dec_2012.docx

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

PART - A

1. a.

ANS -1. a.

Solution:

‘f’ in Hz 0 10 20 50 75 100 150 180 190 200

G(f) = 1- |f|200

1 0.95 0.9 0.75 0.625 0.5 0.25 0.10 0.05 0

WKT Gδ (f) = fs ∑n=−∞

∞

G¿¿s)

i) fs= 300Hz

Therefore, Gδ (f) = 300 ∑n=−∞

∞

G¿¿)

ii) fs = 400Hz

Therefore, Gδ (f) = 400 ∑n=−∞

∞

G¿¿)


1. b.

ANS -1. b.

Sampling theorem for band-pass signal states that, “The bandpass signal g(t) whose maximum bandwidth is 2W can

be completely represented into & recovered from its samples if it is sampled at the minimum rate of twice the

bandwidth.”

In this scheme, the band pass signal is split into two components, one is in-phase component and other is

quadrature component. These two components will be low–pass signals and are sampled separately. This

form of sampling is called quadrature band-pass sampling.

In quadrature sampling of bandpass signal, bandpass signals are converted to lowpass signals by

multiplying bandpass signals with two sinusoidal signals which are phase Quadrature (900) with each other.

The signals are sampled using low pass sampling theorem.

At the receiving end, low-pass signals are reconstructed from low-pass samples & then it is converted to

bandpass signals by multiplying the low-pass signals by 2 sinusoidal which are phase Quadrature (900) with

each other.

Proof:

Consider a band pass signal g(t) whose spectrum is limited to a bandwidth 2W, centered around the

frequency ‘fc’ where fc ¿ W (See Fig 1(a) and Fig 1(b)).

(a)


(b)

Fig 1: (a) Spectrum of band-pass signal g(t). (b) Spectrum of low-pass in-phase component g I(t) and quadrature

component gQ(t).

Any band-pass signal can be represented in terms of its In-phase & Quadrature components as:

g(t) = gI(t) cos2 πfct - gQ(t) sin 2π fct

Where, gI(t) -----¿ In-phase Component

gQ(t) -----¿ Quadrature Component

Fig 2: (a) Generation of In-phase and Quadrative samples from band-pass signal g(t).

The in-phase component, gI(t) is obtained by multiplying g(t) with cos(2πfct) and then filtering out the high

frequency components. Parallelly a quadrature phase component is obtained by multiplying g(t) with

sin(2πfct) and then filtering out the high frequency components.

The in-phase, gI(t) and quadrature phase gQ(t) signals are low–pass signals, having band limited to (-W < f

< W). Accordingly each component may be sampled at the rate of 2W samples per second. Then, the o/p of

sampler will be 12

gI(nTs) & 12

gQ(nTs).

RECONSTRUCTION:


From the sampled signals gI(nTs) and gQ(nTs), the signals gI(t) and gQ(t) are obtained. To reconstruct the

original band pass signal, multiply the signals gI(t)and gQ(t) by cos(2πfct) and sin(2πfct) respectively and

then add the results.

Fig 2: (b) Reconstruction of band-pass signal g(t).

The o/p is the required bandpass signal g(t) given by:

g(t) = gI(t) cos2 πfct - gQ(t) sin 2π fct

1. c.

ANS -1. c.

Advantages of Digital Communication

1. The effect of distortion, noise and interference is less in a digital communication system. This is because

the disturbance must be large enough to change the pulse from one state to the other.

2. Regenerative repeaters can be used at fixed distance along the link, to identify and regenerate a pulse

before it is degraded to an ambiguous state.

3. Digital circuits are more reliable and cheaper compared to analog circuits.

4. The Hardware implementation is more flexible than analog hardware because of the use of

microprocessors, VLSI chips etc.

5. Signal processing functions like encryption, compression can be employed to maintain the secrecy of the

information.

6. Error detecting and Error correcting codes improve the system performance by reducing the probability

of error.

7. Combining digital signals using TDM is simpler than combining analog signals using FDM. The

different types of signals such as data, telephone, TV can be treated as identical signals in transmission and

switching in a digital communication system.

8. We can avoid signal jamming using spread spectrum technique.

Disadvantages of Digital Communication:


1. Large System Bandwidth:- Digital transmission requires a large system bandwidth to communicate the

same information in a digital format as compared to analog format.

2. System Synchronization:- Digital detection requires system synchronization whereas the analog signals

generally have no such requirement.

3. System complexity is increased.

2. a.

ANS -2. a.

i)

Solution:

In One frame, Total number of pulses = 25.

Given, Sampling frequency = fs = 8 kHz.

Time duration for one time frame = Ts = 125μ seconds.

Time duration utilized by pulses = 25μ sec.

Time spacing between successive pulses = (125- 25)/25 = 4μ sec.

ii)

Solution:

In One frame, Total number of pulses = 25.

Given the highest frequency component, fm = 3.4 kHz.

If sampled at the Nyquist rate, then fs = 2 fm = 2 * 3.4 kHz = 6.8 kHz.

Time duration for one time frame = Ts = 147μ seconds.

Time duration utilized by pulses = 25μ sec.

Time spacing between successive pulses = (147- 25)/25 = 4.88μ sec.

2. b.

ANS -2. b.

Consider a binary-encoded PCM wave s(t) that uses NRZ unipolar format:

s(t) = {s1 ( t )=√ EmaxTb

,0 ≤t ≤ Tb for symbol 1

s 2 (t )=0 ,0 ≤ t ≤ Tb for symbol0 } -------------¿ 1

Where Tb = bit duration

Emax = maximum or peak signal energy.


Fig 1: Receiver for baseband transmission of binary-encoded PCM wave.

The channel noise w(t) is modeled as additive white guassian noise (AWGN) with mean and power spectral

density (PSD) No2

, then the received signal

x(t) = s(t) + w(t) 0 ≤ t ≤ Tb

then equation 1 becomes:

s(t) = {s1 ( t )=√ Emax∅ 1 ( t )s2 ( t )=0 }

An ON – OFF PCM system is characterized by having a signal space that is one-dimensional and with two

message points.


Fig 2: (a) Signal space diagram for on – off PCM system. (b) Likelihood function, given that symbol 0 was sent.

(c) Likelihood function, given that symbol 1 was sent.

The co-ordinates of the two message points are:

S11 = ∫0

Tb

s1 (t )∅ 1 (t ) . dt

S11 = √ Emax and


S21 = ∫0

Tb

s2 (t )∅ 1 ( t ) . dt

S21 = 0

For s1(t), μ=√Emax∧σ2= No2

and for s2(t), μ=0∧σ2=No2

A threshold of √Emax2

is used, which is half way point b/w two message points.

i.e. If x(t) ¿ √Emax2

, then decision is taken in favour of symbol ‘1’

If x(t) ¿ √Emax2

, then decision is taken in favour of symbol ‘0’.

The received signal point is calculate by sampling the matched filter o/p at time t = Tb.

x1 = ∫0

Tb

x (t )∅ 1 (t ) . dt

To calculate Pe of 1 st kind :

Symbol ‘0’ is sent but received as symbol ‘1’ called as error of 1st kind.

The decision region Z1 : √Emax2

¿ x1<∞

Since x1 has guassian distribution, it is defined by

Therefore, probability of symbol error, when symbol ‘0’ is transmitted

Substituting eq 2 in eq 3, we get


Let Z = x 1

√No , x1 = Z √No and dx1 = dz √No

Therefore, limits

When, x1 = √Emax2

Z√No = √Emax2

Z = √Emax2√No

Z = √ EmaxN 0

.12

When, x1 = ∞

Z√No = ∞

Z = ∞

√No

Z = ∞

Therefore equation 4 becomes

WKT

erfc (u) = 2

√π ∫

u

∞

e−u2

. du ------------¿ 6

From eq 6, we can write eq 5 as


Where u = 12

√ EmaxNo

Similarly Pe of 2nd kind is

Pe (1) =12

erfc¿)

Average probability of error:

Let probability of sending ‘0’ is P0 = ½

Let probability of sending ‘1’ is P1 = ½

Then average probability of error

The ratio Emax

No represents peak signal energy to noise spectral density ratio, it can be represented as peak

signal to noise power ratio.

WKT Emax = Pmax Tb

Emax

no =

Pmax TbNo

= Pmax{ No|Tb }

Pe decreases very rapidly as this ratio is increased.


Fig 3: Probability of error in a PCM receiver.

2. c.

ANS -2. c.

ROBUST QUANTIZATION

Features of an uniform Quantizer

– Variance is valid only if the input signal does not overload Quantizer.

– SNR Decreases with a decrease in the input power level.

A Quantizer, whose SNR remains essentially constant for a wide range of input power levels. A quantizer

that satisfies this requirement is said to be robust.

To keep signal to quantization noise ratio high, we must use a signal which is large in comparison with step

size. This requirement is not satisfied when signal is small i.e. we need smaller step size for low magnitude

signal samples & higher step-size for higher magnitude signals.

Changing step-size according to signal magnitude is not preferable one. Instead, change the characteristics

of the signal such that lower amplitudes are amplified without changing the maximum value of the signal.


Maintaining of constant SNR throughout the signal range is called “Robust Quantization”.

To achieve this, the signal is passed through a combination of compressor-expander circuit respectively at

the transmitter & receiver. This technique is known as COMPANDING.

To achieve Robust Quantization, the signal is passed through a network which has an I/p – O/p

characteristics as shown in fig (a) & (b).

The signal is changed such that small amplitude signals are boosted up without altering the maximum

amplitude of the signal, small amplitude signals range through more quantization levels.

Any signal when passed through such a network gets compressed leading to signal distortion.

To remove this distortion, the signal is passed through an Inverse network at the receiver called as

Expander.

The complete process of compressing & expanding the signal is referred to as COMPANDING.

There are two types of Companding:

1) μ−Law Companding .

2) A−Law Companding.

3. a,

ANS -3. a.

Solution:

Given: Signal Power x(t) = a0cos¿fot)

The maximum Slope of the signal x(t) is given by


|dx ( t )dt |max = | d

dta o cos(2 π f o t)|max = |−ao 2πfo sin(2πfot )|max

|dx ( t )dt |max = ao 2 πfo -------¿ 1 Since, Sin(2 πfot ) =1 at t = 900

The condition for no slope overload is

∂Ts

≥|d [x ( t )]dt |max ----------¿ 2

Substituting eq (1) in eq (2), we get

∂Ts

≥ ao 2 πfo

ao 2 πfo≤∂Ts

ao= ∂2 πfoTs

ao= ∂ fs2πfo

----------¿ 3

The maximum average power of the signal x(t) is given by

Pmax = V 2

R

Here ‘V’ is the rms value of the signal. ∴V =ao

√2

Normalized signal power is obtained by taking R = 1

Pmax = ¿

Pmax = ao2

2 ----------¿ 4


Pmax = ∂2 fs2

4 π2 fo2 . 12

Pmax = ∂2 fs2

8 π 2 fo2 ---------¿ 5

Eq 5 is the signal power in the delta modulation.

Noise Power:

When there is no slope overload, the maximum quantization error is± ∂. Assuming that the quantization error is

uniformly distributed,

Consider the PDF of the quantization error i.e.


fQ(q) = { 12 ∂

for−∂ ≤ q ≤ ∂

0 ot h erwise }

The variance of Quantization error i.e.

σ Q2 = ∫

−∂

∂

q2 . fQ(q). dq

= ∫−∂

∂

q2 .1

2∂.dq =

12 ∂

∫−∂

∂

q2 dq

σ Q2 =

∂2

3 --------¿ 7

The DM receiver contains a LPF whose cut – off frequency fc = W Hz.

Assuming that the average power of the quantization error is uniformly distributed over a frequency

interval extending from −1Ts

¿+ 1Ts

, we get

Average o/p Noise power ‘No’ = ( fcfs )∂2/3 = WTs( ∂2

3) -------¿ 8

Signal to noise power ratio at the o/p of DM receiver is


(SNR)o = Pmax

No =

∂2 fs2

8π 2 fo2

W ∂2Ts3

= 3

8 π 2 fo2 Ts2WTs

Hence, (SNR)o = 3

8 π 2 fo2 W Ts3 --------¿ 9

Eq 9 shows that, under the assumption of no-slope overload distortion, the maximum o/p signal to noise ratio of a

delta – modulator is proportional to the sampling rate cubed.

3. b.

ANS -3. b.

The Polar octal format employs eight distinct symbols formed by 3-bit binary numbers. Gray and natural codes are

employed. The input binary sequence 011010110 is viewed as a new sequence of 3-bit binary numbers, {011, 010,

110}. Each 3-bit binary number is assigned a level in accordance with the natural code described in Table 1. The

polar octal format of the NRZ type for natural and gray code versions is shown in figure 4.

Table 1: Natural and Gray codes

Level Natural code Gray code

-7 000 000

-5 001 001

-3 010 011

-1 011 010

1 100 110

3 101 111

5 110 101

7 111 100


Figure 4: Polar Octal Format. (a)Natural-encoded. (b)Gray-encoded.

3. c.


ANS -3. c.

Generally, digital data is represented by electrical pulse, communication channel is always band limited. Such a

channel disperses or spreads a pulse carrying digitized samples passing through it. When the channel bandwidth is

greater than bandwidth of pulse, spreading of pulse is very less. But when channel bandwidth is close to signal

bandwidth, i.e. if we transmit digital data which demands more bandwidth which exceeds channel bandwidth,

spreading will occur and cause signal pulses to overlap. This overlapping is called Inter Symbol Interference. In

short it is called ISI. Similar to interference caused by other sources, ISI causes degradations of signal if left

uncontrolled. This problem of ISI exists strongly in Telephone channels like coaxial cables and optical fibers.

The effect of sequence of pulses transmitted through channel is shown in fig 5. The Spreading of pulse is greater

than symbol duration, as a result adjacent pulses interfere. i.e. pulses get completely smeared, tail of smeared pulse

enter into adjacent symbol intervals making it difficult to decide actual transmitted pulse. There are different formats

of transmitting digital data. In base band transmission best way is to map digits or symbols into pulse waveform.

This waveform is generally termed as Line codes.

Figure 5: Effect of sequence of pulses transmitted through the channel.

The solution to overcome the degradation of waveform or signal is by properly shaping pulse. Nyquist Pulse

Shaping Criterion suggests the method for constructing band limited function to overcome effect of ISI.

Ideal Nyquist filter that achieves best spectral efficiency and avoids ISI is designed to have bandwidth as suggested

B0 = 1/2Tb (Nyquist bandwidth) = Rb/2

Practical Filter such as raised cosine filter can also be used to eliminate ISI which has transfer function consists of a

flat portion and a roll off portion which is of sinusoidal form.


4. a.

ANS -4. a.

By adding ISI to the transmitted signal in a controlled manner, it is possible to achieve a bit rate of 2B o bits per

second in a channel of bandwidth Bo Hz. Such a scheme is correlative coding or partial- response signaling scheme.

One such example is Duo binary signaling.

Duobinary Encoding:

Duo means transmission capacity of system is doubled.

Consider binary I/p Sequence {bk} having duration Tb seconds, with symbol ‘1’ represented by a pulse of

amplitude +1V & symbol ‘0’ by a pulse of amplitude -1V.

When this sequence is applied to a duobinary encoder, it is converted into a three-level o/p namely, -2V,

0V, +2V.

To produce this transformation, the Scheme is shown in fig 6.

Figure 6: Duobinary signaling encoding scheme

The binary sequence {bk} is first passed through a simple filter involving a single delay element.

The o/p of duobinary coder Ck is the sum of the present binary digit bk & its previous value bk-1 i.e.

Now, I/p sequence {bk} of uncorrelated digits are transformed into {Ck} of correlated digits.

The symbol {bk} takes ±1 level, thus Ck takes one of three possible values -2, 0, +2. The duo binary code

results in a three level output. In general, for M-ary transmission, we get 2M-1 levels.


Duobinary decoding:

Figure 7: Duobinary decoding

At the receiver, original sequence {bk} may be detected by subtracting the previous decoded binary digit from the

presently received digit Ck This demodulation technique shown in Fig 7 (known as nonlinear decision feedback

equalization) is essentially an inverse of the operation of the digital filter at the transmitter. Thus, the original {b k}

may be detected from the duobinary sequence {Ck} as shown below.

Where, is estimate of original sequence bk.

Transfer function of Duo-binary Filter:

The ideal delay element used produce delay of Tb seconds for impulse & will have transfer function e -j 2π f Tb.

Overall transfer function of the filter H(f)

H(f) = Hc(f) + Hc(f) e -j 2π f Tb

H(f) = Hc(f) [1+ e -j 2π f Tb]

H(f) = 2Hc(f) [(e j π f Tb + e -j π f T

b)/2] e -j π f Tb

H(f) = 2Hc(f) Cos(πfTb) e -j 2π f Tb

As ideal channel transfer function

Thus overall transfer function


H(f) which has a gradual roll off to the band edge, can also be implemented by practical and realizable analog

filtering. Figure 8 shows Magnitude and phase plot of Transfer function.

Figure 8: Frequency response of duobinary conversion filter. (a) Amplitude response. (b) Phase response.

Advantage of obtaining this transfer function H(f) is that practical implementation is easy.

Disadvantage:

If Ck and previous estimate is received properly without error then we get correct decision and current estimate.

Otherwise error made it tends to propagate, because of decision feedback {bk} depends on previous digit bk-1.

This error can be avoided by use of precoder before the duobinary coding.

Example: Demonstration of duobinary encoding and decoding:


4. b.

ANS -4. b.

a) Alternating 1s and 0s.

b) A long sequence of 1s followed by a long sequence of 0s.

After applying to filters response is


4. c.

ANS -4. c.

Raised Cosine Spectrum:

To overcome the practical difficulties encountered with the ideal Nyquist channel by extending the bandwidth from

the minimum value W=Rb/2 to an adjustable value between W and 2W this technique is used.

Its frequency response P(f) consists of a flat portion and a roll off portion that has a sinusoidal form as follows:

The frequency parameter f1 and bandwidth W are related by

The parameter α is called the roll-off factor; it indicates the excess bandwidth over the ideal solution, W.

Specifically, the transmission bandwidth BT is defined by

BT = 2W – f1

= W(1 + α )

The time response p(t) is the inverse Fourier transform of the frequency response P(f).

For α = 1, we have the most gradual rolloff factor in that the amplitudes of the oscillatory tails of p(t) are smallest.

Thus the amount of ISI resulting from timing error decreases as the rolloff factor α is increased from zero to unity.

The special case with α=1 (i.e. f1 = 0) is known as the full-cosine rolloff characteristic, for which the frequency

response simplifies to


Correspondingly, the time response p(t) simplifies to

This time response exhibits two interesting properties:

1. At t = ±Tb/2 = ±1/4W, we have p(t) = 0.5; that is, the pulse width measured at half amplitude is exactly equal to

the bit duration Tb.

2. Thare are sero crossings at t = ±3Tb/2, ±5Tb/2,… in addition to the usual zero crossings at the sampling times t =

±Tb, ±2Tb,…

Thses two properties are extremely useful in extracting a timing signal from the received signal for the purpose of

synchronization. However, the price paid for this desirable property is the use of a channel bandwidth double that

required for the ideal Nyquist channel corresponding to α = 0.

PART - B

5. a.

ANS -5. a.

Binary Phase Shift Keying (BPSK) Generation:

Fig (a): Block Diagram of BPSK Transmitter.

In a binary PSK system a sinusoidal carrier wave of fixed amplitude & frequency ‘fc’ is used to represent

both symbol ‘1’ & ‘0’ except that the carrier phase of each symbol differs by a phase of 180o.

i.e.

S1(t) = √ 2 EbTb

cos 2π fct ---------------------- for symbol ‘1’

S2(t) = √ 2 EbTb

cos ¿fct +π ¿ = - √ 2 EbT b

cos 2π fct --------------- for symbol ‘0’


Where Eb= Average energy transmitted per bit Eb = (Eb0 + Eb1)/2

Fig (b): Waveforms of BPSK signal.

In this case of PSK, there is only one basic function of unit energy which is given by

∅ 1(t) = √ 2Tb

cos2 π fct 0≤ t ≤ Tb

Therefore the transmitted signals are given by

S1(t) = √ Eb∅ 1(t) 0≤ t ≤ Tb for Symbol ‘1’

S2(t) = -√ Eb∅ 1(t) 0 ≤ t ≤ Tb for Symbol ‘0’

A Coherent BPSK is characterized by having a signal space that is one dimensional (N=1) with two

message points (M=2)

S11 = ∫0

Tb

S1(t)∅ 1(t) dt = +√ Eb

S21 = ∫0

Tb

S2(t)∅ 1(t) dt = -√ Eb

The message point corresponding to S1(t) is located at S11 = +√ Eb and S2(t) is located at S21 = -√ Eb

To generate a binary PSK signal we have to represent the input binary sequence in polar form with symbol

‘1’ and ‘0’ represented by constant amplitude levels of +√ Eb & -√ Eb respectively.

This signal transmission encoding is performed by a NRZ level encoder. The resulting binary wave [in

polar form] and a sinusoidal carrier ∅ 1(t) [whose frequency fc = ncTb

] are applied to a product modulator.

The desired BPSK wave is obtained at the modulator output.

Thus in BPSK, Symbol ‘1’ is represented by transmitting a carrier wave with a 0o phase shift. Symbol ‘0’ is

represented by transmitting a carrier wave with a 180o phase shift.

Coherent detection of BPSK:


Fig (c): Coherent binary PSK receiver.

Let x(t) be the received signal which is the sum of AWGN noise and transmitted BPSK signal i.e. x(t) =

S(t) + w(t).

Let ∅ 1(t) be the local carrier signal which is synchronized in terms of frequency & phase.

To detect the original binary sequence of 1’s and 0’s we apply the noisy PSK signal x(t) to a Correlator,

which is also supplied with a locally generated coherent reference signal ∅ 1(t) as shown in fig (c). The

correlator output x1 is compared with a threshold of zero volt.

If x1 > 0, the receiver decides in favour of symbol 1.

If x1 < 0, the receiver decides in favour of symbol 0.

Drawbacks of BPSK: Ambiguity in Output Signal

To regenerate the carrier in the receiver, we start by squaring √ Eb∅ 1(t). If the received signal is - √ Eb∅ 1(t) then

the squared signal remains same as before. Therefore the recovered carrier is unchanged even if the input signal has

changed its sign. Therefore it is not possible to determine whether the received signal is equal to √ Eb or - √ Eb.

This result in ambiguity in the output signal. The problem can be removed if we use differential phase shift keying

(DPSK).

5. b

ANS -5. b.

.Probability of error in QPSK:

The Signal points S1, S2, S3 & S4 are located symmetrically in two dimensional space diagram as shown in

Fig 9.


Fig 9: Signal space diagram for coherent QPSK system.

∴ Computing probability for one message point, is remains same for other three points.

Consider transmission of symbol S4(t) then received signal x(t) will be

X(t) = S4(t) + w(t) 0 ≤ t ≤ T

The samples x1 & x2 are computed as follows

x1 = ∫0

T

x (t )∅ 1(t) dt = ∫0

T

[S 4 ( t )+w(T )]∅ 1(t) dt

x1 = ∫0

T

S 4(t )∅ 1(t) dt + ∫0

T

w(t )∅ 1(t) dt

x1 = S41 + w2

x1 = √ E2

+ w1

Similarly


x2 = ∫0

T

x (t )∅ 2(t) dt = ∫0

T

[S 4 ( t )+w(T )]∅ 2(t) dt

x2 = ∫0

T

S 4(t )∅ 2(t) dt + ∫0

T

w(t )∅ 2(t) dt

x2 = S42 + w2

x2 = √ E2

+ w2

x1 & x2 are the guassian random variables with mean ‘μ’ = √ E2

w1 & w2 are also guassian random variables with variance σ 2=No2

When signal S4(t) is transmitted, the received signal point lies in the decision region ‘Z4’.

If x1¿0 & x2 ¿0, leading to correct decision.

The conditional PDF is given by

Let us assume S4(t) is transmitted. If the received signal ‘x’ should fall in region Z 4 i.e. both x1 & x2 should

be +ve.


Fig 10: Illustrating the region of correct decisions and the region of erroneous decisions, given that signal

S4(t) was transmitted.

Probability of correct decision ‘Pc’ is equal to the product of conditional probabilities of events x 1 > 0 & x2

> 0, both given that S4(t) was transmitted.

Region Z4: 0 ≤ x1 ≤ ∞

0 ≤ x2 ≤ ∞

Let

Z = x1−√ E

2√No

dZ = dx1−0

√No

dx1 = √ No dZ

Z = x2−√ E

2√No

dZ = dx2−0

√No

dx2 = √ No dZ


Limits

x1 = ∞

then, Z = ∞

x1 = 0, Z = x1−√ E

2√No

Z = 0−√ E

2√No

Z = - √ E2 No

x2 = ∞

then, Z = ∞

x2 = 0, Z =

x2−√ E2

√No

Z = 0−√ E

2√No

Z = - √ E2 No

From the definition of the complementary error function we have

1√π

∫−√ E

2No

∞

e−z2

. dz=1−12

erfc (√ E3 No

)

Substituting eq 5 in eq 4


Pc = 1 + 14

erfc2(√ E2 No

)−erfc (√ E2 No

)

Thus average probability of symbol error

Pe = 1 - Pc

Pe = 1 – [1 + 14

erfc2(√ E2 No

)−erfc (√ E2 N o

)]

Pe = erfc (√ E2 No

) – 14

erfc2¿

In the region Z4: (E

2 N 0)≫1, hence we can ignore second term

Pe ≈ erfc (√ E2 No )

In QPSK two bits are transmitted per symbol

Thus E = 2Eb

6. a.

ANS -6. a.

Consider the signals s1(t), s2(t), s3(t), and s4(t) shown in Fig. 11a. Use the Gram-Schmidt orthogonalization procedure

to find an orthonormal basis for this set of signals.


Fig 11(a)

Step 1 We note that the energy of signal s1(t) is

E1 = ∫0

T

s21(t) dt = ∫

0

T /3

(1)2 dt

= T3

The first basis function ∅ 1(t) is therefore

∅ 1(t) = s1 ( t )√E 1

= {√ 3T

, 0 ≤ t ≤T3

0 , ot h erwise}Step 2 Evaluating the projection of s2(t) onto ∅ 1(t), we find that

s21 = ∫0

T

s2(t)∅ 1(t) dt

= ∫0

T /3

(1 )(√ 3T

¿)¿ dt

= √ 3T

The energy of signal s2(t) is

E2 = ∫0

T

s22(t) dt

= ∫0

2 T /3

(1)2 dt

= 2T3

The second basis function ∅ 2(t) is therefore


Step 3 Evaluating the projection of s3(t) onto ∅ 1(t),

S31 = ∫0

T

s3(t)∅1(t) dt

= 0

and the coefficient s32 equals

S32 = ∫0

T

s3(t)∅2(t) dt

= ∫T /3

2 T /3

(1 )(√ 3T

¿)¿ dt

= √ 3T

The corresponding value of the intermediate function gi(t), with i = 3, is therefore

g3(t) = s3(t) – s31∅ 1(t) – s32 ∅ 2(t)

= {1 ,2T3

≤ t ≤ T

0 , elsewhere}Hence, the third basis function ∅ 3(t) is

∅ 3(t) =

g 3 ( t )

√∫0

T

g 32 ( t ) dt

= {√ 3T

,2 T3

≤ t ≤T

0 , elsewhere }The orthogonalization process is now complete.

The three basis functions ∅ 1(t), ∅ 2(t) and ∅ 3(t) form an orthonormal set, as shown in Fig. 11b. In this example, we

thus have M = 4 and N = 3, which means that the four signals s1(t), s2(t), s3(t), and s4(t) described in Fig. 11a do not

form a linearly independent set. This is readily confirmed by noting that s4(t) = s1 (t) + s3(t). Moreover, we note that

any of these four signals can be expressed as a linear combination of the three basis functions, which is the essence

of the Gram-Schmidt orthogonalization procedure.


Figure 11(b)

6. b.

ANS -6. b.

Given: Rb = 2.5 Mbps, No2

= 10-20 W/Hz, Ac = 1× 10-6 V, No = 2×10−20 W/Hz, erfc(√5) = 1.7

Solution:

The bit duration is

Tb = 1

Rb =

1

2.5× 106 = 0.4 ×10−6 s

The signal energy per bit is

Eb = 12

Ac2Tb

= 12

(10-6)2 × 0.4 ×10−6 = 2 ×10−19 joules

The average probability of error for Coherent Binary FSK is

Pe = 12

erfc(√ Eb/2 No)

=12

erfc(√2×10−19 /4 ×10−20)

= 12

erfc(√5)

= 12

×1.7

Pe = 0.85.

6. c.

ANS -6. c.

Optimum Receiver or Maximum likelihood Receiver or Correlative Receiver:

For an AWGN channel, when the transmitted signals S1(t), S2(t), ….. , SM(t) are equally likely, the

optimum receiver consisits of two subsystems as shown in fig 12.


(a)

1. The detector part of the reciver us as shown in fig (a). It consists of a bank of correlators supplied with a

corresponding set of orthonormal basis function ∅ 1(t), ∅ 2(t), …. , ∅N(t) that are generated locally.

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

(b)

Figure 12: (a) Detector. (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 m̂ of the transmitted symbol mi, where i = 1, 2, …. , M to

minimize the average probability of symbol error.

The observation vector x is multiplied by M signal vectors S1, S2, …. , SM & the resulting products are

successively summed in accumulators.

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

transmitted signal.

The optimum receiver is commonly referred to as a correlation receiver.

7. a.

ANS -7. a.

Matched filter properties:

NOTE:

1) hopt(t) = ∅ (T – t)

2) Hopt(f) = ∅*(f) e− j2 πfT

PROPERTY 1

The spectrum of the output signal of a matched filter with the matched signal as input is, except for a time delay

factor, proportional to the energy spectral density of the input signal.

Let ∅ o(f) denote the FT of the filter o/p ∅ o(t), then

∅ o(f) = Hopt(f) ∅ (f) ------>

= ∅*(f) e− j2 πfT ∅ (f)

= ∅*(f) ∅ (f) e− j2 πfT

∅ o(f) = |∅ ( f )|2e− j2 πfT

PROPERTY 2

The output signal of a Matched Filter is proportional to a shifted version of the autocorrelation function of the input

signal to which the filter is matched.

WKT ∅ o(f) = |∅ ( f )|2e− j2 πfT

Taking IFT of the above eqn, we get

∅ o(f) = R∅ (t – T)

Where R∅ (τ ) is the autocorrel ation function of the I/p ∅ (t) for lag τ {i.e. R∅ (τ ) = ∅ (t) * ∅ (t – T)}

We have ∅ o(T) = R∅ (0) = E

Where ‘E’ is the signal energy.

PROPERTY 3


The output Signal to Noise Ratio of a Matched filter depends only on the ratio of the signal energy to the power

spectral density of the white noise at the filter input.

WKT the average power of the o/p noise n(t) is

E[n2(t)] = No2

∫−∞

∞

|H (f )|2 df

E[n2(t)] = No2

∫−∞

∞

|∅ ( f )|2 df

(Because, H(f) = ∅*(f) e− j2 πfT)

From 2nd property, the maximum value of filter o/p at time t = T, is proportional to the signal energy ‘E’.

∴E ¿(f)] = No2

E <------- (multiply and divide by E)

So, the o/p SNR will have the maximum value as

(SNR)o,max = E2

No2

E =

2 ENo

(SNR)o,max = 2 ENo

2 ENo

is a dimensionless as E is in Joules & No2

is in W/Hz.

E

No is termed as the signal energy to noise density ratio.

PROPERTY 4

The Matched Filtering operation may be separated into two matching conditions; namely spectral phase matching

that produces the desired output peak at time T, and the spectral amplitude matching that gives this peak value its

optimum signal to noise density ratio.

In polar form,

∅ ( f )=¿ |∅ ( f )|e jθ(f )

Where |∅ ( f )| is the amplitude spectrum &

θ( f ) is the phase spectrum of the signal.

i.e. θ( f ) = 2 πfT

For Spectral amplitude matching, the amplitude response |H ( f )| of the filter to shape the o/p for best

SNR at t = T by using |H ( f )| = |∅ ( f )|


7. b.

ANS -7. b.

Adaptive Equalization:

An equalizer is a filter that compensates for the dispersion effects of a channel.

Adaptive equalizer can adjust its coefficients continuously during the transmission of data.

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 continuosly 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.

Figure 13: Apative equalizing filter

Fig 13 shows the block diagram of an adaptive equalizing filter. It is adaptive in nature because it is

capable of adjusting its co – efficients ω0, ω1, …. , ωM-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

ωi 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. ωi 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:

ω̂i (nT + T) = ω̂i (nT) + μe(nT ) x(nT – iT)

Where,

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

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

μis the adaptation constant

x (nT – iT) is the filter I/p &

e (nT) is the error signal

Methods of implementing adaptive equalizer

i) Analog

ii) Hard wired digital

iii) Programmable digital

Analog method

Charge coupled devices [CCD’s] are used.

CCD- FET’s are connected in series with drains capacitively coupled to gates.

The set of adjustable tap widths are stored in digital memory locations, and the multiplications of the

analog sample values by the digitized tap weights done in analog manner.

Suitable where symbol rate is too high for digital implementation.

Hard wired digital technique

Equalizer input is first sampled and then quantized in to form that is suitable for storage in shift registers.

Set of adjustable lap weights are also stored in shift registers. Logic circuits are used for required digital

arithmetic operations.

Widely used technique of equalization.

Programmable method


Digital processor is used which provide more flexibility in adaptation by programming.

Advantage of this technique is same hardware may be timeshared to perform a multiplicity of signal

processing functions such as filtering, modulation and demodulation in modem.

8. a.

ANS -8. a.

A feedback shift register is said to be Linear when the feed back logic consists of entirely mod-2-address ( Ex-or

gates). In such a case, the zero state is not permitted. The period of a PN sequence produced by a linear feedback

shift register with ‘n’ flip flops cannot exceed 2n-1. When the period is exactly 2n-1, the PN sequence is called a

‘maximum length sequence’ or ‘m-sequence’.

Figure 14: Maximum-length sequence generator for n=3

Example1: Consider the linear feed back shift register as shown in fig 14 involve three flip-flops. The input S o is

equal to the mod-2 sum of S1 and S3. If the initial state of the shift register is 100. Then the succession of states will

be as follows.

100,110,011,011,101,010,001,100 . . . . . .

The output sequence (output S3) is therefore. 00111010 . . . . .

Which repeats itself with period 23–1 = 7 (n=3)

Maximal length codes are commonly used PN codes

In binary shift register, the maximum length sequence is

N = 2m-1 chips, where m is the number of stages of flip-flops in the shift register.

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 sequence , number of binary ones differ from binary zeros by at most one digit .(i.e. number of

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

Ex:- For 3 – Stage Shift register,

N = 23 – 1

N = 7 i.e. 0010111

Number of 1’s ------> 4

Number of 0’s -------> 3

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 in ML - sequence, 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.

Ex:- 0010111

Total number of runs = (N+1)

2 =

(7+1)2

= 4 runs.

00, 1, 0, 111 = 4 runs.

i) 1,0 -----> Two runs are of length one

ii) 00 ------> One run is of length two

iii) 111 ----> One run is of length three.

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

Rc(k) = 1N∑n=1

N

Cn Cn-k

Where,

N is the length or period of the PN sequence &

k is the lag of the autocorrelation sequence &

The given sequence is

The given sequence has a period L = 7, because after every seven bits, the pattern repeats. This is also in agreement

with L = 2n – 1, when n = 3.


Balance Property: In one period of the sequence, we find that

(i) the number of 1’s = 4 (which is in agreement with number of 1’s = 2n-1).

(ii) the number of 0’s = 3 (which is in agreement with number of 0’s = 2n-l - 1 ). Thus, we

find that number of 0’s and number of l’s differ by one.

Run Property:

Total number of runs = L+1

2=

7+12

= 4 runs. ---------> 1

If we take, the sequence as 0101110, we get five runs, which is not in agreement with equation 1. Hence, we give a

circular shift to right and get a new sequence: 0010111.

Now, there are a total of four runs in one period of the sequence. Reading them from left to right, the four runs are

00, 1, 0, 111. Two of the runs (a half of the total) are of length one, and one run (a quarter of the total) is of length

two, which satisfies the run property.

Autocorrelation Property:

One period of the sequence:

{ C1, C2, C3, C4, C5, C6, C7} = {0, 1 , 0, 1, 1, 1, 0}

Let us represent symbol 0 by + 1 volt and symbol 1 by -1 volt.

Recall:

Rc(k) = 1N∑n=0

N −1

Cn Cn-k = 17∑n=0

6

C n Cn-k

To compute Rc(0):

Cn : C0 C1 C2 C3 C4 C5 C6

Cn-0 : C0 C1 C2 C3 C4 C5 C6

Hence,

Rc(0) = 17

(C02 + C1

2 + C22 + C3

2 + C42 + C5

2 + C62) = 1

Note: C0= C2 = C6 = -1 volt; C1 = C3 = C4 = C5 = +1 volt.

To compute Rc(1):

Cn : C0 C1 C2 C3 C4 C5 C6

Cn-1 : C1 C2 C3 C4 C5 C6 C0

Which is equal to Cn circular shifted right by one index.

Hence,

Rc(1) = 17

(C0C1 + C1C2 + C2C3 + C3C4 + C4C5 + C5C6 + C6C0)

= - 17

To compute Rc(2):

Cn : C0 C1 C2 C3 C4 C5 C6


Cn-2 : C2 C3 C4 C5 C6 C0 C1

Which is equal to Cn circular shifted right by two indices.

Hence,

Rc(2) = 17

(C0C2 + C1C3 + C2C4 + C3C5 + C4C6 + C5C0 + C6C1)

= - 17

Proceeding in a similar manner, we find that,

Rc(3) = Rc(4) = Rc(5) = Rc(6) = - 17

.

Thus ACR function has only two values. Also ACR function is periodic with a period equal to 7.

Summarizing, we have

R(τ ) = { 1 , τ=0 , ± 7 , ±14 ,± 21 , ….−17

, τ ≠ 0 , ± 7 , ±14 ,± 21 , ….}

8. b.

ANS -8. b.

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

Figure 15: Model of direct – sequence spread binary PSK system


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

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.

Model for Analysis:

The spread spectrum i.e. m(t) = b(t) . c(t) can also be performed prior to phase modulation.

For analysis purpose, the spectrum spreading & the BPSK are interchanged because both are linear

operation.

In this model, it is assumed that the interference j(t) limits performance, so that the effect of channel noisy

may be ignored.

From fig 15, channel o/p is given by:

y(t) = x(t) + j(t)

y(t) = c(t) s(t) + j(t) -------> 1

where,

s(t) is the binary PSK signal &

c(t) is the PN sequence.

In the receiver, the received signal y(t) is 1st multiplied by the PN sequence c(t).

Thus u(t) = y(t) c(t) ------> 2


u(t) = [c(t) . s(t) + j(t)] c(t)

u(t) = c2(t) s(t) + j(t).c(t)

Because c2(t) = 1

u(t) = s(t) + c(t) j(t)

This u(t) is fed to coherent detector to detect the information signal b(t).

8. c.

ANS -8. c.

Applications of Spread Spectrum Techniques:

1. Most important application of SS – technique is in the field of secure communication where it has to

provide immunity against jamming & low probability of interception

2. Code Division Multiple Access

3. Multipath Suppression

4. Range determination

5. In Bluettoth devices which is able to establish ad-hoc connections

dec_2012.docx

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