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I'm trying to Fourier transform the following function:

$$ y(t) = 2\cos(2\pi Bt) \sum_{k=-\infty}^\infty rect(2Bt-2k) $$

It's a normalized cosine with the negative lobes zeroed out.
I'm not sure I got the expression right, so here's how I got $y(t)$: I fed $x(t) = \cos(2\pi Bt)$ to $y=g(x):g(x\ge 0) = 2x, g(x\lt 0) = 0$. To transform it, I figured I could just do this:

$$ y(t) - y(t - \frac{1}{2B}) = 2\cos(2\pi Bt) $$

Which seems true to me, then transform both members and isolate $Y(f)$:

$$ Y(f) - Y(f)e^{-\frac{j\pi f}{B}} = \delta(f - B) + \delta(f + B) $$ $$ Y(f) = \frac{\delta(f - B) + \delta(f + B)}{1 - e^{-\frac{j\pi f}{B}}} $$

And this boggles me. I see just two Fourier coefficients, $Y_1$ and $Y_{-1}$, both skyrocketing to $\infty$. I guess I did something invalid, somewhere, I just don't know where. I know I could find $Y_k$ through the integral definition, but I would like to understand why this method doesn't work.

Thank you for your attention.

user480113
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  • It is a $1/B$-periodic function, so it is enough to look at the Fourier series $\sum_n c_n e^{2i \pi n t B}$ of a single period $[0,1/B]$, you'll obtain $y(t) = \sum_n c_n e^{2i \pi n t B}$ (here the series converges pointwise and in $L^2$) ie. $Y(f) = \sum_n c_n \delta(f- n/B)$. $$c_n = B\int_0^{1/B} y(t) e^{-2i \pi n t B} dt = ?$$ – reuns Sep 13 '17 at 02:09
  • Sorry, I'm using the phone and I posted instead of inserting a new line. Using the definition, like you suggested, gives me: $$c_n= \frac{2\cos(\frac{\pi n}{2})}{\pi (1-n^2)}$$ Which seems far more plausible, it's just an awful lot of work compared to my method (more calculations, chances of errors increase). That's why I would like to know why my way does not work! – user480113 Sep 13 '17 at 11:01
  • $y(t) - y(t - \frac{1}{2B}) = 2\cos(2\pi Bt)$ is not correct – reuns Sep 13 '17 at 18:46

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