As an engineer analyzing a system (whether it be a circuit or an audio sample), you should know when to apply the analysis tools you've been given--such as Discrete Time Fourier Transform and Z-Transform. I understand that a Fourier Transform will tell me how the amplitude of the system changes with respect to inputting different frequencies. For example, I believe that performing the Fourier Transform on a LC circuit would give me information as to what the resonant frequency is. However, I don't know when I should apply the Z-Transform. It's usefulness is not apparent to me. I don't understand what information I would get from the Z-Transform that I wouldn't get from a Fourier Transform.

I realize that the Z-Transform and Fourier transform are very similar, except that the Z-Transform has an 'r' term in it. I don't know the significance of this or of the result coming from a Z-Transform compared to the result coming from a Fourier Transform.

What additional information are we getting by doing the Z-transform, that we are not getting from doing the DTFT?

Edit: It has been suggested that my question is a duplicate of this one: Mathematical advantages of the ZT, DTFT and DT?

My question is different from this one because I am asking from the perspective of an engineer, not from the perspective of a mathematician. I would like to understand the intuition and real world physical meaning behind this. These techniques are all developed for the purpose of solving some real world problem and I want to understand what that is.

  • 2
    $\begingroup$ One difference is that with DTFT you can only find the steady-state response to sinusoidal inputs, but with $z$-transform you can find both transient and steady-state response to arbitrary inputs. See this answer for more details. $\endgroup$
    – msm
    Dec 7 '16 at 23:47
  • $\begingroup$ @msm I read your answer to the question, but I don't have enough exposure to DSP to be able to interpret what you mean when you say transient response vs steady state. It would really help me to have it related to a real life example. Right now, what I understand is that, if you change then input (only the amplitude), then you could know what the output is going to be because you have done the Z-transform. It's all very theoretical and I would be very happy to have some kind of physical link to a real engineering scenario. $\endgroup$
    – Klik
    Dec 8 '16 at 1:12
  • 1
    $\begingroup$ @Klik this might get a little complicated. DTFT and z-Transform are DSP concepts, and explaining them without throwing math at you is pretty hard. And hence, I'll have to agree with msm: this has already been answered in the other question. $\endgroup$ Dec 8 '16 at 1:15
  • 1
    $\begingroup$ Puh, I really think it's ok to presume that at some point, you don't get more intuition – as said in the answers there and the answer here, the z Transform works for all z, whereas when you set $z= e^{-j2\pi f}$, you get the DFT – so one is nice for discrete systems, the other is extremely handy for harmonic signals $\endgroup$ Dec 8 '16 at 2:12
  • 1
    $\begingroup$ Difference equations are not the same thing as differential equations! AFAIK analog electronic circuits cannot be described by difference equations; they're strictly discrete. At this point I think you really need a good book; I recommend "Signal Processing for Communications" by Prandoni and Vetterli; it's free online. There are many free books around but I think this particular one approaches the problem in the way you're looking for. $\endgroup$
    – MBaz
    Dec 8 '16 at 3:06

For a signal x(t), with the timescale $t$ sampled at $dt$:

The Z Transform (bilateral): $$\mathbf{Z}(x)=\sum_{t=-\infty}^{\infty}x(t)z^{-t}$$

when evaluated for the orbit $z=e^{-i2\pi f}$ (the complex unit circle, for angular parameter $f$), becomes into the Discrete Fourier Transform (scaled by $dt$): $$\mathbf{F}_{DT}(x)=\sum_{t=-\infty}^{\infty}x(t)e^{-i2\pi ft}$$

Just like the discrete space, in the continuous space, the Laplace Transform (bilateral): $$\mathbf{L}(x)=\int_{-\infty}^{\infty}x(t)e^{-st}dt$$

when evaluated for the orbit $s=i2\pi f$ (the vertical complex axis, with linear parameter $y=2\pi f$), becomes into the Fourier Transform: $$\mathbf{F}(x)=\int_{-\infty}^{\infty}x(t)e^{-i2\pi ft}dt$$

Hence, the DTFT and the FT are essencially special cases of the ZT and the LT respectively, when taking only periodical signals.

Note that, in both cases, the $f$ (or $\omega$) variable in the DTFT and FT is real, but in the ZT and LT the $z$ and $s$ parameters are complex ($s=\alpha+i\omega$), exposing the simplicity of the Fourier Transforms compared with the Laplace / Z versions...

Indeed, the Fourier Transforms has a ROC, which expresses as their integrability conditions for the FT to converge.

  • 1
    $\begingroup$ I'm confused when you say talk about the parameters for the ZT and LT being complex, because in the fourier transform, we also have complex exponentials. Looking at the math, the difference between the ZT and DTFT is 'r' since z = rexp(jwn). Whereas the input for the DT Fourier Transform is exp(jw*n). I am having difficulty understanding how to interpret what the ZT means. I understand the DTFT gives me a frequency response. But, I don't know what information the ZT gives me in comparison. I don't know when an engineer would use DTFT over ZT. $\endgroup$
    – Klik
    Dec 8 '16 at 1:40
  • 2
    $\begingroup$ One uses the FT for checking periods. One uses LT|ZT for studying systems.... $\endgroup$
    – Brethlosze
    Dec 8 '16 at 1:55
  • 1
    $\begingroup$ I follow you. I read your answer to the post that was mentioned and it is a little more advanced than I can understand. Could you elaborate on the difference between FT and ZT as if you were writing one of those "for dummies" books? You say that FT is for checking periods, but isn't it also for determining how the system will respond to a given frequency? $\endgroup$
    – Klik
    Dec 8 '16 at 2:12
  • 2
    $\begingroup$ really, your notation is atrocious. again, you seem to like to use "$i$" for both an index and the imaginary unit at the same time. $\endgroup$ Dec 8 '16 at 4:35
  • 2
    $\begingroup$ Please check your first equation, there are several errors in it. $\endgroup$
    – Matt L.
    Dec 8 '16 at 7:57

Not the answer you're looking for? Browse other questions tagged or ask your own question.