2
$\begingroup$

Okay so I have this question for my homework which I am asked to find the fourier transform of a signal. However I am unable to obtain the correct answer, the question and my working is shown below.

Question( The answer is the equation of Xp(f) ):

Picture of Question

My working(Look from b) onwards):

Working

As can be seen, I am unable to get the answer. But I don't feel that I am wrong.

Thanks.

$\endgroup$
2
$\begingroup$

If $x(t)$ is the (non-periodic) function in the interval $[-2,2]$, then the periodic function $x_p(t)$ is given by

$$x_p(t)=x(t)\star \sum_{n=-\infty}^{\infty}\delta(t-6n)\tag{1}$$

where $\star$ denotes convolution, and, consequently, its Fourier transform is

$$X_p(f)=X(f)\cdot\frac16\sum_{k=-\infty}^{\infty}\delta\left(f-\frac{k}{6}\right)= \frac16\sum_{k=-\infty}^{\infty}X\left(\frac{k}{6}\right)\delta\left(f-\frac{k}{6}\right)\tag{2}$$

where $X(f)$ is the Fourier transform of $x(t)$. It remains to find $X(f)$. The function $x(t)$ can be written as

$$x(t)=\sin\left(\frac{\pi t}{2}\right)\cdot\left[\text{rect}\left(\frac{t-1}{2}\right)-\text{rect}\left(\frac{t+1}{2}\right)\right]\tag{3}$$

With the Fourier transform pairs

$$\begin{align}\sin\left(\frac{\pi t}{2}\right)&\Longleftrightarrow \frac{1}{2j}\left[\delta\left(f-\frac14\right)-\delta\left(f+\frac14\right)\right]\\\text{rect}\left(\frac{t\pm 1}{2}\right)&\Longleftrightarrow 2e^{\pm j2\pi f}\text{sinc}(2f)\end{align}\tag{4}$$

we obtain from $(3)$

$$\begin{align}X(f)&=\frac{1}{2j}\left[\delta\left(f-\frac14\right)-\delta\left(f+\frac14\right)\right]\star 2\;\text{sinc}(2f)(e^{-j2\pi f}-e^{j2\pi f})\\&=2\left[\delta\left(f+\frac14\right)-\delta\left(f-\frac14\right)\right]\star\text{sinc}(2f)\sin(2\pi f)\\ &=2\left[\text{sinc}\left(2f+\frac12\right)\sin\left(2\pi f+\frac{\pi}{2}\right)-\text{sinc}\left(2f-\frac12\right)\sin\left(2\pi f-\frac{\pi}{2}\right)\right]\end{align}\tag{5}$$

The final expression for $X_p(f)$ results from combining $(5)$ and $(2)$.

$\endgroup$
0
$\begingroup$

OK. Re-looking at it... $X_p(f)$ in the screenshot doesn't depend on $f$ except in the argument of the $\delta$ function... so something is missing!

Oh. Does your equation start with $\sin^2$ ? That's going to be different from $|\sin|$ which is what the question says the signal is.

To write that out as equation I'd try something like: $$ \cos(0.5\pi t)\mbox{rect}(t/2) $$ and then move it by $+1$ and $-1$ $$ \cos(0.5\pi (t-1))\mbox{rect}((t-1)/2) + \cos(0.5\pi (t+1))\mbox{rect}((t+1)/2) $$ though I probably have the scalings wrong; please check!

$\endgroup$
  • $\begingroup$ Yes. Which is why I used the rect(.) function to isolate the signal from -2 to 2 by multiplying it by the sin^2 function. I then convoluted it with a comb function to get the replication effect. $\endgroup$ – John Nov 27 '15 at 16:05
  • $\begingroup$ @John Yes, I see that now. But $X_p(f)$ seems strange: the trig terms don't have $f$ as part of their arguments. Are you leaving something out? $\endgroup$ – Peter K. Nov 27 '15 at 16:14
  • $\begingroup$ Oh. Assuming that I had |sin| , how would I expand it? So that I can take the fourier transform of it. $\endgroup$ – John Nov 27 '15 at 16:28
  • $\begingroup$ @John Answer updated. Not sure it's quite correct, but might be better than $\sin^2$. $\endgroup$ – Peter K. Nov 27 '15 at 16:37
  • $\begingroup$ I think it might be easier to use |sin| , but I am just unsure on how to expand it or get the fourier transform of it $\endgroup$ – John Nov 28 '15 at 6:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.