In a broadcast superheterodyne receiver having no RF amplifier, the loaded Q of the antenna coupling circuit is 150. If the IF is 455 kHz, calculate the image frequency and its rejection ratio at 1400 kHz.

Mumbai University > Electronics Engineering > Sem4 > Fundamentals of Communication Engineering

Marks: 5M

Year: Dec2014

Given Data:

Quality factor, Q = 150

Intermediate frequency, $f_{IF}$ = 455 kHz

Signal frequency, $f_s$=1400 kHz

To find :

Image signal frequency, $f_{si}$ = ?

Rejection ratio, α = ?

Solution:

Image frequency $f_{si}$ is given by:

$f_{si} = f_s + 2 f_{IF} \\ \ \ \ = 1400 + 2(455) \\ f_{si} = 2310 kHz$

Image frequency rejection ratio (α) is given by:

$α = \sqrt{(1+Q^2 ρ^2 )} \\ ρ = f_\frac{si}{f_s} - \frac{f_s}{f_{si}} \\ = \frac{2310}{1400} – \frac{1400}{2310} = 1.043 \\ α = \sqrt{(1+(150^2 ×1.043^2))} = 156.453$

α in dB is given as:

α in dB=20 log⁡ α

α = 20 log⁡156.452 = 43.887 dB

Thus the image frequency is 2310 kHz and its rejection ratio is 43.887