Question

A superheterodyne receiver is used to receive an FM signal in the the range of 88 MHz-108 MHz with IF frequency 10 MHz. In order to avoid image frequency, the maximum and minimum capacitance ratio at the local oscillator section is (high side tuned is used)

• A. 6.2 : 1
• B. 1.449 : 1
• C. 1 : 1.449
• D. 1 : 6.2

Answer 1

The correct option is B 1.449 : 1

$f_I=f_c+2\left(IF\right)IF=10MHz\left(\frac{C_{max}}{C_{min}}\right)={\left(\frac{f_{L{o}_2}}{f_{L{o}_1}}\right)}^2$
In order to avoid image frequency
$f_{L{o}_1}=f_{0_1}+IF=88+10=98MHz{f}_{L{o}_2}=f_{o_2}+IF=108+10=118MHz\frac{C_{max}}{C_{min}}={\left(\frac{118}{98}\right)}^2=1.449:1$