Question

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:

• A. \(995\ \text{Hz}, 1005\ \text{Hz}\) and \(25\ \text{V}\)
• B. \(995\ \text{Hz}, 1005\ \text{Hz}\) and \(50\ \text{V}\)
• C. \(995\ \text{Hz}, 1005\ \text{Hz}\) and \(75\ \text{V}\)
• D. \(995\ \text{Hz}, 1005\ \text{Hz}\) and \(0\ \text{V}\)

Hint:

Sidebands lie at \(f_c\pm f_m\); each has amplitude \(mA_c/2\).

Answer 1

The Correct Option is A

Step 1: An amplitude-modulated wave is \[v(t) = A_c\left[1 + m\cos(\omega_m t)\right]\cos(\omega_c t),\] with carrier amplitude \(A_c = 100\ \text{V}\), modulation index \(m = 0.5\), carrier \(f_c = 1\ \text{MHz}\) and modulating \(f_m = 5\ \text{kHz}\).
Step 2: Expanding the product gives three lines: the carrier, plus two sidebands at \(f_c \pm f_m\): \[f_{\text{LSB}} = f_c - f_m = 1000 - 5 = 995\ \text{kHz},\] \[f_{\text{USB}} = f_c + f_m = 1000 + 5 = 1005\ \text{kHz}.\]
Step 3: Each sideband has amplitude \[\frac{m A_c}{2} = \frac{0.5 \times 100}{2} = 25\ \text{V}.\]
Step 4: So both sidebands sit at \(995\ \text{kHz}\) and \(1005\ \text{kHz}\), each of amplitude \(25\ \text{V}\). This is option (A). \[\boxed{f = 995,\ 1005\ \text{kHz},\quad \text{amplitude} = 25\ \text{V}}\]