Question:
Two analog signals \(x_1(t)\) and \(x_2(t)\) (\(t\) in second) are sampled at a rate \(F_s=40\) Hz, where
\[ x_1(t)=\cos(20\pi t),\ t\geq0,\qquad x_2(t)=\cos(100\pi t),\ t\geq0. \] The first ten samples (starting from \(t=0\)) are considered for the analysis. Which of the following statements is TRUE?
Two analog signals \(x_1(t)\) and \(x_2(t)\) (\(t\) in second) are sampled at a rate \(F_s=40\) Hz, where
\[ x_1(t)=\cos(20\pi t),\ t\geq0,\qquad x_2(t)=\cos(100\pi t),\ t\geq0. \] The first ten samples (starting from \(t=0\)) are considered for the analysis. Which of the following statements is TRUE?
\[ x_1(t)=\cos(20\pi t),\ t\geq0,\qquad x_2(t)=\cos(100\pi t),\ t\geq0. \] The first ten samples (starting from \(t=0\)) are considered for the analysis. Which of the following statements is TRUE?
Show Hint
Sample both cosines at t_n = n/40 seconds and simplify cos(2.5*pi*n) using the fact that 2*pi*n is always a whole number of cycles.
Updated On: Jul 20, 2026
- All of the first three samples of \(x_1(t)\) are greater than the corresponding samples of \(x_2(t)\).
- All of the last three samples of \(x_1(t)\) are greater than the corresponding samples of \(x_2(t)\).
- All of the samples of \(x_2(t)\) are greater than the corresponding samples of \(x_1(t)\).
- All of the fourth to seventh samples of \(x_1(t)\) are equal to the corresponding samples of \(x_2(t)\).
Show Solution
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The Correct Option is D
Solution and Explanation
Step 1: Write the sampling instants.
Sampling at \(F_s=40\) Hz means the sampling period is \(T_s=1/40=0.025\) s, so samples are taken at \(t_n=nT_s=0.025n\) for \(n=0,1,2,\ldots,9\).
Step 2: Write the sampled sequence for \(x_1(t)\).
\[ x_1[n]=\cos(20\pi\cdot0.025n)=\cos\left(\frac{\pi n}{2}\right) \]
Step 3: Write the sampled sequence for \(x_2(t)\).
\[ x_2[n]=\cos(100\pi\cdot0.025n)=\cos(2.5\pi n) \] Since \(2.5\pi n=2\pi n+\dfrac{\pi n}{2}\) and \(2\pi n\) is always a whole number of full cycles for integer \(n\), the cosine repeats:
\[ x_2[n]=\cos\left(2\pi n+\frac{\pi n}{2}\right)=\cos\left(\frac{\pi n}{2}\right) \]
Step 4: Compare the two sequences.
Both simplify to the exact same formula, \(\cos(\pi n/2)\), so \(x_1[n]=x_2[n]\) for every integer \(n\). This happens because the \(50\) Hz frequency of \(x_2\) aliases down to \(50-40=10\) Hz when sampled at \(40\) Hz, exactly matching the \(10\) Hz frequency of \(x_1\).
Step 5: List the first ten sample values.
\[ n=0,1,2,\ldots,9:\quad \cos\left(\frac{\pi n}{2}\right)=1,\,0,\,-1,\,0,\,1,\,0,\,-1,\,0,\,1,\,0 \]
Step 6: Check each option against these values.
(A) The first three samples of both signals are \(1,0,-1\); they are equal, not greater, so (A) is false. (B) The last three samples (\(n=7,8,9\)) are \(0,1,0\) for both signals, again equal, so (B) is false. (C) Since every sample of \(x_2\) equals the corresponding sample of \(x_1\), none is strictly greater, so (C) is false. (D) The fourth to seventh samples are the same list of values, \(0,1,0,-1\), for both \(x_1[n]\) and \(x_2[n]\), so they are indeed equal. This statement is TRUE.
Step 7: Final conclusion.
\[ \boxed{\text{All of the fourth to seventh samples of }x_1(t)\text{ are equal to the corresponding samples of }x_2(t).} \]
Sampling at \(F_s=40\) Hz means the sampling period is \(T_s=1/40=0.025\) s, so samples are taken at \(t_n=nT_s=0.025n\) for \(n=0,1,2,\ldots,9\).
Step 2: Write the sampled sequence for \(x_1(t)\).
\[ x_1[n]=\cos(20\pi\cdot0.025n)=\cos\left(\frac{\pi n}{2}\right) \]
Step 3: Write the sampled sequence for \(x_2(t)\).
\[ x_2[n]=\cos(100\pi\cdot0.025n)=\cos(2.5\pi n) \] Since \(2.5\pi n=2\pi n+\dfrac{\pi n}{2}\) and \(2\pi n\) is always a whole number of full cycles for integer \(n\), the cosine repeats:
\[ x_2[n]=\cos\left(2\pi n+\frac{\pi n}{2}\right)=\cos\left(\frac{\pi n}{2}\right) \]
Step 4: Compare the two sequences.
Both simplify to the exact same formula, \(\cos(\pi n/2)\), so \(x_1[n]=x_2[n]\) for every integer \(n\). This happens because the \(50\) Hz frequency of \(x_2\) aliases down to \(50-40=10\) Hz when sampled at \(40\) Hz, exactly matching the \(10\) Hz frequency of \(x_1\).
Step 5: List the first ten sample values.
\[ n=0,1,2,\ldots,9:\quad \cos\left(\frac{\pi n}{2}\right)=1,\,0,\,-1,\,0,\,1,\,0,\,-1,\,0,\,1,\,0 \]
Step 6: Check each option against these values.
(A) The first three samples of both signals are \(1,0,-1\); they are equal, not greater, so (A) is false. (B) The last three samples (\(n=7,8,9\)) are \(0,1,0\) for both signals, again equal, so (B) is false. (C) Since every sample of \(x_2\) equals the corresponding sample of \(x_1\), none is strictly greater, so (C) is false. (D) The fourth to seventh samples are the same list of values, \(0,1,0,-1\), for both \(x_1[n]\) and \(x_2[n]\), so they are indeed equal. This statement is TRUE.
Step 7: Final conclusion.
\[ \boxed{\text{All of the fourth to seventh samples of }x_1(t)\text{ are equal to the corresponding samples of }x_2(t).} \]
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