Homework 2 solution

Homework #1
Chapter 3
1. A TV channel has a bandwidth of 6 MHz. If we send a digital signal using one
channel, what are the data rates if we use one harmonic, three harmonic, and five
harmonics?
Ans BW = 6 MHz
1) BW from 0 Hz to f1st harmonic = 6 MHz;
Bit rate = 2* f1st harmonic = 2 * 6 = 12 Mbps
2) BW from 0 Hz to f3rd harmonic = 6 MHz;
f3rd harmonic = 3 * f1st harmonic
f1st harmonic = 6 MHz / 3 = 2 MHz
Bit rate = 2 * f1st harmonic = 2 * 2 = 4 Mbps
3) BW from 0 Hz to f5th harmonic = 6 MHz;
f1st harmonic = 6 MHz / 5 = 1.2 MHz
Bit rate = 2 * f1st harmonic = 2 * 1.2 = 2.4 Mbps
2. The attenuation of a signal is -10 dB. What is the final signal power if it was
originally 5 W?
Ans –10 = 10 log10 (P2 / 5) → log10 (P2 / 5) = −1 → (P2 / 5) = 10−1 → P2 = 0.5 W
3. We measure the performance of a telephone line (4KHz of bandwidth). When the
signal is 10 V, the noise is 5 mV. What is the maximum data rate supported by
this telephone line?
Ans 4,000 log2 (1 + 10 / 0.005) = 43,866 bps
4. What is the total delay (latency) for a frame of size 5 million bits that is being sent
on a link with 10 routers each having a queuing time of 2 us and a processing time
of 1 us. The length of the link is 2000 Km. The speed of light inside the line is 2 x
108 m/s. The link has a bandwidth of 5 Mbps. Which component of the total delay
is dominant? Which one is negligible?
Ans Propagation time = distance / propagation speed
= 2000 Km / 2 x 108 m/s = 10 ms
Transmission time = Message size / Bandwidth
= 5 x 106 bits/ 5 Mbps = 1 s
Queuing time = 10 routers * 2 us = 20 us
Processing Delay = 10 routers * 1 us = 10 us
Total delay (latency) = 10 ms + 1 s + 20 us + 10 us
= 1010.03 ms = 1.01003 s  1 s
Chapter 4
1. Assume a data stream is ‘1101000000000010’ s.. Encode this stream, using the
following code schemes. How many changes (vertical line) can you find for each
scheme?
a. Unipolar: 4 changes between bit
b. NRZ-L:
4 changes between bit
c. NRZ-I:
4 changes between bit
d. RZ: 16 changes at the middle of each bit + 15 changes between bit
e. Manchester: 16 changes at the middle of each bit + 11 changes between bit
f. Diff. Manchester: 16 changes at the middle of each bit + 11 changes between bit
g.
h.
i.
AMI:
2B1Q:
MLT-3:
6 changes between bit
3 changes between bit
4 changes between bit
2. What is the sampling rate for PCM if the frequency ranges from 1000 to 4000 Hz?
Ans Sampling rate >= 2 x fhighest = 2 x 4000 = 8000 samples / s
3. Using the Nyquist theorem, calculate the sampling rate for the following analog
signals.
a. An analog signal with bandwidth of 2000 Hz
Ans Sampling rate >= 2 x fhighest = 2 x (fLowest + BW)
>= 2 x (fLowest + 2000)
b. An analog signal with frequencies from 2000 to 6000 Hz
Ans Sampling rate >= 2 x fhighest
>= 2 x 6000 = 12,000 samples /s
c. A signal with a horizontal line in the time-domain representation
Ans Sampling rate >= 2 x fhighest
>= 2 x 0 = 0 samples /s
d. A signal with a vertical line in the time-domain representation
Ans Sampling rate >= 2 x fhighest
>= 2 x ∞ = ∞ samples /s
4. A signal is sampled. Each sample represents one of four levels. How many bits are
needed to represent each sample? If the sampling rate is 8000 samples per second,
what is the bit rate?
Ans Quantization 2 bits/sample;
Bit rate = 8000 * 2 = 16,000 bps
5. We have sampled a low-pass signal with a bandwidth of 200 KHz using 1024
levels of quantization. Calculate the bit rate of the digitized signal.
Ans Low pass signal:
frequency between 0 – 200 KHz
BW = 200 KHz
Sampling rate >= 2 x fhighest = 2 x 200 KHz
>= 400,000 samples /s
Quantization 10 bits/sample;
Bit rate = 400,000 x 10 = 4 Mbps