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I am reading a chapter on digital filter design from analog filter design using difference equations. What they do first of all is that they map $s$ (Laplace variable) to $z$ ($z$-transform) by the following relation, $$z = \frac{1}{1-sT}$$ where $T$ is the sampling period.Now there is a complete paragraph on how a derivative in continuous time can be approximated by a difference equation in the discrete time. I have two doubts. The first doubt is regarding the mapping done above. If I replace $s=j\Omega$, where $\Omega$ is the continuous time angular frequency, I get $$z=\frac{1}{1-j\Omega T}$$. The book says that this can be further simplified to $$z=\frac{1}{2}(1+e^{j2\tan^{-1}(\Omega T)})$$ That's fine, I have no problem with the simplification, but now it says that it is not a unit circle but a circle with center at $z=1/2$ and radius equal to $1/2$. I didn't understood this.

My second doubt is regarding this statement from the book.

If a bandlimited analog signal is sampled at the Nyquist rate, then the spectrum is non-zero over the entire unit cicle. If sampling period $T$ is sufficiently small, then the response of the digital filter will be concentrated on the small circle in the vicinity of $z=1$.

I am unable to prove this statement mathematically and I am even unable to understand it intuitively. The second line of the quoted text could be proved easily by the same mapping equation, but the first line is still not clear. Please help.

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Let me show you a quick and easy way to see that $z=\frac12+\frac12 e^{j\phi}$ is a circle in the complex plane with center $\frac12$ and radius $\frac12$. First, note that $z=e^{j\phi}$ is the unit circle, i.e., a circle centered at $z=0$ with radius $r=1$. Changing the radius is easy, just multiply the expression for the unit circle with some positive number $r$: $z=re^{j\phi}$. If you want to change the center of the circle just shift it by adding a complex number $z_0$; this number is the new center: $z=z_0+re^{j\phi}$. Comparing this general equation for a circle in the complex plane with the given expression gives $z_0=\frac12$ (i.e., the circle is centered at $z_0=\frac12$), and $r=\frac12$, i.e., the circle has a radius of $\frac12$.

Concerning the second part of your question, I think you should not try to directly relate that statement to the given mapping (that's maybe where your confusion comes from). The spectrum $X_d(e^{j\omega})$ of a discrete-time signal $x_d[n]$ that is obtained from sampling a continuous-time signal $x(t)$ is given by the aliased spectrum $X(\omega)$ of the continuous-time signal:

$$X_d(e^{j\omega})=\frac{1}{T}\sum_kX\left(\omega-\frac{2\pi k}{T}\right)\tag{1}$$

where $T$ is the sampling interval (i.e., the inverse of the sampling frequency $f_s$). Now if $x(t)$ is sampled at the Nyquist rate, which is the minimum rate that avoids aliasing, there will be no gap between the shifted spectra in the sum of Eq. $(1)$, i.e., the discrete-time frequency axis will be completely filled with images of the continuous-time signal. On the other hand, if the sampling frequency is higher than the Nyquist rate, there will we gaps between the images. If the sampling frequency is very high compared to the upper frequency limit of the continuous-time signal then the gaps between the images will be very large, and the spectrum of the discrete-time signal will be concentrated around frequency zero, i.e., around $z=1$.

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  • $\begingroup$ Thanks @Matt. I found your answer very much explanatory. $\endgroup$ – Himanshu Sharma Nov 12 '18 at 2:36
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Consider the mapping you have given:

$$z = \frac{1}{1 - j \Omega T } = 0.5 ( 1 + e^{j 2 \tan^{-1}(\Omega T)} ) $$

Assume the angle $\phi$ such that $\tan(\phi) = \Omega T$, and hence with $r = \sqrt{ 1 + (\Omega T ) ^2 }$, $\cos(\phi) = 1/r$ and $\sin(\phi) = \Omega T / r$, then :

$$\begin{align} z &= 0.5 ( 1 + e^{j 2 \tan^{-1}(\Omega T)} ) \\ & = 0.5 ( 1 + e^{j 2 \phi} ) \\ & = 0.5 ( 1 + \cos(2\phi) + j \sin(2 \phi) ) \\ & = 0.5 ( 1 + \cos^2(\phi) - \sin^2(\phi) + j 2\sin(\phi)\cos(\phi) \\ & = 0.5 ( 2\cos^2(\phi) + j 2\sin(\phi)\cos(\phi) \\ & = \cos^2(\phi) + j \sin(\phi)\cos(\phi) \\ & = \frac{1}{r^2} + \frac{j \Omega T}{r^2} \\ & = \frac{1 + j \Omega T}{r^2} = \frac{1 + j \Omega T}{1 + (\Omega T ) ^2} =\frac{1}{1 - j \Omega T } \\ \end{align} $$

indeed.

Now, to see the mapping range on the complex variable $z$ ; consider some values of $\Omega T$ such as $\Omega T = 0, 1, \pm \infty , -1$ for which the corresponding angles will be $\phi = 0, \pi/4 , \pm \pi/2, -\pi/4$ respectively. And when $\phi$ takes on those values, then the mapping on $z$ will take on the following:

$$ \Omega T = 0 \implies \phi = 0 \implies z = 0.5(1 + 1) = 1 $$ $$ \Omega T = 1 \implies \phi = \pi/4 \implies z = 0.5(1 + e^{j2 \pi/4}) = 0.5 + j0.5 $$ $$ \Omega T = \pm \infty \implies \phi = \pm \pi/2 \implies z =0.5(1 + e^{j2 \pm \pi/2}) = 0.5(1 - 1) = 0$$ $$ \Omega T = -1 \implies \phi = -\pi/4 \implies z = 0.5(1 + e^{-j2 \pi/4}) = 0.5 - j0.5 $$

Hence this shows that the for the values of $\Omega T$ that span the whole range of $0 \leq \Omega T \leq \infty$, the corresponding $z$ will lie on a circle centered at $z = 0.5$ and a radius of $0.5$ on the complex z-plane.

For the last paragraph, when a signal is critically sampled, its DTFT spectrum will be nonzero on the entire unit circle on the z-plane, for the range of angles $-\pi < \omega < \pi$, i think which is pretty clear from the frequency domain understanding of sampling operation;i.e., the shifted spectra are cosited without a spectral gap in between, which results in a fully occupied unit circle.

Finally, when the sampling period $T$ is sufficiently small (sampling freuqency $F_s$ is sufficiently high = oversmalping) then the spectrum of the signal will be condensed about $\omega = 0$ of the unit circle on the z-plane, which is about a small circle around $z=1$ on that paragraphs convention.

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  • $\begingroup$ I am still not able to understand the last paragraph. Could you please elaborate? $\endgroup$ – Himanshu Sharma Nov 11 '18 at 4:37
  • $\begingroup$ @HimanshuSharma Updated the answer, as now it includes an explanation of every formula and statement in that excerpt that you are working on... $\endgroup$ – Fat32 Nov 11 '18 at 10:31

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