
Find out the image frequency of an FM radio receiver which has been tuned to $91.6MHz$ at the broadcast station?
$\begin{align}
& A.111.8MHz \\
& B.113MHz \\
& C.121.6MHz \\
& D.132MHz \\
\end{align}$
Answer
571.2k+ views
Hint: First of all find the oscillator frequency. Image frequency is the sum of the oscillator frequency and the intermediate frequency. Therefore add the oscillator frequency into the intermediate frequency. This information will help you in solving this question.
Complete answer:
First of all let us mention what is given in the question. The tuned frequency of the FM radio station is given by the equation,
${{f}_{tuned}}=91.6MHz$
The intermediate frequency is typically given as,
${{f}_{IF}}=10.7MHz$
Therefore oscillation frequency with respect to the tuned frequency can be taken as,
$\begin{align}
& {{f}_{tuned}}+{{f}_{IF}}={{f}_{osc}} \\
& \Rightarrow 91.6MHz+10.7MHz=102.3MHz \\
\end{align}$
${{f}_{osc}}=102.3MHz$
The image frequency is given as the sum of the oscillator frequency and difference between the oscillation frequency and the tuned frequency which is the intermediate frequency. This can be written as,
${{f}_{osc}}+\left( {{f}_{osc}}-{{f}_{tuned}} \right)={{f}_{image}}$
Substituting the values in it will give,
$\begin{align}
& {{f}_{image}}=102.3+\left( 102.3-91.6 \right) \\
& \Rightarrow {{f}_{image}}=102.3+10.7=113MHz \\
\end{align}$
So, the correct answer is “Option B”.
Note:
An image frequency is defined as a random input frequency which is equivalent to the station frequency plus or minus twice the intermediate frequency. The image frequency will cause the two stations to obtain at the identical time. This will result in the production of interference patterns. The Image frequencies can be avoided by requiring attenuation on the incoming signal by the radio frequency amplifier filter of the super heterodyne receiver. An intermediate frequency (IF) is a frequency defined as the frequency at which a carrier wave is shifted as an intermediate step in transmission or reception. This is used in communications and electronic engineering. These frequencies will be slightly higher than the radio frequencies and lower than the extremely low frequencies.
Complete answer:
First of all let us mention what is given in the question. The tuned frequency of the FM radio station is given by the equation,
${{f}_{tuned}}=91.6MHz$
The intermediate frequency is typically given as,
${{f}_{IF}}=10.7MHz$
Therefore oscillation frequency with respect to the tuned frequency can be taken as,
$\begin{align}
& {{f}_{tuned}}+{{f}_{IF}}={{f}_{osc}} \\
& \Rightarrow 91.6MHz+10.7MHz=102.3MHz \\
\end{align}$
${{f}_{osc}}=102.3MHz$
The image frequency is given as the sum of the oscillator frequency and difference between the oscillation frequency and the tuned frequency which is the intermediate frequency. This can be written as,
${{f}_{osc}}+\left( {{f}_{osc}}-{{f}_{tuned}} \right)={{f}_{image}}$
Substituting the values in it will give,
$\begin{align}
& {{f}_{image}}=102.3+\left( 102.3-91.6 \right) \\
& \Rightarrow {{f}_{image}}=102.3+10.7=113MHz \\
\end{align}$
So, the correct answer is “Option B”.
Note:
An image frequency is defined as a random input frequency which is equivalent to the station frequency plus or minus twice the intermediate frequency. The image frequency will cause the two stations to obtain at the identical time. This will result in the production of interference patterns. The Image frequencies can be avoided by requiring attenuation on the incoming signal by the radio frequency amplifier filter of the super heterodyne receiver. An intermediate frequency (IF) is a frequency defined as the frequency at which a carrier wave is shifted as an intermediate step in transmission or reception. This is used in communications and electronic engineering. These frequencies will be slightly higher than the radio frequencies and lower than the extremely low frequencies.
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