GATE EC 2026 Set 1 — Question 15

The sampling rate is $F_s = 40$ Hz. The sampling interval is $T_s = 1/F_s = 1/40$ seconds.
The samples are taken at $t_k = k T_s = k/40$ for $k = 0, 1, \dots, 9$ (first ten samples).For $x_1(t) = \cos(20\pi t)$, the samples are $x_1[k] = x_1(k/40) = \cos(20\pi k/40) = \cos(\pi k/2)$.
For $x_2(t) = \cos(100\pi t)$, the samples are $x_2[k] = x_2(k/40) = \cos(100\pi k/40) = \cos(5\pi k/2)$.Let's evaluate the samples for $k = 0, 1, \dots, 9$:$k=0$: $x_1[0] = \cos(0) = 1$, $x_2[0] = \cos(0) = 1$
$k=1$: $x_1[1] = \cos(\pi/2) = 0$, $x_2[1] = \cos(5\pi/2) = \cos(2\pi + \pi/2) = \cos(\pi/2) = 0$
$k=2$: $x_1[2] = \cos(\pi) = -1$, $x_2[2] = \cos(10\pi/2) = \cos(5\pi) = -1$
$k=3$: $x_1[3] = \cos(3\pi/2) = 0$, $x_2[3] = \cos(15\pi/2) = \cos(6\pi + 3\pi/2) = \cos(3\pi/2) = 0$
$k=4$: $x_1[4] = \cos(2\pi) = 1$, $x_2[4] = \cos(20\pi/2) = \cos(10\pi) = 1$
$k=5$: $x_1[5] = \cos(5\pi/2) = 0$, $x_2[5] = \cos(25\pi/2) = \cos(12\pi + \pi/2) = \cos(\pi/2) = 0$
$k=6$: $x_1[6] = \cos(3\pi) = -1$, $x_2[6] = \cos(30\pi/2) = \cos(15\pi) = -1$
$k=7$: $x_1[7] = \cos(7\pi/2) = 0$, $x_2[7] = \cos(35\pi/2) = \cos(16\pi + 3\pi/2) = \cos(3\pi/2) = 0$
$k=8$: $x_1[8] = \cos(4\pi) = 1$, $x_2[8] = \cos(40\pi/2) = \cos(20\pi) = 1$
$k=9$: $x_1[9] = \cos(9\pi/2) = 0$, $x_2[9] = \cos(45\pi/2) = \cos(22\pi + \pi/2) = \cos(\pi/2) = 0$From the calculations, we observe that $x_1[k] = x_2[k]$ for all $k = 0, 1, \dots, 9$.Let's check the given options:
(A) "All of the first three samples of $x_1(t)$ are greater than the corresponding samples of $x_2(t)$." This is false, as $x_1[k] = x_2[k]$.
(B) "All of the last three samples of $x_1(t)$ are greater than the corresponding samples of $x_2(t)$." This is false, as $x_1[k] = x_2[k]$.
(C) "All of the samples of $x_2(t)$ are greater than the corresponding samples of $x_1(t)$." This is false, as $x_1[k] = x_2[k]$.
(D) "All of the fourth to seventh samples of $x_1(t)$ are equal to the corresponding samples of $x_2(t)$." The fourth sample corresponds to $k=3$, and the seventh sample corresponds to $k=6$. For $k=3, 4, 5, 6$, we have:
$x_1[3]=0, x_2[3]=0$ (Equal)
$x_1[4]=1, x_2[4]=1$ (Equal)
$x_1[5]=0, x_2[5]=0$ (Equal)
$x_1[6]=-1, x_2[6]=-1$ (Equal)
This statement is TRUE.The final answer is $\boxed{D}$