GATE 2018 EE Q.38: solution and answer

GATE 2018 · EEQ.38 · 2 marks · MCQ

Consider the two continuous-time signals defined below:

x_{1} (t) = {∣ t ∣, 0, - 1 \leq t \leq 1 otherwise, x_{2} (t) = {1 - ∣ t ∣, 0, - 1 \leq t \leq 1 otherwise

These signals are sampled with a sampling period of $T = 0.25$ seconds to obtain discrete-time signals $x_{1} [n]$ and $x_{2} [n]$ , respectively. Which one of the following statements is true?

(A)
The energy of $x_{1} [n]$ is greater than the energy of $x_{2} [n]$ .
(B)
The energy of $x_{2} [n]$ is greater than the energy of $x_{1} [n]$ .
(C)
$x_{1} [n]$ and $x_{2} [n]$ have equal energies.
(D)
Neither $x_{1} [n]$ nor $x_{2} [n]$ is a finite-energy signal.

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Answer and solution

ANSWER

(A) The energy of $x_{1} [n]$ is greater than the energy of $x_{2} [n]$ : $E_{1} = 3.75$ and $E_{2} = 2.75$ .

Official answer: ASource: Official answer key, GATE 2018 EE (IIT Guwahati)

Sampling: $x [n] = x (n T) = x (0.25 n)$ . Both signals are nonzero only for $- 1 \leq t \leq 1$ , so only $- 4 \leq n \leq 4$ can give nonzero samples. Both are finite-length sequences with finite values, so both have finite energy, which rules out (D).
$x_{1} [n] = ∣0.25 n ∣$ for $∣ n ∣ \leq 4$ . At $n = 0, \pm 1, \pm 2, \pm 3, \pm 4$ the values are $0, 0.25, 0.5, 0.75, 1$ . The endpoints $t = \pm 1$ lie inside the closed interval $- 1 \leq t \leq 1$ , so $x_{1} [\pm 4] = 1$ .
$x_{2} [n] = 1 - ∣0.25 n ∣$ for $∣ n ∣ \leq 4$ . At $n = 0, \pm 1, \pm 2, \pm 3, \pm 4$ the values are $1, 0.75, 0.5, 0.25, 0$ .
Discrete-time energy is $E = \sum_{n} ∣ x [n] ∣^{2}$ . For $x_{1}$ : $E_{1} = 0 + 2 (0.2 5^{2} + 0. 5^{2} + 0.7 5^{2} + 1^{2}) = 2 (0.0625 + 0.25 + 0.5625 + 1) = 2 (1.875) = 3.75$
For $x_{2}$ : $E_{2} = 1^{2} + 2 (0.7 5^{2} + 0. 5^{2} + 0.2 5^{2} + 0^{2}) = 1 + 2 (0.5625 + 0.25 + 0.0625) = 1 + 2 (0.875) = 2.75$
$E_{1} = 3.75 > E_{2} = 2.75$ . The two sequences take the same set of values, but the largest value $1$ occurs twice in $x_{1} [n]$ (at $n = \pm 4$ ) and only once in $x_{2} [n]$ (at $n = 0$ ), while the value $0$ occurs once in $x_{1} [n]$ and twice in $x_{2} [n]$ . So the energy of $x_{1} [n]$ is greater, which is option (A).

KEY IDEA

Sample at $t = n T$ , list the finite set of samples (including the endpoints $t = \pm 1$ , which are inside the closed interval), and add their squares: $E = \sum_{n} ∣ x [n] ∣^{2}$ . Here $x_{1} [n] + x_{2} [n] = 1$ for $∣ n ∣ \leq 4$ . Both sequences take the values $0, 0.25, 0.5, 0.75, 1$ , but $x_{1} [n]$ reaches the peak value $1$ at two samples ( $n = \pm 4$ ) and $x_{2} [n]$ at only one ( $n = 0$ ).

WATCH OUT FOR

Choosing (C) by reasoning from the continuous-time signals. Their energies are equal, $\int ∣ x_{1} (t) ∣^{2} d t = \int ∣ x_{2} (t) ∣^{2} d t = \frac{2}{3}$ , but sampling does not keep that equality: the samples at the interval edges $t = \pm 1$ are where $x_{1}$ is largest and $x_{2}$ is zero, so $x_{1} [n]$ gets two peak samples and $x_{2} [n]$ only one. Dropping the endpoint samples $n = \pm 4$ (treating the interval as open) gives the same kind of miscount, with $E_{1} = 1.75$ against $E_{2} = 2.75$ , which points to (B).

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GATE 2018 EE Q.38: solution and answer

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