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1. Al-Israa University College
Computer Techniques Engineering Department
Digital Communication
Chapter 5 part (1)
Digital Pulse Modulation
BY
Assist. Lec. Karrar S. Faraj

2. Types of Modulation

3. Pulse Code Modulation (PCM)
Most current digital audio systems (computers, compact discs, digital telephony etc.) use multi-bit Pulse
Code Modulation (PCM) to represent the sound signal. Analog to PCM conversion consists of three steps:
Sampling, Quantizing and Encoding.
PCM Transmitter
PCM Receiver
Block Diagram of PCM system

4. Pulse Code Modulation (PCM)

5. PCM Advantages and Disadvantages
• Advantages of PCM:
1- Low channel noise and interference.
2- Easy multiplexing of various PCM signals.
• Disadvantages of PCM:
1- Complex systems.
2- Wide bandwidth is required.

6. Q (1):- A compact dis (CD) records audio signals digitally by using PCM. Assume the audio signal 15KHz.
1- What is the Nyquist rate.
2- If the samples are quantized into L = 65536 levels and then binary coded, determine the number of
binary digits required to encoded a sample.
3- Determine the number of binary digits per second (bit rate).
1- 𝑓𝑁 = 𝑓𝑠 = 2𝑓𝑚
𝑓𝑁 = 2 ∗ 15 𝐾 = 30 𝐾𝐻𝑧
2- 𝐿 = 2𝑏
𝑏 =
𝐿𝑜𝑔 𝐿
𝐿𝑜𝑔 2
=
𝐿𝑜𝑔 (65536)
𝐿𝑜𝑔 2
= 16 𝑏𝑖𝑡
3- 𝑅𝑏 = 𝑏𝑓𝑠
𝑅𝑏 = 16 ∗ 30 𝐾 = 480 𝐾 (b/s)
Solution:-

7. Q (2):- A television has band limited 4.5 MHz. This signal is sampled, quantized and binary coded to obtain
a PCM signal.
1- Determine the sample rate if the signal is to be sampled at a rate 20% above the Nyquist rate .
2- If the samples are quantized into L = 1024 levels and then binary coded, determine the number of
binary digits required to encoded a sample.
3- Determine the number of binary digits per second (bit rate) and the minimum bandwidth required to
transmit this signal.
Solution:-
1- 𝑓𝑁 = 2𝑓𝑚= 2* 4.5 M = 9 MHz
∴ 𝑓𝑠 > 20% 𝑓𝑁
𝑓𝑠 > 20% ∗ 9 𝑀𝐻𝑧 =1.8 MHz
𝑓𝑠 = 9 M + 1.8 M =10.8 MHz
2- 𝐿 = 2𝑏
𝑏 =
𝐿𝑜𝑔 𝐿
𝐿𝑜𝑔 2
=
𝐿𝑜𝑔 (1024)
𝐿𝑜𝑔 2
= 10 𝑏𝑖𝑡
3- 𝑅𝑏= 𝑏𝑓𝑠
𝑅𝑏 = 10 ∗ 10.8 𝑀 = 108 𝑀 (b/s)
𝐵𝑃𝐶𝑀 ≥
1
2
𝑅𝑏
𝐵𝑃𝐶𝑀 ≥
1
2
108 𝑀
𝐵𝑃𝐶𝑀 ≥ 54 MHz

8. Q (3):- Five telemetry signals, each of bandwidth 1KHz, are to be transmitted by binary PCM. The
maximum error in sample amplitude is 0.2% of the peak signal amplitude. This signal must be sampled
rate at least 20% above the Nyquist rate. Determine the minimum possible data rate that must be
transmitted (for both PCM and TDM) and the minimum bandwidth required to transmit this signal.
N=5 signals
f = 1 KHz
Eq = 0.2% R
𝒇𝒔 > 𝟐𝟎% 𝒇𝑵
𝑹𝒃 ? 𝐟𝐨𝐫 𝐏𝐂𝐌
𝑹𝒃= 𝒃𝒇𝒔 &
𝑹𝒑 ? ? 𝐟𝐨𝐫 𝐓𝐃𝐌
𝑹𝒑 = 𝑵𝒇𝒔
𝑩𝑷𝑪𝑴? ? ?
Solution:-
𝑓𝑁 = 2𝑓𝑚= 2* 1 K = 2 KHz
∴ 𝑓𝑠 > 20% 𝑓𝑁
𝑓𝑠 > 20% ∗ 2 𝐾𝐻𝑧 =0.4 KHz
𝑓𝑠 = 2 K + 0.4 K =2.4 KHz
𝒇𝒐𝒓 𝑷𝑪𝑴
𝑅𝑏= 𝑏𝑓𝑠= 8 * 2.4 K = 19.2 K (b/s)
𝐵𝑃𝐶𝑀 ≥
1
2
𝑅𝑏 ≥
1
2
∗ 19.2 𝐾 ≥ 9.6 KHz
𝒇𝒐𝒓 𝑻𝑫𝑴
𝑅𝑝 = 𝑁𝑓𝑠 = 5 ∗ 2.4 𝐾 = 12 𝐾(b/s)

9. Q (4):- Ten telemetry signals, each of bandwidth 100Hz, are to be transmitted by binary PCM. This
signal are sampled, quantized, binary coded and then time-division multiplexer. The maximum error in
sample amplitude is 0.5% of the peak signal amplitude. This signal rate must be sampled at least twice
the Nyquist rate. Determine the minimum possible data rate that must be transmitted and the minimum
bandwidth required to transmit this signal (for both PCM and TDM).
N=10 signals
f = 100 Hz
Eq = 0.5% R
𝒇𝒔 > 𝟐 𝒇𝑵
𝑹𝒃 & 𝑩𝑷𝑪𝑴? 𝐟𝐨𝐫 𝐏𝐂𝐌
𝑹𝒑 & 𝑩𝑻𝑫𝑴? ? 𝐟𝐨𝐫 𝐓𝐃𝐌
Solution:-
𝑓𝑁 = 2𝑓𝑚= 2* 100 = 200 Hz
∴ 𝑓𝑠 = 2 𝑓𝑁
𝑓𝑠 = 2 ∗ 200 = 400 Hz
𝒇𝒐𝒓 𝑷𝑪𝑴
𝑅𝑏= 𝑏𝑓𝑠= 7* 400= 2800 (b/s)
𝐵𝑃𝐶𝑀 ≥
1
2
𝑅𝑏 ≥
1
2
∗ 2800 ≥ 1400 Hz
𝒇𝒐𝒓 𝑻𝑫𝑴
𝑅𝑝 = 𝑁𝑓𝑠 = 10 ∗ 400 = 4 𝐾(b/s)
𝐵𝑇𝐷𝑀 ≥ N*𝑓𝑚 ≥ 10* 100≥ 1 KHz

10. Q (5):- A signal x(t) band-limited to 3 KHz is sampled using PCM system. The signal power is 50 mW
and uses a 8-levels quantization. This signal rate must be sampled at least twice the Nyquist rate. Find
channel capacity?
f = 3 KHz
𝑷𝒙 = 𝟓𝟎 𝒎𝑾
L = 8
𝒇𝒔 > 𝟐 𝒇𝑵
𝑪 ?
Solution:-
𝐶 = 𝐵 ∗ 𝐿𝑜𝑔2(1 +
12𝑷𝒙
∆
2 )
𝑓𝑁 = 2𝑓𝑚= 2* 3 K = 6 KHz
∴ 𝑓𝑠 = 2 𝑓𝑁
𝑓𝑠 = 2 ∗ 6 𝐾 = 12 KHz
𝑅𝑏 = 𝑏𝑓𝑠= 3* 12 K= 36 K (b/s)
𝐵𝑃𝐶𝑀 ≥
1
2
𝑅𝑏 ≥
1
2
∗ 36 𝐾 ≥ 18 KHz
∆ =
𝑅
𝐿
𝑃𝑥 =
𝑅2
16
50 m =
𝑅2
16
𝑅2 = 800 m
R = 800 𝑚 = 0.9
∆ =
𝑅
𝐿
=
0.9
8
= 0.1125
𝐶 = 𝐵 ∗ 𝐿𝑜𝑔2(1 +
12𝑃𝑥
∆
2 )
C = 18 K ∗ 𝐿𝑜𝑔2(1 +
12∗50 𝑚
(0.1125)
2)
C = 100.74 K (b/s)

12. Thank
You