7: Baseband Transmission of Digital Signals · 7: Baseband Transmission of Digital Signals Y. Yoganandam, Runa Kumari, and S. R. Zinka Department of Electrical & Electronics Engineering

Digital Transmission Systems Line Coding Pulse Shaping Summary

7: Baseband Transmission of Digital Signals

Y. Yoganandam, Runa Kumari, and S. R. Zinka

Department of Electrical & Electronics EngineeringBITS Pilani, Hyderbad Campus

Sep 28 – Oct 07, 2015

Baseband Transmission of Digital Signals Communication Systems, Dept. of EEE, BITS Hyderabad


Outline

1 Digital Transmission Systems

2 Line Coding

3 Pulse Shaping

4 Summary



Outline


2 Line Coding

3 Pulse Shaping

4 Summary



A Typical Digital Transmission System

MultiplexerSourceencoder

Basebandmodulation(line coding)

Digitalcarrier

modulationChannel

Regenerativerepeater

Othersignals

1011000...



A Typical Digital Transmission System



Digitalcarrier

modulationChannel


Othersignals

1011000...



Source



Digitalcarrier

modulationChannel


Othersignals

1011000...



Bits, Symbols, and PCM Word

x(t)

t




x(t)

t




x(t)

1

2

3

4

5

6

t

7

0




x(t)

1

2

3

4

5

6

t000 100 101 100 011 100 110 111 101 011 011 100 100

7

0




x(t)

1

2

3

4

5

6

t000 100 101 100 011 100 110 111 101 011 011 100 100

Pulse code (PCM word)

7

0




x(t)

1

2

3

4

5

6

t000 100 101 100 011 100 110 111 101 011 011 100 100

Pulse code (PCM word)

7

0

8-ary symbols (M = 8, k = 3)



Line Coding (Transmission Coding)



Digitalcarrier

modulationChannel


Othersignals

1011000...



Line Coding (Transmission Coding)



Digitalcarrier

modulationChannel


Othersignals

1011000...



Line Coding – Nonreturn-to-Zero (NRZ) Schemes

1 1 1 1 1 100000NRZ - L 1 : + V

0 : - V




1 1 1 1 1 100000NRZ - L 1 : + V

0 : - V




1 1 1 1 1 100000

1 1 1 1 1 100000

NRZ - L

NRZ - M

1 : + V0 : - V

1 : Level change0 : No level change




1 1 1 1 1 100000

1 1 1 1 1 100000

1 1 1 1 1 100000

NRZ - L

NRZ - M

NRZ - S

1 : + V0 : - V


1 : No level change0 : Level change



Line Coding – Return-to-Zero (RZ) Schemes

1 1 1 1 1 100000Unpolar - RZ 1 : Half bit wide pulse

0 : No pulse




1 1 1 1 1 100000Unpolar - RZ 1 : Half bit wide pulse

0 : No pulse




1 1 1 1 1 100000

1 1 1 1 1 100000

Unpolar - RZ

Bipolar - RZ 1 : Half bit wide +V0 : Half bit wide -V

1 : Half bit wide pulse0 : No pulse




1 1 1 1 1 100000

1 1 1 1 1 100000

1 1 1 1 1 100000

Unpolar - RZ

Bipolar - RZ

AMI - RZ1 : Half bit wide +V/-V0 : No pulse

1 : Half bit wide +V0 : Half bit wide -V




Digital Carrier Modulation



Digitalcarrier

modulationChannel


Othersignals

1011000...



Digital Carrier Modulation



Digitalcarrier

modulationChannel


Othersignals

1011000...



Multiplexer



Digitalcarrier

modulationChannel


Othersignals

1011000...



Regenerative Repeater



Digitalcarrier

modulationChannel


Othersignals

1011000...



Noise Immunity of Digital Signals

tTransmittedsignal




tTransmittedsignal




t

t

Transmittedsignal

Recieved distortedsignal (without noise)




t

t

t

Transmittedsignal


Recieved distortedsignal (with noise)




t

t

t

t

Transmittedsignal


Recieved distortedsignal (with noise)

Regenerated signal(delayed)



Outline


2 Line Coding

3 Pulse Shaping

4 Summary



A Few Line Codes

1 1 1 1 1 100000

1 1 1 1 1 100000

1 1 1 1 1 100000

NRZ - L

NRZ - M

NRZ - S

1 : + V0 : - V





A Few Line Codes

1 1 1 1 1 100000

1 1 1 1 1 100000

1 1 1 1 1 100000

NRZ - L

NRZ - M

NRZ - S

1 : + V0 : - V





A Few Line Codes

1 1 1 1 1 100000

1 1 1 1 1 100000

1 1 1 1 1 100000

Unpolar - RZ

Bipolar - RZ

AMI - RZ1 : Half bit wide +V/-V0 : No pulse

1 : Half bit wide +V0 : Half bit wide -V




Choosing an Appropriate Line Code

• Transmission bandwidth (should be as small as possible)

• Power efficiency (for a given bandwidth and a specified detection errorprobability)

• Error detection and correction capability

• Favorable power spectral

• Adequate timing content

• Transparency









• Transparency









• Transparency









• Transparency









• Transparency









• Transparency









• Transparency



Extracting Timing/Clock Information & Transparency

Tb

At

On-OffCoding

If there are too many zeros in sequence, we can’t extract timing information.So, on-of coding is not transparent.




Tb

At

On-OffCoding





tA/2

Tb

At

On-OffCoding





tA/2

tA/2

Tb

At

On-OffCoding





tA/2

tA/2

+

Tb

At

On-OffCoding

If there are too many 0’s in sequence, we can’t extract timing information. So,on-of coding is not transparent.



Power Spectral Density of Any Line Code

x(t)t

(k-1)Tb

(k+1)TbkTb

ak+1ak

ak-1




x(t)t

(k-1)Tb

(k+1)TbkTb

ak+1ak

ak-1




x(t)t

(k-1)Tb

(k+1)TbkTb

ak+1ak

ak-1

What are thepossible values of ak ?




x(t)t

(k-1)Tb

(k+1)TbkTb

ak+1ak

ak-1


p(t)

t




x(t)t

(k-1)Tb

(k+1)TbkTb

ak+1ak

ak-1


p(t)

t

What are thepossible shapes of p(t)?




y(t)t

(k-1)Tb

(k+1)TbkTb

x(t)t

(k-1)Tb

(k+1)TbkTb

ak+1ak

ak-1


p(t)

t





h(t) = p(t)x(t) y(t)

y(t)t

(k-1)Tb

(k+1)TbkTb

x(t)t

(k-1)Tb

(k+1)TbkTb

ak+1ak

ak-1


p(t)

t






y(t)t

(k-1)Tb

(k+1)TbkTb

x(t)t

(k-1)Tb

(k+1)TbkTb

ak+1ak

ak-1


p(t)

t


Sy(ω)=|P(ω)|2Sx(ω)





y(t)t

(k-1)Tb

(k+1)TbkTb

x(t)t

(k-1)Tb

(k+1)TbkTb

ak+1ak

ak-1


p(t)

t


Sy(ω)=|P(ω)|2Sx(ω)



Autocorrelation function of an Impulse Train

x(t)t

(k-1)Tb

(k+1)TbkTb

ak+1ak

ak-1




x(t)t

(k-1)Tb

(k+1)TbkTb

ak+1ak

ak-1




x(t)t

(k-1)Tb

(k+1)TbkTb

ak+1ak

ak-1

x(t)t

(k-1)Tb

(k+1)TbkTb

hk+1hk

hk-1

>ϵ




x(t)t

(k-1)Tb

(k+1)TbkTb

ak+1ak

ak-1

x(t)t

(k-1)Tb

(k+1)TbkTb

hk+1hk

hk-1

>ϵ

tkTb

hk

τ




x(t)t

(k-1)Tb

(k+1)TbkTb

ak+1ak

ak-1

x(t)t

(k-1)Tb

(k+1)TbkTb

hk+1hk

hk-1

>ϵ

tkTb

hk

τ

τ

R0/ϵTb

-ϵ +ϵ




x(t)t

(k-1)Tb

(k+1)TbkTb

ak+1ak

ak-1

x(t)t

(k-1)Tb

(k+1)TbkTb

hk+1hk

hk-1

>ϵ

tkTb

hk

τ

t(k+1)TbkTb

hk+1hk

τ

τ

R0/ϵTb

-ϵ +ϵ

R1/ϵTb

Tb




x(t)t

(k-1)Tb

(k+1)TbkTb

ak+1ak

ak-1

x(t)t

(k-1)Tb

(k+1)TbkTb

hk+1hk

hk-1

>ϵ

tkTb

hk

τ

t(k+1)TbkTb

hk+1hk

τ

τ

R0/ϵTb

-ϵ +ϵ

R1/ϵTb

R2/ϵTb

R3/ϵTb

Tb

2Tb

3Tb-Tb

-2Tb

-3Tb

R1/ϵTb

R2/ϵTb

R3/ϵTb




x(t)t

(k-1)Tb

(k+1)TbkTb

ak+1ak

ak-1

x(t)t

(k-1)Tb

(k+1)TbkTb

hk+1hk

hk-1

>ϵ

tkTb

hk

τ

t(k+1)TbkTb

hk+1hk

τ

τ

R0/TbR1/Tb

R2/Tb

R3/Tb

Tb

2Tb

3Tb-Tb

-2Tb

-3Tb

R1/Tb

R2/Tb

R3/Tb

0




Each impulse of x (t) can be approximated by finite pulses of width ε→ 0 andheight hk = ak

ε . If we designate the corresponding rectangular pulse train byx̂ (t), then by definition

Rx̂ (τ) = limT→∞

1T

ˆ T/2

−T/2x̂ (t) x̂ (t− τ) dt.

When τ < ε,

Rx̂ = limT→∞

1T ∑

kh2

k (ε− τ)

= limT→∞

1T ∑

ka2

k

(ε− τ

ε2

)=

R0εTb

(1− τ

ε

)=

R0εTb

(1− |τ|

ε

), (1)

whereR0 = lim

T→∞

TbT ∑

ka2

k = limN→∞

1N ∑

ka2

k = a2k . (2)







1T

ˆ T/2

−T/2x̂ (t) x̂ (t− τ) dt.

When τ < ε,

Rx̂ = limT→∞

1T ∑

kh2

k (ε− τ)

= limT→∞

1T ∑

ka2

k

(ε− τ

ε2

)=

R0εTb

(1− τ

ε

)=

R0εTb

(1− |τ|

ε

), (1)

whereR0 = lim

T→∞

TbT ∑

ka2

k = limN→∞

1N ∑

ka2

k = a2k . (2)







1T

ˆ T/2

−T/2x̂ (t) x̂ (t− τ) dt.

When τ < ε,

Rx̂ = limT→∞

1T ∑

kh2

k (ε− τ)

= limT→∞

1T ∑

ka2

k

(ε− τ

ε2

)=

R0εTb

(1− τ

ε

)

=R0εTb

(1− |τ|

ε

), (1)

whereR0 = lim

T→∞

TbT ∑

ka2

k = limN→∞

1N ∑

ka2

k = a2k . (2)







1T

ˆ T/2

−T/2x̂ (t) x̂ (t− τ) dt.

When τ < ε,

Rx̂ = limT→∞

1T ∑

kh2

k (ε− τ)

= limT→∞

1T ∑

ka2

k

(ε− τ

ε2

)=

R0εTb

(1− τ

ε

)=

R0εTb

(1− |τ|

ε

), (1)

whereR0 = lim

T→∞

TbT ∑

ka2

k = limN→∞

1N ∑

ka2

k = a2k . (2)







1T

ˆ T/2

−T/2x̂ (t) x̂ (t− τ) dt.

When τ < ε,

Rx̂ = limT→∞

1T ∑

kh2

k (ε− τ)

= limT→∞

1T ∑

ka2

k

(ε− τ

ε2

)=

R0εTb

(1− τ

ε

)=

R0εTb

(1− |τ|

ε

), (1)

whereR0 = lim

T→∞

TbT ∑

ka2

k

= limN→∞

1N ∑

ka2

k = a2k . (2)







1T

ˆ T/2

−T/2x̂ (t) x̂ (t− τ) dt.

When τ < ε,

Rx̂ = limT→∞

1T ∑

kh2

k (ε− τ)

= limT→∞

1T ∑

ka2

k

(ε− τ

ε2

)=

R0εTb

(1− τ

ε

)=

R0εTb

(1− |τ|

ε

), (1)

whereR0 = lim

T→∞

TbT ∑

ka2

k = limN→∞

1N ∑

ka2

k

= a2k . (2)







1T

ˆ T/2

−T/2x̂ (t) x̂ (t− τ) dt.

When τ < ε,

Rx̂ = limT→∞

1T ∑

kh2

k (ε− τ)

= limT→∞

1T ∑

ka2

k

(ε− τ

ε2

)=

R0εTb

(1− τ

ε

)=

R0εTb

(1− |τ|

ε

), (1)

whereR0 = lim

T→∞

TbT ∑

ka2

k = limN→∞

1N ∑

ka2

k = a2k . (2)




Repeating the earlier argument, we can prove that

Rn = limT→∞

TbT ∑

kakak+n = lim

N→∞

1N ∑

kakak+n = akak+n. (3)

Thus, in the limit as ε→ 0, the triangular pulses becomes impulses and

Rx̂ (τ) =1

Tb

∞

∑n=−∞

Rnδ (τ − nTb) . (4)

Since Sx (ω) is the Fourier transform ofRx̂ (τ), and R−n = Rn,

Sx (ω) =1

Tb

∞

∑n=−∞

Rne−jnωTb =1

Tb

(R0 + 2

∞

∑n=1

Rn cos nωTb

). (5)





Rn = limT→∞

TbT ∑

kakak+n = lim

N→∞

1N ∑



Rx̂ (τ) =1

Tb

∞

∑n=−∞

Rnδ (τ − nTb) . (4)


Sx (ω) =1

Tb

∞

∑n=−∞

Rne−jnωTb =1

Tb

(R0 + 2

∞

∑n=1

Rn cos nωTb

). (5)





Rn = limT→∞

TbT ∑

kakak+n = lim

N→∞

1N ∑



Rx̂ (τ) =1

Tb

∞

∑n=−∞

Rnδ (τ − nTb) . (4)


Sx (ω) =1

Tb

∞

∑n=−∞

Rne−jnωTb =1

Tb

(R0 + 2

∞

∑n=1

Rn cos nωTb

). (5)





Rn = limT→∞

TbT ∑

kakak+n = lim

N→∞

1N ∑



Rx̂ (τ) =1

Tb

∞

∑n=−∞

Rnδ (τ − nTb) . (4)

Since Sx (ω) is the Fourier transform ofRx̂ (τ),

and R−n = Rn,

Sx (ω) =1

Tb

∞

∑n=−∞

Rne−jnωTb =1

Tb

(R0 + 2

∞

∑n=1

Rn cos nωTb

). (5)





Rn = limT→∞

TbT ∑

kakak+n = lim

N→∞

1N ∑



Rx̂ (τ) =1

Tb

∞

∑n=−∞

Rnδ (τ − nTb) . (4)

Since Sx (ω) is the Fourier transform ofRx̂ (τ),

and R−n = Rn,

Sx (ω) =1

Tb

∞

∑n=−∞

Rne−jnωTb

=1

Tb

(R0 + 2

∞

∑n=1

Rn cos nωTb

). (5)





Rn = limT→∞

TbT ∑

kakak+n = lim

N→∞

1N ∑



Rx̂ (τ) =1

Tb

∞

∑n=−∞

Rnδ (τ − nTb) . (4)


Sx (ω) =1

Tb

∞

∑n=−∞

Rne−jnωTb =1

Tb

(R0 + 2

∞

∑n=1

Rn cos nωTb

). (5)




So, PSD of any line code is given by

Sy (ω) =|P (ω)|2

Tb

∞

∑n=−∞

Rne−jnωTb =|P (ω)|2

Tb

(R0 + 2

∞

∑n=1

Rn cos nωTb

),

(6)where

Rn = limN→∞

1N ∑




PSD of Polar Line Coding

1 1 1 1 1 100000+V

- V

Since ak is either +1 or -1 and are equally likely,

R0 = limN→∞

1N ∑

ka2

k = limN→∞

1N

(N) = 1. (8)

Since ak and ak+n are either +1 or -1 and are equally likely,

Rn = limN→∞

1N ∑

kakak+n = lim

N→∞

1N ∑

k

[N2(1) +

N2(−1)

]= 0. (9)

So,

Sy (ω) =|P (ω)|2

Tb

(R0 + 2

∞

∑n=1

Rn cos nωTb

)=|P (ω)|2

Tb. (10)




1 1 1 1 1 100000+V

- V


R0 = limN→∞

1N ∑

ka2

k = limN→∞

1N

(N) = 1. (8)


Rn = limN→∞

1N ∑

kakak+n = lim

N→∞

1N ∑

k

[N2(1) +

N2(−1)

]= 0. (9)

So,

Sy (ω) =|P (ω)|2

Tb

(R0 + 2

∞

∑n=1

Rn cos nωTb

)=|P (ω)|2

Tb. (10)




1 1 1 1 1 100000+V

- V


R0 = limN→∞

1N ∑

ka2

k

= limN→∞

1N

(N) = 1. (8)


Rn = limN→∞

1N ∑

kakak+n = lim

N→∞

1N ∑

k

[N2(1) +

N2(−1)

]= 0. (9)

So,

Sy (ω) =|P (ω)|2

Tb

(R0 + 2

∞

∑n=1

Rn cos nωTb

)=|P (ω)|2

Tb. (10)




1 1 1 1 1 100000+V

- V


R0 = limN→∞

1N ∑

ka2

k = limN→∞

1N

(N) = 1. (8)


Rn = limN→∞

1N ∑

kakak+n = lim

N→∞

1N ∑

k

[N2(1) +

N2(−1)

]= 0. (9)

So,

Sy (ω) =|P (ω)|2

Tb

(R0 + 2

∞

∑n=1

Rn cos nωTb

)=|P (ω)|2

Tb. (10)




1 1 1 1 1 100000+V

- V


R0 = limN→∞

1N ∑

ka2

k = limN→∞

1N

(N) = 1. (8)


Rn = limN→∞

1N ∑

kakak+n

= limN→∞

1N ∑

k

[N2(1) +

N2(−1)

]= 0. (9)

So,

Sy (ω) =|P (ω)|2

Tb

(R0 + 2

∞

∑n=1

Rn cos nωTb

)=|P (ω)|2

Tb. (10)




1 1 1 1 1 100000+V

- V


R0 = limN→∞

1N ∑

ka2

k = limN→∞

1N

(N) = 1. (8)


Rn = limN→∞

1N ∑

kakak+n = lim

N→∞

1N ∑

k

[N2(1) +

N2(−1)

]= 0. (9)

So,

Sy (ω) =|P (ω)|2

Tb

(R0 + 2

∞

∑n=1

Rn cos nωTb

)=|P (ω)|2

Tb. (10)




1 1 1 1 1 100000+V

- V


R0 = limN→∞

1N ∑

ka2

k = limN→∞

1N

(N) = 1. (8)


Rn = limN→∞

1N ∑

kakak+n = lim

N→∞

1N ∑

k

[N2(1) +

N2(−1)

]= 0. (9)

So,

Sy (ω) =|P (ω)|2

Tb

(R0 + 2

∞

∑n=1

Rn cos nωTb

)

=|P (ω)|2

Tb. (10)




1 1 1 1 1 100000+V

- V


R0 = limN→∞

1N ∑

ka2

k = limN→∞

1N

(N) = 1. (8)


Rn = limN→∞

1N ∑

kakak+n = lim

N→∞

1N ∑

k

[N2(1) +

N2(−1)

]= 0. (9)

So,

Sy (ω) =|P (ω)|2

Tb

(R0 + 2

∞

∑n=1

Rn cos nωTb

)=|P (ω)|2

Tb. (10)



PSD of Polar Line Coding – Example

1 1 1 1 1 100000 +V

- V

Since

p (t) = rect(

tTb/2

), (11)

the corresponding Fourier transform is given by

P (ω) =Tb2

sinc(

ωTb4

). (12)

So, PSD of the polar line code (whose pulse width is Tb/2) is given by

Sy (ω) =|P (ω)|2

Tb=

Tb4

sinc2(

ωTb4

). (13)




1 1 1 1 1 100000 +V

- V

Since

p (t) = rect(

tTb/2

), (11)


P (ω) =Tb2

sinc(

ωTb4

). (12)


Sy (ω) =|P (ω)|2

Tb=

Tb4

sinc2(

ωTb4

). (13)




1 1 1 1 1 100000 +V

- V

Since

p (t) = rect(

tTb/2

), (11)


P (ω) =Tb2

sinc(

ωTb4

). (12)


Sy (ω) =|P (ω)|2

Tb=

Tb4

sinc2(

ωTb4

). (13)




1 1 1 1 1 100000 +V

- V

Since

p (t) = rect(

tTb/2

), (11)


P (ω) =Tb2

sinc(

ωTb4

). (12)


Sy (ω) =|P (ω)|2

Tb=

Tb4

sinc2(

ωTb4

). (13)




1 1 1 1 1 100000 +V

- V

Since

p (t) = rect(

tTb/2

), (11)


P (ω) =Tb2

sinc(

ωTb4

). (12)


Sy (ω) =|P (ω)|2

Tb=

Tb4

sinc2(

ωTb4

). (13)



Properties of Polar Line Coding

Sy(ω)

2π

Tb

4π

Tb

6π

Tb

8π

Tb

2π

Tb

4π

Tb

6π

Tb

8π

Tb

0ω



PSD of ON-OFF Line Coding

1 1 1 1 1 100000+V

0

Since ak is either 1 or 0 and are equally likely,

R0 = limN→∞

1N ∑

ka2

k = limN→∞

1N

[N2(1) +

N2(0)]=

12

. (14)

Since ak and ak+n are equally likely to be 1 or 0, the product akak+n equallylikely to be 1× 1, 1× 0, 0× 1, or 0× 0. So,

Rn = limN→∞

1N ∑

kakak+n = lim

N→∞

1N ∑

k

[N4(1) +

3N4

(0)]=

14

, and (15)

Sy (ω) =|P (ω)|2

Tb

[14+

2π

4Tb

∞

∑n=−∞

δ

(ω− 2πn

Tb

)]. (prove yourself) (16)



PSD of ON-OFF Line Coding1 1 1 1 1 100000

+V

0


R0 = limN→∞

1N ∑

ka2

k = limN→∞

1N

[N2(1) +

N2(0)]=

12

. (14)


Rn = limN→∞

1N ∑

kakak+n = lim

N→∞

1N ∑

k

[N4(1) +

3N4

(0)]=

14

, and (15)

Sy (ω) =|P (ω)|2

Tb

[14+

2π

4Tb

∞

∑n=−∞

δ

(ω− 2πn

Tb





+V

0


R0 = limN→∞

1N ∑

ka2

k

= limN→∞

1N

[N2(1) +

N2(0)]=

12

. (14)


Rn = limN→∞

1N ∑

kakak+n = lim

N→∞

1N ∑

k

[N4(1) +

3N4

(0)]=

14

, and (15)

Sy (ω) =|P (ω)|2

Tb

[14+

2π

4Tb

∞

∑n=−∞

δ

(ω− 2πn

Tb





+V

0


R0 = limN→∞

1N ∑

ka2

k = limN→∞

1N

[N2(1) +

N2(0)]=

12

. (14)


Rn = limN→∞

1N ∑

kakak+n = lim

N→∞

1N ∑

k

[N4(1) +

3N4

(0)]=

14

, and (15)

Sy (ω) =|P (ω)|2

Tb

[14+

2π

4Tb

∞

∑n=−∞

δ

(ω− 2πn

Tb





+V

0


R0 = limN→∞

1N ∑

ka2

k = limN→∞

1N

[N2(1) +

N2(0)]=

12

. (14)

Since ak and ak+n are equally likely to be 1 or 0, the product akak+n equallylikely to be 1× 1, 1× 0, 0× 1, or 0× 0.

So,

Rn = limN→∞

1N ∑

kakak+n = lim

N→∞

1N ∑

k

[N4(1) +

3N4

(0)]=

14

, and (15)

Sy (ω) =|P (ω)|2

Tb

[14+

2π

4Tb

∞

∑n=−∞

δ

(ω− 2πn

Tb





+V

0


R0 = limN→∞

1N ∑

ka2

k = limN→∞

1N

[N2(1) +

N2(0)]=

12

. (14)


Rn = limN→∞

1N ∑

kakak+n

= limN→∞

1N ∑

k

[N4(1) +

3N4

(0)]=

14

, and (15)

Sy (ω) =|P (ω)|2

Tb

[14+

2π

4Tb

∞

∑n=−∞

δ

(ω− 2πn

Tb





+V

0


R0 = limN→∞

1N ∑

ka2

k = limN→∞

1N

[N2(1) +

N2(0)]=

12

. (14)


Rn = limN→∞

1N ∑

kakak+n = lim

N→∞

1N ∑

k

[N4(1) +

3N4

(0)]=

14

, and (15)

Sy (ω) =|P (ω)|2

Tb

[14+

2π

4Tb

∞

∑n=−∞

δ

(ω− 2πn

Tb




PSD of ON-OFF Line Coding – Example

1 1 1 1 1 100000 +V

0

Since

p (t) = rect(

tTb/2

), (17)


P (ω) =Tb2

sinc(

ωTb4

). (18)

So, PSD of the ON-OFF line code (whose pulse width is Tb/2) is given by

Sy (ω) =Tb4

sinc2(

ωTb4

)[14+

2π

4Tb

∞

∑n=−∞

δ

(ω− 2πn

Tb

)]. (19)




1 1 1 1 1 100000 +V

0

Since

p (t) = rect(

tTb/2

), (17)


P (ω) =Tb2

sinc(

ωTb4

). (18)


Sy (ω) =Tb4

sinc2(

ωTb4

)[14+

2π

4Tb

∞

∑n=−∞

δ

(ω− 2πn

Tb

)]. (19)




1 1 1 1 1 100000 +V

0

Since

p (t) = rect(

tTb/2

), (17)


P (ω) =Tb2

sinc(

ωTb4

). (18)


Sy (ω) =Tb4

sinc2(

ωTb4

)[14+

2π

4Tb

∞

∑n=−∞

δ

(ω− 2πn

Tb

)]. (19)




1 1 1 1 1 100000 +V

0

Since

p (t) = rect(

tTb/2

), (17)


P (ω) =Tb2

sinc(

ωTb4

). (18)


Sy (ω) =Tb4

sinc2(

ωTb4

)[14+

2π

4Tb

∞

∑n=−∞

δ

(ω− 2πn

Tb

)]. (19)




1 1 1 1 1 100000 +V

0

Since

p (t) = rect(

tTb/2

), (17)


P (ω) =Tb2

sinc(

ωTb4

). (18)


Sy (ω) =Tb4

sinc2(

ωTb4

)[14+

2π

4Tb

∞

∑n=−∞

δ

(ω− 2πn

Tb

)]. (19)



Properties of ON-OFF Line Coding

Sy(ω)

2π

Tb

4π

Tb

6π

Tb

8π

Tb

2π

Tb

4π

Tb

6π

Tb

8π

Tb

0ω



Properties of ON-OFF Line Coding

Sy(ω)

2π

Tb

4π

Tb

6π

Tb

8π

Tb

2π

Tb

4π

Tb

6π

Tb

8π

Tb

0ω



PSD of AMI Line Coding

1 1 1 1 1 100000+V

- V

AMI line code has 3 symbols 1, 0, -1. On average, half aks are zeros and otherhalf are either +1 or -1. Hence,

R0 = limN→∞

1N ∑

ka2

k = limN→∞

1N

[N2(±1)2 +

N2(0)]=

12

. (20)

Since ak and ak+1 are equally likely to be 1 or 0, the product akak+1 equallylikely to be ±1×∓1, 1× 0, 0× 1, or 0× 0. So,

R1 = limN→∞

1N ∑

kakak+1 = lim

N→∞

1N ∑

k

[N4(−1) +

3N4

(0)]= −1

4, and (21)

Sy (ω) =|P (ω)|2

Tbsin2

(ωTb

2

). (Rn = 0 for n > 1) (22)



PSD of AMI Line Coding1 1 1 1 1 100000

+V

- V


R0 = limN→∞

1N ∑

ka2

k = limN→∞

1N

[N2(±1)2 +

N2(0)]=

12

. (20)


R1 = limN→∞

1N ∑

kakak+1 = lim

N→∞

1N ∑

k

[N4(−1) +

3N4

(0)]= −1

4, and (21)

Sy (ω) =|P (ω)|2

Tbsin2

(ωTb

2

). (Rn = 0 for n > 1) (22)




+V

- V


R0 = limN→∞

1N ∑

ka2

k =

limN→∞

1N

[N2(±1)2 +

N2(0)]=

12

. (20)


R1 = limN→∞

1N ∑

kakak+1 = lim

N→∞

1N ∑

k

[N4(−1) +

3N4

(0)]= −1

4, and (21)

Sy (ω) =|P (ω)|2

Tbsin2

(ωTb

2

). (Rn = 0 for n > 1) (22)




+V

- V


R0 = limN→∞

1N ∑

ka2

k = limN→∞

1N

[N2(±1)2 +

N2(0)]=

12

. (20)


R1 = limN→∞

1N ∑

kakak+1 = lim

N→∞

1N ∑

k

[N4(−1) +

3N4

(0)]= −1

4, and (21)

Sy (ω) =|P (ω)|2

Tbsin2

(ωTb

2

). (Rn = 0 for n > 1) (22)




+V

- V


R0 = limN→∞

1N ∑

ka2

k = limN→∞

1N

[N2(±1)2 +

N2(0)]=

12

. (20)

Since ak and ak+1 are equally likely to be 1 or 0, the product akak+1 equallylikely to be ±1×∓1, 1× 0, 0× 1, or 0× 0.

So,

R1 = limN→∞

1N ∑

kakak+1 = lim

N→∞

1N ∑

k

[N4(−1) +

3N4

(0)]= −1

4, and (21)

Sy (ω) =|P (ω)|2

Tbsin2

(ωTb

2

). (Rn = 0 for n > 1) (22)




+V

- V


R0 = limN→∞

1N ∑

ka2

k = limN→∞

1N

[N2(±1)2 +

N2(0)]=

12

. (20)


R1 = limN→∞

1N ∑

kakak+1

= limN→∞

1N ∑

k

[N4(−1) +

3N4

(0)]= −1

4, and (21)

Sy (ω) =|P (ω)|2

Tbsin2

(ωTb

2

). (Rn = 0 for n > 1) (22)




+V

- V


R0 = limN→∞

1N ∑

ka2

k = limN→∞

1N

[N2(±1)2 +

N2(0)]=

12

. (20)


R1 = limN→∞

1N ∑

kakak+1 = lim

N→∞

1N ∑

k

[N4(−1) +

3N4

(0)]= −1

4, and (21)

Sy (ω) =|P (ω)|2

Tbsin2

(ωTb

2

). (Rn = 0 for n > 1) (22)




+V

- V


R0 = limN→∞

1N ∑

ka2

k = limN→∞

1N

[N2(±1)2 +

N2(0)]=

12

. (20)


R1 = limN→∞

1N ∑

kakak+1 = lim

N→∞

1N ∑

k

[N4(−1) +

3N4

(0)]= −1

4, and (21)

Sy (ω) =|P (ω)|2

Tbsin2

(ωTb

2

). (Rn = 0 for n > 1) (22)



PSD of AMI Line Coding – Example

1 1 1 1 1 100000+V

- V

Since

p (t) = rect(

tTb/2

), (23)


P (ω) =Tb2

sinc(

ωTb4

). (24)

So, PSD of the AMI line code (whose pulse width is Tb/2) is given by

Sy (ω) =Tb4

sinc2(

ωTb4

)sin2

(ωTb

2

). (25)




1 1 1 1 1 100000+V

- V

Since

p (t) = rect(

tTb/2

), (23)


P (ω) =Tb2

sinc(

ωTb4

). (24)


Sy (ω) =Tb4

sinc2(

ωTb4

)sin2

(ωTb

2

). (25)




1 1 1 1 1 100000+V

- V

Since

p (t) = rect(

tTb/2

), (23)


P (ω) =Tb2

sinc(

ωTb4

). (24)


Sy (ω) =Tb4

sinc2(

ωTb4

)sin2

(ωTb

2

). (25)




1 1 1 1 1 100000+V

- V

Since

p (t) = rect(

tTb/2

), (23)


P (ω) =Tb2

sinc(

ωTb4

). (24)


Sy (ω) =Tb4

sinc2(

ωTb4

)sin2

(ωTb

2

). (25)




1 1 1 1 1 100000+V

- V

Since

p (t) = rect(

tTb/2

), (23)


P (ω) =Tb2

sinc(

ωTb4

). (24)


Sy (ω) =Tb4

sinc2(

ωTb4

)sin2

(ωTb

2

). (25)



Properties of AMI Line Coding

Sy(ω)

2π

Tb

4π

Tb

ω

Polar




Sy(ω)

2π

Tb

4π

Tb

ω

Polar




Sy(ω)

2π

Tb

4π

Tb

ω

AMI

Polar




Sy(ω)

2π

Tb

4π

Tb

ω

AMI

Manchester(Split phase)

Polar



Self Study

1 High density bipolar signaling (HDB)

2 Binary with N-zero substitution signaling (BNZS)



Self Study





Self Study





Outline


2 Line Coding

3 Pulse Shaping

4 Summary



Why Pulse Shaping?

Sy (ω) = |P (ω)|2 Sx (ω) =|P (ω)|2

Tb

∞

∑n=−∞

Rne−jnωTb

In addition to the line code which decides Sx (ω), pulse shape p (t) is also animportant factor in influencing the overall PSD Sy (ω).



Why Pulse Shaping?

Sy (ω) = |P (ω)|2 Sx (ω) =|P (ω)|2

Tb

∞

∑n=−∞

Rne−jnωTb




Why Pulse Shaping?

Sy (ω) = |P (ω)|2 Sx (ω) =|P (ω)|2

Tb

∞

∑n=−∞

Rne−jnωTb




Why Pulse Shaping?

For polar line code, we have seen that

Sy(ω)

2π

Tb

4π

Tb

2π

Tb

4π

Tb

0ω



Why Pulse Shaping?

If the polar coded signal is transmitted through a channel of bandwidth of Rb,

Sy(ω)

2π

Tb

4π

Tb

2π

Tb

4π

Tb

0ω



Why Pulse Shaping?

Pulses get stretched if the channel of bandwidth is limited to Rb leading to ISI.

Sy(ω)

2π

Tb

4π

Tb

2π

Tb

4π

Tb

0ω



Why Pulse Shaping?

If we want to avoid ISI, we need to use infinite bandwidth.

If we want to restrict the bandwidth, we end up in inter symbol interference(ISI).

Then what is the solution ?

Since we are dealing with digital information, we only need the amplitudeinformation at a single instant.

If the amplitude information at that instant is not corrupted by ISI, we shouldbe fine.



Why Pulse Shaping?








Why Pulse Shaping?








Why Pulse Shaping?








Why Pulse Shaping?








Why Pulse Shaping?








Nyquist Criterion for Zero ISI

t




t

0 Tb




t

0 Tb 2Tb




t

0 Tb 2Tb




t

0 Tb 2Tb




t

0 Tb 2Tb

0 Rb/2f

-Rb/2



Problems with Sinc Pulses

• Not realistic as the time starts at −∞

• Truncation can not guarantee Rb/2 Hz BW

• Decays too slowly at a rate 1/t
















t

0 Tb 2Tb




t

0 Tb 2Tb

If pulse rate changesat the transmittter




t

0 Tb 2Tb




t

0 Tb 2Tb

If sampling rate changesat the receiver



Nyquist Pulse

Nyquist showed that a pulse satisfying

p (t) =

{1 t = 00 t = ±nTb

and decaying faster than 1/t would require a bandwidth kRb/2, where 1 <

k < 2.

|P(ω)|

ωb-ωb/2 ωb/2-ωbω



Nyquist Pulse


p (t) =

{1 t = 00 t = ±nTb


k < 2.

|P(ω)|




Nyquist Pulse


p (t) =

{1 t = 00 t = ±nTb


k < 2.

|P(ω)|




Nyquist Pulse


p (t) =

{1 t = 00 t = ±nTb

and decaying faster than 1/t would require a bandwidth kRb/2, where 1 <k < 2.

|P(ω)|


P(ω) = 0 P(ω) = 0



Nyquist Pulse

Let p(t) be sampled at Tb instants. Then the samples signal p̄ (t) and the cor-responding Fourier transform are given as

p̄ (t) =∞

∑n=−∞

p (t) δ (t− nTb) = p (t) δTb (t)

P̄ (ω) =1

Tb

∞

∑n=−∞

P (ω− nωb) .

Since p (t) = 0 at t = ±nTb,

p̄ (t) =∞

∑n=−∞

p (t) δ (t− nTb) = p (0) δ (t) = δ (t) . (assuming p (0) = 1)

Hence,

∞

∑n=−∞

P (ω− nωb) = Tb. (26)



Nyquist Pulse


p̄ (t) =∞

∑n=−∞


P̄ (ω) =1

Tb

∞

∑n=−∞

P (ω− nωb) .


p̄ (t) =∞

∑n=−∞


Hence,

∞

∑n=−∞

P (ω− nωb) = Tb. (26)



Nyquist Pulse


p̄ (t) =∞

∑n=−∞


P̄ (ω) =1

Tb

∞

∑n=−∞

P (ω− nωb) .


p̄ (t) =∞

∑n=−∞


Hence,

∞

∑n=−∞

P (ω− nωb) = Tb. (26)



Nyquist Pulse


p̄ (t) =∞

∑n=−∞


P̄ (ω) =1

Tb

∞

∑n=−∞

P (ω− nωb) .


p̄ (t) =∞

∑n=−∞


Hence,

∞

∑n=−∞

P (ω− nωb) = Tb. (26)



Nyquist Pulse

|P(ω)|

-ωb/2 ωb/2ω

Tb



Nyquist Pulse

|P(ω)|


Tb



Nyquist Pulse

|P(ω)|


Tb



Nyquist Pulse

|P(ω)|

ωbωb/2ω

Tb



Nyquist Pulse

|P(ω)|

ωbωb/2ω

Tb

P(ω) + P(ω - ωb) = Tb

0 < ω < ωb



Nyquist Pulse

Since P (ω) + P (ω−ωb) = Tb, assuming ω = x + ωb/2 gives

P(

x +ωb2

)+ P

(x− ωb

2

)= Tb |x| < ωb

2. (27)

The above equation can be rewritten as

P(ωb

2+ x)+ P∗

(ωb2− x)= Tb |x| < ωb

2. (28)

If we assume P (ω) = |P (ω)| e−jωtd then only |P (ω)| needs to satisfy theabove equation. So,∣∣∣P(ωb

2+ x)∣∣∣+ ∣∣∣P (ωb

2− x)∣∣∣ = Tb |x| < ωb

2. (29)



Nyquist Pulse


P(

x +ωb2

)+ P

(x− ωb

2

)= Tb |x| < ωb

2. (27)


P(ωb

2+ x)+ P∗

(ωb2− x)= Tb |x| < ωb

2. (28)


2+ x)∣∣∣+ ∣∣∣P (ωb

2− x)∣∣∣ = Tb |x| < ωb

2. (29)



Nyquist Pulse


P(

x +ωb2

)+ P

(x− ωb

2

)= Tb |x| < ωb

2. (27)


P(ωb

2+ x)+ P∗

(ωb2− x)= Tb |x| < ωb

2. (28)


2+ x)∣∣∣+ ∣∣∣P (ωb

2− x)∣∣∣ = Tb |x| < ωb

2. (29)



Nyquist Pulse


P(

x +ωb2

)+ P

(x− ωb

2

)= Tb |x| < ωb

2. (27)


P(ωb

2+ x)+ P∗

(ωb2− x)= Tb |x| < ωb

2. (28)


2+ x)∣∣∣+ ∣∣∣P (ωb

2− x)∣∣∣ = Tb |x| < ωb

2. (29)



Nyquist Pulse

|P(ω)|

αω

x x

ωx ωx

α

Tb-α

ωb

00

Tb

Tb/2

ωb/2

Vestigialspectrum

Roll-off factor r is defined as

r =excess bandwidth

theoretical minimum bandwidth=

ωx

ωb/2=

2ωx

ωb. (30)



Nyquist Pulse

|P(ω)|

αω

x x

ωx ωx

α

Tb-α

ωb

00

Tb

Tb/2

ωb/2

Vestigialspectrum


r =excess bandwidth


ωx

ωb/2=

2ωx

ωb. (30)



Nyquist Pulse

|P(ω)|

αω

x x

ωx ωx

α

Tb-α

ωb

00

Tb

Tb/2

ωb/2

Vestigialspectrum


r =excess bandwidth


ωx

ωb/2=

2ωx

ωb. (30)



Nyquist Pulse – Raised Cosine Pulse

|P(ω)|

αω

x x

ωx ωx

α

Tb-α

ωb

00

Tb

Tb/2

ωb/2

Vestigialspectrum

P (ω) =

12

{1− sin

[π(ω− ωb

2 )2ωx

]} ∣∣ω− ωb2

∣∣ < ωx

0 |ω| > ωb2 + ωx

1 |ω| < ωb2 −ωx




|P(ω)|

αω

x x

ωx ωx

α

Tb-α

ωb

00

Tb

Tb/2

ωb/2

Vestigialspectrum

P (ω) =

12

{1− sin

[π(ω− ωb

2 )2ωx

]} ∣∣ω− ωb2

∣∣ < ωx

0 |ω| > ωb2 + ωx

1 |ω| < ωb2 −ωx




|P(ω)|

αω

x x

ωx ωx

α

Tb-α

ωb

00

Tb

Tb/2

ωb/2

Vestigialspectrum

P (ω) =

12

{1− sin

[π(ω− ωb

2 )2ωx

]} ∣∣ω− ωb2

∣∣ < ωx

0 |ω| > ωb2 + ωx

1 |ω| < ωb2 −ωx




r r

ωbωb/2ω

0

1

Tb 2Tb-Tb-2Tb

|P(ω)| p(t)

Rb

0t




r r

ωbωb/2ω

0

1

Tb 2Tb-Tb-2Tb

|P(ω)| p(t)

Rb

0t




rr

rr

ωbωb/2ω

0

1

Tb 2Tb-Tb-2Tb

|P(ω)| p(t)

Rb

0t




r

rr

r

rr

ωbωb/2ω

0

1

Tb 2Tb-Tb-2Tb

|P(ω)| p(t)

Rb

0t




r

rrr

r

rrr

ωbωb/2ω

0

1

Tb 2Tb-Tb-2Tb

|P(ω)| p(t)

Rb

0t




Nyquist pulse for r = 1:

P (ω) = cos2(

ω

4Rb

)rect

(ω

4πRb

)(31)

p (t) = Rbcos (πRbt)1− 4R2

bt2sinc (πRbt) (32)




Nyquist pulse for r = 1:

P (ω) = cos2(

ω

4Rb

)rect

(ω

4πRb

)(31)

p (t) = Rbcos (πRbt)1− 4R2

bt2sinc (πRbt) (32)



Signaling with Controlled ISI

Nyquist pulse results in BW slightly higher than the theoretical minimum BW

BT =(

1+r2

)Rb.

Is it possible to reduce the BW below the theoretical minimum (Rb/2 Hz)?

Widening of the pulse is the only way; However we can’t avoid ISI.

Tb

p(t)

0 2Tb 3Tb 4Tb-Tb-2Tb-3Tb 5Tb-4Tb

1

t

3Tb

duobinarypulse





BT =(

1+r2

)Rb.



Tb

p(t)


1

t

3Tb

duobinarypulse





BT =(

1+r2

)Rb.



Tb

p(t)


1

t

3Tb

duobinarypulse





BT =(

1+r2

)Rb.



Tb

p(t)


1

t

3Tb

duobinarypulse





BT =(

1+r2

)Rb.



Tb

p(t)


1

t

3Tb

duobinarypulse



Signaling with Controlled ISI (with Polar Line Code)

Tb

p(t)

0 2Tb 3Tb 4Tb-Tb-2Tb-3Tb 5Tb-4Tbt

0 0




Tb

p(t)


1

t

0 1




Tb

p(t)


1

t

1 0




Tb

p(t)


1

t

1 1



Decision Rule for Duo-binary Pulses

1 If the sample value is + ve, the present bit is 1 and the previous bit isalso 1.

2 If the sample value is - ve, the present bit is 0 and the previous bit is also0.

3 If the sample value is 0, the present bit is compliment of the previous bit.





















By the Way, How do we do Pulse Shaping?

Nyquist Pulse is generated by driving a ’Nyquist filter’ with the line codes tobe conveyed.

Nyquist Filter is expected to shape the pulse such that at the receiver onedoes not encounter ISI at the sampling point.













Square Root Nyquist Pulse

r = 1r = 0.5r = 0

Though a single Square Root Nyquist pulse does not have nulls at thesampling points, 2 such filters in cascade (Tx & Rx) does satisfy Nyquist

Pulse properties




r = 1r = 0.5r = 0


Pulse properties




r = 1r = 0.5r = 0


Pulse properties




Since the path, between information source and destination, involvestransmitter & receiver (assuming channel to be ideal), the transmit & receive

filters together should have the Nyquist Filter property.

On the receiver side, the channel response, if it can be accurately estimated,can also be taken into account so that the overall response is Raised-cosine

filter.

This can be achieved by having 2 symmetric Square Root Nyquist Filters, oneat the Tx side and the other at the Rx side.







filter.








filter.








filter.




Outline


2 Line Coding

3 Pulse Shaping

4 Summary



Summary

• Sy (ω) =|P(ω)|2

Tb∑∞

n=−∞ Rne−jnωTb = |P(ω)|2Tb

(R0 + 2 ∑∞n=1 Rn cos nωTb)

• Rn = limN→∞1N ∑k akak+n = akak+n

• Polar Line Coding:

Rn =

{1, n = 00, n 6= 0

• ON-OFF Coding:

Rn =

{12 , n = 014 , n 6= 0

• AMI Coding:

Rn =

12 , n = 0− 1

4 , n = 10, |n| > 1

• r = ωxωb/2 ⇒ Transmission bandwidth = (1 + r) ωb

2


Documents

7: Baseband Transmission of Digital Signals · 7: Baseband Transmission of Digital Signals Y. Yoganandam, Runa Kumari, and S. R. Zinka Department of Electrical & Electronics Engineering