# Solving a tough integral

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.

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

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.

• I've read the FAQ about homework questions, I tried to provide some context but I'm not sure how many signal theory guys hang around here, if something is wrong with my question please comment it and I'd be glad to edit it. – Vladimir Cravero Sep 12 '14 at 12:59
• I would first attempt to substitute such that the exponent is not so long and ugly. Note that $e^{-ix} = \overline{e^{ix}}$ for $x\in\mathbb R$ so the $e$-term in the denominator is the complex conjugate of the numerator. – AlexR Sep 12 '14 at 13:05
• yep I've tried that, please note that the numerator exp contains the l parameter. That does not make the integral any easier though... – Vladimir Cravero Sep 12 '14 at 13:07
• This surely involves the use of hypergeometric functions, maybe the parametrization trick will do it? – Hakim Sep 12 '14 at 13:09
• I am sorry @Hakim, I have no idea of what that function is. Actually this exercise was given as an old test so I thought that solving it would actually be possible. Keep in mind this was for an EE course, not math folks. – Vladimir Cravero Sep 12 '14 at 13:11

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.

• FYI, the OP has added a bit in there stating that $0<\alpha<1/2$ – Semiclassical Sep 12 '14 at 13:26
• This is awesome. I've done some calculations on a piece of paper, is it ok if I take a picture and add it to OP? I promise my writing is clear. – Vladimir Cravero Sep 12 '14 at 13:36
• The result I get is $(-\alpha)^\ell$ anyway – Vladimir Cravero Sep 12 '14 at 13:37
• Sure, why not? You can edit your own post. If you want to, you could also add it to my post, but I think that adding it to your own post would make more sense. @Semiclassical: I know, I wrote that $|\alpha| < 1$ is needed mostly to state that slightly more general assumptions would also work. – PhoemueX Sep 12 '14 at 13:46
• @VladimirCravero: Your calculation is correct. The only thing I would have done differently (I like to be pedantic) is that in getting from line 2 to line 3, I would simply note that the integral vanishes for $\ell \neq n$ (why?) and equals $(-\alpha)^n$ for $\ell = n$. Otherwise, the $\ell - n$ in the denominator does not "feel right", at least for $\ell = n$. Also, the $\rm{sinc}$ is defined at $0$ (or at least, $\rm{sinc}$ extends continously to $0$) with $\rm{sinc}(0) = 1$. – PhoemueX Sep 12 '14 at 14:20

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 $\ell<0$: $$p_l=\begin{cases}0,&\ell<0\\(-\alpha)^\ell,&\ell\ge 0\end{cases}.$$

• Thanks for your insight. I have never studied complex analysis, you use some tools I don't understand, your result surprises me though. Where is the error in my calculation? – Vladimir Cravero Sep 12 '14 at 15:44
• Wait, I got it: the sum is over n from 0 to +inf, so $\ell$ can only be greater or equal to zero. Just distraction. – Vladimir Cravero Sep 12 '14 at 15:45