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The signal in question is:

$$x(t)= \sum_{n=-\infty}^{+\infty} \text{rect}\left(\frac{t-4nT}{2T}\right)+\text{trian}\left(\frac{t-n4T}{T}\right)\quad\text{with}\quad T=1\,\textrm{ms}$$

Passes through a filter with frequency response filter:

$$H(f)= \text{rect}\left(\frac{f-375}{400}\right) + \text{rect}\left(\frac{f+375}{400}\right)$$

I must find the output signal $y(t)$ from the filter. I have calculated the Fourier transform of $x$:

$$X(f)=\sum\limits_nC_n\delta\left(f-\frac{n}{4T}\right)\quad\text{and}\quad C_n= \frac{1}{2}\text{sinc}\left(\frac n2\right)+ \frac{1}{4}\text{sinc}\left(\frac n4\right)$$

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then I have post:

\begin{align} Y(f)&=X(f)H(f)\\ &=C_1(\delta(f-250)+\delta(f+250))+C_2(\delta(f+500)+\delta(f-500)) \end{align}

and

$$y(t)= 2C_1\cos(2\pi250)+2C_2\cos(2\pi500)$$

According to my prof the correct answer is:

$$Y(f)=\frac{1}{4}C_1(\delta(f-250)+\delta(f+250))+\frac{1}{4}C_2(\delta(f+500)+\delta(f-500))$$

why?

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