7: Baseband Transmission of Digital Signals · 7: Baseband Transmission of Digital Signals Y....

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

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

Pulse code (PCM word)

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

Pulse code (PCM word)

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

NRZ - L

NRZ - M

1 : + V0 : - V

1 : Level change0 : No level change

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

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

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

Transmittedsignal

Recieved distortedsignal (without noise)

Transmittedsignal

Recieved distortedsignal (with noise)

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

NRZ - L

NRZ - M

NRZ - S

1 : + V0 : - V

A Few Line Codes

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

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

Extracting Timing/Clock Information & Transparency

On-OffCoding

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

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

(k-1)Tb

(k+1)TbkTb

ak+1ak

(k-1)Tb

(k+1)TbkTb

ak+1ak

(k-1)Tb

(k+1)TbkTb

ak+1ak

What are thepossible values of ak ?

(k-1)Tb

(k+1)TbkTb

ak+1ak

(k-1)Tb

(k+1)TbkTb

ak+1ak

What are thepossible shapes of p(t)?

(k-1)Tb

(k+1)TbkTb

(k-1)Tb

(k+1)TbkTb

ak+1ak

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

(k-1)Tb

(k+1)TbkTb

(k-1)Tb

(k+1)TbkTb

ak+1ak

(k-1)Tb

(k+1)TbkTb

(k-1)Tb

(k+1)TbkTb

ak+1ak

Sy(ω)=|P(ω)|2Sx(ω)

(k-1)Tb

(k+1)TbkTb

(k-1)Tb

(k+1)TbkTb

ak+1ak

Sy(ω)=|P(ω)|2Sx(ω)

Autocorrelation function of an Impulse Train

(k-1)Tb

(k+1)TbkTb

ak+1ak

(k-1)Tb

(k+1)TbkTb

ak+1ak

(k-1)Tb

(k+1)TbkTb

ak+1ak

(k-1)Tb

(k+1)TbkTb

hk+1hk

(k-1)Tb

(k+1)TbkTb

ak+1ak

(k-1)Tb

(k+1)TbkTb

hk+1hk

(k-1)Tb

(k+1)TbkTb

ak+1ak

(k-1)Tb

(k+1)TbkTb

hk+1hk

R0/ϵTb

-ϵ +ϵ

(k-1)Tb

(k+1)TbkTb

ak+1ak

(k-1)Tb

(k+1)TbkTb

hk+1hk

t(k+1)TbkTb

hk+1hk

R0/ϵTb

-ϵ +ϵ

R1/ϵTb

(k-1)Tb

(k+1)TbkTb

ak+1ak

(k-1)Tb

(k+1)TbkTb

hk+1hk

t(k+1)TbkTb

hk+1hk

R0/ϵTb

-ϵ +ϵ

R1/ϵTb

R2/ϵTb

R3/ϵTb

3Tb-Tb

R1/ϵTb

R2/ϵTb

R3/ϵTb

(k-1)Tb

(k+1)TbkTb

ak+1ak

(k-1)Tb

(k+1)TbkTb

hk+1hk

t(k+1)TbkTb

hk+1hk

R0/TbR1/Tb

3Tb-Tb

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→∞

ˆ T/2

−T/2x̂ (t) x̂ (t− τ) dt.

When τ < ε,

Rx̂ = limT→∞

1T ∑

k (ε− τ)

= limT→∞

1T ∑

(ε− τ

R0εTb

(1− τ

R0εTb

(1− |τ|

), (1)

whereR0 = lim

T→∞

TbT ∑

k = limN→∞

1N ∑

k = a2k . (2)

ˆ T/2

−T/2x̂ (t) x̂ (t− τ) dt.

When τ < ε,

Rx̂ = limT→∞

1T ∑

k (ε− τ)

= limT→∞

1T ∑

(ε− τ

R0εTb

(1− τ

R0εTb

(1− |τ|

), (1)

whereR0 = lim

T→∞

TbT ∑

k = limN→∞

1N ∑

k = a2k . (2)

ˆ T/2

−T/2x̂ (t) x̂ (t− τ) dt.

When τ < ε,

Rx̂ = limT→∞

1T ∑

k (ε− τ)

= limT→∞

1T ∑

(ε− τ

R0εTb

(1− τ

=R0εTb

(1− |τ|

), (1)

whereR0 = lim

T→∞

TbT ∑

k = limN→∞

1N ∑

k = a2k . (2)

ˆ T/2

−T/2x̂ (t) x̂ (t− τ) dt.

When τ < ε,

Rx̂ = limT→∞

1T ∑

k (ε− τ)

= limT→∞

1T ∑

(ε− τ

R0εTb

(1− τ

R0εTb

(1− |τ|

), (1)

whereR0 = lim

T→∞

TbT ∑

k = limN→∞

1N ∑

k = a2k . (2)

ˆ T/2

−T/2x̂ (t) x̂ (t− τ) dt.

When τ < ε,

Rx̂ = limT→∞

1T ∑

k (ε− τ)

= limT→∞

1T ∑

(ε− τ

R0εTb

(1− τ

R0εTb

(1− |τ|

), (1)

whereR0 = lim

T→∞

TbT ∑

= limN→∞

1N ∑

k = a2k . (2)

ˆ T/2

−T/2x̂ (t) x̂ (t− τ) dt.

When τ < ε,

Rx̂ = limT→∞

1T ∑

k (ε− τ)

= limT→∞

1T ∑

(ε− τ

R0εTb

(1− τ

R0εTb

(1− |τ|

), (1)

whereR0 = lim

T→∞

TbT ∑

k = limN→∞

1N ∑

= a2k . (2)

ˆ T/2

−T/2x̂ (t) x̂ (t− τ) dt.

When τ < ε,

Rx̂ = limT→∞

1T ∑

k (ε− τ)

= limT→∞

1T ∑

(ε− τ

R0εTb

(1− τ

R0εTb

(1− |τ|

), (1)

whereR0 = lim

T→∞

TbT ∑

k = limN→∞

1N ∑

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

∑n=−∞

Rnδ (τ − nTb) . (4)

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

Sx (ω) =1

∑n=−∞

Rne−jnωTb =1

(R0 + 2

∑n=1

Rn cos nωTb

). (5)

Rn = limT→∞

TbT ∑

kakak+n = lim

N→∞

1N ∑

Rx̂ (τ) =1

∑n=−∞

Rnδ (τ − nTb) . (4)

Sx (ω) =1

∑n=−∞

Rne−jnωTb =1

(R0 + 2

∑n=1

Rn cos nωTb

). (5)

Rn = limT→∞

TbT ∑

kakak+n = lim

N→∞

1N ∑

Rx̂ (τ) =1

∑n=−∞

Rnδ (τ − nTb) . (4)

Sx (ω) =1

∑n=−∞

Rne−jnωTb =1

(R0 + 2

∑n=1

Rn cos nωTb

). (5)

Rn = limT→∞

TbT ∑

kakak+n = lim

N→∞

1N ∑

Rx̂ (τ) =1

∑n=−∞

Rnδ (τ − nTb) . (4)

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

and R−n = Rn,

Sx (ω) =1

∑n=−∞

Rne−jnωTb =1

(R0 + 2

∑n=1

Rn cos nωTb

). (5)

Rn = limT→∞

TbT ∑

kakak+n = lim

N→∞

1N ∑

Rx̂ (τ) =1

∑n=−∞

Rnδ (τ − nTb) . (4)

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

and R−n = Rn,

Sx (ω) =1

∑n=−∞

Rne−jnωTb

(R0 + 2

∑n=1

Rn cos nωTb

). (5)

Rn = limT→∞

TbT ∑

kakak+n = lim

N→∞

1N ∑

Rx̂ (τ) =1

∑n=−∞

Rnδ (τ − nTb) . (4)

Sx (ω) =1

∑n=−∞

Rne−jnωTb =1

(R0 + 2

∑n=1

Rn cos nωTb

). (5)

So, PSD of any line code is given by

Sy (ω) =|P (ω)|2

∑n=−∞

Rne−jnωTb =|P (ω)|2

(R0 + 2

∑n=1

Rn cos nωTb

(6)where

Rn = limN→∞

1N ∑

PSD of Polar Line Coding

1 1 1 1 1 100000+V

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

R0 = limN→∞

1N ∑

k = limN→∞

(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 ∑

[N2(1) +

N2(−1)

]= 0. (9)

Sy (ω) =|P (ω)|2

(R0 + 2

∑n=1

Rn cos nωTb

)=|P (ω)|2

Tb. (10)

1 1 1 1 1 100000+V

R0 = limN→∞

1N ∑

k = limN→∞

(N) = 1. (8)

Rn = limN→∞

1N ∑

kakak+n = lim

N→∞

1N ∑

[N2(1) +

N2(−1)

]= 0. (9)

Sy (ω) =|P (ω)|2

(R0 + 2

∑n=1

Rn cos nωTb

)=|P (ω)|2

Tb. (10)

1 1 1 1 1 100000+V

R0 = limN→∞

1N ∑

= limN→∞

(N) = 1. (8)

Rn = limN→∞

1N ∑

kakak+n = lim

N→∞

1N ∑

[N2(1) +

N2(−1)

]= 0. (9)

Sy (ω) =|P (ω)|2

(R0 + 2

∑n=1

Rn cos nωTb

)=|P (ω)|2

Tb. (10)

1 1 1 1 1 100000+V

R0 = limN→∞

1N ∑

k = limN→∞

(N) = 1. (8)

Rn = limN→∞

1N ∑

kakak+n = lim

N→∞

1N ∑

[N2(1) +

N2(−1)

]= 0. (9)

Sy (ω) =|P (ω)|2

(R0 + 2

∑n=1

Rn cos nωTb

)=|P (ω)|2

Tb. (10)

1 1 1 1 1 100000+V

R0 = limN→∞

1N ∑

k = limN→∞

(N) = 1. (8)

Rn = limN→∞

1N ∑

kakak+n

= limN→∞

1N ∑

[N2(1) +

N2(−1)

]= 0. (9)

Sy (ω) =|P (ω)|2

(R0 + 2

∑n=1

Rn cos nωTb

)=|P (ω)|2

Tb. (10)

1 1 1 1 1 100000+V

R0 = limN→∞

1N ∑

k = limN→∞

(N) = 1. (8)

Rn = limN→∞

1N ∑

kakak+n = lim

N→∞

1N ∑

[N2(1) +

N2(−1)

]= 0. (9)

Sy (ω) =|P (ω)|2

(R0 + 2

∑n=1

Rn cos nωTb

)=|P (ω)|2

Tb. (10)

1 1 1 1 1 100000+V

R0 = limN→∞

1N ∑

k = limN→∞

(N) = 1. (8)

Rn = limN→∞

1N ∑

kakak+n = lim

N→∞

1N ∑

[N2(1) +

N2(−1)

]= 0. (9)

Sy (ω) =|P (ω)|2

(R0 + 2

∑n=1

Rn cos nωTb

=|P (ω)|2

Tb. (10)

1 1 1 1 1 100000+V

R0 = limN→∞

1N ∑

k = limN→∞

(N) = 1. (8)

Rn = limN→∞

1N ∑

kakak+n = lim

N→∞

1N ∑

[N2(1) +

N2(−1)

]= 0. (9)

Sy (ω) =|P (ω)|2

(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

p (t) = rect(

), (11)

the corresponding Fourier transform is given by

P (ω) =Tb2

). (12)

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

Sy (ω) =|P (ω)|2

sinc2(

). (13)

1 1 1 1 1 100000 +V

p (t) = rect(

), (11)

P (ω) =Tb2

). (12)

Sy (ω) =|P (ω)|2

sinc2(

). (13)

1 1 1 1 1 100000 +V

p (t) = rect(

), (11)

P (ω) =Tb2

). (12)

Sy (ω) =|P (ω)|2

sinc2(

). (13)

1 1 1 1 1 100000 +V

p (t) = rect(

), (11)

P (ω) =Tb2

). (12)

Sy (ω) =|P (ω)|2

sinc2(

). (13)

1 1 1 1 1 100000 +V

p (t) = rect(

), (11)

P (ω) =Tb2

). (12)

Sy (ω) =|P (ω)|2

sinc2(

). (13)

Properties of Polar Line Coding

Sy(ω)

PSD of ON-OFF Line Coding

1 1 1 1 1 100000+V

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

R0 = limN→∞

1N ∑

k = limN→∞

[N2(1) +

N2(0)]=

. (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 ∑

[N4(1) +

, and (15)

Sy (ω) =|P (ω)|2

∑n=−∞

(ω− 2πn

)]. (prove yourself) (16)

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

R0 = limN→∞

1N ∑

k = limN→∞

[N2(1) +

N2(0)]=

. (14)

Rn = limN→∞

1N ∑

kakak+n = lim

N→∞

1N ∑

[N4(1) +

, and (15)

Sy (ω) =|P (ω)|2

∑n=−∞

(ω− 2πn

R0 = limN→∞

1N ∑

= limN→∞

[N2(1) +

N2(0)]=

. (14)

Rn = limN→∞

1N ∑

kakak+n = lim

N→∞

1N ∑

[N4(1) +

, and (15)

Sy (ω) =|P (ω)|2

∑n=−∞

(ω− 2πn

R0 = limN→∞

1N ∑

k = limN→∞

[N2(1) +

N2(0)]=

. (14)

Rn = limN→∞

1N ∑

kakak+n = lim

N→∞

1N ∑

[N4(1) +

, and (15)

Sy (ω) =|P (ω)|2

∑n=−∞

(ω− 2πn

R0 = limN→∞

1N ∑

k = limN→∞

[N2(1) +

N2(0)]=

. (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.

Rn = limN→∞

1N ∑

kakak+n = lim

N→∞

1N ∑

[N4(1) +

, and (15)

Sy (ω) =|P (ω)|2

∑n=−∞

(ω− 2πn

R0 = limN→∞

1N ∑

k = limN→∞

[N2(1) +

N2(0)]=

. (14)

Rn = limN→∞

1N ∑

kakak+n

= limN→∞

1N ∑

[N4(1) +

, and (15)

Sy (ω) =|P (ω)|2

∑n=−∞

(ω− 2πn

R0 = limN→∞

1N ∑

k = limN→∞

[N2(1) +

N2(0)]=

. (14)

Rn = limN→∞

1N ∑

kakak+n = lim

N→∞

1N ∑

[N4(1) +

, and (15)

Sy (ω) =|P (ω)|2

∑n=−∞

(ω− 2πn

PSD of ON-OFF Line Coding – Example

1 1 1 1 1 100000 +V

p (t) = rect(

), (17)

P (ω) =Tb2

). (18)

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

Sy (ω) =Tb4

sinc2(

∑n=−∞

(ω− 2πn

)]. (19)

1 1 1 1 1 100000 +V

p (t) = rect(

), (17)

P (ω) =Tb2

). (18)

Sy (ω) =Tb4

sinc2(

∑n=−∞

(ω− 2πn

)]. (19)

1 1 1 1 1 100000 +V

p (t) = rect(

), (17)

P (ω) =Tb2

). (18)

Sy (ω) =Tb4

sinc2(

∑n=−∞

(ω− 2πn

)]. (19)

1 1 1 1 1 100000 +V

p (t) = rect(

), (17)

P (ω) =Tb2

). (18)

Sy (ω) =Tb4

sinc2(

∑n=−∞

(ω− 2πn

)]. (19)

1 1 1 1 1 100000 +V

p (t) = rect(

), (17)

P (ω) =Tb2

). (18)

Sy (ω) =Tb4

sinc2(

∑n=−∞

(ω− 2πn

)]. (19)

Properties of ON-OFF Line Coding

Sy(ω)

Properties of ON-OFF Line Coding

Sy(ω)

PSD of AMI Line Coding

1 1 1 1 1 100000+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 ∑

k = limN→∞

[N2(±1)2 +

N2(0)]=

. (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 ∑

[N4(−1) +

(0)]= −1

4, and (21)

Sy (ω) =|P (ω)|2

Tbsin2

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

PSD of AMI Line Coding1 1 1 1 1 100000

R0 = limN→∞

1N ∑

k = limN→∞

[N2(±1)2 +

N2(0)]=

. (20)

R1 = limN→∞

1N ∑

kakak+1 = lim

N→∞

1N ∑

[N4(−1) +

(0)]= −1

4, and (21)

Sy (ω) =|P (ω)|2

Tbsin2

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

R0 = limN→∞

1N ∑

limN→∞

[N2(±1)2 +

N2(0)]=

. (20)

R1 = limN→∞

1N ∑

kakak+1 = lim

N→∞

1N ∑

[N4(−1) +

(0)]= −1

4, and (21)

Sy (ω) =|P (ω)|2

Tbsin2

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

R0 = limN→∞

1N ∑

k = limN→∞

[N2(±1)2 +

N2(0)]=

. (20)

R1 = limN→∞

1N ∑

kakak+1 = lim

N→∞

1N ∑

[N4(−1) +

(0)]= −1

4, and (21)

Sy (ω) =|P (ω)|2

Tbsin2

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

R0 = limN→∞

1N ∑

k = limN→∞

[N2(±1)2 +

N2(0)]=

. (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.

R1 = limN→∞

1N ∑

kakak+1 = lim

N→∞

1N ∑

[N4(−1) +

(0)]= −1

4, and (21)

Sy (ω) =|P (ω)|2

Tbsin2

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

R0 = limN→∞

1N ∑

k = limN→∞

[N2(±1)2 +

N2(0)]=

. (20)

R1 = limN→∞

1N ∑

kakak+1

= limN→∞

1N ∑

[N4(−1) +

(0)]= −1

4, and (21)

Sy (ω) =|P (ω)|2

Tbsin2

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

R0 = limN→∞

1N ∑

k = limN→∞

[N2(±1)2 +

N2(0)]=

. (20)

R1 = limN→∞

1N ∑

kakak+1 = lim

N→∞

1N ∑

[N4(−1) +

(0)]= −1

4, and (21)

Sy (ω) =|P (ω)|2

Tbsin2

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

R0 = limN→∞

1N ∑

k = limN→∞

[N2(±1)2 +

N2(0)]=

. (20)

R1 = limN→∞

1N ∑

kakak+1 = lim

N→∞

1N ∑

[N4(−1) +

(0)]= −1

4, and (21)

Sy (ω) =|P (ω)|2

Tbsin2

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

PSD of AMI Line Coding – Example

1 1 1 1 1 100000+V

p (t) = rect(

), (23)

P (ω) =Tb2

). (24)

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

Sy (ω) =Tb4

sinc2(

). (25)

1 1 1 1 1 100000+V

p (t) = rect(

), (23)

P (ω) =Tb2

). (24)

Sy (ω) =Tb4

sinc2(

). (25)

1 1 1 1 1 100000+V

p (t) = rect(

), (23)

P (ω) =Tb2

). (24)

Sy (ω) =Tb4

sinc2(

). (25)

1 1 1 1 1 100000+V

p (t) = rect(

), (23)

P (ω) =Tb2

). (24)

Sy (ω) =Tb4

sinc2(

). (25)

1 1 1 1 1 100000+V

p (t) = rect(

), (23)

P (ω) =Tb2

). (24)

Sy (ω) =Tb4

sinc2(

). (25)

Properties of AMI Line Coding

Sy(ω)

Manchester(Split phase)

Self Study

1 High density bipolar signaling (HDB)

2 Binary with N-zero substitution signaling (BNZS)

Self Study

Outline

2 Line Coding

3 Pulse Shaping

4 Summary

Why Pulse Shaping?

Sy (ω) = |P (ω)|2 Sx (ω) =|P (ω)|2

∑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

∑n=−∞

Rne−jnωTb

Why Pulse Shaping?

Sy (ω) = |P (ω)|2 Sx (ω) =|P (ω)|2

∑n=−∞

Rne−jnωTb

Why Pulse Shaping?

For polar line code, we have seen that

Sy(ω)

Why Pulse Shaping?

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

Sy(ω)

Why Pulse Shaping?

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

Sy(ω)

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?

Nyquist Criterion for Zero ISI

0 Tb 2Tb

0 Rb/2f

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

0 Tb 2Tb

If pulse rate changesat the transmittter

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

∑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

∑n=−∞

P (ω− nωb) .

p̄ (t) =∞

∑n=−∞

Hence,

∑n=−∞

P (ω− nωb) = Tb. (26)

Nyquist Pulse

p̄ (t) =∞

∑n=−∞

P̄ (ω) =1

∑n=−∞

P (ω− nωb) .

p̄ (t) =∞

∑n=−∞

Hence,

∑n=−∞

P (ω− nωb) = Tb. (26)

Nyquist Pulse

p̄ (t) =∞

∑n=−∞

P̄ (ω) =1

∑n=−∞

P (ω− nωb) .

p̄ (t) =∞

∑n=−∞

Hence,

∑n=−∞

P (ω− nωb) = Tb. (26)

Nyquist Pulse

|P(ω)|

-ωb/2 ωb/2ω

Nyquist Pulse

|P(ω)|

Nyquist Pulse

|P(ω)|

Nyquist Pulse

|P(ω)|

ωbωb/2ω

Nyquist Pulse

|P(ω)|

ωbωb/2ω

P(ω) + P(ω - ωb) = Tb

0 < ω < ωb

Nyquist Pulse

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

x +ωb2

(x− ωb

)= Tb |x| < ωb

2. (27)

The above equation can be rewritten as

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

x +ωb2

(x− ωb

)= Tb |x| < ωb

2. (27)

2+ x)+ P∗

(ωb2− x)= Tb |x| < ωb

2. (28)

2+ x)∣∣∣+ ∣∣∣P (ωb

2− x)∣∣∣ = Tb |x| < ωb

2. (29)

Nyquist Pulse

x +ωb2

(x− ωb

)= Tb |x| < ωb

2. (27)

2+ x)+ P∗

(ωb2− x)= Tb |x| < ωb

2. (28)

2+ x)∣∣∣+ ∣∣∣P (ωb

2− x)∣∣∣ = Tb |x| < ωb

2. (29)

Nyquist Pulse

x +ωb2

(x− ωb

)= Tb |x| < ωb

2. (27)

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)|

ωx ωx

Vestigialspectrum

Roll-off factor r is defined as

r =excess bandwidth

theoretical minimum bandwidth=

ωb/2=

ωb. (30)

Nyquist Pulse

|P(ω)|

ωx ωx

Vestigialspectrum

r =excess bandwidth

ωb/2=

ωb. (30)

Nyquist Pulse

|P(ω)|

ωx ωx

Vestigialspectrum

r =excess bandwidth

ωb/2=

ωb. (30)

Nyquist Pulse – Raised Cosine Pulse

|P(ω)|

ωx ωx

Vestigialspectrum

P (ω) =

{1− sin

[π(ω− ωb

2 )2ωx

]} ∣∣ω− ωb2

∣∣ < ωx

0 |ω| > ωb2 + ωx

1 |ω| < ωb2 −ωx

|P(ω)|

ωx ωx

Vestigialspectrum

P (ω) =

{1− sin

[π(ω− ωb

2 )2ωx

]} ∣∣ω− ωb2

∣∣ < ωx

0 |ω| > ωb2 + ωx

1 |ω| < ωb2 −ωx

|P(ω)|

ωx ωx

Vestigialspectrum

P (ω) =

{1− sin

[π(ω− ωb

2 )2ωx

]} ∣∣ω− ωb2

∣∣ < ωx

0 |ω| > ωb2 + ωx

1 |ω| < ωb2 −ωx

ωbωb/2ω

Tb 2Tb-Tb-2Tb

|P(ω)| p(t)

ωbωb/2ω

Tb 2Tb-Tb-2Tb

|P(ω)| p(t)

ωbωb/2ω

Tb 2Tb-Tb-2Tb

|P(ω)| p(t)

ωbωb/2ω

Tb 2Tb-Tb-2Tb

|P(ω)| p(t)

ωbωb/2ω

Tb 2Tb-Tb-2Tb

|P(ω)| p(t)

Nyquist pulse for r = 1:

P (ω) = cos2(

p (t) = Rbcos (πRbt)1− 4R2

bt2sinc (πRbt) (32)

Nyquist pulse for r = 1:

P (ω) = cos2(

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

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.

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

duobinarypulse

Signaling with Controlled ISI (with Polar Line Code)

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

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.

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:

{1, n = 00, n 6= 0

• ON-OFF Coding:

{12 , n = 014 , n 6= 0

• AMI Coding:

12 , n = 0− 1

4 , n = 10, |n| > 1

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

7: Baseband Transmission of Digital Signals · 7: Baseband Transmission of Digital Signals Y....

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