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):


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



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 $(2)$.

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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!

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  • $\begingroup$ 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

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