If the IF is 455KHz calculate : i)The frequency and its rejection ratio for the tuning at 1044MHz ii) The image frequency and its rejection ratio for the tuning at 30MHz.

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

Given :

Super Hetrodyne receiver loaded Q of antenna coupling circuit = 100

Image frequency (f i) = 455 KHz

To Find :

i) Image frequency and rejection ratio for tunning 1044 MHz

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

= 1044 + 2 x 0.455 = 1044.91

$f = \frac{f_{si}}{f_s} - \frac{f_s}{f_{si}} = \frac{1044.91}{1044} - \frac{1044}{1044.91}$ = 1.7425 X $10^{-3}$

$\alpha$ = $\sqrt{1 + 100^2 X (1.7425 X 10^{-3})^2}$ = $\sqrt{1.03036}$ = 1.0150

ii)

Image frequency and rejection ratio for tunning 30 MHz

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

= 30 + 2 x 0.455 = 30.91

$f = \frac{f_{si}}{f_s} - \frac{f_s}{f_{si}} = \frac{30.91}{30} - \frac{30}{30.091}$ = 1.0303 - 0.9705 = 0.0598

$\alpha = \sqrt{1 + Q^2 f^2}$ = $\sqrt{1 + (100)^2 (0.0598)^2}$ = 6.063