What is the theoretical capacity of a channel in each of the following cases: a.
Bandwidth: 20 KHz SNRdB =40
b.
Bandwidth: 200 KHz SNRdB =4
c.
Bandwidth: 1 MHz SNRdB =20

The answer is below

Explanation:

The theoretical capacity of a channel (C) is the maximum data rate at which data can be transmitted in a channel. It is measured in bits per second (bps) and is given by the formula:

C = bandwidth * (SNR dB)/3

a) Bandwidth = 20 KHz = 20000 Hz, SNR dB =40

C = 20000 * (40)/3 = 26666.7 bps

C ≅ 267 Kbps

b) Bandwidth = 200 KHz = 200000 Hz, SNR dB =40]

C = 200000 * (4)/3 = 26666.7 bps

C ≅ 267 Kbps

c) Bandwidth = 1 MHz = 1000000 Hz, SNR dB =20

C = 1000000 * (20)/3 = 6666666.7 bps

C ≅ 6667 Kbps

for this, you need the pythagorean theorem, which is a^{2}+ b^{2}= c^{2}

your a should be 12, and your b should be 35. so now you can solve for c, which equals 37 miles.

so the answer is 37 miles

i just did so your welcome!