Question:medium
Let \(x_1(t)=\cos(2\pi nt)\) and \(x_2(t)=2\sin(4\pi nt)\) represent two sinusoids for a positive integer \(n\) and \(-\infty<t<\infty\). Which of the following statements about \(x_1(t)\) and \(x_2(t)\) is/are valid?
Let \(x_1(t)=\cos(2\pi nt)\) and \(x_2(t)=2\sin(4\pi nt)\) represent two sinusoids for a positive integer \(n\) and \(-\infty<t<\infty\). Which of the following statements about \(x_1(t)\) and \(x_2(t)\) is/are valid?
Show Hint
Turn the product of the two sinusoids into a sum of sines, then integrate over each interval to test orthogonality.
Updated On: Jul 20, 2026
- \(x_1(t)\) and \(x_2(t)\) are orthogonal to each other over \(0\leq t<1/n\).
- \(x_1(t)\) and \(x_2(t)\) are orthonormal to each other over \(0\leq t<1/n\).
- \(x_2(t)\) is a harmonic of \(x_1(t)\).
- \(x_1(t)\) and \(x_2(t)\) are non-orthogonal to each other over \(0\leq t<1/(2n)\).
Show Solution
The Correct Option is A, C, D
Solution and Explanation
Step 1: Understanding the Concept:
Two signals are orthogonal on an interval when the integral of their product over that interval is zero, and they are orthonormal when they are additionally orthogonal and each one has unit energy over the same interval. A harmonic simply means an integer multiple of a base frequency.
Step 2: Key Formula or Approach:
Turn the product $x_1(t)x_2(t)=2\cos(2\pi nt)\sin(4\pi nt)$ into a sum of sines using $2\cos A\sin B=\sin(A+B)+\sin(B-A)$, then integrate term by term over each interval asked about.
Step 3: Detailed Explanation:
With $A=2\pi nt$ and $B=4\pi nt$, the product becomes $\sin(6\pi nt)+\sin(2\pi nt)$.
Over the full period $0\leq t<1/n$, both $\sin(6\pi nt)$ and $\sin(2\pi nt)$ complete a whole number of cycles (three and one), so each integrates to zero. The total is zero, so $x_1(t)$ and $x_2(t)$ are orthogonal here, confirming (A).
To test orthonormality, look at the energy of $x_1(t)$ alone over this period: $\int_0^{1/n}\cos^2(2\pi nt)\,dt=\frac{1}{2n}$, which equals $1$ only for a special value of $n$, not generally. So the pair is not orthonormal in general, and (B) fails.
Since $x_1(t)$ oscillates at frequency $n$ and $x_2(t)$ oscillates at frequency $2n$, and $2n$ is a whole number times $n$, $x_2(t)$ is by definition a harmonic of $x_1(t)$, confirming (C).
Now shrink the window to $0\leq t<1/(2n)$, only half the earlier period. Here $\sin(6\pi nt)$ completes one and a half cycles and $\sin(2\pi nt)$ completes half a cycle, so neither term returns to zero net area. Direct evaluation gives $\int_0^{1/(2n)}\sin(6\pi nt)\,dt=\frac{1}{3\pi n}$ and $\int_0^{1/(2n)}\sin(2\pi nt)\,dt=\frac{1}{\pi n}$, and their sum $\frac{4}{3\pi n}$ is not zero. So the two signals are non-orthogonal on this shorter window, confirming (D).
Step 4: Final Answer:
\[ \boxed{\text{(A), (C), (D)}} \]
Two signals are orthogonal on an interval when the integral of their product over that interval is zero, and they are orthonormal when they are additionally orthogonal and each one has unit energy over the same interval. A harmonic simply means an integer multiple of a base frequency.
Step 2: Key Formula or Approach:
Turn the product $x_1(t)x_2(t)=2\cos(2\pi nt)\sin(4\pi nt)$ into a sum of sines using $2\cos A\sin B=\sin(A+B)+\sin(B-A)$, then integrate term by term over each interval asked about.
Step 3: Detailed Explanation:
With $A=2\pi nt$ and $B=4\pi nt$, the product becomes $\sin(6\pi nt)+\sin(2\pi nt)$.
Over the full period $0\leq t<1/n$, both $\sin(6\pi nt)$ and $\sin(2\pi nt)$ complete a whole number of cycles (three and one), so each integrates to zero. The total is zero, so $x_1(t)$ and $x_2(t)$ are orthogonal here, confirming (A).
To test orthonormality, look at the energy of $x_1(t)$ alone over this period: $\int_0^{1/n}\cos^2(2\pi nt)\,dt=\frac{1}{2n}$, which equals $1$ only for a special value of $n$, not generally. So the pair is not orthonormal in general, and (B) fails.
Since $x_1(t)$ oscillates at frequency $n$ and $x_2(t)$ oscillates at frequency $2n$, and $2n$ is a whole number times $n$, $x_2(t)$ is by definition a harmonic of $x_1(t)$, confirming (C).
Now shrink the window to $0\leq t<1/(2n)$, only half the earlier period. Here $\sin(6\pi nt)$ completes one and a half cycles and $\sin(2\pi nt)$ completes half a cycle, so neither term returns to zero net area. Direct evaluation gives $\int_0^{1/(2n)}\sin(6\pi nt)\,dt=\frac{1}{3\pi n}$ and $\int_0^{1/(2n)}\sin(2\pi nt)\,dt=\frac{1}{\pi n}$, and their sum $\frac{4}{3\pi n}$ is not zero. So the two signals are non-orthogonal on this shorter window, confirming (D).
Step 4: Final Answer:
\[ \boxed{\text{(A), (C), (D)}} \]
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