Lab/Tutorial :
SEHH2238 : Computer Networking
Session 3 : PCM and Error Detection (Solution)
1. We have sampled a low-pass signal with a bandwidth of 200 kHz using 1024 levels of quantization.
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a. Calculate the number of bits per sample.
no. of bits per sample (nb) = Log2L
= Log21024
b. Calculate the bit rate of the digitized signal
Sampling frequency = 2 x highest frequency component = 2 x 200 kHz
Bit rate = Sampling freq. x no. of bits per sample = 400 x 10
c. Calculate the SNRdB for this signal.
SNRdB = 6.02 (nb) + 1.76 = 6.02 x 10 + 1.76
2. An analog signal has voltage level in the range of 0 to 5 V. The signal is digitized using PCM
with the signal-to-noise ratio due to quantization confined to 55 dB.
a. Determine the minimum number of bits required.
SNRdB = 6.02 (nb) + 1.76 55 ≤ 6.02 (nb) + 1.76 nb ≥ 8.84
Therefore 9 bits are required.
b. Suppose that all “0”s represents the lowest signal voltage level and all “1”s represents the highest signal voltage. If the quantization value is round-up and assigned linearly to each signal level, what is the binary code for 1.75 V?
1111111112 = 51110 represents 5V
For 1.75V, the value = 511 x 1.75 / 5
= 178.85 = 179
The binary code is 010110011
SEHH2238 Computer Networking Tutorial 3 Page 1
2) Error Detection
1. What is the maximum effect of a 2-ms burst of noise on data transmitted at the following rates? a. 1500 bps
no. of affected bits
b. 100 kbps
no. of affected bits
= Data rate x burst duration = 1500 x 2 x 10-3
= Data rate x burst duration = 100 x 103 x 2 x 10-3
= 200 bits
2. Assuming even parity, find the parity bit for each of the following data units.
a. 1011011
b. 0001100
3. 01001 01101 11000 10001 00101 is received using two dimensional even parity bit. The first 4 blocks are data with the parity bit in the rightmost bit, while the last block is all parity. Assume that no more than 2 bits contain error. Find the error bit(s).
01001 01101 11000 10001 00101
4. Given the dataword 1010011110 and the polynomial x4 + x2 + x + 1.
a. Show the generation steps of the codeword at the sender site.
x4 + x2 + x + 1 (degree 4) represents the divisor 10111 (5 bits). Append 4 “0”s at the end of the dataword, i.e. 10100111100000 ** Division Step **
Get the reminder 1010
Final codeword transmitted is 10100111101010
b. Assuming no error, show the checking of the codeword at the receiver site.
Use the same divisor 10111 with the received codeword ** Division Step **
If no error, the reminder should be 0.
SEHH2238 Computer Networking Tutorial 3 Page 2
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