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Question

Consider a real, narrowband signal $x(t) = A(t)\cos[2\pi f_c t + \theta(t)]$ where the maximum frequency components of $A(t)$ and $\theta(t)$ are $f_M$ and $f_c \ (= 1000 f_M)$, respectively. 

Which of the following statements is/are correct for $-\infty < t < \infty$?

Narrowband Signal Representation Analysis

The signal is defined as a real, narrowband signal: $x(t) = A(t)\cos[2\pi f_c t + \theta(t)]$. Key parameters include the maximum frequency component of $A(t)$ as $f_M$, and that of $\theta(t)$ as $f_c$, with the condition $f_c \ (= 1000 f_M)$. We examine each statement regarding its potential representation.

PSK Signal Representation

A Phase Shift Keying (PSK) signal encodes information in the phase shifts of a carrier wave. The given signal form $x(t) = A(t)\cos[2\pi f_c t + \theta(t)]$ can represent PSK if the amplitude $A(t)$ remains constant, and the phase term $\theta(t)$ is modulated to carry the information. Thus, for suitable choices of $A(t)$ and $\theta(t)$, $x(t)$ can be a PSK signal. This statement is correct.

Amplitude Modulation Signal Representation

Amplitude Modulation (AM) encodes information by varying the amplitude of a carrier wave. If the phase term $\theta(t)$ is held constant (e.g., $\theta(t) = \theta_0$), the signal becomes $x(t) = A(t)\cos[2\pi f_c t + \theta_0]$. In this scenario, the amplitude $A(t)$ carries the message signal. Therefore, for suitable choices of $A(t)$ and a constant $\theta(t)$, $x(t)$ represents an AM signal. This statement is correct.

Band-limited Gaussian Noise Representation

A narrow-band Gaussian noise process centered around a carrier frequency $f_c$ can be mathematically described using envelope $A(t)$ and phase $\theta(t)$. In this representation, $A(t)$ and $\theta(t)$ are random processes. The condition $f_c \gg f_M$ aligns with the characteristics of a narrow-band signal where the modulating components' bandwidths are much smaller than the carrier frequency. Thus, $x(t)$ can represent a band-limited Gaussian noise process. This statement is correct.

Narrowband FM Signal Representation Analysis

Frequency Modulation (FM) involves varying the instantaneous frequency of a carrier signal. The instantaneous frequency ($f_i$) of $x(t)$ is derived from its phase: $f_i(t) = f_c + \frac{1}{2\pi}\frac{d\theta(t)}{dt}$. For $x(t)$ to represent a narrowband FM signal, the amplitude $A(t)$ must be constant, and the frequency deviation term $\frac{1}{2\pi}\frac{d\theta(t)}{dt}$ must be small relative to $f_c$ and proportional to a message signal. Since the general form allows for such conditions, $x(t)$ can represent a narrowband FM signal. The statement that it never represents one is therefore incorrect.

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Important Questions from Modulation

  1. Which of the following modulations is used in India for radio transmission?

  2. Which of the following is a result of over-modulation?
  3. Which of the following statements are correct?

    A. DSB‐SC modulation is well suited for point to point communication involving one transmitter and one receiver.

    B. VSB modulation is a linear modulation scheme.

    C. SSB is a non‐linear modulation scheme.

    D. FM is a linear modulation scheme.

    Choose the correct answer from the options given below:

  4. The condition for achieving distortion-less demodulation of an amplitude-modulated signal using an envelope detector is

  5. When a broadcast AM transmitter is 50 % modulated, its antenna current is 12 A. What will be the current when the modulation depth is increased to 90%?

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