about qualification of rajya sabha memeber

1. Dr. AMBEDKAR INSTITUTE OF TECHNOLOGY
(An Autonomous Institution, Affiliated to Visveswaraya Technological University, Belagavi)
Near Jnana Bharathi Campus, Mallathahalli, Bengaluru-560056
DEPARTMENT OF ELECTRONICS AND COMMUNICATION ENGINEERING
Group Activity Report
on
“INTERSYMBOL INTERFERENCE,NYQUIST’S CRITERION FOR
BASEBAND,IDEAL SOLUTION FOR ZERO (ISI)”
Submitted in partial fulfillment of the curriculum
DIGITAL COMMUNICATION-18EC53
In
ELECTRONICS AND COMMUNICATION ENGINEERING
For
BACHELOR’S IN ENGINEERING
Submitted By:
AKSHATHA B R 1DA20EC009
ANUSHA CHOWDARY D 1DA20EC016
ARADHANA B 1DA20EC017
CHINMAY D 1DA20EC032
DARSHAN D 1DA21EC403

2. Dr. AMBEDKAR INSTITUTE OF TECHNOLOGY
(An Autonomous Institution, Affiliated to Visvesvaraya Technological University, Belagavi)
Near Jnana Bharathi Campus, Mallathahalli, Bengaluru-560056
DEPARTMENT OF ELECTRONICS AND COMMUNICATION ENGINEERING
CERTIFICATE
Certified that the group activity entitled “INETRSYMBOL INTERFERENCE,NYQUIST”S CRITERRION FOR
BASEBAND,IDEAL SOLUTION FOR ZERO(ISI)” carried out by Ms. ,AKSHATHA B R bearing USN
1DA20EC009, Ms.,ANUSHA CHOWDARY D bearing USN 1DA20EC016, Ms.,ARADHANA B
bearing USN 1DA20EC017, Mr. CHINMAY D, bearing USN 1DA20EC032, Mr.DARSHAN bearing
USN 1DA21EC403 a bonafide student of Dr. Ambedkar Institute of Technology, Bengaluru-560056 in
partial fulfillment of curriculum in “DIGITAL COMMUNICATION” in Bachelor of Engineering in
Electronics and Communication Engineering of Dr. Ambedkar Institute of Technology during the year
2022-23.
The intergroup Activity report has been approved as it satisfies the academic requirements in respect
of the subject prescribed for the said subject.
Signature of Co-ordinator
SPOORTHI P A
Assistant professor

3. CONTENTS :
1). INTERSYMBOL INTERFERNCE (ISI):
2).NYQUIST CRITERION FOR DISTORTIONLESS BASEBAND
(FOR BINARY TRANSMISSION):
3).IDEAL SOLUTION OR NYQUIST SDOLUTION FOR ZERO
ISI:

4. 1).Intersymbol Interfernce (ISI):
Intersymbol Interferance:
 Inter symbol Interference is a form of a distortion of a signal, in
which one or more symbols interfere with subsequent signals,
causing noise or delivering a poor output.
Block diagram:
Fig: Block Diagram Of Intersymbol Interferance
Let us assume that the channel is free from noise. Let the input to the
transmitting filter be represented in time domain form as
x(t)= ∑ 𝐴𝑘𝑔(𝑡 − 𝑘𝑇𝑏)
∞
𝑘=−∞ --> (1)
where:
 g(t) is a rectangular pulse.
 A discrete random variable taking the value of +a for symbol 1
and -a for symbol 0.
 T is the time allocated for one bit and is known as the bit
duration.

5. Hence, the fourier transform of equation gives
X(f)=∑ 𝐴𝑘𝐺(𝑓)𝑒−𝑗2𝜋𝑓𝑘𝑇𝑏
∞
𝑘=−∞ --> (2)
Let us denote the output of the receiving filter by
Y(t)=∑ 𝜇𝐴𝑘𝑝(𝑡 − 𝑘𝑇𝑏)
∞
𝑘=−∞ --> (3)
Where:
p(t) is the pulse shaping function of y(t).
𝜇 is the scaling factor.
let the FT of y(t) is
Y(f)=∑ 𝜇𝐴𝑘𝑃(𝑓)𝑒−𝑗2𝜋𝑓𝑘𝑇𝑏
∞
𝑘=−∞ --> (4)
As a second step, the output of the receiving filter in frequency domain
is given by
Y(f)=X(f)Ht(f)Hc(f)Hr(f) -->(5)
Substituting equations 2 and 4 in equation 5, we get
∑ 𝜇𝐴𝑘𝑃(𝑓)𝑒−𝑗2𝜋𝑓𝑘𝑇𝑏
∞
𝑘=−∞ =∑ 𝐴𝑘𝐺(𝑓)𝑒−𝑗2𝜋𝑓𝑘𝑇𝑏
∞
𝑘=−∞ Ht(f)Hc(f)Hr(f)
𝜇P(f) = G(f)Ht(f)Hc(f)Hr(f) -->(6)

6. Taking inverse FT on both sides of equation 6 ,we can determine the
shape of p(t).
y(t) = ∑ 𝜇𝐴𝑘𝑝(𝑡 − 𝑘𝑇𝑏)
∞
𝑘=−∞
Let t=iTb,
Where
i=0, +1 or -1, +2 or -2
y(iTb)=∑ 𝜇𝐴𝑘𝑝(𝑖𝑇𝑏 − 𝑘𝑇𝑏)
∞
𝑘=−∞
=∑ 𝜇𝐴𝑘𝑝[(𝑖 − 𝑘)𝑇𝑏]
∞
𝑘=−∞
Therefore,
y(iTb)=𝜇Aip(0) + ∑ 𝜇𝐴𝑘𝑝[(𝑖 − 𝑘)𝑇𝑏]
∞
𝑘=−∞
𝑘≠𝑖
-->(7)
Since p(0)=1,
y(iTb)=𝜇Ai + ∑ 𝜇𝐴𝑘𝑝[(𝑖 − 𝑘)𝑇𝑏]
∞
𝑘=−∞
𝑘≠𝑖
--> (8)
Specification:
 The first term on the right-hand side of equation 8 represents the
contribution of ith transmitted symbol.
 The second term represents the unwanted contribution of all other
transmitted bits on the detection of ith transmitted bit.
 This unwanted contribution is called intersymbol interference.

7. 2).NYQUIST’SCRITERION FOR DISTORTIONLESS BASEBAND BINARY
TRANSMISSION (OR ZERO ISI):
 The pulse shaping function p(t) with Fourier transform given by
P(f),
∑ 𝑃(𝑓 − 𝑛𝑅𝑏)
∞
𝑘=−∞ =Tb
has,
p(iTb-Ktb)={
1, 𝑖 = 𝑘
0, 𝑖 ≠ 𝑘
This condition is known as Nyquist criterion for zero ISI.
Proof:
>>> Let us sample p(t) by using a Dirac comb with a period equal to
Tb. The process of ideal sampling may be defined mathematically by
P𝛿(t)=p(t)S𝛿(𝑡)
Applying FT on both sides,
P𝛿(f)=P𝛿(f)*S𝛿(f)
P𝛿(f)=P(f)*fs∑ 𝛿(𝑓 − 𝑛𝑓𝑠)
∞
𝑛=−∞
Let fs = (1/T) = Rb
then above equation

8. therefore,
P𝛿(f)=P(f)*Rb∑ 𝛿(𝑓 − 𝑛𝑅𝑏)
∞
𝑛=−∞
Applying the convolution property of an impulse function,
P𝛿(f)=Rb∑ 𝑃(𝑓 − 𝑛𝑅𝑏)
∞
𝑛=−∞ -->(1)
Using the defining equation of FT,
P𝛿(f)=Rb∫ 𝑝𝛿(𝑡)
∞
−∞
𝑒−𝑗2𝜋𝑓𝑡
dt
Using,
P𝛿(t)=p(t)S𝛿(𝑡)
=𝑝(𝑡) ∑ 𝛿(𝑡 − 𝑚𝑡𝑏)
∞
𝑚=−∞
then,
P𝛿(t)=∑ 𝑝(𝑚𝑇𝑏)𝛿(𝑡 − 𝑚𝑇𝑏)
∞
𝑚=−∞
Then above equation becomes,
P𝛿(f)=∫ ∑ [𝑝(𝑚𝑇𝑏)𝛿(𝑡 − 𝑚𝑇𝑏)]𝑒−𝑗2𝜋𝑓𝑡
∞
𝑚=−∞
∞
−∞
dt -->(2)
Let the integer, m = i-k,
Then,
i = k corresponds to m = 0, and similarly i ≠ k corresponds to m ≠ 0.
Using the condition:
p[(i-kTb)] = p[mTb]

9. then,
p[(i-kTb)] ={
1, 𝑖 = 𝑘
0, 𝑖 ≠ 𝑘
Equation 2 bcomes,
P𝛿(f)=∫ 𝑝(0)𝛿(𝑡)
∞
−∞
𝑒−𝑗2𝜋𝑓𝑡
dt
=p(0) 𝑒−𝑗2𝜋𝑓𝑡
|t=0 (using shifting property)
=p(0)
Since p(0) = 1, we get P𝛿(f)= 1,
 As a consequence of this, equation 1 gives P𝛿(f) = 1 only when
∑ 𝑃(𝑓 − 𝑛𝑅𝑏)
∞
𝑛=−∞ =Tb
Hence the proof,
 Because of the significance of this theorem in baseband
transmission, the above equation or equivalently
p(iTb-Ktb)={
1, 𝑖 = 𝑘
0, 𝑖 ≠ 𝑘
Finally,it constitutes that Nyquist's criterion for distortionless
baseband transmission (zero ISI).

10. 3).IDEAL SOLUTION OR NYQUIST SOLUTION
FOR ZERO ISI:
 The ISI can be minimized by controlling p(t) in time-domain and
P(f) in frequency domain.
One of the functions that gives zero ISI is
>> p(t) = sinc(2Bot) shown in fig 1
And spectrum of same signal is shown in fig 2
Where:
Bo = 1/2Tb is called Nyquist bandwidth.
“Nyquist bandwidth is defined as the minimum transmission bandwidth
for zero ISI”.
The FT of p(t) gives
P(f)={
1
2𝐵𝑜
, |𝑓| < 𝐵𝑜
0, |𝑓| > 𝐵𝑜
 1
 The above equation implies that frequencies of absolute value
greater than half the bit rate are not needed.

11. Equation 1:
 Suggests that P(f) is the frequency response of an ideal low pass
filterand p(t) = sinc(2Bot) is the impulse response of an ideal low
pass filter.
 Since p(t) is a sinc function, it goes through zero at integer
multiples of Tb.
Thus if,
Tb = 1/2Bo.
 Then,it is clear that p(t — kTb) = sinc[2Bo(t— kTb)] for integer
values of k will appear as shown in fig 3.
Also fig 3 implies that if y(t) is sampled at instants of time t= 0, +Tb
or –Tb , +2Tb or -2Tb….., will have zero ISI.
Fig 1: Impulse response of an ideal Low pass filter
Fig 2: Frequency response of an ideal low pass filter

12. Fig 3: Demonstration of Sampling Instants For Zero ISI
But,we know that

13.  The first term on the right–hand side of the above equation gives the
desired symbol.
 The second term represents the ISI caused by timing error ∆(𝑡) due to
inaccurate synchronisation of the clock in receiver sampling circuit.