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How can I find the frequency response of the following linear system enter image description here

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    $\begingroup$ This looks like homework, which is OK, but you should show some evidence of having tried yourself and ask some more concrete question than just "please solve it for me". $\endgroup$ – Matt L. Oct 22 '14 at 8:50
  • $\begingroup$ I don't know if i am doing this right. I am proceeding in frequency domain from the start($X(w)$). $X(w)-X(w)e^{-j \omega T}$ would come out of first sigma block..... $\endgroup$ – user2332665 Oct 22 '14 at 9:02
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    $\begingroup$ That sounds like a good start ... :) And note that you're almost done. Just find out what an ideal integrator is in the frequency domain, and note that you have the same system twice in a row. What does that mean for the frequency response? $\endgroup$ – Matt L. Oct 22 '14 at 9:12
  • $\begingroup$ the final answer is $T^2 sinc^2(fT) e^{-j 2\pi fT}$, going like this I am not getting there. 'same system twice in a row' i think i need to find transfer function upto the integrator and square it. $\endgroup$ – user2332665 Oct 22 '14 at 9:38
  • $\begingroup$ You're right about squaring. I think you got the correct result, but you just don't see how to get the sinc function in there, which doesn't matter because it's just another way to write out the solution. Show us your result and I can tell you if it's correct. $\endgroup$ – Matt L. Oct 22 '14 at 9:47
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OK, in the comments you've shown that you can actually do it by yourself. This answer just summarizes the steps and shows you how to write your (correct) result in the form that you were given as an answer.

First of all, note that you have two identical systems in series. If $G(\omega)$ is the frequency response of the system up to and including the first integrator (from the left), then the total frequency response is

$$H(\omega)=G^2(\omega)\tag{1}$$

The response $G(\omega)$ can be seen as a multiplication of two subsystems, the system up to and including the adder, and the integrator. The first of the two has a frequency response

$$G_1(\omega)=1-e^{-j\omega T}\tag{2}$$

Since $G_1(\omega)$ has a zero at $\omega=0$, we don't need to bother with any problems of the integrator at $\omega=0$ (I'm talking about the delta impulse). So for calculating $G(\omega)=G_1(\omega)G_2(\omega)$, we can simply use

$$G_2=\frac{1}{j\omega}\tag{3}$$

which gives

$$G(\omega)=\frac{1-e^{-j\omega T}}{j\omega}\tag{4}$$

and, from (1),

$$H(\omega)=\frac{(1-e^{-j\omega T})^2}{(j\omega)^2}\tag{5}$$

In order to get from (5) the answer that you were given, use this trick to rewrite $G_1(\omega)$:

$$G_1(\omega)=e^{-j\omega T/2}(e^{j\omega T/2}-e^{-j\omega T/2})=2je^{-j\omega T/2}\sin(\omega T/2)\tag{6}$$

Using (6) you can rewrite $G(\omega)$ as

$$G(\omega)=2e^{-j\omega T/2}\frac{\sin(\omega T/2)}{\omega}= Te^{-j\omega T/2}\frac{\sin(\omega T/2)}{\omega T/2}\tag{7}$$

Squaring (7), using $\omega=2\pi f$ and $\text{sinc}(x)=\sin(\pi x)/(\pi x)$ will give you the final form of the result.

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Neat. This one you could also do in the time domain. Let's make x(t) a dirac impulse. At t=0 that gets into the integrate and the output of the integrator is 1 for t >=0. Now at T=t, the negative impulse gets into the integrator the result for this is -1 for t >=T. This cancels the original impulse and the output becomes zero again. So the impulse response of the first stage is basically a rectangle: h(t) = 1, for 0<= t < T, 0 otherwise.

This corresponds, of course, to the sinc function in the frequency domain. Cascading two systems is done by convolving the two impulse response, so we have to convolve two rectangles which simply results in a triangle.

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