Question:easy

A sinusoidal carrier voltage of frequency \(1\ \text{MHz}\) and amplitude \(100\ \text{Volts}\) is amplitude modulated by a sinusoidal voltage of frequency \(5\ \text{kHz}\) producing \(50\%\) modulation. The frequency and amplitude of the lower and upper sideband terms will be:

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Sidebands lie at \(f_c\pm f_m\); each has amplitude \(mA_c/2\).
Updated On: Jul 2, 2026
  • \(995\ \text{Hz}, 1005\ \text{Hz}\) and \(25\ \text{V}\)
  • \(995\ \text{Hz}, 1005\ \text{Hz}\) and \(50\ \text{V}\)
  • \(995\ \text{Hz}, 1005\ \text{Hz}\) and \(75\ \text{V}\)
  • \(995\ \text{Hz}, 1005\ \text{Hz}\) and \(0\ \text{V}\)
Show Solution

The Correct Option is A

Solution and Explanation

Modulating a carrier with a single tone produces a spectrum of exactly three frequencies: the carrier itself and a symmetric pair of sidebands offset by the message frequency.

The sideband frequencies are the sum and difference of carrier and message:
\[f_{\pm} = f_c \pm f_m = 1000\ \text{kHz} \pm 5\ \text{kHz},\]
giving a lower sideband at $995\ \text{kHz}$ and an upper sideband at $1005\ \text{kHz}$.

For the amplitude, expand $A_c[1 + m\cos\omega_m t]\cos\omega_c t$ with the product-to-sum identity. Each sideband carries a coefficient of $mA_c/2$. Plugging in $m = 0.5$ and $A_c = 100\ \text{V}$:
\[\frac{mA_c}{2} = \frac{0.5 \times 100}{2} = 25\ \text{V}.\]
Both sidebands therefore have equal amplitude $25\ \text{V}$, which points to option (A). \[\boxed{995,\ 1005\ \text{kHz};\ 25\ \text{V each}}\]
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