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Time division multiplexing
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EEEN 4329 Examples for TDM Problems
1. Twenty-three analog signals, each with a bandwidth of 3.4 KHz, are sampled at an 8 KHz rate and multiplexed together with a synchronization channel (8 KHz) into a TDM PAM signal. Draw a block diagram for the system indicating the frequency of the commutator and the overall pulse rate of the TDM PAM signal. Solution:
The frequency of the commutator = 8 KHz. The period of each cycle =
1 8 KHz
Within 125 ms, 24 channels are multiplexed together. 125 1 1 × The time interval between two adjacent channels = ms = 8 KHz 24 24
=
=
125 ms.
1
(8
×
24 ) KHz
The overall pulse (baud) rate of the TDM PAM signal = 8 × 24 2 4 KHz = 192 KHz Note: The baud rate for a TDM PAM signal is a signal level.
2. Rework problem 1 for a TDM PCM system where an 8-bit quantizer is used to generate the PCM words for each of the analog inputs and an 8-bit synchronization word is used in the synchronization channel. Solution:
The TDM PAM is at 192 KHz, that is the same as 192 × 103 samples/sec. The baud rate is at ( 8 × 24 KHz ) 8 = 1.536 Mbits/sec Note: The symbol for a TDM PCM signal is a binary bit
3. Design a TDM PCM system that will accommodate four 300 bits/s (synchronous) digital inputs and one analog input that has a bandwidth of 500 Hz. Assume that the analog samples will be encoded into 4-bit PCM words. Solution:
The first commutator is to multiplex the four 300 bits/s digital inputs. The baud rate = 4 × 300 = 1200 bits/sec The analog input has a bandwidth of 500 Hz. We can pick the sampling frequency to be 1200 Hz or 1200 samples/sec. Each analog sample will be encoded into 4-bit PCM word. The analog input converted into 4 × 1200 = 4800 bits/sec The second commutator will multiplex the 1200 bits/sec and the 4800 bits/sec. The commutator frequency = 1200 Hz. Each rotation will pick one bit from the 1200 bits/sec and four bits from 4800 bits/sec. The baud rate = 5 × 1200 = 6000 Hz.
4. Design a TDM system that will accommodate two 2400 bits/s synchronous digital inputs and an analog input that has a bandwidth of 2700 Hz. Assume that the analog input is sampled at 1.11111 times the Nyquist rate and converted into 4-bit PCM words. Draw a block diagram for your design and indicate the data rate at various points on your diagram. Solution:
The sampling frequency of the analog input = 1.11111 × Nyquist rate = 1.11111 × 2 × 2700 = 6000 Hz The bit rate for the PCM signal converted from the analog signal = 4 × 6000 = 24 KHz The bit rate for the two digital inputs = 2400 bits/sec = 2.4 KHz The commutator will have 12 contacts, two out twelve are for the digital inputs, and 10 out of 12 are the PCM signals. The overall baud rate = 12 × 2.4 = 28.8 Kbits/sec
5. Design a time-division multiplexer that will accommodate 11 sources. Assume that the sources have the following specifications: Source # 1: Source # 2: Source # 3: Source # 4-11:
Assume the analog sources will be converted into 4-bit PCM words and, for simplicity, that frame sync will be provided via a separate channel and synchronous TDM lines are used. To satisfy the Nyquist rate for the analog sources, sources # 1, 2, and 3 need to be sampled at 4, 8, and 4 KHz, respectively. As shown in the following figure, the desoign is to rotate the first commutator at f 1 = 4 KHz and sampling source # 2 twice on each revolution. This produces a 16 ksamples/s TDM PAM signal on commutator output. Each of the analog sample values is converted into a 4-bit PCM word, so that the rate of the TDM PCM signal on the A/D converter output is 64 kbits/s. The digital data on the A/D converter output may be merged with the data from the digital sources by using a second commutator rotating at f 2 = 8 KHz and wired so that the 64 kbits/s PCM signal is present on 8 out of 16 terminals. This provides an effective sampling rate of 64 kbits/s. On the other eight terminals the digital sources are connected to provide a data transfer rate of 8 kbits/s for each source. Since the digital sources are supplying a 7.2 kbits/s data stream, pulse stuffing is used to raise the source rate to 8 kbits/s.
