Question

A modulated signal ${C_m}(t)$has the form ${C_m}(t) = 30\sin 300\pi t + 10(\cos 200\pi t - \cos 400\pi t)$. The carrier frequency ${f{'}_c}$ the modulating frequency( message frequency) ${f_\omega }$ and the modulation index $\mu $ are respectively given by
A) ${f{'}_c}$=$200Hz;{f_\omega } = 50Hz;\mu = \dfrac{1}{2}$
B) ${f{'}_c} = 150Hz;{f_\omega } = 50Hz;\mu = \dfrac{2}{3}$
C) ${f{'}_c} = 150Hz;{f_\omega } = 30Hz;\mu = \dfrac{1}{3}$
D) ${f{'}_c} = 200Hz;{f_\omega } = 30Hz;\mu = \dfrac{1}{2}$

Answer 1

Hint: We can approach the solution to this question as recalling the generalized equation of an EM wave because this question is from EM waves. By remembering the generalized equation of an EM wave and then comparing it with what is given in the question, we can reach the solution

Complete step by step solution:
Before solving the question let us first understand what all these terms actually mean. Let's go through them one by one
${C_m}(t)$ is the equation of the wave. We can understand its nature by comparing it with the general equation for an AM wave
A general equation for a AM wave is \[\
y(t) = Asin(2\pi {{f{'}}_c}t) + A\mu sin(2\pi {{f{'}}_c}t)sin(2\pi {f_w}t) \\
= Asin(2\pi {{f{'}}_c}t) + 2A\mu [cos(2\pi ({f_c}^\prime - {f_w})t) - cos(2\pi ({f_c}^\prime + {f_w})t)] \\
\ \]
So, if we compare this equation to the equation of the wave in the question, we can easily determine that ${C_m}(t)$ is an equation of an AM wave
${f{'}_c}$ is the carrier frequency of the AM wave. The carrier frequency of a wave is the frequency at which it is sent into the atmosphere. The signal is converted to this frequency to prevent the transmission losses due to disturbances.
${f_\omega }$ is the modulating frequency. For modulation of a wave, it is mixed with another wave of a certain frequency. This frequency is called the modulating frequency.
$\mu $ is called the modulation index. It describes how much the modulated variable of the signal varies around its unmodulated level. In this case, the amplitude is modulated. So $\mu $ tells us about how the amplitude of the wave varies with the original unmodulated amplitude.
Now, let us come back to the original question. We have been asked to determine the carrier frequency, the modulation frequency, and the modulation index.
For that, we can compare the given equation with the general equation of an AM wave.
So, form the comparison, we can say that
$\
  A = 30 \\
  2\pi {f_c}^\prime = 300\pi \\
  2\pi ({f_c}^\prime - {f_\omega }) = 200\pi \\
  \dfrac{{A\mu }}{2} = 10 \\
    \\
   \Rightarrow {{f{'}}_c} = 150Hz \\
  {f_{\omega = }} = 50Hz \\
  A = 30 \\
  \mu = \dfrac{2}{3} \\
\ $

Hence the correct answer is option (B).

Note:
This question tests your concepts on EM waves. Many students do this question wrong because they are careless once they see a question from EM waves because it is actually an easy question, the most common mistake made is an incorrect comparison with the general equation, incorrect writing of the general equation. If you can avoid these mistakes, this question is pretty easy to solve.