Solutions manual to accompany introduction to wireless systems 1st edition 9780471321675

Chapter 3 Problem Solutions

Problem 3.1

A. Equation (3.25) gives r dB = Pt dB + Gt dB + Gr + 20log(ht ) + 20log(hr ) Lsys 4 10log(d ).

Substituting gives

= 87.5 dBm.

B. The wavelength is given by  = 3 108 1.2 109 = 0.250 m.

The free-space path loss is then  4d   4 20 103 

path dB = 20log   = 20log 0.250  = 120 dB.

The received power in free space is given by the range equation:

Pr dB = Pt dB + Gt dB + Gr dB Lsys dB Lpath dB = 10log 20  + 6 + 2 2 120

 0.001 = 71.0 dBm.

C. Equation (3.16) gives

= 2A sin 2 ht hr  . dir  d 

The argument of the sine function is 2hthr d = 2  20  3 = 0.0754 0.250  (20 103 )

Then sin(0.0754) = 0.0753 That is, the argument is about 0.1% larger than the sine.

Problem 3.2

From the geometry of Figure 3.2,

tan( i ) = ht x

x

which gives x = 1.15 103 m. Similarly,

tan(1 ) = 20 , tan( r ) = hr d x

tan(1 ) = 3 , d x

which gives d x = 172 m Adding gives d = 1.32103 m

Now Equation (3.16) gives

Equation (3.17) gives

Relative to the value obtained from Equation (3.16), the approximation of Equation (3.17) is 20log  2.29Adir  = 1.99 dB

Problem 3.3

We start by calculating the thermal noise floor. The noise figure is F = 9 dB or F = 7.94. The noise power referred to the receiver input is

In decibels this is Pn dB = 120 dBm. Then for a SNR of 17 dB we need a signal power of Pr dB = 120 + 17 = 103 dBm.

A. From Equation (3.25) we have 103 dBm = 10log 30

= 84.4 dBm 40log(d )

Solving gives d = 48.4 km.

B. In free space the range equation gives 103 dBm = 10log 30  + 6 + 3 3 L .

 0.001 path dB

This gives a maximum path loss of Lpath = 154 dB Now L = 20log 4d  At 850 MHz the wavelength is  = 0.353 m path dB   



Substituting gives d = 1.41106 m.

Problem 3.4

Since the noise and interference powers add, we will have to work with power rather than with decibels. From Problem 3.3 we know that the 18 noise floor is at Pn = 957 10 W The signal power is 17 dB higher, or P = 48.0 10 15 W . The signal power exceeds the sum of interference and noise by 16 dB, or by a factor of 39.8. That is, P 48.0 10 15 r = = 39.8. P + P 957 10 18 + P n i i

r

 0.001

= 84.4 dBm 40log(di ), is the distance from base station B to the hand-held receiver.

Solving gives di = 182 km The separation between base stations is then dAB = d + di = 48.4 + 182 = 230 km.

Problem 3.5

A. Equation (3.34) gives a(hre ) = 3.2(log(11.75h ))2 4.97 dBfor f  400 MHz

re 0

=3.2(log(11.751.6))2 4.97 dB

= 0.225 dB.

Then Equation (3.33) gives the median path loss as

i

L50 (urban)

= 69.55 + 26.16log( f0 ) 13.82log(hte )

+ (44.9 6.55log(hte ))log(d ) a (hre )

= 69.55 + 26.16log(1200) 13.82log(30)

+ (44.9 6.55log(30))log(12) 0.225

= 167 dB.

B. The received signal power is given by P = 10log 15  dBm + 7 dB + 2 dB 4 dB 167 dB

 

r dB  0.001

= 121 dBm.

C. Using Equation (3.25), P = 10log 15  + 7 + 2 + 20log(30) + 20log(1.6) 4 4 10log(12 103 )

r dB  0.001

= 82.8 dBm.

The flat Earth model gives an overly optimistic estimate of the received signal power.

Problem 3.6

We start by calculating the thermal noise floor. The noise figure is F = 6 dB or F = 3.98 The noise power referred to the receiver input is

Pn = kT0BF

= (1.38 10 23 )(290)(30 103 )(3.98)

= 478  10 18 W.

In decibels this is Pn dB = 123 dBm. Then for a SNR of 18 dB we need a signal power of Pr dB = 123 + 18 = 105 dBm.

A. Using the parameters from part B of Problem 3.5 we have 105 dBm = 10log 15  dBm + 7 dB + 2 dB 4 dB L (urban) .  0.001 50 db

The maximum path loss is gives L50 (urban) = 152 dB The Hata model

dB

152 dB = 69.55 + 26.16log( f0 ) 13.82log(hte ) + (44.9 6.55log(hte ))log(d ) a(hre )

= 69.55 + 26.16log(1200) 13.82log (30) + (44.9 6.55log(30))log(d ) 0.225.

Solving gives d = 4.35 km .

B. Using Equation (3.25), 105 dBm = 10log 15  + 7 + 2 + 20log(30) + 20log(1.6) 4 4 10log(d )

 0.001

Solving gives d = 43.1km .

The flat Earth model overestimates the maximum range by an order of magnitude.

Problem 3.7

Equation (3.35) gives

a(hre ) = (1.1log( f0 ) 0.7)hre (1.56log( f0 ) 0.8)

= (1.11log(1200) 0.7)1.6 (1.56log(1200) 0.8)

= 0.345 dB.

Then Equation (3.33) gives the median path loss in urban areas as

L50 (urban) = 69.55 + 26.16log( f0 ) 13.82log(hte )

+ (44.9 6.55log(hte ))log(d ) a (hre )

= 69.55 + 26.16log(1200) 13.82log(30)

+ (44.9 6.55log(30))log(12) 0.345

= 167 dB.

In suburban areas we have from Equation (3.36) L50 suburban dB

In rural areas Equation (3.37) gives

( ) ( ) ( ) ( )

Problem 3.8

= 138 dB.

In the Lee model, distances are measured in feet and miles. Converting the parameters given in Problem 3.5 gives ht

ft

r = 5.25 ft , and d = 7.46 mi Now Equation (3.50) gives

= 9.77 dB.

Table 3.1 gives P1 mile dB = 77 dBm and

= 4.8 for New York. Equation (3.49), modified to include system losses, gives P = 77 dBm 10  4.8log

,50 dB

= 133 dBm.

Problem 3.9

In the solution to Problem 3.6 we showed that the minimum received signal power is Pr dB = 105 dBm . Using the numbers from Problem 3.8, we have

105 dBm = 77 dBm 10  4.8log

4 dB.

Solving gives d = 1.98 mi , or 3.18 km. This is lower than the 4.35 km obtained in Problem 3.6 using the Hata model. Part of the difference may be a consequence of the large value of path-loss exponent that the Lee model uses for New York City.

Problem 3.10

From Equation (3.64) we have

where the fade margin fm dB is given by

Problem 3.11

The receiver sensitivity is pr dB = 110 dBm

15

95 dBm. From the solution to Problem 3.10 we have

This is equivalent to

Solving gives fm dB = 7.26 dB.

 

0.15 = Q  7 dB 

Fade margin is defined by fm dB = Pr dB pr dB Substituting gives Pr dB = pr dB + fm dB = 95 dBm + 7.26 dB = 87.7 dBm.

Problem 3.12

Equation (3.50) gives

 = 20log 100  + 20log 5  30log 1800  c      

100   10   850 

+ (44 40) + (4 8.15) + (2 2.15)

= 16.1 dB.

Table 3.1 gives P1 mile dB = 70 dBm and  = 3.68 for Philadelphia.

Equation (3.49), modified to include system losses, gives P = 70 dBm 10  3.68log 10  16.1 dB 2 dB

r ,50 dB

= 125 dBm.

A. A received signal power 10 dB above the median power is -115 dBm. From Equation (3.64) we have

 pr Pr 

Pr P  p  = Q  dB dB 

 r dB r dB   path 

Pr P  115 dBm = Q  115 dBm ( 125 dBm) 

 r dB   8 dB 

= Q (1.25) = 0.106.

B. At 5 mi the median received signal power is P r,50 5 mi dB = 70 dBm 10  3.68log 5  16.1 dB 2 dB  

= 114 dBm.

Then,

Pr P (5 mi)

 115 dBm = Q  115 dBm ( 114 dBm) 

 r dB   8 dB

Problem 3.13

The receiver’s thermal noise floor is given by

P = kT BF = (1.38 10 23 )(290)(200 103 )(3.98) = 3.19 10 15 W, or Pn dB = 115 dBm. The receiver sensitivity is 15 dB above the noise floor, or pr dB = 115 dBm + 15 dB = 100 dBm . Now

 pr Pr   100 dBm Pr  Pr P  p  = 0.90 = Q  dB dB  = Q  dB  .  r dB r dB      8 dB 

 path   

Solving gives P = 89.7 dBm Using the Lee model from Problem dB 3.12, 89.7 dBm = 70 dBm 10  3.68log d  16.1 dB 2 dB.  

r  1 

Solving gives the maximum operating range as d = 1.11mi or 1.78 km

Problem 3.14

A. From Problem 3.10 we have Pr P  p  = Q  f m dB   r dB r dB     0.95 = Q



 path    6 dB   

The required fade margin is fm dB = 9.87 dB.

B. The receiver’s noise figure is F = 4 dB or F = 2.51. We find the receiver’s noise floor to be

P = kT BF = (1.3810 23 )(290)(30 103 )(2.51) = 30210 18 W, or Pn dB = 125 dBm. The receiver sensitivity is 12 dB above this noise floor, or pr dB = 113 dBm Since the fade margin is given by fm dB = Pr dB pr dB , we have the required median received signal level Pr dB = fm dB + pr dB = 9.87 dB 113 dBm = 103 dBm.

C. The transmitted power is 35 W or Pt dB = 45.4 dBm The range equation gives

103 dBm = 45.4 dBm + 6 dB + 3 dB 3 dB Lpath dB ,

For a medium-sized city Equation (3.35) gives a(hre ) = (1.1log(f0 ) 0.7)hre (1.56log(f0 ) 0.8)

= (1.1log(1200) 0.7)(1.6) (1.56 log (1200) 0.8)

= 0.296 dB.

Then Equation (3.33) gives

L50 (urban) = 69.55 + 26.16log( f0 ) 13.82log(hte )

+ (44.9 6.55log(hte ))log(d ) a (hre )

= 69.55 + 26.16log(1200) 13.82log(hte )

+ (44.9 6.55log(hte ))log(8) 0.296

= 190 19.7log(hte ).

Setting 154 = 190 19.7log(hte ), gives the required antenna height as hte = 63.4 m

Problem 3.15

Equation (3.80) gives

The amplitude A is at its peak when

Now Then

Substituting and solving for f gives

Similarly a the amplitude A is at a null

c 1