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sladkih [1.3K]
3 years ago
12

) Assuming different AM regulations; the receiver is using mixer with subtracting format. The frequency selectivity ratio is app

roximately 10 and the AM range is from 750 kHz to 2600 kHz. The intermediate frequeney is most nearly:
Engineering
1 answer:
Zarrin [17]3 years ago
8 0

Answer:

F=710KHZ

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=\frac{B}{2}

 F=\frac{1420}{2}

 F=710KHZ

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PLEASE HELP QUICK!!
ivolga24 [154]

R01= 14.1 Ω

R02=  0.03525Ω

<h3>Calculations and Parameters</h3>

Given:

K= E2/E1 = 120/2400

= 0.5

R1= 0.1 Ω, X1= 0.22Ω

R2= 0.035Ω, X2= 0.012Ω

The equivalence resistance as referred to both primary and secondary,

R01= R1 + R2

= R1 + R2/K2

= 0.1 + (0.035/9(0.05)^2)

= 14.1 Ω

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= 0.035 + (0.05)^2 * 0.1

= 0.03525Ω

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brainly.com/question/17563681

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