Problem 6:
 
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:

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).

 

 

 

Figure (a)

 

 

 

Figure (b)

 

 

(b) To recover from , we do the following two things:

1)     Modulate the signal by .

2)     Apply a low pass filter of bandwidth .

 

 

(c) To recover from , we do the following two things:

1)     Modulate the signal by .
2)     Apply a low pass filter of bandwidth .

 

 

(d) Maximum value for recoverability is as can be seen from the graphs.