The power in the signal $s(t) = 8\cos(20\pi t - \pi/2) + 4\sin(15\pi t)$ is

Question

A periodic signal $x(t)$ of period $T_0$ is given by $ x(t) = \begin{cases} 1, & |t|<T_1 \\ 0, & T_1<|t|<\frac{T_0}{2} \end{cases} $. The dc component of $x(t)$ is

Answer 1

Step 1: Split the signal into its two sinusoidal components: $ 8\cos(20\pi t - \pi/2) $ with amplitude $ 8 $, and $ 4\sin(15\pi t) $ with amplitude $ 4 $.
Step 2: The average power of any sinusoid of amplitude $ A $ is $ P = A^2/2 $, giving $ P_1 = 8^2/2 = 32 $ and $ P_2 = 4^2/2 = 8 $.
Step 3: Since the two angular frequencies $ 20\pi $ and $ 15\pi $ are different, the cross term between them averages to zero over time, so the total power is just the sum of the individual powers. \[ \boxed{P = 32 + 8 = 40} \]

Answer 2

A slightly different way to arrive at the same number is to compute the power using the RMS values of each term directly, rather than the amplitude-squared-over-two shortcut, then check which option matches.

Option 40: The RMS value of $ 8\cos(20\pi t - \pi/2) $ is $ 8/\sqrt{2} $, so its power is $ \left(8/\sqrt{2}\right)^2 = 32 $. The RMS value of $ 4\sin(15\pi t) $ is $ 4/\sqrt{2} $, giving power $ \left(4/\sqrt{2}\right)^2 = 8 $. Because the two sinusoids are orthogonal over one common period (their frequencies are unequal, so their product integrates to zero on average), the total power is the direct sum $ 32 + 8 = 40 $.
Option 41: No manipulation of the RMS values of $ 8 $ and $ 4 $ amplitude sinusoids produces this figure; it does not correspond to any combination of $ 32 $ and $ 8 $.
Option 42: This value is sometimes obtained by an unjustified extra correction to the summed powers, but there is no such correction needed once orthogonality (unequal frequencies) is properly accounted for; the two powers simply add without any adjustment.
Option 82: This would only follow from squaring the sum of amplitudes directly, $ (8+4)^2/2 = 72 $, which itself does not even equal $ 82 $, and in any case is not valid since the two sinusoids do not combine amplitude-wise when their frequencies differ.

Computing each term's power separately via its RMS value and summing (valid because the frequencies are unequal, so cross terms vanish on averaging) gives a clean total.

So the correct answer is 40.

Answer 3

A slightly different way to arrive at the same number is to compute the power using the RMS values of each term directly, rather than the amplitude-squared-over-two shortcut, then check which option matches.

Option 40: The RMS value of $ 8\cos(20\pi t - \pi/2) $ is $ 8/\sqrt{2} $, so its power is $ \left(8/\sqrt{2}\right)^2 = 32 $. The RMS value of $ 4\sin(15\pi t) $ is $ 4/\sqrt{2} $, giving power $ \left(4/\sqrt{2}\right)^2 = 8 $. Because the two sinusoids are orthogonal over one common period (their frequencies are unequal, so their product integrates to zero on average), the total power is the direct sum $ 32 + 8 = 40 $.
Option 41: No manipulation of the RMS values of $ 8 $ and $ 4 $ amplitude sinusoids produces this figure; it does not correspond to any combination of $ 32 $ and $ 8 $.
Option 42: This value is sometimes obtained by an unjustified extra correction to the summed powers, but there is no such correction needed once orthogonality (unequal frequencies) is properly accounted for; the two powers simply add without any adjustment.
Option 82: This would only follow from squaring the sum of amplitudes directly, $ (8+4)^2/2 = 72 $, which itself does not even equal $ 82 $, and in any case is not valid since the two sinusoids do not combine amplitude-wise when their frequencies differ.

Computing each term's power separately via its RMS value and summing (valid because the frequencies are unequal, so cross terms vanish on averaging) gives a clean total.

So the correct answer is 40.

LIST-I	LIST-II
A. \( e^{-at}u(t), a>0 \)	I. \( \pi[\delta(\omega - \omega_0) + \delta(\omega + \omega_0)] \)
B. \( \cos \omega_0 t \)	II. \( \frac{1}{j\omega + a} \)
C. \( \sin \omega_0 t \)	III. \( \frac{1}{(j\omega + a)^2} \)
D. \( te^{-at}u(t), a>0 \)	IV. \( -j\pi[\delta(\omega - \omega_0) - \delta(\omega + \omega_0)] \)

The power in the signal $s(t) = 8\cos(20\pi t - \pi/2) + 4\sin(15\pi t)$ is

Show Hint

The Correct Option is A

Approach Solution - 1

Approach Solution -2

Top Questions on Signals and Systems

Questions Asked in OJEE exam