# Fourier Transform of infinite sum

I'm looking at last year's exams, and I found an exercise I can't solve: (Roughly translated)

Consider $x(t)=\sum\limits_{k=1}^{+\infty}\left(\frac{1}{2}\right)^k \cos(k2\pi t)$ the input to a LTI system, with the transfer function shown in the picture below a) Determine (and sketch) the Fourier transform of x(t) and y(t).

b) Determine the Fourier coefficients of y(t). Justify.

Edit:

The only thing that comes to mind is using the Linearity of the Fourier Transform and calculate the Transform of $\left(\frac{1}{2}\right)^k \cos(k2\pi t)$, k dependant, and $X(jw)=\sum\limits_{k=1}^{+\infty}X_k(jw)$, so it would be something like $$X(jw)=\sum\limits_{k=1}^{+\infty}\left[2^{-k}\pi\left[\delta(\omega-2k\pi) + \delta(\omega+2k\pi)\right]\right]$$

However, I have never seen a Fourier Transform that looked like this, so I'm not quite sure that's correct, or, if it is, if it can be used (efficiently) for calculations, in this form.

After I get $X(jw)$, I can get $Y(jw)=X(jw)H(jw)$, so right now I'm pretty much just stuck in calculating $X(jw)$. Any help is much appreciated, and feel free to ask for clarification, if something I wrote is not clear enough.

• Hint: There is no $u(t)$ in this problem. Also, the Fourier transform of each term in your $x(t)$ involves impulses (also called Dirac deltas by mathematicians and delta functions by engineers) and so you need to look at a different section of your Fourier transform table. – Dilip Sarwate Dec 30 '13 at 6:58
• @DilipSarwate yeah, I figured that was a bit off, only wrote it just in case. If you don't mind me asking, is the last idea in the right path, transforming each term and then do the sum of all of the transforms? – Sampaio Dec 30 '13 at 16:41
• @DilipSarwate Please take a look at the edit – Sampaio Dec 30 '13 at 23:41

I believe you are correct with

$$X(jw)=\sum\limits_{k=1}^{+\infty}\left[2^{-k}\pi\left[\delta(\omega-2k\pi) + \delta(\omega+2k\pi)\right]\right]$$

Now we also have $$H(jw) = \mathbf{1}_{[-5\pi,-3\pi]} + \mathbf{1}_{[3\pi,5\pi]}$$

Then $$Y(jw) = X(jw)\cdot H(jw)$$ $$= \left[2^{-2}\pi\left[\delta(\omega-4\pi) + \delta(\omega+4\pi)\right]\right]$$ $$= \frac{\pi}{4} \left[\delta(\omega-4\pi) + \delta(\omega+4\pi)\right]$$

Transforming back:

$$y(t) = \frac{1}{4} cos(4\pi t)$$

The trick is to see that $X(jw)$ is an infinite sum of dirac deltas, but only two of those deltas lie in the non-zero range of the transfer function.

i think it's pretty obvious that only the component with angular frequency of $4 \pi$ gets through the LTI system to the output. so the output should be $\frac{1}{4} \cos(4 \pi t)$.

• yes, but what is the Fourier Transform of x(t)? – Sampaio Dec 30 '13 at 3:00
• or rather, how to go about calculating it? – Sampaio Dec 30 '13 at 3:11
• what's the fourier transform of one term in $x(t)$? – robert bristow-johnson Dec 30 '13 at 3:32
• I believe it's $2^{-k}\pi\left[\delta(\omega-2k\pi)+\delta(\omega+2k\pi)\right]$. Is $X(jw)$ the infinite sum of that? – Sampaio Dec 30 '13 at 20:18
• yes. that's what linearity means in a the Fourier Transform. and it's only the term where $k=2$ that passes through your LTI filter, the other terms are multiplied by zero. that's $Y(j\omega)$. then inverse transform back to $y(t)$. – robert bristow-johnson Dec 31 '13 at 7:28