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Isi and nyquist criterion
1. Pulse Shaping
Sy(w)=|P(w)|^2Sx(w)
Last class:
– Sx(w) is improved by the different line codes.
– p(t) is assumed to be square
How about improving p(t) and P(w)
– Reduce the bandwidth
– Reduce interferences to other bands
– Remove Inter-symbol-interference (ISI)
EE 541/451 Fall 2006
2. ISI Example
sequence sent 1 0 1
sequence received 1 1(!) 1
Signal received
Threshold
0 t
-3T -2T -T 0 T 2T 3T 4T 5T
Sequence of three pulses (1, 0, 1)
sent at a rate 1/T
EE 541/451 Fall 2006
3. Baseband binary data transmission system.
ISI arises when the channel is dispersive
Frequency limited -> time unlimited -> ISI
Time limited -> bandwidth unlimited -> bandpass channel ->
time unlimited -> ISI
p(t)
EE 541/451 Fall 2006
4. ISI
First term : contribution of the i-th transmitted bit.
Second term : ISI – residual effect of all other transmitted bits.
We wish to design transmit and receiver filters to minimize the
ISI.
When the signal-to-noise ratio is high, as is the case in a
telephone system, the operation of the system is largely limited
by ISI rather than noise.
EE 541/451 Fall 2006
5. ISI
Nyquist three criteria
– Pulse amplitudes can be
detected correctly despite pulse
spreading or overlapping, if
there is no ISI at the decision-
making instants
x 1: At sampling points, no
ISI
x 2: At threshold, no ISI
x 3: Areas within symbol
period is zero, then no ISI
– At least 14 points in the finals
x 4 point for questions
x 10 point like the homework
EE 541/451 Fall 2006
6. 1st Nyquist Criterion: Time domain
p(t): impulse response of a transmission system (infinite length)
p(t)
1
shaping function
0 no ISI !
t
1
=T
2 fN t0 2t0
Equally spaced zeros,
-1 1
interval =T
2 fn
EE 541/451 Fall 2006
7. 1st Nyquist Criterion: Time domain
Suppose 1/T is the sample rate
The necessary and sufficient condition for p(t) to satisfy
1, ( n = 0 )
p( nT ) =
0, ( n ≠ 0 )
Is that its Fourier transform P(f) satisfy
∞
∑ P( f + m T ) = T
m = −∞
EE 541/451 Fall 2006
8. 1st Nyquist Criterion: Frequency domain
∞
∑ P( f + m T ) = T
m = −∞
f
0 fa = 2 f N 4 fN
(limited bandwidth)
EE 541/451 Fall 2006
9. Proof
∞
Fourier Transform p( t ) = ∫ P ( f ) exp( j 2πft ) df
−∞
∞
At t=T p( nT ) = ∫ P( f ) exp( j 2πfnT ) df
−∞
∞ ( 2 m +1)
p( nT ) = ∑ ∫( P( f ) exp( j 2πfnT ) df
2T
2 m −1) 2T
m = −∞
∞
∑∫ P( f + m T ) exp( j 2πfnT ) df
1 2T
=
−1 2T
m = −∞
∞
∑ P( f + m T ) exp( j 2πfnT ) df
1 2T
=∫
−1 2T
m = −∞
∞
=∫
1 2T
B( f ) exp( j 2πfnT ) df B( f ) = ∑ P( f + m T )
−1 2T m = −∞
EE 541/451 Fall 2006
10. Proof
∞ ∞
B( f ) = ∑ P( f + m T ) B( f ) = ∑ b exp( j 2πnfT )
n
m = −∞ n = −∞
B ( f ) exp( − j 2πnfT )
1 2T
bn = T ∫
−1 2T
bn = Tp ( − nT ) T ( n = 0)
bn =
0 ( n ≠ 0)
∞
B( f ) = T ∑ P( f + m T ) = T
m = −∞
EE 541/451 Fall 2006
11. Sample rate vs. bandwidth
W is the bandwidth of P(f)
When 1/T > 2W, no function to satisfy Nyquist condition.
P(f)
EE 541/451 Fall 2006
12. Sample rate vs. bandwidth
When 1/T = 2W, rectangular function satisfy Nyquist
condition
sin πt T πt T , ( f < W )
p( t ) = = sinc P( f ) = ,
πt T 0, ( otherwise )
1
0.8
0.6
0.4
Spectra
0.2
0
-0.2
-0.4
0 1 2 3 4 5 6
S bcarrier N m k
u u ber
EE 541/451 Fall 2006
13. Sample rate vs. bandwidth
When 1/T < 2W, numbers of choices to satisfy Nyquist
condition
A typical one is the raised cosine function
EE 541/451 Fall 2006
14. Cosine rolloff/Raised cosine filter
Slightly notation different from the book. But it is the same
sin(π T ) cos(rπ T )
t t
prc 0 (t ) = ⋅
π Tt
1 − (2 r T ) 2
t
r : rolloff factor 0 ≤ r ≤1
1 f ≤ (1 − r ) 21T
Prc 0 ( j 2πf ) = 1
2
[1 + cos( π
2r ( πTf + r − 1)) ] if 1
2T (1 − r ) ≤ f ≤ 1
2T (1 + r )
0 f ≥ 1
2T (1 + r )
EE 541/451 Fall 2006
15. Raised cosine shaping
Tradeoff: higher r, higher bandwidth, but smoother in time.
P(ω)
π W r=0
r = 0.25
r = 0.50
r = 0.75
r = 1.00
p(t)
W 2w ω
π π
− +
W W
0
0 t
EE 541/451 Fall 2006
16. Cosine rolloff filter: Bandwidth efficiency
Vestigial spectrum
Example 7.1
data rate 1/ T 2 bit/s
β rc = = =
bandwidth (1 + r ) / 2T 1 + r Hz
bit/s 2 bit/s
1 ≤ < 2
Hz (1 + r ) Hz
2nd Nyquist (r=1) r=0
EE 541/451 Fall 2006
17. 2nd Nyquist Criterion
Values at the pulse edge are distortionless
p(t) =0.5, when t= -T/2 or T/2; p(t)=0, when t=(2k-1)T/2, k≠0,1
-1/T ≤ f ≤ 1/T
∞
Pr ( f ) = Re[ ∑( −1) n P ( f + n / T )] = T cos( fT / 2)
n =−∞
∞
PI ( f ) = Im[ ∑( −1) n P ( f + n / T )] = 0
n =−∞
EE 541/451 Fall 2006
19. 3rd Nyquist Criterion
Within each symbol period, the integration of signal (area) is
proportional to the integration of the transmit signal (area)
( wt ) / 2 π
sin( wT / 2) , w ≤ T
P ( w) =
0, π
w >
T
π /T
1 ( wt / 2)
p (t ) = ∫/ T sin( wT / 2) e dw
jwt
2π −π
2 n +1T
1, n=0
A = ∫2 n−1 p(t )dt =
2
2
T
0, n≠0
EE 541/451 Fall 2006
21. Example
Duobinary Pulse
– p(nTb)=1, n=0,1
– p(nTb)=1, otherwise
Interpretation of received signal
– 2: 11
– -2: 00
– 0: 01 or 10 depends on the previous transmission
EE 541/451 Fall 2006
22. Duobinary signaling
Duobinary signaling (class I partial response)
EE 541/451 Fall 2006
23. Duobinary signal and Nyguist Criteria
Nyguist second criteria: but twice the bandwidth
EE 541/451 Fall 2006
24. Differential Coding
The response of a pulse is spread over more than one signaling
interval.
The response is partial in any signaling interval.
Detection :
– Major drawback : error propagation.
To avoid error propagation, need deferential coding (precoding).
EE 541/451 Fall 2006
25. Modified duobinary signaling
Modified duobinary signaling
– In duobinary signaling, H(f) is nonzero at the origin.
– We can correct this deficiency by using the class IV partial
response.
EE 541/451 Fall 2006
