Solution 6
(a) As , by dual of convolution theorem we have . So, we first find the Fourier Transform of as follows:
The Fourier Transform of a periodic function is an impulse train at intervals of , each impulse being of magnitude:
![](Solution_Template4_clip_image010.gif)
Here w e see that the impulses on the axis vanish at even values of k .
Hence, the Fourier Transform of is as shown in figure (a). I n the frequency domain, the output signal Y can be found by multiplying the input with the frequency response . Hence is as shown below in the figure (b).
![](Solution_Template4_clip_image018.gif) ![](Solution_Template4_clip_image020.jpg)
Figure (a)
![](Solution_Template4_clip_image022.jpg)
Figure (b) ![](Top.gif)
(b) To recover from , we do the following two things:
1) Modulate the signal by .
2) Apply a low pass filter of bandwidth .
![](Top.gif)
(c) To recover from , we do the following two things:
1) Modulate the signal by . ![](Solution_Template4_clip_image018_0000.gif)
2) Apply a low pass filter of bandwidth .
![](Top.gif)
(d) Maximum value for recoverability is as can be seen from the graphs.
![](Top.gif)
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