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I try to work out the simplest system, whose response is $H(z) = 1 + cz + c^2z^2 + c^3z^3 + \cdots = 1 / (1-cz)$

Now, Z-transform done quick (Fourier connection) and 4.5 Transfer Function, Poles and Zeros say that we just need to plug $z => e^{iw}$ to get the frequency response. I am getting single pole at $w = i \ln c$. It is complex. What is the frequency response? The second link says that real coefficients must generate complex poles in pairs. Where is the second pair?

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Frequency response is $1/(1−ce^{iw})$, if you want to get amplitude information you just need to get the magnitude of this complex function and its phase will be the inverse of the phase of $1/(1−ce^{iw})$ or

$$-\arctan (-c\sin w/(1-c\cos w))$$

You can compute the magnitude of the transfer function by:

$$ 1 / sqrt( (-c\sin w)^2 + (1-c\cos w)^2 )$$

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    $\begingroup$ +1 for deleting your comment on the poles and correcting your formula for the phase. Just one more advice/hint: you could write you formulas using Latex commands. This makes them looks much nicer. I'll edit it for you and you can see how it works. $\endgroup$ – Matt L. Jun 4 '13 at 13:22
  • $\begingroup$ How do you compute the magnitude of $1/(1-e^{iw})$? $\endgroup$ – Val Jun 4 '13 at 13:40
  • $\begingroup$ How did you get the magnitude formula? $\endgroup$ – Val Jun 4 '13 at 13:49
  • $\begingroup$ just taking the magnitude of a complex number: en.wikipedia.org/wiki/Complex_number $\endgroup$ – RonaldoMessi Jun 4 '13 at 13:50
  • $\begingroup$ Thanks, I have completely forgot that $e^{iw} = cos + isin$. I have got the same magnitude. It looks like there is a pole every $2 \pi$ $\endgroup$ – Val Jun 4 '13 at 14:24
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You need to evaluate $H(z)$ on the unit circle $z=e^{iw}$ in order to get the frequency response (assuming that the system is stable, i.e. the region of convergence contains the unit circle). But your $\mathcal{Z}$-transform looks a bit unusual because for causal signals (or filter impulse responses) you should get negative powers of $z$:

$$H(z)=\sum_{n=0}^{\infty}x(n)z^{-n}$$

In your case $x(n)=c^n$, $n\ge 0$, so you get

$$H(z)=\frac{1}{1-cz^{-1}}=\frac{z}{z-c}\tag{1}$$

From (1) you see that the pole is at $z_{\infty}=c$. Note that you need to determine the poles of $H(z)$ for general $z$, not on the unit circle $z=e^{iw}$ because stable systems cannot have poles on the unit circle (this is also why you got a complex frequency, which does not make sense). The frequency response is

$$H(e^{iw})=\frac{1}{1-ce^{-iw}}$$

It is true that for real filter coefficients you get complex conjugate pairs of poles and zeros. But this is not the whole truth. You can also get real-valued poles and zeros, and for a first-order system like yours, this must be the case. You need at least a second-order system to get a complex pole pair. In your case you simply got one real pole (assuming that $c$ is a real-valued constant).

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  • $\begingroup$ Wait, how is pole $ce{-iw}$ real? I see that it contains complex i in it? It is a line of poles actually because you will have a new pole for every new value of w. $\endgroup$ – Val Jun 4 '13 at 10:54
  • $\begingroup$ Have you read my answer? You need to determine the poles in the z-plane, so the pole is at z=c which is real as long as c is real. $\endgroup$ – Matt L. Jun 4 '13 at 10:57
  • $\begingroup$ But you say that I must avoid $e^{iw}$ and prompt me to evaluate H(z) at that "point". Actually, I think that it is a line, since it is w-dependent. We have "a" pole at $w = -i \ln c$ $\endgroup$ – Val Jun 4 '13 at 12:34
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    $\begingroup$ No, you must compute the points in the z-plane for which the denominator of $H(z)$ becomes zero. If you try to compute the zeros of the denominator for $z=e^{iw}$ then you implicitly assume that there must be poles on the unit circle and you're trying to compute the pole frequencies. This would be OK if there were poles on the unit circle, but since there aren't any (unless $|c|=1$), you get a nonsensical (i.e. complex-valued) result for $w$. Why don't you see that the pole $z_{\infty}=c$ is real if $c$ is real? $\endgroup$ – Matt L. Jun 4 '13 at 12:41
  • $\begingroup$ Actually, I compute the poles from the $1/(1-e^{iw})$ equation. Resolving $1/(1-e^{iw}) = \infty$ gitves me the poles rather than the unit circle. I do not know how to assume the poles on the unit circle. Resolving equation gives me the whole line of poles, $-i \ln c$, but, they all seem imaginary. No matter which $c$ I choose. $\endgroup$ – Val Jun 4 '13 at 17:52

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