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Question

A discrete-time signal $u \in \mathcal{l}^2(\mathcal{Z})$ has DTFT \begin{equation} \hat{u}(\omega) = \frac{5+3\cos(\omega)}{17+8\cos(\omega)} \end{equation} Use complex integration to find $u(k)$ for $k\in\mathcal{Z}$


My Attempt

If I'm not mistaken, to find $u[k]$, I should find the inverse Fourier Transform \begin{equation} u(k) = \frac{1}{2\pi} \int_{-\pi}^{\pi} \frac{5+3\cos(\omega)}{17+8\cos(\omega)} e^{j\omega k} d\omega \end{equation} Is this where I use complex integration? In order to do so, doesn't my equation need poles? I can't think of when $17+8\cos(\omega)$ would equal zero for this to not be analytic.

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  • $\begingroup$ Is is mandatory to use complex integration? It can also be done with real integration (calculus II) It's not that hard at all. Either way, I would first carry out a division $\endgroup$ – imranfat Dec 8 '13 at 23:12
  • $\begingroup$ It is required to use complex integration for this particular question. $\endgroup$ – John Dec 8 '13 at 23:15
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Using the residue theorem (complex integration) on integrals of rational functions of sines and cosines is commonplace. The trick is to let $z=e^{i \omega}$. Then $d\omega = -i dz/z$ and the integral may be expressed as a complex integral over the unit circle:

$$\begin{align}u(k) &= -\frac{i}{2 \pi} \oint_{|z|=1} \frac{dz}{z} \frac{5+\frac{3}{2} (z+z^{-1})}{17+4 (z+z^{-1})} z^k\\ &= -\frac{i}{4 \pi} \oint_{|z|=1} dz \, z^{k-1} \frac{3 z^2+10 z+3}{4 z^2+17 z+4}\end{align}$$

Because $k \in \mathbb{Z}$, we have no branch points. Consider $k \ge 1$. Then the only poles in the integrand are where

$$4 z^2+17 z+4=0 \implies z_{\pm} = \frac{-17 \pm 15}{8} $$

i.e., $-1/4$ or $-4$, respectively. As only $z_+=-1/4$ is the only pole within the unit circle, this is the only pole contributing to the integral.

By the residue theorem, the integral is $i 2 \pi$ times the residue at the pole $z=-1/4$. Thus

$$u(k) = \frac12 \left (-\frac14 \right)^{k-1} \frac{11}{16 \cdot 15} = -\frac{11}{120} \left (-\frac14 \right)^{k} $$

when $k \ge 1$. When $k=0$, there is an additional pole at $z=0$, which contributes $3/8$ to add to the above piece, or

$$u(0) = \frac{3}{8}-\frac{11}{120} = \frac{17}{60} $$

For $k \lt 0$, we can use the fact that

$$u(-k) = u(k)^*$$

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  • $\begingroup$ The numerator as well as the denominator are always positive, yet you get a negative answer for the integral. What am I missing here? $\endgroup$ – imranfat Dec 9 '13 at 18:49
  • $\begingroup$ @imranfat: Drat - I hate when someone else finds a dumb mistake like that. Thanks. $\endgroup$ – Ron Gordon Dec 9 '13 at 19:09

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