Manal Ibrahim Hamammu
4931
Digital Communications
63244
Problems in
Encoding
2026
Problem 1
A human voice signal is sampled at a rate of 8 kHz and converted into a PCM signal using a quantizer with 256 discrete
levels (𝐿 = 256).
1)
Calculate the number of bits required per sample (n).
2)
Determine the resulting bit rate (R) of the encoder.
Solution:
1. Using the codeword length formula:
2. Using the bit rate formula:
𝐿 = 256
𝑅 = 𝑛 = 8 . 8000 = 64,000 𝑏𝑝𝑠 𝑜𝑟 64 𝑘𝑏𝑝𝑠
𝑛 = log2 (256) = 8 𝑏𝑖𝑡𝑠/𝑠𝑎𝑚𝑝𝑙𝑒
Problem 2
A low-pass signal with a bandwidth of 15𝑘𝐻𝑧 is to be converted to PCM. The sampling rate is set to 20% above the Nyquist
rate. If each sample is encoded into an 11 − 𝑏𝑖𝑡 𝑐𝑜𝑑𝑒𝑤𝑜𝑟𝑑:
1. What is the sampling frequency (𝑓𝑠 )?
2. What is the minimum bandwidth (𝐵𝑃𝐶𝑀) required to transmit this signal using rectangular pulses?
Solution
•Part 1 (𝑓𝑠 ≥ 2𝑓𝑚 ):
Nyquist rate = 2 × 15 𝑘𝐻𝑧 = 30 𝑘𝐻𝑧.
𝑓𝑠 = 30𝑘𝐻𝑧 × 1.20 = 36𝑘𝐻𝑧.
•Part 2 (B_PCM = 𝒏 × 𝑓_𝑠):
𝐵𝑃𝐶𝑀 = 11 × 36𝑘 = 396𝑘𝐻𝑧.
Problem 3
A PCM system uses 256 quantization levels. Due to poor weather, the channel bit
error rate (𝑃𝑒 ) increases to 10−4. Calculate the average output Signal-to-Noise Ratio
(SNR).
Solution:
L=256 and 𝑃𝑒 =0.0001
𝑳𝟐
𝟐𝟓𝟔𝟐
𝑺𝑵𝑹 =
=
≈ 𝟐, 𝟒𝟎𝟖. 𝟏𝟕
𝟏 + 𝟒(𝑳𝟐 − 𝟏)𝑃𝑒
𝟏 + 𝟒 𝟐𝟓𝟔𝟐 − 𝟏 0.0001
Problem 4
A 4-bit PCM system (𝑛 = 4) is used to transmit data over a noisy channel. The channel has a bit error rate
(𝑃𝑒 ) of 0.01 (or 10−2). Calculate the output SNR and compare it to the ideal SNR (where 𝑃𝑒 = 0).
Solution
•
Determine L:
𝐿 = 2𝑛 = 24 = 16.
•
Actual SNR with Channel Noise:
𝟏𝟔𝟐
𝑆𝑁𝑅𝑜𝑢𝑡 = 𝟏+𝟒 𝟏𝟔𝟐 −𝟏 0.01
•Ideal SNR (𝑃𝑒 = 0):
𝟐𝟓𝟔
= 22.86
𝟏+𝟒 𝟐𝟓𝟓 0.01
𝑆𝑁𝑅𝑖𝑑𝑒𝑎𝑙 = 𝐿2 = 162 = 256
𝑆𝑁𝑅𝑜𝑢𝑡 =
In dB= 10 log10 256 = 24.082𝑑𝐵
In dB= 10 log10 22.86 = 13.590𝑑𝐵
Problem 5
A high-fidelity audio system uses 12 bits per sample to encode a signal. The transmission channel is slightly noisy, with a bit
error rate (𝑃𝑒 ) of 10−6 (0.000001).
1.Calculate the Peak SNR in dB.
2.Calculate the Average SNR in dB.
3.Determine the "Peak-to-Average" ratio (the difference between the two) and explain what this tells a designer about the
system's headroom.
Solution:
n = 12 bit
L = = 212 = 4096
𝟑𝑳𝟐
𝟑(𝟒𝟎𝟗𝟔)𝟐
=
= 739,000
𝟏+𝟒(𝑳𝟐 −𝟏)𝑃𝑒
𝟏+𝟒 𝟒𝟎𝟗𝟔𝟐 −𝟏 0.000001
In dB: 10 log10 (739,000) ≈ 58.7 𝑑𝐵
𝑳𝟐
𝟒𝟎𝟗𝟔𝟐
=
= 246,333
𝟏+𝟒(𝑳𝟐 −𝟏)𝑃𝑒 𝟏+𝟒 𝟒𝟎𝟗𝟔𝟐 −𝟏 0.000001
In dB: 10 log10 (246,333) = 53.9𝑑𝐵
Difference: 58.7 -- 53.9 = 4.8 dB
Problem 6
A compact disc (CD) records audio signals digitally by using PCM assume the audio signal bandwidth to be 15kHz
• What is the Nyquist rate?
• If the Nyquist samples are quantized int L=65,536 levels and then binary coded, determine the number of binary digits
required to encode a sample
• Determine the number of binary digits per second required to encode the audio signal.
Solution:
Part 1:
Bandwidth (W): 15 kHz
Nyquist Rate (𝒇𝑵 ): 2 × 𝑊 = 2 × 15 𝑘𝐻𝑧 = 30 𝑘𝐻𝑧
Part 2:
𝐿 = 65,536 → 2𝑛 = 65,536 → log 2 65,536 = 16 𝑏𝑖𝑡
Part 3:
𝑅 = 𝑓𝑠 × 𝑛 → 30,000 × 16 = 480,000𝑏𝑖𝑡𝑠/𝑠 𝑜𝑟 480𝑘𝑏𝑠
Problem 7
A television signal has a bandwidth of 4.5MHz, this signal is samples, quantized, and binary coded to obtain a PCM signal.
•
Determine the sampling rate if the signal is to be sampled at a rate 20% above the Nyquist rate.
•
If the samples are quantized into 1024 levels, determine the numbers of binary pulses required to encode each sample.
Solution:
Part 1:
Bandwidth (W): 4.5MHz
Nyquist Rate (𝒇𝑵 ): 2 × 𝑊 = 2 × 4.5𝑀𝐻𝑧 = 9 𝑀𝐻𝑧
𝑓𝑠 = 9𝑀𝐻𝑧 × 1 + 0.20 → 9𝑀𝐻𝑧 + 1.2 = 10.8𝑀𝐻𝑧
Part 2:
𝐿 = 1024 → 2𝑛 = 1024 → log 2 1024 = 10 𝑏𝑖𝑡
Part 3:
𝑅 = 𝑓𝑠 × 𝑛 → 10.8𝑀𝐻𝑧 × 10 = 108𝑀𝐻𝑧
Problem 8
A digital communications system is to carry a single voice signal using linearly quantized PCM. What
PCM bit rate will be required if an ideal anti-aliasing filter with a cut-off frequency of 3.4 kHz is used
at the transmitter and the SNqR is to be kept above 50 dB?
Solution:
SNqR = 4.8 + 6n − 𝛼𝑑𝐵 . For voice signals α = 10 dB, i.e.:
𝑛 =
50 + 10 − 4.8
= 9.2
6
fs = 2.2 × 3.4 kHz = 7.48 kHz.
We used 2.2 instead 2.0 Because is a practical multiplier that includes a for non-ideal filter performance and to prevent
aliasing in real physical components.
The PCM bit rate (or more strictly binary baud rate) is therefore:
𝑅𝑏 = 𝑓𝑠 × 𝑛 → 7.48 × 103 × 10
𝑏𝑖𝑡
= 74.8 𝑘𝑏𝑖𝑡/𝑠
𝑠
Problem 9
Find the overall SNR for the reconstructed analogue voice signal in Example 8 if receiver noise
induces an error rate, on average, of one in every 106 PCM bits.
Solution:
𝑆𝑁𝑞𝑅
4.8 + 6𝑛 + 𝛼𝑑𝐵 → 4.8 + 6 × 10 − 10
1 + 4 𝑆𝑁𝑞𝑅 𝑃𝑒
= 54.8 𝑑𝐵 𝑜𝑟 3.020 × 105
𝑆𝑁𝑅𝑜𝑢𝑡 =
𝑆𝑁𝑅𝑜𝑢𝑡 =
3.020 × 105
= 1.368 × 105 = 51.4 𝑑𝐵
1 + 4(3.020 × 105 )(1 × 10−6 )
References
•For Example, 3 & 5: Refer to Proakis & Salehi, Chapter on "Digital Representation of Memoryless Sources," where they derive
the effects of channel noise on PCM performance. (Inspired)
•For Example, 1 & 2: Refer to B.P. Lathi, Chapter 6 (Sampling and Pulse Code Modulation), which provides many practical
problems on bit rate and bandwidth. (Inspired)
•For Example, 4: Refer to Bernard Sklar, Section 2.8, which discusses the trade-offs between bandwidth, bits per sample, and
quantization levels. (Inspired)
•For Example, 6 & 7: taken from Modern Digital and Analog Communication Systems, Book by B. P Lathi. Chapter 6, Page 291
in the book.
•For Example, 8 & 9: Digital Communications 3rd Edition Ian Glover. Chapter 5, Page 187 in the book, Example 5.3, 5.4