If the IF is 455KHz calculate image frequency and rejection ratio for tunning (i) 100KHz (ii) 20MHz.

Mumbai University > Electronics > Sem 4 > Priniciples of Communication Engineering

Given :

Super Hetrodyne receiver loaded Q of antenna coupling circuit = 80

Image frequency (f i) = 455 KHz

To Find :

i) Image frequency and rejection ratio for tunning 100 KHz

$\rightarrow f_{si} = f_s + 2f_i$

= 100 + 2 x 455 = 1010 KHz

$f = \frac{f_{si}}{f_s} - \frac{f_s}{f_{si}} = \frac{1010}{100} - \frac{100}{1010}$ = 10.1 - 0.09 = 10.01

$\alpha = \sqrt{1 + Q^2 f^2}$ = $\sqrt{1 + (80)^2 (10.01)^2}$ = 800.80

ii)

Image frequency and rejection ratio for tunning 20 MHz

$\rightarrow f_{si} = f_s + 2f_i$

= 20 + 2 x 0.455 = 20.91 MHz

$f = \frac{20.91}{20} - \frac{20}{20.091}$ = 0.0890

$\alpha = \sqrt{1 + Q^2 f^2}$ = $\sqrt{1 + (80)^2 (0.0890)^2}$ = 7.18