Nyquist ISI Criterion
In communications, the Nyquist ISI criterion describes the conditions which, when
satisfied by a communication channel (including responses of transmit and receive
filters), result in no intersymbol interference or ISI. It provides a method for
constructing band-limited functions to overcome the effects of intersymbol
interference.
Nyquist Pulse Shaping Criterion
Case I: (Ideal Case)
When T<1/2W or1/T>2W , B(f) consist of non-overlapping replicas of H(f)
separated by 1/T as shown below
There is no choice for H(f) to ensure B(f) = T in this case and hence there is no way
that we can design a system with no ISI
Case II (Mathematical Case):
When T=1/2W or1/T=2W (the Nyquist rate), the replication of H(f) separated by
1/T as shown below.
Its clear that in this case there exists only one H(f) that results B(f)=T, namely
T , f <w
H(f)= 
0 ,otherwise
Which corresponds to the pulse
h(t)=
sin( t / T )
 t 
 sinc  
t /T
T 
This means that the smallest value of T for which transmission with zero ISI is
possible is T=1/2W, and for this value h(t) has to be sinc function.
ISI is minimized by controlling P(t) in time domain or P(f) to be rectangular function
in frequency domain.
Ideal Nyquist signaling - Notice zero ISI at sampling points
As we know, this is non-causal and physically non-realizable. To be physically
realizable a delayed version of it is used, i.e., sinc[(t - t0)/T] is used and t0 is chosen
such that for
t < 0, sinc[(t - t0)/T]  0. The sampling time should also be shifted to “nT + t0”
Another difficulty is that rate of convergence to zero is slow. The tails of h(t) decay
as 1/t, thus a small timing “jitter” at the receiver or mistiming error in sampling could
produce an infinite series of ISI components.
Another thing is that the amplitude characteristic of H(f) should be flat from -W to
W and zero elsewhere. This is physically unrealizable because of the abrupt
transitions at the band edges W.
Case III (Practical Case):
When T>1/2W or1/T>2W , B(f) consist of overlapping replicas of H(f) separated
by 1/T as shown below.
The practical difficulties encountered with the ideal Nyquist channel can be
overcome by reducing the signaling rate from the maximum value 2W to an
adjustable value between W and 2W. A particular pulse spectrum, for the T > 1/2W
case, that has desirable
First Nyquist Criterion for Zero ISI
In the first method, Nyquist achieves zero ISI by choosing a pulse shape that has a
non-zero amplitude at its center (say t=0) and zero amplitudes ±nTb (n=1, 2, 3…).
There is exists one (and only one) pulse that meets Nyquist 1st criterion and has a
bandwidth of Rb/2 Hz. This pulse is
We can write such a pulse as
p(t) 
sin(2 B0 t)
 sinc(2B0 t)
2 B0 t
The Fourier transform of this is
P(f)=
 f 
1
rect 

2B0
2B
 0
 1
 2B if f <B0
 0
P(f)= 
 0 if f >B
0


Using this pulse, we can transmit at rate of Rb pulses per seconds without ISI over
a bandwidth of Rb/2
1st Nyquist criterion pulse shape: a) Pulse in time domain b) Pulse in frequency
domain c) Stream of pluses in time domain
Limitations of first criterion pulse
1- Unfortunately, this pulse is impractical because it starts at 
2- It has undesirable feature that it decays too slowly at rate 1/t, this causes a
serious practical problem.
Second Nyquist Criterion
The solution is to find a pulse p(t) that satisfies the condition specified above but
decays faster than 1/t. The solution is to design raised cosine filter which has transfer
function consists of a flat portion and a roll off portion which is of sinusoidal form
and the bandwidth to be adjustable between B0 and 2B0
Where
B0 (Nyquist Bandwidth)=
bit rate R b

2
2
f1 Signal frequency
For that the Impulse response P(t) is given by
 is called the roll off factor (The ratio of the excess bandwidth to the theoretical
minimum bandwidth)
α=
Excess BW
Theoretical minimum BW
and given by
 1
f1
B0
The following figures illustrates the raised cosine spectral characteristics and the
corresponding pulses for  = 0, 0.5 and 1.
Since we are sampling below the Nyquist rate 2Rb, the shifted transforms opverlap.
The transmission bandwidth required can be obtained from the relation
B=2B0 -f1
Where
B =transmission bandwidth
B 0 = Nyquist bandwidth
But
 1
Which results
f1
B0
B=2B0 -B0 (1   )
B=B0 (1   )