Question

A superheterodyne receiver is designed to receive the signals in the range (88 to 108) MHz. The local oscillator frequency of the mixer is set at the lower of the two possible values. To avoid the problem of image frequency in the receiver, the minimum value of the intermediate frequency $\left(f_{IF}\right)$ should be equal to

• A. 20~MHz
• B. 44~MHz
• C. 88~MHz
• D. 10~MHz

Answer 1

The correct option is D 10~MHz

Given that, $f_{LO}<f_c$
Image frequency $f_i=f_c-2f_{IF}$

To avoid image problem,

$108-2f_{IF}\le 88$

$2f_{IF}\ge 20MHz$

$f_{IF}\ge 10MHz$

$f_{IF(min)}=10MHz$