Solving a tough integral

Question

I am studying telecommunications theory and I was doing an exercise where it's required to find the (infinite) taps of a zero forcing equalizer. Here's the point where I am stuck at:

$$ p_\ell=T\int_{-\frac{1}{2T}}^{\frac{1}{2T}}\frac{e^{j2\pi f\ell T}}{1+\alpha e^{-j2\pi fT}}df $$ Where:

• $\ell\in \mathbb{Z}$
• $0<\alpha<\frac{1}{2}$
• $T>0$
• $T,\alpha\in\mathbb{R}$

That comes out because the channel time domain response is: $$ g(t)=\delta(t)+\alpha\delta(t-T) $$ And its fourier transform of course is: $$ G(f)=1+\alpha e^{-j2\pi fT} $$ In a ZF equalizer it is required that the total f-response of the channel and equalizer is unity, i.e. $ P(f)\cdot G(f)=1 $, so to find the $p_\ell$ sequence one has to anti-transform $\frac{1}{G(f)}$.

It doesn't look to me I've done any errors before the integral but I don't have a clue on how to solve it, if possible. Some help/hints would be very appreciated.

Thanks to PhoemueX answer:

$$ \frac{1}{1 + \alpha e^{-2\pi i f T}} = \frac{1}{1 - (- \alpha e^{-2\pi i f T})} = \sum_{n=0}^{\infty} (-\alpha \cdot e^{-2\pi i f T})^n, $$

So let's start rocking:

$$ p_\ell=T\int_{-\frac{1}{2T}}^{\frac{1}{2T}}e^{j2\pi f\ell T}\sum_{n=0}^{\infty} (-\alpha \cdot e^{-2\pi i f T})^ndf=\\ =T\sum_{n=0}^{\infty}\int_{-\frac{1}{2T}}^{\frac{1}{2T}}(-\alpha)^ne^{j2\pi f T(\ell-n)}df=\\ =\frac{T}{j2\pi T}\sum_{n=0}^{\infty}\frac{(-\alpha)^n}{\ell-n} \left(e^{j\pi(\ell-n)}-e^{-j\pi(\ell-n)}\right)=\\ =\frac{2j}{2j\pi}\sum_{n=0}^{\infty}(-\alpha)^n\frac{\sin[\pi(\ell-n)]}{\ell-n}=\\ =\sum_{n=0}^{\infty}(-\alpha)^n\text{sinc}(\ell-n) $$

That last line equals zero whenever $\ell\neq n$, while when $\ell=n$ the sinc is not defined. We can not compute the limit because that is nonsense in $\mathbb{Z}$ but looking at the second equation we can see that when $\ell=n$ the integral becomes trivial and that sum equals $(-\alpha)^\ell$

To sum up: $$ p_\ell=(-\alpha)^\ell $$

Math is awesome.

Answer 1

Expand the denominator into a geometric series like this:

$$ \frac{1}{1 + \alpha e^{-2\pi i f T}} = \frac{1}{1 - (- \alpha e^{-2\pi i f T})} = \sum_{n=0}^{\infty} (-\alpha \cdot e^{-2\pi i f T})^n, $$

where the series converges uniformly as long as $|\alpha|<1$ (this is the case in your question).

Hence, we can interchange summation and integration.

Why does that help you?

Answer 2

You can also solve this problem with the standard tools of complex analysis. If you make a substitution $z=\exp(2\pi i fT)$, then the integral becomes $$p_\ell=\frac{1}{2\pi i}\int dz\ \frac{z^\ell}{z+\alpha}$$ where the contour of integration is the unit circle oriented counter-clockwise.

Because $\alpha\in(0,\tfrac{1}{2})$, there is a pole at $z=-\alpha$ contained in the contour. Note that, when $\ell<0$, there is a second pole at $z=0$. The integral therefore evaluates to $$p_\ell={\rm Res}\left(\frac{z^\ell}{z+\alpha};z=-\alpha\right)+{\rm Res}\left(\frac{z^\ell}{z+\alpha};z=0\right).$$ These residues can be evaluated easily: $${\rm Res}\left(\frac{z^\ell}{z+\alpha};z=-\alpha\right)=(-\alpha)^\ell,$$ $${\rm Res}\left(\frac{z^\ell}{z+\alpha};z=0\right)=\begin{cases}0,&\ell\ge 0\\-(-\alpha)^\ell,&\ell<0\end{cases}.$$

The result is therefore slightly different from what you found. Namely, the integral vanishes for $\ell<0$: $$p_l=\begin{cases}0,&\ell<0\\(-\alpha)^\ell,&\ell\ge 0\end{cases}.$$