Time division multiplexing tdm
This presentation is the property of its rightful owner.
Sponsored Links
1 / 26

Time Division Multiplexing (TDM) PowerPoint PPT Presentation


  • 97 Views
  • Uploaded on
  • Presentation posted in: General

Time Division Multiplexing (TDM). TDM is a technique used for transmitting several message signals over a single communication channel by dividing the time frame into slots, one slot for each message signal. TDM-PAM: Transmitter. TDM-PAM : Receiver. time. 0. Ts. 2Ts. g 1 (t).

Download Presentation

Time Division Multiplexing (TDM)

An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.


- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Presentation Transcript


Time division multiplexing tdm

Time Division Multiplexing (TDM)

  • TDM is a technique used for transmitting several message signals over a single communication channel by dividing the time frame into slots, one slot for each message signal


Tdm pam transmitter

TDM-PAM: Transmitter


Tdm pam receiver

TDM-PAM : Receiver


Samples of signal 1

time

0

Ts

2Ts

g1(t)

Samples of Signal -1


Samples of signal 2

Ts

Ts

Samples of signal - 2

g2(t)


Multiplexing of two signals

0

Ts

2Ts

Multiplexing of TWO signals


Tdm pam for 3 signals

TDM-PAM for 3 signals

2

2

1

1

2

3

3

1

3

Time


Tdm pam for 4 signals

4

4

4

1

1

1

2

2

2

3

3

3

Time

TDM-PAM for 4 signals.


Commutator arrangement

g1(t)

g2(t)

g3(t)

g4(t)

Commutator Arrangement


Decommutator

g1(t)

g2(t)

2

1

3

4

g3(t)

g4(t)

Decommutator


Tdm pam

TDM-PAM

  • Types of TDM:

  • Synchronous TDM

  • Asynchronous TDM


Synchronous tdm

  • Same Sampling rate for all signals.

  • Minimum Sampling rate = twice the maximum frequency of all the signals.

  • Total number of samples transmitted per second is equal to N times the sampling rate, Fs plus sync pulses.

  • Transmission Bandwidth = N. Fs/2

Synchronous TDM


Asynchronous tdm

  • Different Sampling rate for different. signals.

  • Sampling rate of a signal = twice the maximum frequency of that signal.

  • Total number of samples transmitted per second is equal to Sum of samples of all the signals plus sync pulses.

Asynchronous TDM


Asynchronous tdm1

  • 4. Transmission Bandwidth = Half the total number of samples transmitted.

  • 5. Bandwidth is less for Asynchronous TDM.

  • 6. Design of Commutator / Decommutator is difficult.

Asynchronous TDM


Problem 1

  • Two low-pass signals of equal bandwidth are sampled and time division multiplexed using PAM. The TDM signal is passed through a Low-pass filter & then transmitted over a channel with a bandwidth of 10KHz.

  • Continued….

Problem-1


Problem 1 continued

Problem –1 continued

  • What is maximum Sampling rate for each Channel?

  • What is the maximum frequency content allowable for each signal?


Problem 1 solution

  • Channel Bandwidth = 10 KHz.

  • Number of samples that can be transmitted through the channel = 20K

  • Maximum Sampling rate for each channel = 10K Samples/sec.

  • Maximum Frequency for each Signal = 5KHz

Problem –1 : Solution


Problem 2

Problem –2

  • Two signals g1(t) and g2(t) are to transmitted over a common channel by means of TDM. The highest frequency of g1(t) is 1KHz and that of g2(t) is 1.3KHz. What is the permissible sampling rate?

  • Ans: 2.6K samples/sec and above.

  • { Synchronous TDM}


Problem 3

  • 24 voice signals are sampled uniformly and then time division multiplexed . The sampling operation uses the flat-top samples with 1microsec duration. The multiplexing operation includes provision for Synchronization by adding an extra pulse of sufficient amplitude and also 1micro second.

  • Contd…

Problem-3


Problem 31

  • Contd..

  • Assuming a sampling rate of 8KHz, calculate the spacing between successive pulses of the multiplexed signal.

Problem-3


Problem 3 solution

  • In One frame, Total number of pulses = 25.

  • Time duration for one time frame = Ts = 125µ seconds.

  • Time duration utilized by pulses = 25µ sec

  • Time spacing between successive pulses

  • = (125- 25)/25 = 4µ sec.

Problem-3: Solution


Problem 4

  • Three independent message signals of bandwidths 1KHz, 1KHz and 2KHz respectively are to be transmitted using TDM scheme. Determine

  • Commutator segment arrangement

  • Speed of the commutator if all the signals are sampled at its Nyquist rate.

  • Minimum Transmission bandwidth.

Problem-4


Problem 4 solution

g1(t)

g2(t)

2

1

g3(t)

4

3

  • Commutator Segment arrangement.

Problem-4: Solution


Problem 4 solution1

3

3

3

1

1

1

3

3

3

2

2

2

Time

Problem-4 : Solution.


Problem 4 solution2

  • Total number of Samples to be transmitted per second = 8K samples/sec.

  • Number of Commutator segments = 4

  • Speed of Commutator = 2000 rotations/sec.

  • c) Transmission Bandwidth = 4000 Hz.

Problem-4: Solution


Problem 5

  • Q. Eight independent message signals are sampled and time multiplexed using PAM. Six of the message signals are having a bandwidth of 4KHz and other two have bandwidth of 12KHz. Compare the transmission bandwidth requirements of Synchronous TDM and Asynchronous TDM.

Problem-5.


  • Login