Question

In the tuned circuit the oscillator in a simple AM transmitter employs a $\text{50}\mu \text{H}$ coil and a $\text{1nF}$ capacitor. If the oscillator output is modulated by audio frequencies up to $\text{10 kHz}$, calculate the range occupied by the sidebands
(A) $\text{922 - 802 kHz}$
(B) $\text{722 - 702 kHz}$
(C) $\text{722 - 802 kHz}$
(D) $\text{122 - 202 kHz}$

Answer 1

We have to find the value of the sidebands in a tuned circuit for AM modulation. In radio communications, a sideband is a band of frequencies higher than or lower than the carrier frequency, that is the result of the modulation process. The sidebands carry the information transmitted by the radio signal. Sidebands are produced in all types of modulation processes.

Formula Used:
$\text{LSB = }{{\text{f}}_{\text{L}}}\text{ = }{{\text{f}}_{\text{c}}}\text{ - }{{\text{f}}_{\text{m}}}$ $\text{USB = }{{\text{f}}_{\text{U}}}\text{ = }{{\text{f}}_{\text{c }}}\text{+ }{{\text{f}}_{\text{m}}}$

Complete step by step answer:
We have been given the value of the inductance and the capacitance used in the circuit and we have been given the modulation frequencies of the signal. We first need to find out the carrier wave frequency to find the frequency of the upper and the lower sidebands.
The inductance of the circuit $(L)=\text{ 50 }\mu \text{H = 50}\times \text{1}{{\text{0}}^{\text{-6}}}H\text{ (}\because \text{1}\mu \text{=1}{{\text{0}}^{\text{-6}}})$
The capacitance of the circuit $(C)=\text{ 1 nF = 1}\times \text{1}{{\text{0}}^{\text{-9}}}F\ \text{(}\because \text{1nF=1}{{\text{0}}^{\text{-9}}}F)$
The angular frequency of oscillation of the circuit will be $\omega =\dfrac{1}{\sqrt{LC}}$
Substituting the values, we can find the value of angular frequency

& \omega =\dfrac{1}{\sqrt{\text{50}\times \text{1}{{\text{0}}^{\text{-6}}}H\times \text{1}\times \text{1}{{\text{0}}^{\text{-9}}}F}} \\\ & \Rightarrow \omega =\dfrac{1}{\sqrt{5\times {{10}^{-14}}}}=\dfrac{1}{\sqrt{5}}\times {{10}^{7}} \\\ & \Rightarrow \omega =4.47\times {{10}^{6}}\text{ rad/s} \\\ \end{aligned}$$ The frequency of oscillation of the circuit is given as $$f=\dfrac{\omega }{2\pi }$$ where $$\omega $$ is the angular frequency of oscillation of the circuit, which we have calculated above Substituting the values, we can get the value of frequency as $$\begin{aligned} & f=\dfrac{4.47\times {{10}^{6}}\text{ rad/s}}{2\pi } \\\ & \Rightarrow f=711.762\times {{10}^{3}}Hz \\\ & \Rightarrow f\simeq 712kHz\text{ (}\because \text{1kHz=1}{{\text{0}}^{\text{3}}}Hz) \\\ \end{aligned}$$ This is the value of carrier frequency. Thus $${{\text{f}}_{\text{c}}}=712kHz$$ We have been given the value of modulation frequency, that is $${{\text{f}}_{\text{m}}}=10kHz$$ Now applying the formula for lower side band, we get $$\begin{aligned} & \text{LSB = }{{\text{f}}_{\text{L}}}\text{ = }{{\text{f}}_{\text{c}}}\text{ - }{{\text{f}}_{\text{m}}} \\\ & \Rightarrow {{\text{f}}_{\text{L}}}\text{ =(}712-10)kHz=702kHz \\\ \end{aligned}$$ Similarly, for upper side band, we have $$\begin{aligned} & \text{USB = }{{\text{f}}_{\text{U}}}\text{ = }{{\text{f}}_{\text{c }}}\text{+ }{{\text{f}}_{\text{m}}} \\\ & \Rightarrow {{\text{f}}_{\text{U}}}\text{ =}(712+10)kHz=722kHz \\\ \end{aligned}$$ The values of the LSB and USB obtained above form the range of the sideband. Hence option (B) is the correct answer. **Note:** The value of the angular frequency of the circuit can also be obtained using the resonance approach. When an inductor and a capacitor are used in a circuit and the value of their reactance becomes equal, the circuit becomes an oscillatory circuit and this phenomenon is known as resonance. Resonance occurs when a circuit can store and transfer energy from one mode to another.