0
$\begingroup$

Having taken some courses in Systems and Stability dealing mostly with continuous time signals, I am accustomed to thinking about the Laplace and Fourier transforms dealing with complex and pure imaginary signals respectively. For example, if I transformed $x(t)$ and evaluated its magnitude and phase at 0.5+0.5j, I would looking at the response of the system having injected it with $\exp(.5t)\exp(j0.5 \pi t)$

Recently, I have been working in the discrete domains of the DTFT (continuous in the transform domain), DFT, and Z. Here, we have complex exponentials mapping to the unit circle. Now I am struggling at a high level to reconcile these continuous and discrete domains. If I look at a DTFT of a discrete time signal and evaluate it for some input $\exp(0.5 j \hat{\omega} \pi n)$ where $n$ is some index and $\hat{omega}$ is a relative frequency based on the sampling rate of the continuous time signal, I end up somewhere on the unit circle (not on the imaginary axis) even though the signal is a purely imaginary exponential.

The questions I have are

1.) How does a complex signal (real and imaginary)$ x[n] = \exp(0.5n) \cos(j .5 \pi \hat{\omega}n)$ map into the DTFT/DFT and Z frequency domain? I am guessing somehow real part does not map to DTFT/DFT but does to Z?

2.) Why the concept of a multidimensional space for these discrete frequency signals (DTFT/DFT) when the imaginary axis (Fourier domain) worked fine for continuous signal? Why do we need a unit circle that repeats when we could map the same function to a straight line that repeats every $-\pi$ to $\pi$?

$\endgroup$
1
$\begingroup$

Simple answer: that's how the math works out.

Better answer: In the discrete domain, frequency is periodic with the sample rate. If you sample at 48 kHz, the range of frequencies that you can represent range from -24 kHz to +24 kHz. Let's say so you have a 100 kHz signal. You can sample this at 48 kHz but it would look exactly like a 4 kHz signal. In fact the two are indistinguishable, so it makes no sense to treat them differently.

That's why a circle makes a lot of sense here. Frequency is represented as the angle between the origin at any point on the circle. If you go around the circle an integer number of times, you end up exactly where you started.

So in the continuous domain 4 kHz and 100 kHz are different points. In the discrete domain (when sampled at 48 kHz) they are not.

$\endgroup$
1
$\begingroup$

Consider the following $$x(t)=\cos(2\pi f t) $$

$$\frac{d}{dt} x(t)= -2\pi f \sin(2\pi f t) $$ you can take derivative with respect to time. In the $s$ domain $s X(s)$ corresponds to the time derivative of $X(s)$ (with zero initial condition)

For discrete time, $$ x[n]=\cos(2\pi f n) $$ $n$ is a discrete variable $$ \frac{d}{dn} x[n]= \lim_{\Delta \rightarrow 0} \frac{x[n+\Delta]-x[n]} {\Delta} \quad \text{is nonsense} $$ Instead we have the $z$ domain where $zX(z)$ corresponds to $x[n+1]$.

Since poles and zeros can lay in an arbitrary location, (not restricted to $s=j\omega$ or $z$ on the unit circle), using the $s$ domain just doesn't work.

The Z Transform is a special case of the Laurent Series

https://en.wikipedia.org/wiki/Laurent_series

from the first paragraph:

In mathematics, the Laurent series of a complex function f(z) is a representation of that function as a power series which includes terms of negative degree. It may be used to express complex functions in cases where a Taylor series expansion cannot be applied.

Since $(d^k /dn^k) x[n]$ would be necessary for a Taylor Series, and they don't exist, we use the Z transform.

But wait, Isn't $e^{-sT}$ a delay operator in the $s$ plain ? Yes it is, so we can describe signals like $ y(t)= a_0 x(t) + a_1x(t-T) + a_2 x(t-2T)$ in the $s$ domain as $(a_0+ e^{-sT}a_1 + e^{-2sT}a_2) X(s)$. The problem is that a LTI continuous time filter is described as a (finite) ratio of polynomials in $s$.

$$ H(s)=\frac{a_0 + a_1 s + a_2 s^2 +\dots + a_n s^{n}}{b_0 + b_1 s + b_2 s^2 +\dots + b_n s^{m}} $$ The transfer function for $e^{-sT}$ has infinite terms. This can be approximated by a Pade expansion, and often is in some applications, but you can see that the $s$ plain get very complicated and is only approximate. $e^{sT}$ and approximations do come in handy when relating $s$ to $z$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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