Frequency MUXFrequency DEMUX

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Omar Siddiqui

Department of Electrical Engineering College of Engineering

Taibah University Madinah

Email:[email protected]

EE 372 – Communication Theory and Systems I Lecture 11: Digital Modulations

ةبيط ةعماج

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Advantages of Digital Signals

1. Noise Immunity

Digital Systems perform better in noisy environment.

 Recovering exact pulse shape is not important

 The receiver can detect the

amplitude of the signal and make a decision based on thresholds

Transmitted digital signal

Received signal without noise

Received signal with noise

Regenerated signal

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2. Regeneration at repeaters

The digital signal can be amplified and regenerated from the distorted signal

 If analog signals are amplified, more noise is added and therefore cannot be regenerated

 Long haul communications are possible

channel

TX RX

Distortion + Noise

A Decision

Repeater

(4)

3. Hardware Implementation is flexible because availability of digital microprocessors, switching, and integrated circuits

4. Digital coding provides lower error rates 5. Multiplexing is possible

6. Exchange between SNR and bandwidth 7. Cheap storage devices

(5)

Types of Pulse Modulation

Pulse Modulation

Analog Pulse Modulation

PAM, PWM, and PPM Digital Pulse Modulation

The information is transmitted in analog form but transmitted at discrete times Analog Pulse Modulation

Digital Pulse Modulation

The information is transmitted in digital form and also transmitted at discrete times. Therefore, it requires the analog to digital conversion

(6)

Steps in Analog Pulse Modulation

T_s

1. Analog Signal Continuous Time Continuous Amplitude

2. Sampled Signal Discrete Time Continuous Amplitude

T_s

3. Coding (PAM)

T_s

Analog signal is sampled at times T_s

Sampled Signal is represented in the form of the pulses Types of Analog Pulse Modulation:

1. PAM (Pulse Amplitude Modulation) 2. Pulse position modulation (PPM) 3. Pulse width Modulation (PWM)

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Types Analog Pulse Modulation

2. PWM (Pulse width modulation):

The Pulse Amplitudes are same. The width of the pulse is proportional to the amplitude of the signal.

3. PPM (Pulse Position modulation):

The Pulse Amplitudes and widths are the same. The position of the pulse from the sample time is proportional to the amplitude of the signal.

PAM

PWM 1. PAM (Pulse Amplitude Modulation) :

The Pulse Widths are same but

amplitudes are equal to the sample’s amplitude

T_s

PPM Analog

Signal

W₁ W₂ W₃ W₄

P₁ P₂ P₃

P₄

t

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Analog Pulse Modulation

Generation of the PWM and PPM

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1. Sampling

2. Triangular Pulse clocking

3. Adding clock and Sampled signal

t

t 4. Threshold

Comparison

5. Draw PWM

6. Draw PPM VREF

Analog Signal

W₁ W₂ W₃ W₄

t T_s

(10)

1. Sampling

2. Triangular Pulse clocking

3. Adding clock and Sampled signal

t

t 4. Threshold

Comparison

5. Draw PWM

6. Draw PPM Analog Signal

W₁ W₂ W₃ W₄

P₁ P₂ P₃ P₄

VREF

t T_s

(11)

3. Coding: (to change the quantized samples to binary codes)

Steps of A/D Conversion

Analog Signal

L

Digital Pulse Modulation

The information is transmitted in digital form and also transmitted at discrete times. Therefore, it requires the analog to digital conversion

_s

_s 1. Sampling

(Perpendicular to t-axis)

2. Quantization:

Perpendicular to the y-axis (amplitude)

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Analog to Digital Conversion

T_s

Sampling interval

t

L

  

 



 T_s

L

T_s    

     



     

Sampled Signal

Quantized Signal

3. Coding

For every sample 4 bits are transmitted

0 1 2 4 3 5 6 7

1. Sampling 2. Quantization

0 0 0 1

0 1 1

4 5

0 0

1 1

0

6

6 7

1

1 1 1

7

1 1

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Analog to Digital Conversion

t

Sampling interval

t

Sampling interval

L Sampling

Quantization

Coding

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Sampling Time

-In the first step of A/D conversion, what should be the sampling time Ts?

- This question is answered by the Sampling Theorem

_s 1. Sampling

(Perpendicular to t-axis)

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Similarity between Modulation and Sampling

g(t)

t

g(t)cos₀t cos₀t





Modulation is multiplication of a low frequency signal with a

sinusoidal high frequency carrier

Sampling is multiplication of a continuous- time signal by an impulse train so that it can be converted to a discrete time signal.

Time Domain (Multiplication)

0 Ts

Ts

 2T_s

Ts

2 1 1

 

^







n

nTs

 t

0 Ts

Ts

 2T_s Ts

2

 

^









n

nTs

t t

g t

g_( ) ( ) 

T_sis called the sampling interval and f_s = 1/T_s is the sampling rate

t t

g(t) t



 ^t

Modulation is convolution of the low frequency signal with the the FT of cosine (two delta functions).

Frequency Domain (Convolution)

) ( f ) G

0 ( G

0 B

B

 

²

) 0 ( G

0 f₀ B f₀ B



f0

2 / 1 f0



f0



) 2 (

) 1 2 (

2 1 cos )

(t f₀t G f f₀ G f f₀

g     

Sampling is convolution of a continuous-time signal by FT of an impulse train (which is also an impulse train)

The low frequency signal is shifted to high frequency

) ( f ) G

0 ( G

0 B

B



0 Ts

1

 

s s

T1 f

^



 



 



 

n s

s f nT

T

1

1 

Ts

2 Ts

1 Ts

2

 ^ ^

^



 n

nTs

 t

 

^







n

s

s f nf

f 

 ^ ^

^



 n

nTs

 t

fs 2f_s fs

 fs

2



^{G )}^T⁽^s⁰

fs

2  f_s 2fs

 

0

B f_s  B

f₀ 

   

 

(16)

The Minimum Sampling Time

0 T^s Ts

 2T_s

Ts

2 1 1

 

^







n

nTs

 t

 

^









n

nTs

t t

g t

g_ ( ) ( )  g(t)

t



0 Ts

Ts

 2T_s Ts

2

t

) ( f ) G

0 ( G

0 B

 B



0 Ts

1

 

s s

T1f

^



 



 



 

n s

s f n T

T

1

1  ^T^s

2 Ts

1 Ts

2

 



^







n

s

s f nf

f 

) 0 1 ( T_s G

fs

0

B f_s  B

f₀ 

   

 

   

^^







 ⁿ

n s

s G f T

f T

G 1 )

1 ( )

 ( ^^







 ⁿ

n

s

s G f f

f ( )

fs

2 3fs

fs

2  f_s

To avoid overlap

Ts

1

T1_sB T1_sB

B

T1_s B

B



^f^s ^f^s ²^f^s

s  f

2

T1 _S 2B

  f_s 2B

T_S B

2

 1

 ^or



(17)

Sampling Theorem

1. A band-limited signal of finite energy (having no frequency component above B Hz) is completely described by specifying the values of the signal separated by 1/2B seconds

2. A band-limited signal of finite energy (having no frequency component above B Hz) may be completely reconstructed from a knowledge of its samples taken at the rate of 2B samples per seconds

The rate 2B is called the Nyquist Rate and the interval 1/2B is called the Nyquist interval

Nyquist Rate and Nyquist Interval

) ( f ) G

0 ( G

0 B

 B

A signal limited to bandwidth ‘B’

(18)

Sampling theorem

_s=1/2B g_(t)

0 t

Sampling Rate: What is the minimum sampling rate?

^









n

B n f G W f

G_( ) 2 ( 2 )

Time domain Frequency domain

f_s=2B

) 2 / (

) ( )

(t g t t n B

g

n^







 



What happens if the rate is above or below f_s = 2W ?

Sampling Rate

_s

g_(t)

0 t

) ( f G_ ) 0 ( G f_s

f B B



f_s>2B

) ( f G_ ) 0 ( 2BG

f

B 3B 5B

B

5 3B B 7B

fs 2fs

fs

 f_s B f_s B

fs 2fs 3f_s fs

s  f

2 0

0

) ( f G_ ) 0 ( G f_s

f

F_s<2B

fs

3 ⁰

_s

g_(t)

0 t

_s<1/2B

_s>1/2B

fs

2 3f_s 4f_s fs

s  f

2

(19)

Back to Analog to PCM Conversion

Sampling Quantization Encoding

m(t) m_s(t) m_(t)

0 t

m_s(t)

0 t

m_(t)

0 t

Sampled Signal Quantized Signal

(20)

Quantization Error

The sampled levels are not equal to quantization levels, so there is an error in the signal amplitude

(21)

Quantization (Example)

1.3 1.5

5 101

3.6 3.5

7 111

2.3 2.5

6 110

0.7 0.5

4 100

-0.7 -0.5

3 011

-2.4 -2.5

1 001

-3.4 -3.5

0 000

(22)

(Uniform Quantization)

 The quantization levels are uniformly spaced.

 Two types are shown below

Midtread Midrise

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The Signal Multiplexing

Multiplexing is done to increase the number of users who can communicate at the same time on the same medium

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Multiple signals with same frequency spectrum cannot be transmitted on the same channel. (Interference)

 Modulation allows several signals having different frequencies to be transmitted on the same channel

Frequency MUX

f₁ f₂ f₃ f₄ f

f₁ f

f₂ f

f₃ f f f

4

Channel

Frequency DEMUX

f₁ f

f₂ f

f₃ f f f

4

Frequency Division Multiplexing Frequency Domain Multiplexing (FDM)

(25)

Time Division Multiple Access

g₂(t)

0 t

g₁(t)

0 t

Rate = R

Rate = 2R

TDM Signal with higher rate g₁(t)

g₂(t)

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LPF LPF

LPF

Pulse LPF

modulator channel

Pulse De- modulator Message

inputs

1 2

N

synchronized

Commutator Decommutator

Message outputs

1 2

N Time Division Multiplexing Schematic Diagram

(27)

LPF LPF

LPF

Pulse LPF

inputs

1 2

N

synchronized

1 2

 LPF removes all the unwanted frequencies from the message signal and restrict the bandwidth so that sampling may be done at twice the BW

(28)

LPF LPF

LPF

Pulse LPF

inputs

1 2

N

synchronized

1 2

 Commutator has two functions: (1) samples each signal at the Nyquist rate or more than the Nyquist rate and (2) sequentially interleave the ‘N’ samples within the sampling interval T_s

(29)

LPF LPF

LPF

Pulse LPF

inputs

1 2

N

synchronized

1 2

 The pulse modulator modulates the sampled signal to make it suitable for the channel

(30)

LPF LPF

LPF

Pulse LPF

inputs

1 2

N

synchronized

1 2

 The pulse demodulator and decommutator reverse the operation performed by the pulse modulator and commutator

(31)

LPF LPF

LPF

Pulse LPF

inputs

1 2

N

synchronized

1 2

 The pulse demodulator and decommutator reverse the operation performed by the pulse modulator and commutator

 The LPF reconstructs the message signal

(32)

LPF LPF

LPF

Pulse LPF

inputs

1 2

N

synchronized

1 2

Example: Three voice channel having a bandwidth of 3.3 KHz and two data channels of 10 KHz bandwidth are to be TDM. Suggest the TDM frame scheme using uniform sampling. Calculate the frame rate and channel and transmission bandwidth.

(33)

Time Division Multiplexing Schematic Diagram

Example: Three voice channel having a bandwidth of 3.3 KHz and two data channels of 10 KHz bandwidth are to be TDM. Suggest the TDM frame scheme using uniform sampling. Calculate the frame rate and channel and transmission bandwidth.

Solution: For uniform sampling, all the channels are sampled at the same rate which is the highest Nyquist rate among all the channels. There will be total 5 voice and data channels (V1, V2, V3, D1, D2) and one sync (S) channel

V1 V2 V3 D1 D2 S V1 V2 V3 D1 D2 S

Frame frequency = f_fr= 1/T_fr = 1/T_s=20KHz

T_s= T_fr

TDM channel BW = 1/T_slot = N/T_s = Nf_s = Nf_fr=6x20 = 120 KHz Slot Time = T_slot= T_s/N = 50s/6=8.33s

V1 f_s=2xhighest BW = 2 x10KHz, = 20KHz

therefore T_s = 1/f_s = 5x10^-5 sec=50s= T_fr

Number of slots per frame N = 6 (1 for each channel and 1 sync)

(34)

Time Division Multiplexing Schematic Diagram

V1 V2 V3 D1 D2 S V1 V2 V3 D1 D2 S

T_s= T_fr

V1 TDM Frame

TDM Commutator

V1

V2

D1

D2 V₃

S

(35)

Time Division Multiplexing-Non Uniform sampling

In the previous example voice channels had a bandwidth of 3.3 KHz but still they were sampled at 20KHz.

V1 V2 V3 D1 D2 S

Uniform Sampling

Sampling rate f_s = 2 x highest BW Frame time T_fr = T_s

No of slots per channel per frame N = 1 Transmission bandwidth B_ch = Nf_s = Nf_fr

T_s= T_fr

Bandwidth can be saved if every channel is sampled at its Nyquist rate. This is called non-uniform sampling. So higher BW channels will have more slots in one frame

(36)

Time Division Multiplexing-Non Uniform sampling Non-uniform Sampling

Calculation of slots per frame per channel

1 S

Total = 10 slots per frame

20/6.6 = 3 20 KHz

10KHz D2

20/6.6 = 3 20 KHz

10KHz D1

6.6/6.6 = 1 6.6 KHz

3.3 KHz V3

6.6/6.6 = 1 6.6 KHz

3.3 KHz V2

6.6/6.6 = 1 6.6 KHz

3.3 KHz V1

slots per channel per frame (Nyquist rate/ minimum sampling rate) Nyquist Rate

Bandwidth Channels

(37)

Time Division Multiplexing-Non Uniform sampling

TDM Frame Scheme: T_fr=1/f_smin

V1 D1 D2 V2 D1 D2 V3 D1 D2 S

Calculation of slot time and transmission BW 1/f_smin

Sampling rate = Nyquist rate

Frame rate = f_fr= f_smin =2 X lowest BW T_fr= 1/ f_fr= 1/6.6 KHz = 1.5x10^-4sec

Number of slots per channel per frame = depend on rate of each channel

Total slots per frame N = 10

T_slot = T_s/N= 1/Nf_smin=1/(10x6.6KHz) = 15.15 s B_tr = 1/T_slot =66KHz

V1

V2

D1

D2

V3

S

D1

D2

D1

D2

TDM Commutator

(38)

Comparison between Uniform and non Uniform sampling for the previous example

V1 D1 D2 V2 D1 D2 V3 D1 D2 S

T_fr = 1/f_smin

Sampling rate = Nyquist rate

Frame rate = f_fr = f_smin =2 X lowest BW T_fr= 1/ f_fr= 1/6.6 KHz = 1.5x10^-4sec

Number of slots per channel per frame = depend on rate of each channel

Total slots per frame = 10

T_slot = T_s/N= 1/Nf_smin=1/(10x6.6KHz) = 15.15 s B_tr = 1/T_slot =66KHz

V1 V2 V3 D1 D2 S

Uniform Sampling

Sampling rate f_s= 2 x highest BW Frame rate = sampling rate

Frame time T_fr = T_s=1/ f_{s =}5x10^-5sec No of slots per channel per frame = 1 Total slots per frame N = 6

Slot Time = T_s/ N = 8.33 s

Transmission bandwidth B_ch = Nf_s= Nf_fr

= 120 KHz

T_fr= 1/f_s

Non-Uniform Sampling

(39)

Quantization Levels (Uniform Quantization)

 The quantization levels are uniformly spaced.

 Two types are shown below

Midtread Midrise

(40)

The Process of Encoding

 The sampling and quantization processes render a discrete-time discrete-amplitude signal

 But this form is not best-suited for the transmission channel

 To make the signal more robust for transmission, noise, and other channel impairments, the encoding is done

Code: The arrangement of discrete events is called a code

Code Element or symbol: One of the discrete events is called a symbol

Code Word: A particular arrangement to represent a single value of the discrete set is called a code word

0001

Code word

Symbol

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The Process of Encoding-Line Codes

Line code: The electrical representation of the binary digits is called a Line code. Consider 5 types of line codes representing the binary data stream 0 1 1 0 1 0 0 1

Unipolar NRZ Signaling

Polar NRZ Signaling

Unipolar RZ Signaling

Bi-Polar RZ Signaling Split-Phase or Manchester Code

(42)

Split-Phase or Manchester Code

 Symbol 1 is represented by a positive pulse of amplitude +A followed by a negative pulse of amplitude –A

 Symbol 0 is represented by a pulse of amplitude –A followed by another one of amplitude +A

 The Manchester code suppresses dc component and has relatively insignificant low frequency components

0 1 1 0 1 0 0 1

(43)

Reading from the book

 Chapter 6: Sampling and Analog to Digital Conversion, Pages 302 - 324 B.P. Lathi, “Modern Digital and analog Communication Systems”, 4th Edition

College of Engineering, Taibah University

EE 372 College of Engineering, Taibah University Communication TheoryCommunication TheoryCommunication Theory EE 372 College of Engineering, Taibah University Communication Theory EE 372 College of Engineering, Taibah University Communication Theory EE 372 College of Engineering, Taibah UniversityCollege of Engineering, Taibah University Communication TheoryCommunication TheoryCommunication Theory EE 372 College of Engineering, Taibah University Communication Theory