So I'm trying to decide whether the cosine part is intended to be plugged in for $z$ or whether it is strictly part of $h[n]$. (the number a lies in the open unit disk)

I mean I was pretty sure it was all part of $h[n]$ but then upon performing the z-transform I get this rational function

$$\frac{1 - a\cos(2\pi\frac{f_0}{F_s})z^{-1}}{1-2a\cos(2\pi\frac{f_0}{F_s})z^{-1} + a^2z^{-2}}$$

The thing is then I'm supposed to evaluate the poles and zeros and if you just ignore the cosine parts you get this really nice rational expression which factors and simplifies down to $\displaystyle\frac{z}{z-a}$.

So that has gotten me thinking that maybe I'm not understanding things correctly and the cosine portion is supposed to be plugged in for $z$ or something. Can anyone clarify this for me?

  • 2
    $\begingroup$ Hint: Use Euler's identity to express $\cos(2\pi n/F_0 f_0)$ as the sum of two complex exponential functions and then sum the resulting geometric series. Reading my answer to your other question might help in figuring out what geometric series is meant. $\endgroup$ – Dilip Sarwate Mar 17 '12 at 19:16
  • $\begingroup$ I did all that, that's how I got the rational expression above. Since I posted this I was actually able to factor it and obtain the poles and zeros, thanks for your help either way though. Actually could you do me a solid and tell me the matlab code required to plot the frequency response of this system with a=0.8, F_s=128 and f_0=32? Thanks. $\endgroup$ – Zaubertrank Mar 17 '12 at 19:23
  • $\begingroup$ Did you get two complex conjugate poles at locations on the circle of radius $|a|$? As far as MATLAB is concerned, I am sorry that I cannot help you since I am not familiar with MATLAB syntax. Just wait a while and I am sure that someone else will help you out. $\endgroup$ – Dilip Sarwate Mar 17 '12 at 19:31
  • $\begingroup$ yup that's where I got em. $\endgroup$ – Zaubertrank Mar 17 '12 at 19:34
  • $\begingroup$ @Zaubertrank "freqz" works very nicely for filter performance analysis in Matlab. $\endgroup$ – Jim Clay Mar 19 '12 at 3:31

The time domain signal (or impulse response)

$$h(n)=a^{n}\cos n\theta_0,\quad \theta_0=2\pi\frac{f_0}{f_s},\; n\ge 0$$

is very common: it is a damped sinusoidal function (assuming $|a|<1$) which occurs frequently, because it is one possible response of a second order linear time-invariant system. So concerning your doubt, the cosine part is definitely meant to be part of the time domain signal. The poles and zeros can be found by rewriting $H(z)$:

$$\begin{align}H(z)&=\frac{1-az^{-1}\cos \theta_0}{1-2az^{-1}\cos\theta_0+a^2z^{-2}}\\&= \frac{z(z-a\cos \theta_0)}{z^2-2az\cos\theta_0+a^2}\end{align}\tag{1}$$

From (1) it is easy to determine the zeros of $H(z)$: $$z_{0,0}=0\quad z_{0,1}=a\cos\theta_0$$

For determing the poles, we write $H(z)$ as a partial fraction expansion:

$$H(z)=\frac{1}{2}\left [ \frac{1}{1-ae^{j\theta_0}z^{-1}} + \frac{1}{1-ae^{-j\theta_0}z^{-1}} \right ]\tag{2}$$

From (2) we see that the poles are given by $$z_{\infty,0}=ae^{j\theta_0}\quad z_{\infty,1}=ae^{-j\theta_0}$$ We have complex conjugate poles because $h(n)$ is real-valued. Assuming that $h(n)$ is an impulse response, we can see from the poles that the system is stable if $|a|<1$ because then the poles are inside the unit circle of the complex plane.

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The Z-transform of $x(n) = a^ncos(nθ)u(n)...$ will be: enter image description here Hope its helpful

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  • $\begingroup$ Would you mind putting it in TeX, instead of picture? $\endgroup$ – jojek Jun 24 '15 at 10:02

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