Fourier, Laplace and Z transforms

[Jag1972]

Hello All,
I am experiencing some difficulties with Laplace transform and Fourier transforms. I am summarized my understanding of my conceptual understanding would really appreciate if someone could comment on it:

A continuous time domain signal can be transferred into the complex frequency domain by using frequency domain by using the Fourier transform (multiplying the input signal by sine and cosine waves: x(t) e-(j*pi*t)). However some signals/functions do not converge in the frequency domain, for those signals/functions a convergence factor can be used which is multiplied to the Fourier transform e-[tex]\sigma[/tex]*t. When the Fourier transform is multiplied by this convergence factor (exponential decay or growth) the transform is referred to a Laplace transform.

A discrete time domain signal can be transferred into the complex frequency domain by using frequency domain by using the discrete Fourier transform (multiplying the input signal by sine and cosine waves: x(t) e-(j*pi*n)). However some signals/functions do not converge in the frequency domain, for those signals/functions a convergence factor can be used which is multiplied to the Fourier transform r-n. When the discrete Fourier transform is multiplied by this convergence factor (exponential decay or growth) the transform is referred to a Z transform.



What is convergence: I have used the Fourier series on functions before and convergence was when the amplitude of the harmonics was starting to be equal i.e. there would be no point continuing after a point as all amplitudes would be the same.
Example: For a square wave, the Fourier series is sine waves with odd multiplies of the fundamental frequency and convergence occurs pretty quickly.
Am I right to assume that this is the same convergence if so could someone please let me know what type of function would not converge.
My particular interest is in Electronics however any function example will be extremely beneficial as books just point it out and don’t give examples.

Jag.

 

[marcusl]

I've never seen convergence used as to categorize these transforms, and am pretty sure that your usage of convergence is wrong. (For one thing, if the amplitude of harmonics in a Fourier series became constant, the series diverges.) The question of convergence is subtle and involves concepts like Lebesgue integration for the FT. We won't go into this here.

In any case this obscures the useful differences between them. A more useful view is by application and domain. The Laplace transform (LT) operates over the entire complex plane. As you point out, it is useful for characterizing functions that are both oscillatory and damped. it has other uses in complex analysis as well.

The FT is just the LT evaluated along the imaginary axis [tex]\sigma=0[/tex] or [tex]s=i\omega[/tex]. Along this axis one gets the oscillatory part, called the frequency response.

The z transform is a purely discrete construct. In digital systems, z^(-1) is a delay operator corresponding to one sample delay. There are various ways to map the Laplace transform of the entire plane to the z plane. In one simple case, bend the iw axis around to the left to form a circle such that s=+ and - infinity map to z=+/-pi. The left half plane maps to the interior of the unit circle, the right to the exterior, and the frequency response is exactly on the circle.

There are many EE books that cover these concepts. The one I have is out of print, but you might find it at your university library:
Gupta, Transform and State Variable Methods in Linear Systems.

I have not used this book but it is a well-respected standard text and I see in the table of contents that all three transforms are covered:
Oppenheim and Willsky, Signals and Systems (1996)

 

[Jag1972]

Marcus: Thank you very much for taking the time to reply, very usefull. I will order that book from the library. I have also been watching the DSP lectures on youtube from NPTEL (Indian university). There are fantastic.

 

[marcusl]

You are welcome. Please come back if you have more questions.

 

[Jag1972]

Marcus/All

I am sorry but I have another question what does convergence mean in English in the context of Fourier/Laplace and Z transforms. Does it simply mean that it exists for example from zero to a pole?
Zero is when the transform is 0 and a pole when the transform is infinite.
Thanks in advance for your help.

Jag.

 

[marcusl]

You'll see discussion of convergence and existence, though they are just two sides of the same thing. In general, transforms are finite (exist) for some functions, over some ranges of parameters. For instance, every function f(t) that is absolutely integrable, that is

[tex]\int_{-\infty}^{\infty}{|f(t)|dt}[/tex]

is finite, has a finite well-defined Fourier transform. This is an existence condition. In other cases, existence of the transform of a function occurs only for certain ranges of parameters. For example, the function f(t)=exp(-|t|) possesses a Fourier transform by the property above, but has a convergent Laplace transform only in the region of the s plane defined by [tex]-1 < \sigma < 1[/tex]. This is a convergence condition. (Of course, the Laplace transform evaluated at [tex]\sigma=0[/tex] is the Fourier transform, so both of these conditions are the same there.)

You can find these properties in textbooks on transforms, and by searching the web.

 

[Jag1972]

I am sorry about the delay in response hope you don’t think I was being rood; I was just really busy over the last couple of day. Thank you very much for the reply it was extremely helpful.
I started it off with the Fourier series and thought for some reason that converges was when the amplitudes of the harmonics were the same and it seemed to make sense at the time.
I realize now what you mean and to treat convergence as a function that is summable and does not go to infinity as your example shows.
This leads me on to another question to do with Region of Convergence (ROC) for a Z transform, I know that is normally expressed in annular form i.e. on a circle there will be a small circle and around that a larger circle the ROC is between the larger circle and smaller circle. The larger circle can extend to infinity and the smaller one can be 0 (I am only really concerned with casual functions).
Am I right in my following statement, if the transfer function of a system in the ‘Z’ plane was:

H(Z) =[1]\frac{}{}[1-Z^-1]

This means that, [Z^-1] has to be less then 1 and [Z] has to be grater then 1 for convergence to occur and it stretches to infinity. So would that mean the ROC is from 1 to infinity? So in annular form the small circle would have distance 1 and larger one would have distance infinity from the origin. Also this function has only one pole at 1, hopefully that is correct and assuming it is I would like to go for another example and assumption.


H(Z) =[1-Z]\frac{}{}[1+0.5Z^-1]

I have just made this one up, this transfer function has 1, zero at 1 and one pole when [Z]>2 or [Z^-1] <2. Does the zero play any part in the ROC as the Transfer function I would think not and the ROC is from 2 to infinity.

PS: I find these forums extremely helpful and find the people on them who help others for no self gain extremely endearing :)

Jag.

 

[marcusl]

Am I right in my following statement, if the transfer function of a system in the ‘Z’ plane was:

H(Z) =[1]\frac{}{}[1-Z^-1]

Your Latex script is not written correctly to render. Click on my equation to see how to do it:

[tex]H(z) =\frac{1}{1-z^{-1}}[/tex]

This means that, [Z^-1] has to be less then 1 and [Z] has to be grater then 1 for convergence to occur and it stretches to infinity.

Almost, but z is complex so the ROC is actually |z|>1.

So would that mean the ROC is from 1 to infinity?

Yes, if you mean an annulus of inner radius 1.

H(Z) =[1-Z]\frac{}{}[1+0.5Z^-1]

I have just made this one up, this transfer function has 1, zero at 1 and one pole when [Z]>2 or [Z^-1] <2.

No, the pole is at z=-0.5, so the ROC is |z|>0.5.

Does the zero play any part in the ROC

No, as you surmised.
BTW, Oppenheim and Willsky has a useful and concise chapter on the z transform.

PS: I find these forums extremely helpful and find the people on them who help others for no self gain extremely endearing :)

Jag.

We're glad to help!

 

[Jag1972]

Hello,
I have been revisiting Discrete Fourier Transform by watching some excellent you tube videos, however there is one part I don’t understand would really appreciate it if you could give me a hand. I have attached a screen shot of the Fourier Transform.

The discrete function is: x(n) = (1/2)^n u(n)

The DFT of this is equal to summation with limits 0 to infinity, that takes care of u(n). The remaining part of x(n) is (1/2)^n multiplied by e-jwn.

This is equal to (e-jw/2)^n

This function converges and the closed loop form is 1/(1-0.5e-jw)

multiply numerator and denominator by 2 equals:

2/(2-e_jw) or 2/(2-cosw+jsinw)

I don't know how the polar form is calculated to:

2/(sqrt(5-4 cosw))

I am sorry I have not used the latex code, have not got used to it yet, I have however attached screen shots of the video I was watching with all these equations.

Your help will be appreciated.

 

[Jag1972]

Latex code version.

x(n)=(1/2)[tex]^{n}[/tex] u(n)

X(w)= [tex]\Sigma[/tex](1/2)[tex]^{n}[/tex] e[tex]^{-jwn}[/tex] (The limits are 0 - infinity)

=[tex]\Sigma[/tex] [tex]\frac{e^{-jw}}{2}[/tex][tex]^{n}[/tex]

= [tex]\frac{1}{1-(0.5) e^{-jw}}[/tex]

= [tex]\frac{2}{2-cosw+jsinw}[/tex]
I don't know how the magnitude was calculated.

[tex]\frac{2}{\sqrt{5-4 cos w}}[/tex]

 

[marcusl]

The last step looks like a mistake. I get

[tex]
\frac{2}{\sqrt{5-2cos w}}
[/tex]

 

[Jag1972]

Hello Marcus,
Could I be cheeky and ask how you work out the magintidude from the rectunglar coordinates, I don't mind if you don't give the answer just a hint will do. I thought all I would have to do is to multiply the numerator and deniomenator with the complex conjugate of the deniominator. Is that correct, thanks in advance.

 

[marcusl]

No, that just puts it over a real denominator. To find the magnitude, multiply by the complex conjugate and take the square root. The magnitude of f is
[tex]|f|=\sqrt{ff^*}[/tex]

 

[Jag1972]

Modulus of F is [tex]\sqrt{F.F^*}[/tex]
[tex]F=\frac{1}{2 - cosw + jsin w}[/tex]

[tex]|F|= \sqrt{\frac{4}{((2-cosw + jsinw)(2-cosw - jsinw))}[/tex]}

denominator only with square root taken out, will replace in the end

2[tex]^{2}[/tex]- 2cosw - j2sinw -2cosw +cos[tex]^{2}[/tex]w +cosw.jsinw + j2sinw
- jsinw.cosw -j[tex]^{2}[/tex]sin[tex]^{2}[/tex]w

Hope I have not fudged it up too much, if this is correct then I will look at the trig identities to reduce.
Thanks in advance Marcus

 

[Jag1972]

unfortunately I deleted my introductory post by mistake as I was reviewing my post to check my use of latex.

Introductory post:

Hello Marcus,
I am not ashamed to say that my maths is poor, however I have had a bash at multiplying the function by its complex conjugate. I would very much appreciate if you could let me know if I am on the right track if so I will look up the trig identities to reduce down.

Thanks in advance :)

 

[marcusl]

Hello Jag1972,
Your math is perfect. No need to use trig identities, just group terms--you'll see that many cancel--and evaluate.

I see that my post #11 was incorrect--sorry. I should know better than doing math in my head. Your answer in #10 is correct.

 

[Jag1972]

Thanks a lot Marcus, you are a legend :)

 

[Jag1972]

denominator:

2[tex]^{2}[/tex]-2 cosw - j2 sinw - 2 cosw + cos[tex]^{2}[/tex]w + cosw.jsinw + j2sinw
- j2 sinw - cos.jsinw - j[tex]^{2}[/tex] sin[tex]^{2}[/tex]w

Grouping like terms together

4 - 2cosw -2 cosw + cos[tex]^{2}[/tex]w -j2 sinw + j2sin w - j[tex]^{2}[/tex] sin [tex]^{2}[/tex]w + cosw.jsinw - cosw.jsinw

last 2 terms are 0, also j[tex]^{}^{2}[/tex] = -1

4 - 2cosw -2 cosw + cos[tex]^{2}[/tex]w -j2 sinw + j2sin w + sin [tex]^{2}[/tex]w

cos[tex]^{2}[/tex]w + sin[tex]^{}[/tex]w = 1

4 - 2cosw -2 cosw -j2 sinw + j2sin w + 1

-j2 sinw + j2sin w = 0

4 - 2cosw -2 cosw + 1

5 - 2cosw -2 cosw

-2 cos w - 2 cos w = -4 cos w

5 - 4 cosw

[tex]|F|-\sqrt{\frac{4}{5 - 4 cos w}}[/tex]
[tex]|F|=\frac{2}{\sqrt{5 - 4 cos w}}[/tex]

Thanks Marcus, couldn't have done it without your help. Happy new year mate :)

 

[marcusl]

Happy new year to you too!
Best regards.