Question

In a broadcast superheterodyne AM receiver tuned to 1.4 MHz and IF = 455 kHz, which of the followiing will be correct?

• A. If the receiver is to operate in the frequency range of 550 kHz to 1650 kHz then the capacitance ratio of local oscillator is 3 where ratio $=C_{max}/C_{min}$
• B. IRR for $f_L>f_S$ is 52.2
• C. The bandwidth of IF amplifier for Q = 50 is 9.1 kHz.
• D. The image frequency, $f_{Si}$ will be 2310 kHz and 490 kHz for $f_L>f_S$ and $f_L<f_S$ respectively.

Answer 1

Answer: (B), (C), (D)

The image frequency, $f_{Si}$ will be 2310 kHz and 490 kHz for $f_L>f_S$ and $f_L<f_S$ respectively.

$f_{si}=f_s+2IF$ $[\text{for}{f}_L>f_S]$
$=1400+910=2310kHz$
$f_{si}=f_s-2IF$ $[\text{for}{f}_L<f_S]$
$=1400-910=490kHz$
Bandwidth $=\frac{f_r}{Q}=\frac{455K}{50}=9.1kHz$ ($f_r=\text{IF for IF amplifier})$
$\text{IRR}=\sqrt{1+P^2{Q}^2}\Rightarrow P=\frac{f_{si}}{f_s}-\frac{f_s}{f_{si}}$
$=\frac{C_{max}}{C_{min}}={\left(\frac{f_{L_{max}}}{f_{L_{min}}}\right)}^2={\left(\frac{f_{S_{max}+IF}}{f_{S_{min}}+IF}\right)}^2={\left(\frac{2105}{1005}\right)}^2=4.38$