In the system shown in the figure below, two continuous-time signals

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In the system shown in the figure below, two continuous-time signals ??1(??) ?and ??2(??) ?are

convolved to give the signal ??(??)=??1(??)**??2(??). ?The convolved signal is then sampled by a

periodic impulse train ??(??)=???=-88??(??-????). ?The signals ??1(??) ?and ??2(??) ?are band limited

and their respective Fourier transform ??1(??��) ?and ??2(????) ?are sketched below.

(i) (10pts). ?Determine the maximum possible sampling interval ??samp (or equivalently

minimum sampling frequency (??samp ) ?such that ??(??) ?can be completely recovered

from ????(??)=??(??)??(??) ?using an ideal low pass filter.(ii) (5 ?pts.) ?Determine the maximum sampling interval for the following cases:

a) ??1=20??,??2=40??

b) ??1=30??,??2=15??

(iii) (5 ?pts.) ?Consider the discrete-time sequence ??[??]=????(????) ?obtained from ????(??).

Let ??[??]h????????(??) ?from a DTFT pair. Assuming that ??(??) ?is sampled using the

maximum permissible interval ??samp ?determined in part (i), ?sketch the DTFT ??(??)

for the two cases discussed in part (ii).

The Correct Answer and Explanation is :

To answer your question step by step, we need to understand the components and use the key principles of signal processing and sampling theory.

Given Information:

  • Two continuous-time signals ( x_1(t) ) and ( x_2(t) ) are convolved to create the signal ( x(t) = x_1(t) * x_2(t) ).
  • The convolved signal ( x(t) ) is then sampled by a periodic impulse train ( \delta(t) = \sum_{n=-\infty}^{\infty} \delta(t-nT_{samp}) ), where ( T_{samp} ) is the sampling interval.
  • Both ( x_1(t) ) and ( x_2(t) ) are band-limited, meaning their Fourier transforms ( X_1(f) ) and ( X_2(f) ) are bounded in frequency.

Part (i): Maximum Sampling Interval

We are asked to determine the maximum possible sampling interval ( T_{samp} ) or equivalently the minimum sampling frequency ( f_{samp} = \frac{1}{T_{samp}} ) such that the signal ( x(t) ) can be completely recovered from its sampled version using an ideal low-pass filter.

Key Principle:

To ensure that ( x(t) ) can be perfectly reconstructed, the sampling rate must satisfy the Nyquist-Shannon sampling theorem, which states that the sampling frequency must be at least twice the maximum frequency component in the signal. This holds true for ( x(t) ), which is the result of convolving ( x_1(t) ) and ( x_2(t) ).

The Fourier Transform of the Convolution:
The Fourier transform of the convolved signal is the product of the Fourier transforms of the individual signals:
[
X(f) = X_1(f) \cdot X_2(f)
]
Thus, the maximum frequency of ( x(t) ) will be the sum of the maximum frequencies of ( x_1(t) ) and ( x_2(t) ). If ( X_1(f) ) is band-limited to ( f_1 ) (maximum frequency of ( x_1(t) )) and ( X_2(f) ) is band-limited to ( f_2 ) (maximum frequency of ( x_2(t) )), the maximum frequency of ( X(f) ) will be ( f_1 + f_2 ).

Sampling Frequency Condition:

To avoid aliasing, the sampling frequency must be at least twice the highest frequency component of ( x(t) ):
[
f_{samp} \geq 2(f_1 + f_2)
]
Thus, the maximum sampling interval ( T_{samp} ) is:
[
T_{samp} \leq \frac{1}{2(f_1 + f_2)}
]

Part (ii): Maximum Sampling Interval for Specific Cases

We are given two specific cases for the maximum frequencies of ( x_1(t) ) and ( x_2(t) ):

(a) ( f_1 = 20 \text{Hz}, f_2 = 40 \text{Hz} ):

  • The maximum frequency of ( x(t) ) is ( f_1 + f_2 = 20 + 40 = 60 \text{Hz} ).
  • Thus, the minimum sampling frequency is ( f_{samp} = 2 \times 60 = 120 \text{Hz} ).
  • The maximum sampling interval is ( T_{samp} = \frac{1}{120} \text{ seconds} \approx 8.33 \times 10^{-3} \text{ seconds} ).

(b) ( f_1 = 30 \text{Hz}, f_2 = 15 \text{Hz} ):

  • The maximum frequency of ( x(t) ) is ( f_1 + f_2 = 30 + 15 = 45 \text{Hz} ).
  • Thus, the minimum sampling frequency is ( f_{samp} = 2 \times 45 = 90 \text{Hz} ).
  • The maximum sampling interval is ( T_{samp} = \frac{1}{90} \text{ seconds} \approx 11.11 \times 10^{-3} \text{ seconds} ).

Part (iii): Sketching the DTFT

For the discrete-time sequence ( x[n] = x(t) \delta(t-nT_{samp}) ), the Discrete-Time Fourier Transform (DTFT) ( X(\omega) ) will correspond to periodic versions of the continuous-time spectrum ( X(f) ), scaled and shifted in frequency domain according to the sampling interval.

  • The DTFT of ( x[n] ) will have periodic components with period ( f_{samp} ).
  • For the two cases in part (ii), the DTFT of ( X[n] ) will have frequency components spaced by the sampling frequency ( f_{samp} ), and we would expect periodic replicas of the continuous-time spectrum ( X(f) ) centered at multiples of ( f_{samp} ).

In both cases:

  1. For ( f_{samp} = 120 \text{Hz} ), the DTFT will consist of periodic copies of ( X(f) ) spaced by 120 Hz.
  2. For ( f_{samp} = 90 \text{Hz} ), the DTFT will have periodic copies of ( X(f) ) spaced by 90 Hz.

Conclusion:

  • The maximum sampling interval for perfect recovery of ( x(t) ) is determined by the sum of the maximum frequencies of the individual signals.
  • For each case, the corresponding maximum sampling intervals are ( T_{samp} \approx 8.33 \times 10^{-3} ) seconds for the first, and ( T_{samp} \approx 11.11 \times 10^{-3} ) seconds for the second.

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