 2009 at 4:56 am

@kalidoss: You may checkout the post on Frequency offset estimation using 802.11a short preamble
http://www.dsplog.com/2008/03/03/frequency-offset-estimation-using-80211a-short-preamble/

Urbie October 13, 2010 at 10:05 pm

Krishna,
I stand self-corrected. I realized that the term “[0:length(xt)-1]/Fs” corresponds to time, i.e t in the expression
exp(1i*2*pi*freqOffsetkHz_v(ii)*1e3*t), and you are simply shifting the spectrum by multiplying the two time-domain signals.
Sorry, I didn’t know what I was thinking.
Urbie


text:
lilypond-user
[Top][All Lists]

Re: difficulty implementing grob-suicide! for spanned bendAfter

 From: Mark Polesky Subject: Re: difficulty implementing grob-suicide! for spanned bendAfter Date: Thu, 9 Jul 2009 21:47:42 -0700 (PDT)

Mike Solomon wrote:
> Hey lilypond-users,
>     I am working on a piece with lots of beams/events across
> staves, many of which have bendAfter attached to them.  The
> spanned bend after clashes with some other stuff in the work, so
> I wanted to kill it w/ a callback function, but after having
> written the bit of code below I came to realize that this didn't
> exist as I was conceiving it (should have checked the docs
> first...). Taking a peek at the code below, do any of you have
> suggestions for how I could make this work?

Mike, I *do* want to be encouraging, but you really have to format
the scheme code. It's really hard to read, and the good developers
are not going to waste their time rewriting it.

*Please* do something more like this:

#(define (bendcallback grob)
(let* ((orig (ly:grob-original grob))
(siblings (if (ly:grob? orig)
(ly:spanner-broken-into orig)
'())))
(begin
(format #t "I will print if this function is called.\n")
(if (>= (length siblings) 2)
(if (not (eqv? grob (car siblings)))
(ly:grob-suicide! grob))))))

I can understand this in about 30 seconds. It would take me about
2 or 3 minutes to understand it the way you have it formatted. It's
quicker for me to rewrite it than to try to figure it out the way
you have it. But ultimately it's not worth my time to rewrite it.

If you have trouble finding matching parentheses, use an editor
with bracket matching. Personally I use LilyPondTool (there are
others as well) - I can put the cursor to the right of a ")", do
"<ctrl>-]" and it takes me to the matching "(".

Also, the code didn't compile - there's an error on this line:
\bendAfter #-3 c1 \bendAf

1) This:
\score {{ \new Staff {
\relative c'' { ...

can be simplified to this:
\score {
\new Staff \relative c'' {
...

2) This:
\override Voice . BendAfter #'after-line-break =
#(lambda (grob) (bendcallback grob))

can be simplified to this:
\override Voice.BendAfter #'after-line-break = #bendcallback

3) An initial quarter rest followed by whole notes doesn't fit the
time signature. Either remove the "r4" or add "\partial 4".

4) You can automate repetitive input this way:
\repeat unfold 45 { \bendAfter #-3 c1 }

5) The \layout block goes *inside* the \score block.

7) Before you post to the mailing list, always do a final
test compile to make sure what you're sending compiles!

___________________________________

So here's my rewrite:

\version "2.13.0"

#(define (bendcallback grob)
(let* ((orig (ly:grob-original grob))
(siblings
(if (ly:grob? orig)
(ly:spanner-broken-into orig)
'())))
(begin
(format #t "I will print if this function is called.\n")
(if (>= (length siblings) 2)
(if (not (eqv? grob (car siblings)))
(ly:grob-suicide! grob))))))

\score {
\new Staff \relative c'' {
\override Voice.BendAfter #'after-line-break = bendcallback
\repeat unfold 45 { \bendAfter #-3 c1 }
}
\layout {
\context {
\Voice
\remove "Forbid_line_break_engraver"
}
}
}

I'm only doing this so that you see what it should look like.
I understand that a lot of the coding stuff is tricky; there
are a lot of things that are easy for a beginner to forget and
many useful tricks that a beginner just doesn't know yet. Most
of us are more than happy to help out with those things.

But we can't and won't keep rewriting code!

still problems with the code. But I think proper formatting is
more important than you realize. In terms of content, I'm
actually fairly impressed -- you're tackling some ambitious
concepts.

Okay, I'm sure someone can take over where I've left off...
If not, I'll try to help some more later.

- Mark