digital communication vtu paper

Digital Communication 6th Sem B.E. June 2012 Solved

PART - A

1. a.

ANS-1.a.

Figure 1: Basic Digital Communication System

The Block Diagram Shown above consists of 3 main Blocks:

i) Transmitter

ii)Communication Channel

iii) Receiver

Digital Information source:

The source of information is assumed to be digital i.e. symbols, letters etc.

If I/p is analog signal, then it is converted into digital form by using sampler & quantizer.

The Sources of information are human voice, Television picture, Teletype data, atmospheric

temperature & pressure etc.

Source Encoder & Decoder:

Digital information coming out of source consists of lots of redundancy which when transmitted as it is

results in “Improper utilization of Bandwidth”. Hence results in poor efficiency. The objective of source

encoder is to eliminate or reduce redundancy.


Source Decoder:

Source Decoder at the receiver behaves exactly in a reverse way to source encoder. Decoder converts the

codes back to symbols i.e. converts digital information to discrete symbols.

Channel Encoder/Decoder:

Channel Encoder & decoder are used to reduce the channel Noise Effect.

Channel coding is the process of adding controlled redundancy to the data to be transmitted, to

detect and/or correct the errors caused by the channel noise at the receiver.

Addition of redundancy increases bit rate & hence increases bandwidth.

Decoder detects the error in the received data & corrects the error.

E.g.:- Error correcting codes like linear block codes, cyclic codes & convolution codes.

Modulator & Demodulator:

Modulator converts the bitstream into a waveform suitable for transmission over the

communication channels. E.g.:- ASK, FSK, PSK, QPSK etc.

Demodulator converts the waveform into digital data. {optimum detectors are used to minimise

the probability of error.}

Communication Channel:

It is the media through which signal can be transmitted.

E.g.:- Free Space, Twisted wire, Co-axial cable, Waveguide, OFC etc.

1. b.

ANS- 1.b. g (t )=10 cos20 πt .cos 200 πt 1

A=200, B=20

W.K.T CosA.CosB= ½[Cos (A-B) + Cos (A+B)]

g (t )=102 [cos (200−20 ) πt+cos (200+20 ) πt ]

g (t )=5 [ cos180 πt+cos220 πt ] g (t )=5 cos180 πt +5cos220 πt 2

From eqn2,

W1=180π W2=220π 2πf1= 180π 2πf2=220π f1= 90Hz f2=110Hz

fm=max(f1,f2)=f2=110Hz fm=110Hz

g ( t )=5 cos2π (90)t +5 cos2 π (110)t 3

Taking FT on both sides of eqn3,we get


G(f) = 52 [δ ( f −90 )+δ ( f +90 ) ]+ 5

2 [δ ( f −110)+δ (f +110 ) ]

G(f) = 2.5 [δ ( f −90 )+δ ( f +90 ) ]+2.5 [δ ( f −110 )+δ ( f +110 ) ] 4 The Spectrum of the signal g(t) is drawn using eqn4 as shown in figure below:

i) WKT

Gδ (f) = fs ∑n=−∞

∞

G ( f −nfs ) Given: fs=250Hz

Gδ (f) = 2.5 fs ∑n=−∞

∞

[¿δ ( f −250 n−90 )+δ( f −250 n+90)]+2.5 fs ∑n=−∞

∞

¿¿¿

ii)


The cut-off frequencies of the ideal LPF should be more than 110Hz & less than 140Hz for recovering g(t)

from gδ (t).

iii) Nyquist rate for g(t):

Fs=2fm = 2* 110Hz

Fs=220Hz

1. c.

ANS- 1.c. Flap-Top Sampling

As the name itself indicates after sampling, the pulses will have “Flat Top”. It is very easy to

obtain flat of samples compared to Natural samples.

The top of the samples remains constant &equal to instantaneous value of the base band signal g(t) at the

start of sampling.


It is observed from the figure that only starting edge of the pulse represents the instantaneous value of base

band signal g(t).

The Sampled signal gδ (t) is given by

gδ (t) = g(t). Sδ (t)

gδ (t) = g(t) ∑n=−∞

∞

δ (t−nTs )

gδ (t) = ∑n=−∞

∞

g (nTs )δ (t−nTs) 1

Convolving gδ (t) with the pulse h(t), we get

S (t) = gδ (t)*h (t) 2


S (t) = ∫−∞

∞

gδ (t ) . h ( t−τ ) d τ 3

Substituting eqn1 in eqn2, we get

S (t) = ∫−∞

∞

∑n=−∞

∞

g (nTs ) δ (τ−nTs ) h ( t−τ ) dτ 4

{From Shifting Property

∫−∞

∞

x ( t ) . δ (t−¿ )dt=x (¿)

∫−∞

∞

h ( t−τ ) δ (τ−nTs ) dτ=h ( t−nTs )

}

Applying Shifting property in eqn4, we get

S (t) = ∑n=−∞

∞

g (nT s ) h(t−nTs) 5

Eqn5 represents the value of S(t) in terms of sampled value g(nTs) & function h(t-nTs) for flat top sampled

signal.

WKT

S (t) = gδ (t) * h (t)

Taking F.T on both sides of above equation, we get

S (f) = Gδ (f) . H (f)

Substituting Gδ (f) in above equation, we get

S (f) = fs ∑n=−∞

∞

G ( f −nfs ) . H (f ) 6

Eqn6 represents the spectrum of flat Top Sampled Signal.

2. a.

ANS- 2. a.

Let the random variable ‘Q’ denotes the quantization error & ‘q’ its sample value.

Let us assume that the quantization error ‘Q’ is uniformly distributed over a single quantizer

interval ‘∆’


Hence, probability density function (PDF) of Quantization error ‘Q’ is then

The mean quantization error μ=0

The variance of Quantization error is

σ Q 2 = ∫

−∆ /2

∆ /2

(q−μ)2 .fQ (q). dq

=∫−∆

2

∆2

(q−μ )2 . 1∆

. dq


Eqn1 is known as “Mean Squared Quantization Error “or Normalized Noise Power or

Quantization error in terms of power.

Let us Consider ‘N’- bits to represent ‘L’ quantized level, then L=2N


Let ‘p’ denotes the average power of the message signal x(t), then the o/p SNR of a

uniform quantizer is

2. b.

ANS -2. b.

Soln:- Given: μ=100, (SNR)dB = 45dB, L=?

(SNR)dB = 45dB

10 log10 (SNR) =45 dB

(SNR) = log10−1 (

4510

¿

(SNR)o =31622.7766


2. c.

ANS -2. c.

An important feature of TDM is conservation of time i.e. different time intervals (periods) are allocated for

different message signals, so that a common channel is utilized for transmission of these signals without

any interference.


The concept of TDM is illustrated in the block diagram.

The Low Pass pre-alias filters are used to remove high frequency components which may be present in the

message signal.

The o/p of the pre-alias filters are then fed to a commutator, which is usually implemented using electronic

switching circuitry.

The function of Commutator is 2 fold:

1) To take a Narrow Sample of each of the ‘N’ I/p signals at a rate fs ≥ 2W, where ‘W’ is the cut-off

frequency of pre-alias filter.

2) To sequentially interleave these ‘N’ samples inside a sampling interval Ts= 1/fs .

This interleaving is nothing but multiplexing.

The multiplexed signal is applied to a pulse amplitude modulator whose purpose is to transform the

multiplexed signal into a form suitable for transmission over a common channel.

At the receiving end, the pulse amplitude demodulator performs the reverse operation of PAM & the

decommutator distributes the signals to the appropriate low pass reconstruction filters.

The decommutator operates in synchronization with the commutator.


Suppose that the ‘N’ message signals to be multiplexed (Txed) have the same spectral properties (BW). Then

the Sampling rate for each message signal is determined in accordance with the Sampling theorem.

Let ‘Ts’ denotes the Sampling period.

Let ‘Tx’ denote the time spacing between adjacent samples in the TDM signal.

i.e. Tx= Ts¿ N as shown in fig2.

Number of pulses per second = 1/ Tx = 1/(Ts/N) = N/ Ts

Number of pulses per second is also called as Signalling rate ‘r’

i.e. r=Nfs

Since fs ≥ 2fm

r ≥ N2fm

Transmission Bandwidth = Signalling rate

2

3. a.

ANS -3. a.

Delta Modulation transmits only one bit per sample i.e. the present sample value is compared with the

previous sample value & the indication, whether the amplitude is increased or decreased is sent.

The i/p signal x(t) is approximated to step signal by the delta modulator. The difference between I/p signal

x(t) & staircase approximated signal is quantized into only two levels i.e. +δ or –δ .

If the difference is +ve, then approximated signal is increased by one step i.e. +δ∧¿ bit 1 is transmitted.

If the difference is –ve, then approximated signal is reduced by one step i.e. –δ & bit 0 is transmitted.

Thus for each sample only one-bit is transmitted.

DM Transmitter:-


The error between the sampled value x(nTs) & last approximated sample is given by

e(nTs) = x(nTs) - x(nTs) 1 Let u(nTs) be the present sample approximation of Staircase o/p.

From above Fig: x(nTs) = u(n-1)Ts

x(nTs) = u(nTs-Ts) 2

Substituting eqn2 in eqn1, we get

e(nTs) = x(nTs)- u(nTs-Ts) 3

The binary quantity b(nTs) is the algebraic sign of the error e(nTs), except for the scaling factor δ .

i.e. b(nTs) = δsgn[e(nTs)] 4

b(nTs) depends on the sign of error e(nTs), the sign of step-size ‘δ ’ will be decided

i.e. b(nTs) = +δ , if x(nTs) ≥ x(nTs)

b(nTs) = -δ , if x(nTs) ≤ x(nTs)

If b(nTs) = +δ , then binary ‘1’ is transmitted, & if b(nTs) = -δ , then binary ‘0’ is transmitted.

Hence, u(nTs) =u[nTs-Ts] + b(nTs)

The previous Sample approximation u[nTs-Ts] is restored by delaying one sample period ‘Ts’.

DM Receiver:-

Fig above shows the block diagram of DM Receiver.


The accumulator generates the Staircase approximated signal O/p & is delayed by one Sampling

period ‘Ts’. It is then added to the I/p Signal.

If I/p is binary ‘1’ then it adds +δ step to the previous o/p.

If I/p is binary ‘0’ then one step ' δ ' is subtracted from the delayed Signal.

The LPF is used to remove Step variation & to get smooth reconstructed message signal x(t).

DM Systems are subjected to two types of Quantization error

1) Slope-Overload distortion

2) Granular Noise

1) Slope overload distortion:-

Slope overload distortion arises because of the large dynamic range of the I/p Signal.

In Fig above, it can be seen, the rate of rise of I/p signal x(t) is so high that the Staircase Signal cannot

approximate it, the Step Size ‘δ ’ becomes too small for Staircase Signal x(t) to follow the Step Segment of

x(t). Thus large error between the Staircase approximated Signal & the original I/p Signal x(t). This error is

called Slope overload distortion.

To reduce this error, the Step-Size should be increased when slope of the Signal x(t) is high.

i.e. Slope of the Staircase u(t) ≥ Slope of the message Signal.

Granular Noise:-


This noise occurs when the Step Size is too large compared to small variations in the I/p Signal i.e. for very

Small variations in the I/p Signal, the Staircase Signal is changed by large amount because of large Step

Size ‘δ ’.

In Fig above, I/p Signal is almost flat, the Staircase Signal u(t) keeps on oscillating by +δ around the

Signal.

The error between the I/p & approximated Signal is called Granular Noise. The Solution of this problem is

to make Step Size small.

3. b.

ANS -3. b. Polar NRZ Format:-


3. c.

ANS -3. c.

Sr.No.

Parameter PCM Differential Pulse Code

Modulation (DPCM)

1 Number of Bits It can use 4, 8 or 16 bits per sample. Bits can be more than one but

are less than PCM.

2 Levels, step size The number of levels depend on

number of bits. Level size is fixed

Fixed number of levels are

used.

3 Quantization error and

Distortion

Quantization error depends on number

of levels used.

Slope overload distortion and

quantization noise is present.

4 Bandwidth of transmission

channel

Highest bandwidth is required since

number of bits are high.

Bandwidth required is lower

than PCM.

5 Feedback There is no feedback in transmitter or

receiver.

Feedback exists.

6 System Complexity System is complex. Simple.

7 SNR Good. Fair.

8 Applications Audio & Video Telephony. Speech and Video.

4. a.

ANS -4. a.

Nyquist Pulse Shaping Criterion:-

In detection process received pulse stream is detected by sampling at intervals ±KT b, and then in detection

process we will get desired output. This demands sample of i th transmitted pulse in pulse stream at K th sampling

interval should be

------------- (1)If received pulse P(t) satisfy this condition in time domain, then

y(ti) = μai


Let us look at this condition by transform eqn (1) into frequency domain.Consider sequence of samples {P(nTb)} where n=0,±1. . . . . . . by sampling in time domain, we write in frequency domain

Pδ(f) = 1

T b ∑n=−∞

∞

p (f − nTb

) 2

Where Pδ(f) is Fourier transform of an infinite period sequence of delta functions of period T b but Pδ(f) can

be obtained from its weighted sampled P(nTb) in time domain.

Pδ(f) = ∫−∞

∞

∑m=−∞

∞

p(mTb ) δ (t-mTb ) e− j2 πft dt = p(t). δ (t)

Where m = i-k, then i=k, m=0; so

Pδ(f) = ∫−∞

∞

p (0 ) δ (t ) e− j 2 πft dt

Using property of delta function

i.e. ∫−∞

∞

δ(t) dt =1

Therefore Pδ(f) =P(0) =1

Hence, Pδ(f) =1 3

P(0) =1 ,i.e. pulse is normalized (total area in frequency domain is unity)

Comparing (3) and (2)

1 ∑n=−∞

∞

p¿¿/ Tb ) =1

Tb

OR

∑n=−∞

∞

p¿¿/ Tb ) = Tb =1 4

Rb

Where Rb = Bit Rate

Is desired condition for zero ISI and it is termed Nyquist’s first criterion for distortion less base band

transmission. It suggests the method for constructing band limited function to overcome effect of ISI.

4. b.

ANS -4. b.

EYE Pattern:-

The Eye Pattern is used to study the effect of ISI in a PCM or data transmission System.

Eye Pattern can be obtained by applying the received wave to the vertical deflection plates of an

oscilloscope & to apply a sawtooth wave to the horizontal deflection plates at a transmitted Symbol rate

R= 1/T.


The waveforms in Successive Symbol intervals are thereby translated into one interval on the oscilloscope

display as shown in fig 1- a & b.

The resulting display is called Eye Pattern because of its resemblance to the human eye for binary waves.

The interior region of the Eye Pattern is called the Eye Opening.

An Eye Pattern provides information about the performance of the System, as described in Fig.2.


1) The width of the Eye Opening defines the time interval over which the received wave can be sampled

without error from ISI.

2) The Sensitivity of the System to timing error is determined by the rate of closure of the Eye as the

Sampling time is varied.

3) The height of the Eye Opening, at a specified Sampling time defines the margin over Noise.

4) Any non linear transmission distortion would reveal itself in an asymmetric or squinted eye. When the

effected of ISI is excessive, traces from the upper portion of the eye pattern cross traces from lower portion

with the result that the eye is completely closed.

Example of eye pattern:

Binary-PAM Perfect channel (no noise and no ISI)


4. c.

ANS -4. c.

Adaptive Equalization:-

The transmission characteristics of the Channel keep on changing. To compensate this, adaptive

equalization is used.

In Adaptive Equalization, the filters adapt themselves to the dispersive effects of the channel. The co-

efficients of the filters are changed continuously according to the received data in such a way that the

distortion in the data is reduced.

There are two types of equalization:

1) Pre-Channel Equalization


2) Post-Channel Equalization.

Pre- Channel equalization is done at the transmitting side. It requires feedback to know the amount of

distortion in the received data.

In post-channel equalization, feedback is not required. The equalizer is placed after the receiving filter in

the receiver.

Fig Above shows the block diagram of an adaptive equalizing filter. It is adaptive in nature because it is

capable of adjusting its co-efficients w0, w1, ….. , wm-1 by operating on the channel o/p in accordance with

some algorithm.

The adaptive equalizing filter consists of delay elements & adjustable filter co-efficients (Taps).

The Sequence x(nT) is applied to the I/p of the adaptive filter . The O/p y(nT) of the adaptive filter will be:

y(nT) = ∑i=0

M

w i x(nT-iT)

A known sequence {d(nT)} is transmitted 1st . This Sequence is known to the receiver.

An error Sequence is calculated i.e.

e(nT) = d(nT) – y(nT)

If there is no distortion in the channel, then d(nT) & y(nT)will be exactly same producing zero error

Sequence.

If there is distortion in the Channel, then e(nT) exists. The weights of the filter i.e. w i are changed

recursively such that error e(nT) is minimized.

The algorithm used to change the weights of the adaptive filter is Least Mean Square Algorithm (LMS).

The tap weights are adapted by this algorithm as follows:

w i (nT + T) = w i (nT) + μe(nT) x(nT- iT)


Where,

w i (nT) is the present estimate for Tap ‘i’ at time nT

w i (nT + T) is the updated estimate for tap ‘i’ at time nT

μ is the adaptation constant

x(nT – iT) is the filter I/p &

e(nT) is the error Signal.

PART - B

5. a.

ANS -5. a.

Requirements of Passband Transmission Scheme:-

Any passband transmission scheme should satisfy following requirements:-

1. Maximum data transmission rate.

2. Minimum probability of symbol error.

3. Minimum transmitted power.

4. Minimum channel bandwidth.


5. Maximum resistance to interfering signals.

6. Minimum circuit complexity.

5. b.

ANS -5. b.

Power Spectral Density of a BPSK Signal:-

Step 1: Fourier transform of basic NRZ pulse.

We know that the waveform b (t) is NRZ bipolar waveform. In this waveform there are rectangular pulses of

amplitude ± Vb. If we say that each pulse is ± (Tb¿2) around its center as shown in Fig.1, then it becomes easy to

find Fourier Transform of such pulse.

Fig.1 NRZ pulse

The Fourier Transform of this type of pulse is given as,

X(f)= Vb Tb sin ¿¿ Tb) By Standard relations: 1 (πf Tb)

Step 2: PSD of NRZ pulse.

For large number of such positive and negative pulses the power spectral density S(f) is given as

S(f) = ¿ X ( f )∨¿¿2 2 Ts

Here X ( f )denotes average value of X(f) due to all the pulses in b (t). And Ts is symbol duration. Putting value of

X(f) from equation 1 in equation 2 we get,

Step 3: PSD of baseband signal b(t).

For BPSK since only one bit is transmitted at a time, symbol and bit durations are same i.e. Tb = T s. Then above

equation becomes,

The above equation gives the power spectral density of baseband signal b (t).

3


Step 4: PSD of BPSK signal.

The BPSK signal is generated by modulating a carrier by the baseband signal b (t). Because of modulation of the

carrier of frequency f0, the spectral components are translated from f to (f0 + f) and (f0 – f). The magnitude of those

components is divided by half.

Therefore from equation 3 we can write the power spectral density of BPSK signal as,

The above equation is composed of two half magnitude spectral components of same frequency 'f above and below

f0· Let us say that the value of ± Vb = ±√P. That is the NRZ signal is having amplitudes of + √ P and - √P. Then

above equation becomes,

4

The above equation gives power spectral density of BPSK signal for modulating signal b(t) having amplitudes of ±

√P. We know that modulated signal is given by s1(t) = √2P cos¿¿ f0t) and s2(t) = - √2P cos¿¿ f0t) as,

s(t) = ± √2 P cos¿¿ f0t) Since, A= √2PIf b(t) = ±√P, then carrier becomes,

∅ (t) = √2 cos¿¿ f0t) 5

Plot of PSD:-

Equation 3 gives power spectral density of the NRZ waveform. For one rectangular pulse, the shape of S (f) will be

a sinc pulse as given by equation 3. Fig. 2 shows the plot of magnitude of S (f). Below figure shows that the main

lobe ranges from -fb to + fb · Here fb = (1/Tb). Since we have taken ± Vb = ± √ P in equation 3, the peak value of the

main lobe is P Tb,

Fig. 2 Plot of power spectral density of NRZ baseband signal


Now let us consider the power spectral density of BPSK signal given by equation 4. Fig. 3 shows the plot of this

equation. The figure thus clearly shows that there are two lobes; one at f0 and other at -f0. The same spectrum of Fig.

2 is placed at + f0 and - f0· But the amplitudes of main lobes are (PTb/2) in Fig. 3.

Fig. 3 Plot of power spectral density of BPSK signal

Thus they are reduced to half. The spectrums of S (f) as well as SBPSK (f) extends over all the frequencies.

5. c.

ANS -5. c.

Advantages of MSK as Compared to QPSK:-

1. The MSK baseband waveforms are smoother compared to QPSK.

2. MSK signal have continuous phase in all the cases, whereas QPSK has abrupt phase shift of π2

orπ .

3. MSK waveform does not have amplitude variations, whereas QPSK signal have abrupt amplitude

variations.

4. The main lobe of MSK is wider than that of QPSK. Main lobe of MSK contains around 99% of signal

energy whereas QPSK main lobe contains around 90% signal energy.

5. Side lobes of MSK are smaller compared to that of QPSK. Hence interchannel interference is

significantly large in QPSK.

6. To avoid interchannel interference due to sidelobes, QPSK needs bandpass filtering, where as it is not

required in MSK.

7. Bandpass filtering changes the amplitude waveform of QPSK because of abrupt changes in phase. This

problem does not exist in MSK.

The distance between signal points is same in QPSK as well as MSK. Hence the probability of error is also same.

6. a.

ANS -6. a.

Geometric interpretation of signal:-

Using N orthonormal basis functions we can represent M signals as


Si (t) = ∑j=1

n

S ij ∅ j (t) 0 ≤ t ≤T i = 1, 2 ,…….., M 1

Coefficients are given by

Sij = ∫0

T

Si (t) ∅ j (t) dt i = 1, 2,….., M

j = 1, 2,…..., N 2

Given the set of coefficients {sij}, j= 1, 2, ….N operating as input we may use the scheme as shown in fig(a) to

generate the signal si(t) i = 1 to M. It consists of a bank of N multipliers, with each multiplier supplied with its own

basic function, followed by a summer.

Conversely given a set of signals si(t) i = 1 to M operating as input we may use the scheme shown in fig below to

calculate the set of coefficients {sij}, j= 1, 2, ….N.


The vector si is called signal vector

We may visualize signal vectors as a set of M points in an N dimensional Euclidean space, which is also called

signal space.

The squared-length of any vector si is given by inner product or the dot product of si.

Where sij are the elements of si.

Two vectors are orthogonal if their inner product is zero.

The energy of the signal is given by


Ei = ∫0

T

Si2(t) dt

Substituting the value si(t) from

We get

Interchanging the order of summation and integration

Since ∅ j (t) forms an orthonormal set, the above equation reduce to

Ei = ∑j=1

N

S ij2

This shows that the energy of the signal si(t) is equal to the squared-length of the signal vector si.

The Euclidean distance between the points represented by the signal vectors si and sk is

6. b.

ANS -6. b.


Let rearrange the above Figure as shown below.

We first observe that s1(t), s2(t) and s3(t) are not linearly independent because s3(t) = s2(t) + s1(t)

The energy of s1(t) is

E1 = ∫0

2

32 dt = 9 * 2= 18

The first basis function is therefore ∅ 1(t) = s1( t)√ E1

= 3

√18 =

33√2

. =1√2

. For 0 ≤ t ≤ 2

Define: s21 = ∫0

2

s2(t) ∅ 1(t) dt = ∫0

2

0 1√2

dt =0

The Energy of the signal s2(t) is

E2 = ∫2

4

32 dt = 9 * 2 = 18

The second basis function ∅ 2(t) is therefore

= 3−0

√18−0 =

33√2

= 1√2

For 2≤t ≤ 4

Coefficient s31: s31 = ∫0

2

s3(t) ∅ 1(t) dt = ∫0

2 3∗1√2

dt = 3√2

∗2 = 3√2


Coefficient s32: s32 = ∫2

4

s3(t) ∅ 2(t) dt = 3√2

Intermediate function

g3(t) = -3 for 0 ≤ t ≤ 4The third basis function

=

−3

√∫0

4

9dt = -3/6 =

−12

For 0 ≤ t ≤ 4

The corresponding orthonormal functions are shown in the figure below:

Representation of the signals

s1(t) = 3√2 ∅ 1(t)

s2(t) = 3√2 ∅ 2(t)

s3(t) = 3√2 ∅ 1(t) + 3√2 ∅ 2(t) + 6∅ 3(t)

The signal constellation diagram is shown below:


7. a.

ANS -7. a.

Without loss of Generality, assume that the signal s(t) =a0p(t) is normalized so that Ep = ∫−∞

∞

|p (μ)2| dμ = 1 and thus

Es = E[|ao|2]. If a0 is a binary random variable with values a0 ∈ {A, B}, then b0 is a Guassian random variable with

mean either A or B and variance σ b2 = N0/2, i.e.,

The decision rule for a maximum likelihood (ML) receiver upon receiving b0 = β is:

Decide ao = A iff , else set ao = B.

Substituting the conditional pdfs yields:

Decide ao = A iff (β−A)2<( β−B)2 or β< A+B2


Assuming B ¿ A, the probability of a decision error given a0 = A is

where the complementary error function erfc(x) is defined as

erfc (x) = 2√π∫z

∞

e−μ2

dμ → erfc (-x) = 1-erfc(x) = 2√π ∫

−∞

z

e− μ2

dμ.

The probability of a decision error given a0 = B is computed similarly as

The probability of a symbol error for an AWGN channel with matched filter receiver an ML decision rule and a 0 ∈

{A, B} is therefore

,

i.e., it depends only on the distance |B−A| between the possible values of a0 and the (2-sided) noise power

spectral density No /2. If antipodal signaling is used (e.g., for BPSK or QPSK modulation of a carrier) then B = -A

which implies

|B−A|=2|A|=2√Eb whereEb=A2 P (ao=A )+B2 P(ao=B),

That is Eb is the bit energy.

Uncoded antipodal (a0∈{− A ,+ A }) signaling. The probability of bit error on an AWGN channel with SNR

Eb/No and a matched filter receiver with ML decision rule is

,

Where Eb = A2 is the energy per bit.


Coded antipodal (a0∈{−A ,+ A }) signaling. The probability of error between two binary codewords Hamming

distance‘d’ apart on an AWGN channel with SNR Ec/No per code bit and a matched filter receiver with ML decision

rule is

,

Where Ec = A2 is the energy per coded bit.

7. b.

ANS -7. b.

For an AWGN channel and for the case when the transmitted signals are equally likely,

the optimum receiver consists of two subsystems as shown in figure below.

(a) Detector

1. The detector part of the receiver is as shown in fig (a). It consists of a bank of M product-integrator or

correlators supplied with a set of orthonormal basis function Φ1(t) ,Φ2(t) …….ΦM(t) that are generated

locally.

This bank of correlator operate on the received signal x(t) to produce observation vector x.


(b) Vector Receiver

2. The 2nd part of the receiver, namely the vector receiver is as shown in fig (b).

The vector x is used to produce an estimate m of the transmitted symbol mi, where i= 1,2, …. , M to

minimize the average probability of symbol error.

The N elements of the observation vector x are first multiplied by the corresponding N elements of each of

the M signal vectors s1, s2… sM , and the resulting products are successively summed in accumulator to

form the corresponding set of Inner products {(x, sk)} k= 1, 2 ..M. The inner products are corrected for the

fact that the transmitted signal energies may be unequal.

Finally, the largest in the resulting set of numbers is selected and a corresponding decision on the

transmitted message made.

The optimum receiver is commonly referred as a correlation receiver.

8. a.

ANS –8. a.

The definition of SS (Spread Spectrum) may be stated in two parts:

1. It is a means of transmission in which the transmitted data sequence occupies a larger bandwidth then the

minimum bandwidth necessary to send the data.

2. Spreading of data is done before transmission through the channel using a code which is independent of

data sequence. The same code is used at the receiving end to despread the received signal so that original

data may be recovered.

Direct – Sequence Spread Spectrum with coherent binary Phase shift Keying:-

The transmitter involves two stages of modulation


In 1st stage, the data sequence b(t) is modulated with the code sequence c(t). So the Spread signal is

m(t) =b(t). c(t)

In 2nd stage, the Spread signal m(t) is modulated with the binary PSK modulator.

When the polarity of b(t) & c(t) are same, the product b(t). c(t) = 1, hence the phase of the BPSK signal is

(2πfct) radians.

Similarly when b(t) & c(t) are of different polarities, the product b(t). c(t) = -1, hence the phase of the

BPSK signal is (2πfct + π) radians.

Figure.1: Direct-sequence spread coherent phase-shift keying. (a) Transmitter. (b) Receiver.


Figure.2: (a) Product signal m(t) = c(t). b(t). (b) Sinusoidal carrier. (c) DS/BPSK

The receiver consists of two stages of demodulation.

In 1st stage demodulation, the received signal y(t) & a locally generated replica of the PN sequence are

applied to a multiplier.

In 2nd stage, m(t) is despread by multiplying it by c(t) i.e. it consists of a coherent detector, the o/p which

provides an estimate of the original data sequence.

8. b.

ANS –8. b.

Properties of PN Sequence

Randomness of PN sequence is tested by following properties

1. Balance property

2. Run length property


3. Autocorrelation property

1. Balance property

In each Period of the ML (Maximum Length) - sequence, the number of 1’s is always one more than the number of

0’s (i.e. number of 1’s exceeds the number of 0’s by one).

2. Run length property

A run is defined as a subsequence of identical symbols within the ML-Sequence. The length of the Subsequence is

known as the run-length.

The total number of runs = (N+1)

2

Among the runs of ones and zeros in each period, it is desirable that about one half the runs of each type are of

length 1, one- fourth are of length 2 and one-eighth are of length 3 and so-on.

3. Auto correlation property

Auto correlation function of a maximal length sequence is periodic and binary valued.

Autocorrelation sequence of binary sequence in polar format is given by

where,

N is the length or period of the PN sequence &

K is the lag of the autocorrelation sequence &


8. c.

ANS -8. c.

Sl.No Slow FH – SS Fast FH – SS

1) Multiple Symbols are transmitted in one frequency hop Multiple hops are taken to transmit one symbol

2) Symbol rate = Chip rate (Rs = Rc) Hop rate = Chip rate (Rh = Rc)

3) Hop rate is lower than Symbol rate (Rh ¿ Rs) Hop rate is higher than Symbol rate (Rh ¿ Rs)

4) One or more Symbols are transmitted over the same carrier

frequency

One symbol is transmitted over multiple

carriers in different hops

5) Less secure than fast FH-SS More secure than slow FH-SS

6) PG = 2K PG = 2K

7) Several modulation symbols per hop Several frequency hops per modulation

8) Shortest uninterrupted waveform in the system is that of

data symbol

Shortest uninterrupted waveform in the system

is that of hop

digital communication vtu paper

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