Question

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?

• A. \(x_1(t)\) and \(x_2(t)\) are orthogonal to each other over \(0\leq t<1/n\).
• B. \(x_1(t)\) and \(x_2(t)\) are orthonormal to each other over \(0\leq t<1/n\).
• C. \(x_2(t)\) is a harmonic of \(x_1(t)\).
• D. \(x_1(t)\) and \(x_2(t)\) are non-orthogonal to each other over \(0\leq t<1/(2n)\).

Hint:

Turn the product of the two sinusoids into a sum of sines, then integrate over each interval to test orthogonality.

Answer 1

The Correct Option is A, C, D

Step 1: Write the product in a sum of sines.
We need
\[ x_1(t)x_2(t) = \cos(2\pi nt)\cdot 2\sin(4\pi nt) \] Using \(2\cos A\sin B = \sin(A+B)+\sin(B-A)\) with \(A=2\pi nt\) and \(B=4\pi nt\),
\[ x_1(t)x_2(t) = \sin(6\pi nt)+\sin(2\pi nt) \]

Step 2: Test orthogonality over \(0\leq t<1/n\).
The period of \(x_1(t)\) is \(1/n\), so we integrate over one full period,
\[ \int_0^{1/n}\sin(6\pi nt)\,dt = \left[\frac{-\cos(6\pi nt)}{6\pi n}\right]_0^{1/n} = \frac{-\cos(6\pi)+\cos(0)}{6\pi n} = 0 \] \[ \int_0^{1/n}\sin(2\pi nt)\,dt = \left[\frac{-\cos(2\pi nt)}{2\pi n}\right]_0^{1/n} = \frac{-\cos(2\pi)+\cos(0)}{2\pi n} = 0 \] Both terms vanish, so
\[ \int_0^{1/n}x_1(t)x_2(t)\,dt = 0 \] So \(x_1(t)\) and \(x_2(t)\) are indeed orthogonal over \(0\leq t<1/n\), and option (A) is correct.

Step 3: Test orthonormality over the same interval.
Orthonormal needs unit energy on top of orthogonality. The energy of \(x_1(t)\) over one period is
\[ \int_0^{1/n}\cos^2(2\pi nt)\,dt = \frac{1}{2}\cdot\frac{1}{n} = \frac{1}{2n} \] which equals \(1\) only for a special value of \(n\), not for a general positive integer \(n\). So option (B) is incorrect.

Step 4: Check the harmonic relation.
The frequency of \(x_1(t)\) is \(n\) and the frequency of \(x_2(t)\) is \(2n\), an integer multiple of \(n\). A signal whose frequency is an integer multiple of a fundamental frequency is a harmonic of it.
So option (C) is correct.

Step 5: Test the shorter interval \(0\leq t<1/(2n)\).
Integrating the same product over this half period,
\[ \int_0^{1/(2n)}\sin(6\pi nt)\,dt = \frac{-\cos(3\pi)+\cos(0)}{6\pi n} = \frac{1+1}{6\pi n} = \frac{1}{3\pi n} \] \[ \int_0^{1/(2n)}\sin(2\pi nt)\,dt = \frac{-\cos(\pi)+\cos(0)}{2\pi n} = \frac{1+1}{2\pi n} = \frac{1}{\pi n} \] Adding these,
\[ \int_0^{1/(2n)}x_1(t)x_2(t)\,dt = \frac{1}{3\pi n}+\frac{1}{\pi n} = \frac{4}{3\pi n} \neq 0 \] This is nonzero, so \(x_1(t)\) and \(x_2(t)\) are non-orthogonal over \(0\leq t<1/(2n)\), and option (D) is correct.

Step 6: Final conclusion.
\[ \boxed{\text{(A), (C), (D)}} \]