Web for Principle of Communication - B. Tech 4th Sem

Web answers for Principles of Communication Engineering Chapter 2: Amplitude Modulation Example 8 An amplitude-modulated wave has a power content of 800W at its carrier frequency. Calculate the power content of each of the sidebands for 80% modulation. Solution: Pc = 800W m = .8 PLSB  PUSB 

m2 Pc  128W 4

Example 10 A carrier wave Vc = A sin c t is amplitude modulated by an audio wave

B B sin3ωa t  sin5ωa t 3 5 Determine the upper and lower bands. Sketch the complete spectrum of the modulated wave. Va  Bsinωa t 

Estimate the total power in the side bands in terms of the carrier power if the modulated index B/A = 0.6. Solution: After modulated, the frequencies will be c  a, c  3a, and c  5a, respectively. The frequency spectrum is shown below. A MA 2 MA 6

MA

c + 5a

c + 3a

c + a

c a

c - a

c - 3a

c - 5a

10

L

Total power in side bands, 2

2

2

Pm Pm Pm  C 1 C 2 C 3 2 2 2

PSBT

m1 = m = 0.6

m2  and m3 

m 0.6   0.2 3 3

m 0.6   0.12 5 5 2

2

2

P m  m 2 m3  SBT  1 PC 2



0.36  0.04  0.0144 2

= 0.2072 % Power = 0.2072  100 = 20.72% Example 17 Determine the percent increase in the effective value of the antenna current when the modulation of a transmitter is increased to 65% from 0%. Solution: 2

I  m2 We know  t   1  2  Ic  Where It = modulated current (or total) Ic = unmodulated current m = modulation index

 1



(0.65)2 0.4225  1  1.2112 2 2

It  1.2112  1.1 Ic

or It = 1.1 Ic

So increase in antenna current = (It – Ic) = 1.1 Ic – Ic = 0.1 Ic

Example 19 The ac rms antenna current of an AM transmitter is 6.2 A when unmodulated and rises to 6.7 A when modulated. Calculate % age modulation. Solution: 2

 It  m2 We know    1  2  Ic  2

I    t  1  Ic 

m2 2

or

where It = modulated current = 6.7 Ic = unmodulated current = 6.2 2

m 2  6.7    1 2  6.2 



m  2[(1.08)2  1]



 2[(1.16- 1)



 0.32  0.56  56 %

Example 22 A certain transmitter radiates 9 kW with the carrier unmodulated, and 10.125 kW when the carrier is sinusoidally modulated. Calculate the modulation index. If another sine wave, corresponding to 40% modulation, is transmitter Simultaneously, determine the total radiated power. Solution: 1st part: Here, Pt = Total Power (with carrier modulated) 10.125 kW Pc = Power with carrier unmodulated = 9 kW

  

From equation, Pt = Pc 1 

m2 2

  

m2 Pt 10.125  1   1  1.125 -1  0.125 2 Pc 9 m2 = 2  0.125 m = m1 =

2  0.125  0.50

2nd Part: 40% modulation means m2 = 0.4 2

mt =

2

m 1 m 2

So, (Total unmodulation indes)

m  (0.5)  (0.4) 2  0.25  0.16  0.41  0.64 t 2   m P  Pc 1  t  t  2   

 (0.64) 2    9  103 1  2   = 9  103 (1 + 0.205) =10.84 kW. Example 24 An AM broadcast station operates at its maximum allowable output of 50 kW and at 90% modulation. How much of its transmitted power is intelligence? Solution: Maximum allowable output, Pt = 50 kW m = 90% = 0.9

 m2  We know Pt  Pc 1   2  



50 kW = Pc 1 



 

Pc 

(0.9) 2   2 

50  10 3 1  0.405

Pc  35.58 kW

This will be the intelligence power transmitted.

Example 30 When the modulation percentage is 75, an AM transmitter produces 10 kW (a) How much of this is carrier power. (b) What would be the percentage power saving if the carrier and one of the sidebands were suppressed before transmission took place? Solution: (a)

 m2  Pt  Pc 1   2  

 (0.75)2  10  Pc 1   2    Pc  1.28

 Pc 

10  7.81kW 1.28

(b) We know,

Pt  Pc 

m2 m2 Pc  Pc for two side bands 4 4

where power due to carrier and one of the sidebands

 Pc 

m2 Pc 4

 m2   Pc 1   4    0.752  7.811  4 

   

 7.811.14 = 8.9 kW power suppressed, i.e., saved as per condition. = 89% [Since, 10 kW considered 100%, so 8.9 kW in 89%]

Example 31 Refer to the following figure. Determine the effect of a phase error in the local oscillator on synchronous DSB demodulation. xDSB(t)

d(t)

y(t) LPF

~

cosct

Fig: Synchronous demodulator

Solution: Let  be the phase error of the local oscillator in the given figure. Then the local carrier is expressed as cos( c t   ). Now

x DSB (t )  m(t ) cos  c t d (t )  [m(t ) cos  c t ] cos( c t   )

and



1 m(t )[cos  (cos2 c t   )] 2



1 1 m(t ) cos  m(t ) cos(2 c t   ) 2 2

The second term on the R. H. S. is filtered out by the low-pass filter, and we get

y (t )  When 

1 m(t ) cos 2

is constant, this output is proportional to m(t). The output is completely lost when

    2 . Thus, as long as  is constant and not equal to   2 , the phase error in the local carrier causes attenuation of the output signal without any distortion. If the phase error  varies randomly with time, then the output also will vary randomly and is undesirable. Example 32 When a broadcast AM transmitter is 50% modulated, its current is 12 A. What will be the current when the modulation depth is increased to 0.9? Solution: We know It = Ic 1 

m 2 here I = 12 A , 2 m = 0.5 Ic = ?



12  I c 1 

0.5 2 2



12  I c 1.125  I c 1.06



Ic 

12  11.32 1.06

m  0.9, Ic  11.32, I t  ?

Next,

I t  11.32 1 

(0.9) 2 2

= 11.32 1.18  13.35A Example 36 Take the example of an analog baseband communication system with additive white noise. The transmission channel is assumed to be distortionless and the power spectral density of white noise /2 is 10–9 watt per hertz (W/Hz). The signal to be transmitted is an audio signal with 4 kHz bandwidth. At the receiver end, an RC low-pass filter with a 3dB bandwidth of 8 kHz is used to limit the noise power at the output. Determine the output noise power. Solution: The frequency response H() of an RC low-pass filter with a 3 db bandwidth of 8 kHz is given by

H   

1 1  j /  0

where  0  2 8000  . Then the output noise power N0 is





1 N 0  E n t   2 

2 0

 1 2 2





  2 H  

d



1

 1   /  

2



2

d

0

1 1  0  2 109   2  8  103 W 4 4  25.2 W

Chapter 3: Angle Modulation Example 6 A modulating signal 5cos 2 (15 103 )t , angle modulates a carrier A cos  c t . a) Find the modulation index and the bandwidth for FM systems. b) Determine change in the bandwidth and modulation index for FM, if fm is reduced to 5KHz. Solution: a) Here Em  5V ; f m  15KHz So frequency deviation

f  k f E m  15  103  5  75KHz mf 

So

75 5 15

Hence bandwidth

 2(f  f m )  2(75  15)  2  90  180KHz

mf 

b)

75  15 5

Therefore, Bandwidth

 2(f  f m )  2(75  5)  160 KHz Example 22 A FM wave is represented by voltage equation v = 20 sin (5  108t + 4 sin 1500 t). Find the carrier, modulating frequency modulation index, and maximum deviation of the FM. What power will this FM voltage dissipate in a 20  resistor? Solution: A = 20 (Emax), c = 5  108, mf = 4

5  10 8 5  10 8 f    79 .6 MHz. c 2 6.28 m = 1500

f

m



mf 

1500 1500   238.8 Hz. 2 6.28

f f m

 f = mf  fm = 4  238.8 = 955 Hz

E 2rm s (20 / 2 ) 2 20  20    10 W Power  R 20 20  2 Example 27 A FM signal has deviation of 3 kHz and modulating frequency of 1 kHz. Its total power is PT = 5W developed across a 50  resistor. The carrier frequency is 160 MHz. (i) Calculate RMS signal voltage VT. (ii) If J 0 = - 0.26, J 1 = 0.34, J 2 = 0.49 Find the amplitude of carrier and first two side bands. (iii) Find power in side bands (iv) Plot the frequency spectrum. Solution: Df = 3 kHz, PT = 5 W, R = 50 , fm = 1 KHz, FC = 160MHz



Df



fm

3 103 3 1103

2 2 2 We have, P  P [ J (  )  2( J (  )  J (  ) )] T



C

0

1

2

5  P [(0.26) 2  2(0.342  0.49 2 )] C 5 = PC [0.0676 + 0.1156 + 0.2401] 5 = PC (0.779)

PC 

5  6.418 W 0.779

We have,

E2 P  C C R

EC  PC R  6.418  50 EC = 17.91 V = Amplitude of Carrier The amplitudes of sidebands can be determined by En = Jn() EC = 0.26 + 17.91 = 4.65 V E1 = J1() EC = 0.34  17.91 = 6.08 V  E2 = J2 () EC = 0.49  17.91 = 8.77 V

Power in side bands is given by,

Pn  E 2 / R n  P0  E 02 / R  (4.65 ) 2 / 50  0.43W

 P1  E12 2 / R  (6.08 ) 2 / 50  0.739W  P2  E 22 / R  (8.77 ) 2 / 50  1.53W VT2  PT R;VT2  5  50 VT  250  15 .81 V Example 29 In an Armstrong FM transmitter with a carrier frequency of 152 MHz and a maximum deviation of 15 kHz at a minimum audio frequency of 100 Hz, the primary oscillator is a 100 kHz crystal, oscillator and the initial phase modulation deviation is to be kept less than 12º in order to minimize distortion. Estimate the frequency multiplication to give the required deviation and specify the required mixer crystal and multiplier stages. Solution: Here, the maximum phase deviation of the modulator is



max

 120 

 180

 12 rad  0.2094  mP

Now, f max =   max f min = 0.2094  100 = 20.94 Hz = 0.02094 kHz So, the required frequency deviation is, 

Maximum frequency deviation allowed f max (allowed)  Maximum frequency deviation f max



15 103  716.33 0.02094  103

Also 729 = 36, this requires a chain of six tripler stages to give a total deviation of 0.02094  729 = 15.265 KHz at a carrier frequency of 72.9 MHz. The mixer oscillator signal is f0 = (152- 72.9) = 79.1 MHz Here f0 can be contained by employing two tripler stages from a 8.7889 MHz crystal oscillator.

Example 30 The following is a block diagram of an indirect (Armstrong) FM transmitter. Calculate the maximum frequency deviation f of the output of the FM transmitter and the carrier frequency fc if f1 =200 KHz, fLO = 10.8 MHz, f1 = 25 Hz, n1 = 64, and n2 = 48. m(t)

Frequency Multiplier

NBFM

 n1 f1

xc(t)

Frequency Multiplier  n2 f2 = n1f1

f1

f3 ~

fc f

fLO

Solution:

f  (f1 )(n1 )(n2 )  (25)(64)(48) Hz  76.8 KHZ f 2  n1 f1  (64 )( 200 )(10 3 )  12 .8(10 6 ) Hz 12.8 MHz

23.6MHz f 3  f 2  f LO  (12.8  10.8)(106 ) Hz   2.0 MHz Thus, if f3 = 23.6 MHz, then

f c  n2 f 3  (48 )( 23 .6)  1132 .8 MHz If f3 = 2 MHz, then

f c  n2 f 3  (48 )( 2)  96 MHz. Example 32 The crystal oscillator frequency in an Armstrong-type FM generator of the following figure is 200 KHz. To avoid distortion, the maximum phase deviation is limited to 0.2. f m ranges from 50 Hz to 15 KHz. The carrier frequency at the output is 108 MHz, and the maximum frequency deviation is 75 KHz. Select multiplier and mixer oscillator frequencies. m(t)

Frequency Multiplier

NBFM

 n1 f1

 n2 f2 = n1f1

f1

Solution: Referring to the given figure, we have

f1   f m  (0.2)  (50)  10 Hz

f 75 103   7500  n1n2 f1 10

xc(t)

Frequency Multiplier

f3 ~

fLO

fc f

f 2  n1 f1  n1 (2  105 ) Hz Let us assume down conversion. Then, we have

f 2  f LO 

fc n2

Thus,

f LO  n1 f1 

fc 7500(2 105 )  108 106  n2 n2 

1392  106 Hz n2

Let n2 = 150. Then, we get n1 = 50

and

fLO = 9.28 MHz.

Example 35 Determine the resonant frequency of series combination of a 0.001 F capacitor and 2 mH inductor. Solution: The resonant frequency is given by

fr  fr 

1 2 LC 1 2 0.001  10 6  2  10 3

= 112.53  103 Hz = 112.53 KHz.

Chapter 4: A/D & D/A Converter Example 5 Compare the maximum conversion periods of an 8-bit digital ramp ADC and an 8-bit successive approximation ADC if both utilize a 1 MHz clock frequency. Solution: For the digital-ramp converter, the maximum conversion time is (2N  1)  (1 clock cycle) = 255  1 s = 255 s For an 8-bit successive-approximation converter, the conversion time is always 8 clock periods, i.e., 8  1 s = 8 s. Thus, the successive-approximation conversion is about 30 times faster than the digital-ramp conversion.

Chapter 6: Pulse Modulation Question 2: What are the disadvantages for digital communication? Answer: Disadvantages 1. Digital transmission system are incompatible with all practical analog system. 2. Digital transmission requires Synchronization. 3. Digital Systems are expensive. Question 7: Compare TDM and FDM systems. Answer: 1) TDM instrumentation is somewhat simpler than FDM. FDM requires modulators, filters and demodulators. TDM however needs commutators and distributor only. 2) TDM is invulnerable to the usual sources of FDM inter channel cross talk. In TDM virtually no cross talk occur when pulse are completely isolated and are non-overlapping. 3) The bandwidth required for TDM system, for multiplexing n signals, each band limited to fm Hz is nfm and if modulated by a carrier it becomes 2n f m. Now if the ‘n’ signals are multiplexed in PDM, using (SSB) technique, the bandwidth is nfm. 4) Because of the advantages of TDM over FDM, TDM systems are commonly used in long distance telephone communication. Question 9: Calculate SNR for PCM systems. Explain with an example. Answer: Pulse-code modulation (PCM) is a digital representation of an analog signal where the magnitude of the signal is sampled regularly at uniform intervals, then quantized to a series of symbols in a numeric (usually binary) code. PCM has been used in digital telephone systems and 1980s-era electronic musical keyboards. It is also the standard form for digital audio in computers and the compact disc format. There are two major sources of noise in a PCM system (I) Transmission noise introduced outside the transmitter (II) Quantization noise introduced in the transmitter. The quantized staircase waveform is an approximation to the original waveform. The difference between the two waveforms amounts to ‘noise’ added to the signal by the quantizing circuit. The mean square quantization noise voltage has a value of

2 E nq 

S2 12

Where S is the voltage of each step or the sub range voltage span. m7(t) 8 7 6

m(t)

5 4 3 2 1 0

As a result, the number of quantization level must be kept high in order to keep the quantization noise below some acceptable limit, given by power signal-to-noise ratio, which is the ratio of average signal power to average noise power for a sinusoidal signal which occupies the full range, the mean square signal voltages is, 2

E S2 

1 2 1  MS  (MS)2 E peak     2 2 2  8

where

M = Number of steps S = Step height voltage

The signal-to-noise ratio is now given by,

Signal ES2 ( MS ) 2 12 3 2  2   2 M Noise Enq 8 2 S The number of levels M is related to the number n of bits per level by, M = 2n.



Signal S 3 3    (2n )2   22n Noise N 2 2

S 3  or    10 log   2 2 n   N  db 2 

 1.761  6.02ndb In PCM SNR can be controlled by transmission band width. The same feature is also observed for FM or PM. But that case it require double of band width to quadruple the SNR So PCM compare to PM or FM.

Example 9 Suppose that a binary channel with bit rate Rb = 36000 bps is available for PCM voice transmission. What will be the appropriate values of the sampling rate f s, the quantizing level L, and the binary digits n, assuming fM = 3.2 KHz. Solution: It is required that

f s  2 f m  6400

and nf s  Rb  36000 .

So,

n

Rb 36000   5. 6 fs 6400

Therefore,

fs 

36000  7200 Hz = 7.2 KHz 5

Example 10 Consider an analog signal to be quantized and transmitted by using a PCM system. When each system at the receiving end of the system must be known to within 0.5 % of the peak-to-peak full-scale value, calculate the number of binary digits that each sample must contain. Solution: Let 2 m p be the peak-to-peak value of the signal. The peak error is then 0.005( 2 m p )  0.01m p . The peak-to-peak error is 2( 0.01m p )  0.02m p (the maximum step size ). Therefore, the required number of quantizing levels is

L

2m p 



2m p 0.02m p

 100  2 n

Thus, the required number of binary digits is n  7. Example 15 Calculate the roll-off factor  for a communication channel of bandwidth 75 KHz which will be required to transmit binary data at a rate of 0.1 Mb/s using raised-cosine pulses. Solution:

Tb 

1  105 s 6 0.110

f B  75 kHz  75  103 Hz

Then, we have

1    2 f B Tb  2(75 )(10 3 )(10 5 )  1.5 Hence we get the roll-off factor as  = 0.5

Chapter 7: Digital Modulation Question 4: Draw the diagram of Integrate and Dump Receiver. Answer: Integrate and Dump Receiver The circuit diagram of Integrate and Dump filter is shown in figure. White noise, n(t)

x(t)

+

+ + Integrator

Fig: Integrate and dump type filter

The input to the integrator and output is shown below: v(t) x(t) A

t Integrator input variations

0

t Integrator output

Question 5: What is Matched filter? Answer: Matched filter Matched filter is a type of filter in which after a time delay, the transfer function of the optimum filter is the same as the complex conjugate of the spectrum of the input signal. The Block diagram of a matched filter is shown in figure. y(t)

Matched filter

x(t)

Decision device

Threshold

Fig: Matched filter Receiver

The figure clearly depicts that it consists of a filter matched to input signal, a sampler and a decisiondevice. At time t = T, matched filter output is sampled and amplitude of this sample is compared with a present threshold . If threshold is exceeded, receiver decides that known signal S(t) is present, otherwise as shown it will decide that it is absent. Question 10: Draw the block diagram of Band Pass Binary delta transmission scheme. Answer: Band Pass Binary delta transmission scheme Block diagram of a band bass being data transmission scheme using digital modulation is shown below. Transmitter

Local carrier

Clock pulse

Clock pulse

Node Modulator

Channel

+



Demodulator

+ Binary data Binary data output

Fig: Band pass binary data transmission system

Chapter 8: Information Theory & Coding Question 6: Prove that maximum entropy is maximum when all messages are equally likely? Also calculate the average information per message. Answer: Let us consider a memoriless source m emitting messages m1, m2, ……, mn with probabilities P1, P2, ……, Pn respectively. Conditions, P1  P2  ......  Pn  1 We know that, n

H   Pi I i i 1

We need to consider only terms –Pi log Pi and Pn log Pn [because Pn is a function of Pi]

So

dH d  Pi log Pi  Pn log Pn   dPi dPi 1  1   Pi   log e  log Pi  Pn   Pi   Pn P   log  n   Pi 

when Pn = Pi then

  log e  log Pn 

dH 0 dPi

Because this is true for all i so

P1  P2  ......  Pn 

1 n

 (1)

Similarly,

d 2H

1  ve for P1  P2  ...... Pn  . n dPi 2

So maximum entropy occurs when messages are equally likely.

When P1 = 1 and P2 = P3 = … Pn = 0, H(m) = 0, whereas the probabilities in equation (1) gives,

 Average information per message = log n. Example 4 A source generates one of four possible messages during each interval with probabilities

P1 

1 1 1 , P2  , P3  P4  . 2 4 8

Find the information content of each of these messages. Solution: We know that,

I

1 1 log 2   Pi

  

Example 7 A transmission channel has a bandwidth of 4KHs and signal to noise power ratio of 31. a) How much should the bandwidth be in order to have the same channel capacity if reduced to 15. b) What will be the signal power if bandwidth is reduced to 3KHz?

S ratio is N

Solution: a) Channel capacity, S  C  B log 2 1  bits / sec N   4  10 3 log 2 1  31  20K bits / sec . S when   ratio is reduced from 31 to 15 then we can write, N 20  10 3  B log 2 1  15 

or ,20  10 3  B  4

Therefore, Bandwidth, B = 5KHz b) In the second case, B = 3KHz So,

S  20  log 2 1    N 3 

So,

S    90.4 N

So,

S So, required    90.4 N

Example 10 An ideal receiver receives information from a channel with bandwidth B Hz. Assuming the message to be band limited to fm Hz, compare

S   N

ratio at receiver output to that at its input.

Solution: According to Shannon Hartley’s law, channel capacity at receiver input is

 S  C  B log 2 1  in  bits / sec  Nin  Similarly, channel capacity at receiver output is,

 S  C  B log 2 1  o  bits / sec  No  Since at receiver output, B = fm then

 S  C  f m log 2 1  o  bits / sec  No 

According to the problem, there is no information loss in the receiver. So,

  S  S  B log 2 1  in   f m log 2 1  o  N in  No    B

 S  fm  S  or, 1  in   1  o  N in  No    S Since  in  N in

 S   and  o  are very large compared to unity N    o

S So we can write,  o  No

  S in      N in

  

B / fm

Example 16 Consider the parity-check code having the following parity check matrix 1 0 1 1 0 0  H  1 1 0 0 1 0  0 1 1 0 0 1

(i) Find the generator matrix G. (ii) Find the code word that begins 101…. (iii) Let that the received word be 110110. Decode this received word. Solution: (i) Since H is a 63 matrix, n = 6 and k = 3. So,

1 1 0  P  0 1 1 1 0 1 T

Then the generator matrix G is

1 0 0 1 1 0  G [ I 3 P ]  0 1 0 0 1 1 0 0 1 1 0 1 T

(ii)

1 0 0 1 1 0    c = dG = 1 0 1 0 1 0 0 1 1   0 0 1 1 0 1 = 1 0 1 0 1 1

r  1 1 0 1 1 0

(iii)

1 0  1 s  rH T  1 1 0 1 1 0  1 0  0

1 0 1 1 0 1   0 1 0 0 0 1 0  0 1

As s is equal to the second row of HT, an error is at the second bit, the correct code word is 100110 and the data bits are 100. Example 17 Show that if Ci and Cj are two code vectors in an (n, K) linear Block code then their sum is also a code vector. Solution: We know that for any code, CHT = 0 So we can write, Ci H T  0 and C j H T  0 So,

So,

C i  C j  is also a code vector.

Example 22 A signal amplitude X is a randomly variable uniformly distributed in the range (-1,1). This signal is passed through an amplifier of gain 2. The output y is also a random variable, uniformly distributed in the range (-2, 2). Determine the entropies H(x) and H(y). Solution: We have P  x  

1 2

x 1

=0

P  x 

1 2

otherwise

x 1

=0

otherwise

Hence, 1

1 H  x    log 2 dx  1bit 2 1 H  y 

2

1

 4 log 4 dx  2 bits

2

Chapter 9: Satellite Communication

Cross-talk All communication channels are of limited bandwidths. Hence a communication channel can be represented by an RC low pass filter as shown in Figure (a) below whose upper cut-off frequency is

fC 

1 2RC

Now, when a pulse is applied to this channel, the output of the channel will be distorted as shown in Figure (b). This is due to the HF limitations of the channel. It can be seen that the signal pulse allotted to the time slot 1 extends into the time slot 2, resulting in cross-talk. On an average, we can assume that the multiplexed signals are equally strong. Hence during the time slot 2, the area corresponding to the signal, i.e., A2 (not shown in the figure) will be the same as A1. We can define the cross-talk factor, K, as the ratio of the cross talk signal to the desired signal. Thus

K

A12 A12  A2 A1

… (1)

The pulse in time slot 1 is almost rectangular (the figure here is exaggerated for clarity). Hence,

A1  V

… (2) R C Fig: (a)

V Guard time T.S.2

 T.S.1

g

A1 Time constant A12c = RC

V

t Fig: (b)

In the above Fig. Cross Talk due to HF Cut-off of Channel; (a) Low pass Filter Representation of the Channel (b) Output of Channel of (a) for a pulse Input Now,

A12  V c e

 g / c

(1  e / c )

… (3)

The time constant  C must be much smaller than  in order to have minimum cross talk. Therefore,

 c   , and Equation.3 becomes A12  V c e

 g / c

… (4)

From Equation (1), (2) and (4), we get

K

 c  / e  g

… (5)

c

The cross-talk factor K should be as low as possible. Equation 5 suggests that  g should be much larger than  c . Of course, there are other considerations as well. For example, a large  g will result in less number of channels that can be multiplexed, and/or reduced signal strength. This type of cross talk is restricted to the neighboring channel because  c is very small, and hence the pulse ends in the neighboring time slot only.

Question 9: What is ISL (inter satellite link)? Answer: Occasionally, there is an application where it is necessary to communicate between satellites. This is done using satellite cross-links or inters satellite links (ISLs). Cross-link Sat 1

Sat 2

Up-link/down-link

ES 1

Earth

Fig:

Up-link/down-link

ES 2

Inter Satellite link (ISL)

A disadvantage of using an ISL is that both the transmitter and receiver are space bound. Consequently, both the transmitter’s output power and the receiver’s input sensitivity are limited.

Question 15: Define antenna pattern and study how the coverage or footprint depends on beamwidth? Answer: An antenna pattern is a plot of the field strength in the far field of the antenna when the antenna is driven by a transmitter. 3 dB beam width of the antenna is given by,

3dB 

75 degrees D

Thus from then we see that the coverage or footprint depends on beam width. A table below shows the dependence. Beam width ()

Coverage dia (D)

10°

3821 miles

5.7°

2235 miles

2.8°

1117 miles

1.0°

382 miles

0.7°

223 miles

Example 6 A satellite moving in an elliptical eccentric orbit has the semi major axis of the orbit equal to 16000 Km (Fig. Shown below). If the difference between the apogee and the perigee is 30100 Km, determine the orbit eccentricity. Solution: Apogee = a (1 + e) Perigee = a (1 – e) Where, a = semi-major axis of the ellipse e = orbit eccentricity Apogee – Perigee = a(1 + e) – a(1 – e)

Orbit

= 2ae or, Eccentricity, e 

 

Apogee-Perigee 2a

30100 2  16000

30100 32000

 0.9406

b

Earth a

Example 7 The farthest and the closest points in a satellite’s elliptical eccentric orbit from earth’s surface are 30,000 Km and 200 Km respectively. Determine the apogee, the perigee and the orbit eccentricity. Assume radius of earth to be 6370 Km. Solution: Apogee

= 30000 + 6370 = 36370 Km

Perigee

= 200 + 6370 = 6570 Km

Eccentricity



Where

a = semi-major axis of the elliptical orbit

Also,

a

Apogee  Perigee 2a Apogee  Perigee 2

or,2a  Apogee  Perigee Therefore, orbit eccentricity





36370  6570 36370  6570



29800  0.693 42940

Apogee  Perigee Apogee  Perigee

Example 8 Refer to Fig. showing a satellite moving in an elliptical, eccentric orbit. Determine the apogee and perigee distances if the orbit eccentricity is 0.5.

Earth C

O

S

15000 km

Solution: The distance from center of ellipse (o) to the center of earth (c) is given by (a × e) where (a) is the semi-major axis and (e) is the eccentricity.

Therefore, a × e = 15000

a

15000  30000 Km 0.5

Now apogee

 a(1  e)  30000(1  0.5)  45000 Km Perigee

 a(1  e)  30000(1  0.5)  30000 0.5  15000 Km Example 14 The semi-major axis and the semi-minor axis of an elliptical satellite orbit are 20,000 Km and 16000 Km respectively. Determine the apogee and perigee distances. Solution: If (ra)and (rp) are the apogee and perigee distances respectively, then semi-major axis  Semi-minor axis  ra rp

ra  rp 2

 20000 Km  ra  rp  40000 Km

ra  rp  16000  ra r p  256000000 Now

ra  rp  40000 ......( 1 )

ra  r p  256000000 ......( 2) Substituting the value of (rp) from (2) in (1)

ra (40000  ra )  256000000 or,

ra  40000ra  256000000  0 2

ra  

40000  16  10 8  10 .24  10 8 2

40000  5.76  10 8 2

40000  2.4  10 4  2  3.2  10 4 ,1.6  10 4

 32000 Km, 16000 Km

ra  rp 2

ra = 32000 Km as it cannot be 16000 Km if the semi-major axis is 20,000 Km.

r p  40000  32000  8000 Km Example 23 What would be the new maximum coverage angle and the slant range if the minimum possible elevation angle is 5 and not zero as in Example – 22. Solution: The maximum coverage angle, (2 max) is given by

 R e    cos E min  2 max  2 sin 1   R e  H  

2max

where Emin = Minimum elevation angle

 6378   2 max  2sin 1   cos 5  6378  35786   1  2sin [0.1512  0.996]

5

5

 2sin 1 0.1506  2  8.66  17.32 Example 24 The Fig.

(Showing below) a geostationary satellite orbiting earth. Calculate the angle ()

subtended by the arc of the satellite’s footprint at the center of the earth. Solution:

  1  2 1  90  1  E1

( E1  5)

 2  90   2  E2 ( E2  0)

Therefore,

  (90  1  E1 )  (90   2  E2 )  180  (1   2 )  ( E1  E2 )  175  (1   2 ) 35786 km

6378 cos 5 6378  35786 6378cos 5  sin 1 42164  8.66

1  sin 1



B

90

6378 km A

2

6378 cos 0 6378  35786  8.69

1

 2  sin 1

Therefore,   175  (8.66  8.69)  157.65.

Example 28 A satellite is currently in its elliptical transfer orbit (Fig.) with apogee and perigee being at distances of 35786 Km and 300 Km respectively above the surface of earth. If the transfer orbit inclination to the equatorial plane is 0, the incremental velocity to be given to the satellite at the apogee point by the apogee kick motor to circularize the orbit (Assume earth’s radius = 6378 Km). Circular path

35786 km

Transfer orbit

Equator

300 km Perigee

Solution: The velocity (V) at any point along an elliptical orbit is given by: 2 1 V     r a

Where  = GM r = apogee distance from center of earth a = semi – major axis of ellipse Now

r = 35786 + 6378 = 42164 Km

a

35786  6378  6378  300  24421 Km 2

The velocity (Va) at the apogee point can then be computed from 2 1 Va     r a 1  2  Va 2  39.8  1013   1013   10 3  24421  42164  

39.8  1013  6678 42164  24421  10 3

Va 2  2.58  10 6 Va  2.58  10 6  1.61 Km / s

For a circular orbit with a radius of 42164 Km,

 39.8  1013  r 42164  10 3  3.07 Km / s

Vc 

The velocity change (V) required to circularize is given by

V  Va 2  Vc 2  2Va Vc cosi Here

i = 0°

Therefore,

V  Va  Vc  2VaVc 2

2

Example 30 Determine the maximum line of sight distance between two communication satellites moving in a circular orbit at a height (H) above the surface of earth. Orbit

A 

C Earth

O  B

Solution: Maximum line of sight distance between the two satellites would occur when the satellites are so placed that the line joining the two becomes tangent to the earth’s surface as shown in Fig. Maximum line of sight distance = AB = OA + OB = 2 × OA or 2 × OB as OA = OB If (R) us the radius of earth and (H) the height of the orbit, then OA  AC sin   R  H  sin 

Now,

 R   RH 

  cos 1 

 

Therefore, OA  R  H  sin  cos 1

R   RH



 R    R  H 

Maximum line of distance  2R  H  sin cos1 



Example 31 If the two satellites in Example – 30 have an orbital radius that is twice the radius of earth, determine the maximum line of sight distance. Repeat the Example for geostationary satellites having an orbital radius of 42164 Km. (Assume earth’s radius, R = 6370 Km). Solution: Maximum line of sight distance   R   2R  H  sin cos 1    R  H  

It is given that (R + H) = 2R Therefore, maximum line of sight distance

 2  2 R sin cos 1 0.5  4 R sin 60 4 3 R 2  2 3R 

 2 3  6370  22066 Km For geostationary satellite, (R + H) = 42164 Km Therefore, maximum line of sight distance

6370    2  42164sin  cos 1  42164    84328sin(cos 1 81.3)  84328  0.998  83361Km Example 43 Te G1.Te1

G1.Te2

G1.Te3

Fig. shows a cascaded arrangement of three gain blocks. It is given that G1 = 106 and its associated equivalent noise temperature Te1 = 100 deg Kelvin, G2 = 104 and Te2 = 60 deg K, G3 = 1000 and Te3 = 20 deg K. Determine the equivalent noise temperature of the cascaded arrangement. Solution: If (Te) is the equivalent noise temperature of the cascaded arrangement, Then Te  Te1   100 

Te 2 T  e3 G 1 G 1G 2 60 10

6



20 10  10 4 6

 100  60  10 6  20  10 10  100 K

Example 44 Fig. Shows the cascaded arrangement of four gain blocks with their gain and noise figures as G1 = 100, F1 = 2, G2 = 10, F2 = 6 , G3 = 10, F3 = 15, G4 = 10, F4 = 20. Determine the noise figure of the cascaded arrangement. G1=100 F1=2

G2=10 F2= 6

G3=10 F3=15

G4=10 F4=20

Solution: The overall noise figure (F) is given by

F  F1 

F2  1 F3  1 F4  1   G1 G1G2 G1G2G3

6  1 15  1 20  1   100 100 10 100 10 10  2  0.05  0.014  0.0019  2.0659dB  2

Example 45 Fig. shows the receive side of satellite earth station comprising of earth station antenna followed by wave guide that connects the antenna feed point to the low noise amplifier input with the output of the low noise amplifier feeding the down-converter. Assume that the receive antenna has a gain of 66 dB at the received down link frequency of 11.9 GHz. The other parameters characterizing the receive chain are: Antenna noise temperature, TA = 60 K Loss in the wave guide, L1 = 1.075 (0.3 dB) Equivalent noise temperature of low noise amplifier Te2 = 160 K Low noise amplifier gain, G2 = 106 Down converter equivalent noise temperature, Te3 = 10000 K Ambient temperature, T0 = 290 K Determine the earth station system noise temperature and (G/T) referred to the input of the low noise amplifier. TA

Wave guide (L1, Te1)

Low noise Amplifier (G2, Te2)

Down Converter (Te3)

Solution: Earth station system noise temperature referred to the input of the low noise amplifier is given by: T  Ts  Te

where Ts = Noise temperature measured at the output of the wave guide Te = Equivalent noise temperature at the input of low noise amplifier

Now

Ts 

T A L1  1T0  L1 L1

60  1.075  1   290 1.075  1.075   55.81  20.23 

 76.04 K

Te  Te 2   160 

Te3 G2 10000 10 6

 160.01

T  76.04  160.01  236.05 K G  66  0.3  65.7 dB G  G in dB   10 log T T  65.7  10 log 236.05  65.7  23.73  41.97 dB / K .