examples for tdm

7/31/2019 Examples for TDM

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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 KHzrate 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 andthe overall pulse rate of the TDM PAM signal.

Solution:

The frequency of the commutator = 8 KHz. The period of each cycle =1

1258 KHz

= ms.

Within 125 ms, 24 channels are multiplexed together.

The time interval between two adjacent channels =125

24ms =

( )1 1 1

8 24 8 24KHz KHz =

8 24 192KHz KHz = The overall pulse (baud) rate of the TDM PAM signal =

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 thePCM words for each of the analog inputs and an 8-bit synchronization word is used in thesynchronization channel.

Solution:

The TDM PAM is at 192 KHz, that is the same as samples/sec.

The baud rate is at

3192 10

( )8 24 8 1.536KHz = 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) digitalinputs and one analog input that has a bandwidth of 500 Hz. Assume that the analogsamples 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 = 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

4 300 1200 =

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 = Hz.5 1200 6000 =


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4. Design a TDM system that will accommodate two 2400 bits/s synchronous digital inputsand 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

Hz

The bit rate for the PCM signal converted from the analog signal =

1.11111 2 2700 6000= =

4 6000 24 = KHz

The bit rate for the two digital inputs = 2400 bits/sec = 2.4 KHzThe 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 = Kbits/sec

5. Design a time-division multiplexer that will accommodate 11 sources. Assume that thesources have the following specifications:

Source # 1: Analog, 2 KHz bandwidth

Source # 2: Analog, 4 KHz bandwidthSource # 3: Analog, 2 KHz bandwidth

Source # 4-11: Digital, synchronous at 7200 bits/s

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 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

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

12 2.4 28.8 =

1 4f =

converter output is 64 kbits/s.

2 8f = 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 areconnected 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.


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Fig. TDM with analog and digital inputs

examples for tdm

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