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If I have a signal x(t) as follow

x(t)=$\sum\limits_{k=-∞}^∞ R_{T}(t-2kT)$

$R_{T}(t) = \begin{cases}1 ;&[\dfrac{-T}{2},\dfrac{T}{2}] \\ 0; & \text{otherwise} \end{cases}$

$\mathfrak{F}\{R_{T}(t)\}=Tsinc(\pi fT)$

$H(f)= \begin{cases}1 ;&|f| \le \dfrac{3}{4T} \\ 0; & |f| \ge \dfrac{3}{4T} \end{cases} $

$Y(f)=\dfrac{1}{2}\delta(t) + \dfrac{1}{\pi}\delta(f-\dfrac{1}{2T}) +\dfrac{1}{\pi}\delta(f-\dfrac{1}{2T})$

Can Some one Please Help me in how X(f) will Look Like graphically . I am just Poor in the graphical representation of how X(f) multiplied With H(f) to obtain Y(f).
Please this is not a home work question but i really need to know to move forward.i tried a lot but i just don't get graphical representation in frequency domain. thank You.

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  • $\begingroup$ This is very similar to your previous question. Is this finally the correct version? Do you know how to obtain $X(f)$? Otherwise how do you want to get a graphical representation? Please show us what you have done so far and where you're stuck. $\endgroup$ – Matt L. Feb 11 '15 at 9:14
  • $\begingroup$ Yes the last Question is almost the same. Exactly X(f) is my problem. Kindly Matt I need to understand this it is very necessary for me to understand. i know how to find the fourier transform of periodic signal but i simply do not understand how this rectangular function is transformed here. $\endgroup$ – dfeast Feb 11 '15 at 11:22
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The Fourier transform of $x(t)$ can be easily obtained if you realize that $x(t)$ can be written as the convolution of the pulse $R_T(t)$ and a Dirac comb:

$$x(t)=R_T(t)*\sum_{k=-\infty}^{\infty}\delta(t-2kT)\tag{1}$$

Since convolution corresponds to multiplication in the frequency domain, you get for the Fourier transform of $x(t)$

$$X(f)=\mathcal{F}\{R_T(t)\}\cdot\frac{1}{2T}\sum_{k=-\infty}^{\infty}\delta(f-k/2T)\tag{2}$$

because the Fourier transform of a Dirac comb is a Dirac comb (see e.g. this table). And since

$$f(t)\delta(t)=f(0)\delta(t)$$

Eq. (2) can be written as

$$X(f)=\frac{1}{2T}\sum_{k=-\infty}^{\infty}\hat{R_T}\left(\frac{k}{2T}\right) \delta(f-k/2T)\tag{3}$$

where $\hat{R_T}(f)$ is the Fourier transform of $R_T(t)$. Eq. (3) shows that the Fourier transform of $x(t)$ is a weighted sum of Dirac impulses, where the weights are equidistant samples of the Fourier transform of the pulse $R_T(t)$.

Since the low pass filter $H(f)$ only passes frequencies lower than $3/4T$, only the contributions for $k\in\{-1,0,1\}$ of the sum in (3) are passed. This gives the final result for $Y(f)$. Note that there are two typos in your formulation of $Y(f)$. It should be

$$Y(f)=\frac12\delta(f)+\frac{1}{\pi}\delta(f-1/2T)+\frac{1}{\pi}\delta(f+1/2T)\tag{4}$$

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  • $\begingroup$ If There is a heaven and i am sure there is one and i am sure you are going there. Thank You Sir Matt. $\endgroup$ – dfeast Feb 12 '15 at 17:11
  • $\begingroup$ you took me from Confused mode to enlightened mode. $\endgroup$ – dfeast Feb 12 '15 at 17:13
  • $\begingroup$ And One small question if the rect(t) is defined in the interval [0,T] then the simplified x(t) should be like this right? x(t)=Rect(t-T/2) * Train_Of_Deltas? $\endgroup$ – dfeast Feb 12 '15 at 17:34
  • $\begingroup$ @dfeast: Yes, that just shifts the pulse to the right by $T/2$. $\endgroup$ – Matt L. Feb 12 '15 at 17:41

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