# DFT/FFT Transfer function

I want play and record a sine sweep. When i have both signals the recorded one and the send one i can create a Transferfunction. That is what i know so far.

$$H_0 = \frac{OUT}{IN} = \frac{Y}{X}$$

Where i'm stuck is that when i read about the Transfer function it is all about the $Laplace \, transform$ (and sometimes the $Z-Transform$) While i use the $FFT/DFT$. I tried some GNU Octave code to tried an Transfer function and i think that looks what i suspected.
But now i curious

1. if the $FFT/DFT$ can be used for/ as a transferfunction and
2. what is the relation between an $FFT/DFT$, $Laplace \, Transform$ and the $Z-Transform$

Your question is fairly broad, let me answer it step by step.

First of all $H(s)$ is indeed called the transfer function, and is the laplace transform of the impulse response. It's useful for finding the poles and zeroes, which is what the fourier transform can't do alone. $H(\omega)$ is the frequency response of the impulse response, not really the "transfer function".

If the frequency response is what you want to attain, then you can use the frequency response (fourier transform). It does not exist for unstable signals though.

The relation between Fourier Transform, discrete Fourier transform and Laplace transform/z-transform:

Fourier transform and DFT are in principle the same. One is for discrete signals, and the other for continuous signals. Because of that there are some different properties like frequency domain aliasing. What both of these transforms do is decompose the signal into complex exponentials (easily converted into sines and cosines (or sinusoids with magnitude and phase), which is the original idea).

Laplace and z-transform are again in principle the same, z-transform is the discrete equivalent to Laplace. It's simply the Fourier transform/DFT with the signal multiplied by a varying exponential $e^{\sigma t}$, or in discrete case just $r^t$. Then you have multiple fourier transforms/DFTs relating to each different value of the exponential ($r$ or $\sigma$). Point of interests are the poles and zeroes, IE. the points where the impulse response in question multiplied by an exponential has integral that is exactly infinite or exactly zero.

Note that when $r=1$, or $\sigma = 0$, the Laplace/Z-transform is equal to the Fourier transform. In other words, the Laplace/Z-transform contains the Fourier transform in it.

EDIT: I wrote this fairly quickly so please point out any errors.

• The DFT and FFT compute exactly the same thing, the Discrete Fourier Transform. But great answer +1 – johnnymopo Dec 28 '15 at 17:24
• Oke I think i understand the relation between both transforms. But in that case i have one question(if it fits's within this question otherwise i can create a new one) Matlab and Python don't know a discrete Z-transform function (like FFT is for Fourier Transform). That i need to use the FFT for create discrete signals and convert that to Laplace. How do i use the r and $\sigma$ to convert – Jan-Bert Dec 29 '15 at 11:13
• @Jan-Bert Matlab is a computer program and thus operates with discrete variables. It can not perform a FT, but a DFT (which FFT is an algorithm for). As for how to transform the DFT/FFT into Z-transform. Simply multiply the time domain function by $r^n$ and then calculate the DFT. You have one more variable now, which is the value of r – Dole Dec 29 '15 at 11:30

The Fourier transformation and the Laplace transformation defined for continous signals. Fourier is a special case of Laplace for non-negative frequencies. Both are not applicable for dicrete signals directly. FFT name an algorithm which approximates the Fourier transformation in it's discrete form, called DFT. The question about the relation between Fourier and Laplace is further answered in this math-stackexchange topic.

The Z transform is essentially a discrete version of the Laplace transform and, thus, can be useful in solving difference equations, the discrete version of differential equations. The Z transform maps a sequence $f[n]$ to a continuous function $F(z)$ of the complex variable $z=r\cdot e^{jΩ}$

This quote is from this posting.

• I'll add that the DFT is a Z - transform where the magnitude is 1, that is, evaluated along the unit circle in the z plane. The FFT is a fast means to evaluate the DFT. – johnnymopo Dec 28 '15 at 16:26
• uhm, there are several mistakes that need to be corrected before they are reinforced in anyone's understanding:  1. "Fourier is a special case of Laplace for non-negative frequencies." -- that's false. FT specifically has negative frequencies.  2. "Both are not applicable for discrete signals directly." -- both LT and FT can be applied to sampled signals as you get the Z-transform and DTFT.  3. "FFT name an algorithm which approximates the Fourier transformation in it's discrete form, called DFT." -- not an approximation, but an implementation. – robert bristow-johnson Dec 28 '15 at 22:30