Question

The tuned circuit the oscillator in a simple AM transmitter employs a 50 $\mu H$ coil and a 1 nf capacitor. If the oscillator output is modulated by audio frequencies up to 10m KHz, calculate the range occupied by the side bands

• A. 922 to 802 KHz.
• B. 722 to 702 KHz.
• C. 722 to 802 KHz.
• D. 122 to 202 KHz.

Answer 1

The correct option is B 722 to 702 KHz.

The frequency of tuned circuit is
$f=\frac{1}{2\pi \sqrt{LC}}=\frac{1}{2\pi {(5\times {10}^{-5}\times 1\times {10}^{-9})}^{1/2}}$
$=\frac{1}{2\pi {\left(5\right)}^{1/2}\times {10}^{-7}}=712KHz$
Since highest modulating frequency is 10 KHz, the frequency range occupied by the side bands will range from 10 KHz above to 10 KHz below the carrier, extending from 722 to 702 KHz.