Question

) Assuming different AM regulations; the receiver is using mixer with subtracting format. The frequency selectivity ratio is approximately 10 and the AM range is from 750 kHz to 2600 kHz. The intermediate frequeney is most nearly:

Answer 1

`F=710KHZ`

Step-by-step explanation:

From the question we are told that:

Frequency selectivity ratio
`R=10`

AM range 750 kHz to 2600 kHz

Therefore Bandwidth is

`B=2100-680`

`B=1420KHZ`

Generally the equation for The intermediate frequency is mathematically given by

Intermediate frequency=\frac{Bandwidth}{2}

`F=(B)/(2)`

`F=(1420)/(2)`

`F=710KHZ`