Math  /  Calculus

QuestionDrill Problem 5.7 Consider a continuous-time signal defined by g(t)=sin(πt)πtg(t)=\frac{\sin (\pi t)}{\pi t}
The signal g(t)g(t) is uniformly sampled to produce the infinite sequence {g(nTs)}n=\left\{g\left(n T_{s}\right)\right\}_{n=-\infty}^{\infty}. Determine the condition that the sampling period TsT_{s} must satisfy so that the signal g(t)g(t) is uniquely recovered from the sequence {g(nTs)}\left\{g\left(n T_{s}\right)\right\}.

Studdy Solution

STEP 1

1. The signal g(t)=sin(πt)πt g(t) = \frac{\sin(\pi t)}{\pi t} is a sinc function.
2. The sinc function is a bandlimited signal with a specific bandwidth.
3. To recover a continuous-time signal from its samples, the sampling theorem (Nyquist-Shannon) must be satisfied.
4. The sampling period Ts T_s must be chosen such that the sampling frequency fs=1Ts f_s = \frac{1}{T_s} is greater than twice the maximum frequency of the signal.

STEP 2

1. Identify the bandwidth of the signal g(t) g(t) .
2. Apply the Nyquist-Shannon sampling theorem.
3. Determine the condition on the sampling period Ts T_s .

STEP 3

Identify the bandwidth of the signal g(t)=sin(πt)πt g(t) = \frac{\sin(\pi t)}{\pi t} .
The sinc function sin(πt)πt \frac{\sin(\pi t)}{\pi t} is known to be bandlimited with a cutoff frequency of 12 \frac{1}{2} Hz. This is because the Fourier transform of a sinc function is a rectangular function in the frequency domain.

STEP 4

Apply the Nyquist-Shannon sampling theorem.
According to the Nyquist-Shannon sampling theorem, to perfectly reconstruct a bandlimited signal from its samples, the sampling frequency fs f_s must be greater than twice the maximum frequency of the signal.
For g(t) g(t) , the maximum frequency is 12 \frac{1}{2} Hz, so the sampling frequency must satisfy:
fs>2×12=1 Hz f_s > 2 \times \frac{1}{2} = 1 \text{ Hz}

STEP 5

Determine the condition on the sampling period Ts T_s .
The sampling frequency fs f_s is related to the sampling period Ts T_s by fs=1Ts f_s = \frac{1}{T_s} . Therefore, the condition on Ts T_s is:
1Ts>1 \frac{1}{T_s} > 1
Solving for Ts T_s , we get:
Ts<1 T_s < 1
Thus, the sampling period Ts T_s must be less than 1 second to ensure that the signal g(t) g(t) can be uniquely recovered from its samples.
The condition that the sampling period Ts T_s must satisfy is Ts<1 \boxed{T_s < 1} .

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