Question

1. A line has a signal-to-noise ratio of 1000 and a bandwidth of 4000 KHz. What is
the maximum data rate supported by this line?

2. A signal with 200 milliwatts power passes through 10 devices, each with an
average noise of 2 microwatts. What is the SNR? What is the SNRdB?

3. If the peak voltage value of a signal is 20 times the peak voltage value of the
noise, what is the SNR? What is the SNRdB?

4. In a digital transmission, the sender clock is 0.2 percent faster than the receiver
clock. How many extra bits per second does the sender send if the data rate is 1
Mbps?

5. We have a baseband channel with a 1-MHz bandwidth. What is the data rate for
this channel if we use each of the following line coding schemes?
a. NRZ-L b. Manchester c. MLT-3 d. 2B1Q

6. A signal has passed through three cascaded amplifiers, each with a 4 dB gain.
What is the total gain? How much is the signal amplified?

Answer 1

1. The maximum data rate supported by a line is given by the Shannon capacity formula:

C = B * log2(1 + S/N)

where C is the maximum data rate in bits per second, B is the bandwidth in hertz, and S/N is the signal-to-noise ratio.

Substituting the given values, we get:

C = 4000 KHz * log2(1 + 1000) = 53.1 Mbps

Therefore, the maximum data rate supported by this line is 53.1 Mbps.

2. The signal power is 200 milliwatts, or 0.2 watts. The total noise power is the sum of the noise powers of the 10 devices, which is 2 microwatts/device * 10 devices = 20 microwatts, or 0.00002 watts.

The signal-to-noise ratio is therefore:

SNR = signal power / noise power = 0.2 / 0.00002 = 10,000

The SNRdB is:

SNRdB = 10 log10(SNR) = 40 dB

3. If the peak voltage value of a signal is 20 times the peak voltage value of the noise, then the power of the signal is 400 times the power of the noise, since power is proportional to the square of the voltage.

The signal-to-noise ratio is therefore:

SNR = signal power / noise power = 400

The SNRdB is:

SNRdB = 10 log10(SNR) = 26 dB

4. If the sender clock is 0.2 percent faster than the receiver clock, then the sender sends 0.2 percent more bits per second than the receiver can receive.

The extra bits per second sent by the sender is:

0.2 percent of 1 Mbps = 0.002 * 1 Mbps = 2 kbps

Therefore, the sender sends an extra 2 kbps compared to what the receiver can receive.

5. The data rate for a baseband channel with a 1-MHz bandwidth depends on the line coding scheme used.

a. For NRZ-L coding, each bit takes 1 microsecond to transmit, so the data rate is 1/1 microsecond = 1 Mbps.

b. For Manchester coding, each bit is transmitted twice, so the data rate is 0.5 Mbps.

c. For MLT-3 coding, each symbol can represent 2 bits, so the data rate is 2 * 1/2 microseconds = 1 Mbps.

d. For 2B1Q coding, each symbol can represent 2 bits, so the data rate is 2 * 1/4 microseconds = 0.5 Mbps.gain of three c

6. The total ascaded amplifiers, each with a 4 dB gain, is:

total gain = 10 log10(gain1 * gain2 * gain3) = 10 log10(10^(0.4) * 10^(0.4) * 10^(0.4)) = 5.6 dB

The signal is amplified by a factor of:

amplification factor = 10^(total gain / 10) = 10^(5.6/10) = 3.98

Therefore, the signal is amplified by a factor of approximately 3.98.

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