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I am trying to solve these questions. I have attached my solutions. Please I would appreciate if some one can check my approach and tell me if I am correct or not. Also, how can I proceed to find $h_0(t)$?


Question (s) with My Attempts

  1. Given the LTI-System with frequency response $$\LARGE H(j\omega)=T_1e^{-\frac{j\omega T_1}{2}}\cdot\frac{\sin(\frac{\omega T_1}{2})}{\frac{\omega T_1}{2}}$$Find
    a. Impulse Response $h_0(t)$ of this system.
    b. Reaction $y(t)$ of this system to the input signal $\Gamma(t)$

    enter image description here

  1. Determine the Fourier Transform of the Rectangular Pulse shown in the following figure.
    enter image description here

    enter image description here
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  • $\begingroup$ Your calculation is correct, but I think it will not lead to a simple solution (You'd need to calculate inverse FT of (1/jw), which does converge in the common sense, I think.) Instead, I give you two hints: 1) Check out the time-shift property of the FT and apply it to your equation. 2) Remember the what the Fourier transform of a rectangular function is and try to apply it here. Let us know, if you can solve on your own now or need more help, explaining your problems with my hints. $\endgroup$ Mar 31, 2017 at 11:31
  • $\begingroup$ Maximilian, thanks very much for the tips. I am trying to apply the Inverse FT but I am confuse on how to simply it. $\endgroup$
    – Soso
    Mar 31, 2017 at 12:02
  • $\begingroup$ I didn't check your calculations, but $\frac{2}{j\omega}$ is the FT of sgn function. So you actually have a difference of a sgn and a shifted sgn by $T$ which is a box function (width of $T$). However, a DC value is missing (compared to the IFT of the sinc). So some delta should also be there in your result... $\endgroup$
    – msm
    Mar 31, 2017 at 12:10
  • $\begingroup$ I have tried to do some maths manipulation which I have attached above, i don't know if I am on the right track. $\endgroup$
    – Soso
    Mar 31, 2017 at 12:39
  • $\begingroup$ @msm: You are right with the signs, but I think actually solving this integral is not easly. Further, I dont think there is a DC missing, the IFT of a sinc is also just a box. $\endgroup$ Mar 31, 2017 at 12:50

1 Answer 1

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$h_0(t)$ is inverse Fourier transform of $H(jw)$. Your formula is OK, you can continue your calculation to practice your math manipulation, why not. To check the result, you can remark that $H(jw)$ is rotated $sinc$ function. And $sinc$ must be the Fourier transform of a rectangular box function.

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  • $\begingroup$ So it is like a shifted rectangular function: rect(t - T1/2) $\endgroup$
    – Soso
    Mar 31, 2017 at 13:41

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