Using Complex Conjugates to Decompose a Fraction

And the problem is, this is from the solution manual of a book. So it's not like it's a mistake or anything. I'm just missing something.
  • #1
WhiteWolf98
86
5
Homework Statement
How do you decompose the following:

$$\frac{-3z^{-1}+2z^{-2}}{1-z^{-1}+z^{-2}}$$
Relevant Equations
None.
So the original question is from Control Theory, and the topic is the inverse z-transform. This is a part from the solution I just can't understand. The reason it has to be in this form (##z^{-1}##) is because that's the form used in the z-transform table. The question essentially is, how do you get from the left side to the right side:

$$\frac{-3z^{-1}+2z^{-2}}{1-z^{-1}+z^{-2}}=-3z^{-1}\frac{1-0.5z^{-1}}{1-z^{-1}+z^{-2}}+z^{-1}\frac{0.5z^{-1}}{1-z^{-1}+z^{-2}}$$

In the solution, it states: 'Noting that the poles in the quadratic term are complex conjugates'. I have no idea what that means. In the normal (##z##) form, it is:

$$\frac{-3z+2}{z^2-z+1}$$

The solutions of this, I know, are complex conjugates:

$$\frac{1}{2}+\frac{\sqrt{3}}{2}i,\,\frac{1}{2}-\frac{\sqrt{3}}{2}i$$

I don't see how that helps me though. Any help would be appreciated. Thank you.
 
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  • #2
WhiteWolf98 said:
Homework Statement:: How do you decompose the following:

$$\frac{-3z^{-1}+2z^{-2}}{1-z^{-1}+z^{-2}}$$
Relevant Equations:: None.

So the original question is from Control Theory, and the topic is the inverse z-transform. This is a part from the solution I just can't understand. The reason it has to be in this form (##z^{-1}##) is because that's the form used in the z-transform table. The question essentially is, how do you get from the left side to the right side:

$$\frac{-3z^{-1}+2z^{-2}}{1-z^{-1}+z^{-2}}=-3z^{-1}\frac{1-0.5z^{-1}}{1-z^{-1}+z^{-2}}+z^{-1}\frac{0.5z^{-1}}{1-z^{-1}+z^{-2}}$$
In the above, they rewrote the numerator of the first term as ##-3z^{-1} + 1.5z^{-2} + .5z^{-2}##.
They then made two separate fractions, with the first's numerator the first two terms above, and the second's numerator the remaining term. Finally, they factored ##-3z^{-1}## from the first fraction and ##z^{-1}## from the second.
WhiteWolf98 said:
In the solution, it states: 'Noting that the poles in the quadratic term are complex conjugates'. I have no idea what that means. In the normal (##z##) form, it is:

$$\frac{-3z+2}{z^2-z+1}$$
I haven't worked things through, but I suspect that they rewrote the common denominators like so: $$1 - z^{-1} + z^{-2} = 1 - \frac 1 z + \frac 1 {z^2}$$
and then rewrote the last expression as $$\frac{z^2 - z + 1}{z^2}$$
WhiteWolf98 said:
The solutions of this, I know, are complex conjugates:

$$\frac{1}{2}+\frac{\sqrt{3}}{2}i,\,\frac{1}{2}-\frac{\sqrt{3}}{2}i$$

I don't see how that helps me though. Any help would be appreciated. Thank you.
 
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  • #3
Mark44 said:
In the above, they rewrote the numerator of the first term as −3z−1+1.5z−2+.5z−2.
Okay... that makes sense. But then the next part:
Mark44 said:
They then made two separate fractions, with the first's numerator the first two terms above, and the second's numerator the remaining term. Finally, they factored −3z−1 from the first fraction and z−1 from the second.
How can you just split a fraction like that? For example:
$$\frac{x+y}{z}=\frac{x}{z}+\frac{y}{z}$$
You can do that. But what you can't do is:
$$\frac{x+y}{w+z}=\frac{x}{w+z}+\frac{y}{w+z}$$
So how come it's being done there? Is there something I'm missing here?

I also don't see how the fact that the solutions of the denominator are complex conjugates, is being used here. I get it needs to be in a certain form, but it seems pretty random for this statement to be a reason for splitting the fraction like so. Or I'm just missing something/ there's a rule I don't know
 
  • #4
WhiteWolf98 said:
Okay... that makes sense. But then the next part:

How can you just split a fraction like that? For example:
$$\frac{x+y}{z}=\frac{x}{z}+\frac{y}{z}$$
You can do that. But what you can't do is:
$$\frac{x+y}{w+z}=\frac{x}{w+z}+\frac{y}{w+z}$$
No, the above is perfectly valid. You should review the basics of fraction addition.

You could also do something like this, if there's a need for it.
$$\frac{x+y}{z}=\frac{x - y}{z}+\frac{2y}{z}$$
This is similar to what they did in the work you posted.
What you can't do is the following:
$$\frac{z}{x + y}=\frac{z}{x}+\frac{z}{y}$$
WhiteWolf98 said:
So how come it's being done there? Is there something I'm missing here?

I also don't see how the fact that the solutions of the denominator are complex conjugates, is being used here. I get it needs to be in a certain form, but it seems pretty random for this statement to be a reason for splitting the fraction like so. Or I'm just missing something/ there's a rule I don't know
 
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  • #5
Mark44 said:
No, the above is perfectly valid. You should review the basics of fraction addition.
Wow, uh. Right you are, my apologies. That's quite embarrassing lol

Mark44 said:
What you can't do is the following:
$$\frac{z}{x + y}=\frac{z}{x}+\frac{z}{y}$$
THAT is what I meant to say! Seems I've got it all muddled up. Thanks a lot Mark, that's made it quite clear how that fraction is split.
 
  • #6
I think it is simpler than that. It is easy to see the identity by rewriting the numerator as ##-3z^{-1}+1.5z^{-2}-.5z^{-2}## and separating the fraction into two fractions. I do not see how any of this has anything to do with the poles being complex conjugates. You should complete the statement in the solution. What does it say after 'Noting that the poles in the quadratic term are complex conjugates'? (Or perhaps this is the motivation for deciding to rewrite the fraction as they did?)
 
  • #7
FactChecker said:
I think it is simpler than that. It is easy to see the identity by rewriting the numerator as ##-3z^{-1}+1.5z^{-2}-.5z^{-2}## and separating the fraction into two fractions. I do not see how any of this has anything to do with the poles being complex conjugates. You should complete the statement in the solution. What does it say after 'Noting that the poles in the quadratic term are complex conjugates'? (Or perhaps this is the motivation for deciding to rewrite the fraction as they did?)
1610665171561.png

That's the solution as I saw it; everything after is using the inverse z-transform. And I agree, I still have absolutely no clue what complex conjugates have to do with this
 
  • #8
consider the familiar Z-transform pair
$$(1-r e^{\pi i t}z^{-1})^{-1}\leftrightarrow r^n e^{n\pi i t}\mathrm{u}(n)$$
if a real to real function has a complex singularity it must have the conjugate as well. When the above pair appears so to will its conjugate
$$(1-r e^{-\pi i t}z^{-1})^{-1}\leftrightarrow r^n e^{-n\pi i t}\mathrm{u}(n)$$
the sum of the above two pairs divided by 2 being
$$\frac{1-r\cos(\pi t)z^{-1}}{1-2r \cos(\pi t)z^{-1}+r^2z^{-2}}\leftrightarrow r^n \cos(n\pi t)\mathrm{u}(n)$$
and the difference divided by 2
$$\frac{r\sin(\pi t)z^{-1}}{1-2r \cos(\pi t)z^{-1}+r^2z^{-2}}\leftrightarrow r^n \sin(n\pi t)\mathrm{u}(n)$$
of course in you example ##t=3^{-1}##
so
$$\cos(\pi i t)=\frac{1}{2}$$
$$\sin(\pi i t)=\frac{\sqrt{3}}{2}$$
 
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  • #9
My apologies, maybe I should have actually listened and shared the rest of the solution. I sort of get what you're doing there, and recognise two formulas from the z-transform table. This is the rest of the solution:
1610999063624.png

Could it possibly be that the statement, 'Noting that the poles in the quadratic term are complex conjugates' has absolutely nothing to do with the decomposition, and in-fact, is aimed at what you've shown/ solution I've shared above?

When I went through the rest of the solution after breaking down the fraction, I was able to do the rest of the problem without thinking of complex conjugates. The two formulas I got straight from the tables, and after that, it was just a matter of identifying what's what. There was only one thing that confused me, which was why ##1/\sqrt{3}## was taken out. I assumed this was the case so the two terms in the final solution have the same term in the fraction##(\frac{n\pi}{3})##. I think if ##\frac{1}{\sqrt{3}}## isn't taken out, it would end up being ##\sin(\frac{n\pi}{6})##.
1610999661502.png
 

Related to Using Complex Conjugates to Decompose a Fraction

1. What are complex conjugates?

Complex conjugates are pairs of complex numbers that have the same real part but opposite imaginary parts. For example, the complex conjugates of 2+3i are 2-3i and vice versa.

2. How can complex conjugates be used to decompose a fraction?

By multiplying the numerator and denominator of a fraction by the complex conjugate of the denominator, the fraction can be rewritten in a form where the denominator is a real number. This makes it easier to simplify and solve the fraction.

3. Can any fraction be decomposed using complex conjugates?

Yes, any fraction with a complex number in the denominator can be decomposed using complex conjugates. However, this method may not always be the most efficient way to solve the fraction.

4. Are there any limitations to using complex conjugates to decompose a fraction?

One limitation is that the fraction must have a complex number in the denominator. If the fraction only has real numbers, using complex conjugates will not be necessary. Additionally, this method may not always result in the simplest form of the fraction.

5. Why is it useful to decompose a fraction using complex conjugates?

Decomposing a fraction using complex conjugates can make it easier to simplify and solve the fraction, especially when dealing with complex numbers. It also allows for the use of other algebraic methods that may not be possible with a fraction containing complex numbers.

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