San Jose State University
EE255
Spring 2017
Prof. Kamali
Solution to Problem Set #4
1. Problem 6.1
1
1
1

 50  sec
, thus Ts 
2 B 20 KHz
2
PT
2Es
P
 2 s   s  5 SNR per symbol
b) SNR = 10dB = 10 = r  r s 
N 0 B N 0 BTs
N0
s
5
  2.5 SNR per bit
For 4-QAM,  b 
log 2 M 2
s
5
  1.25
c)  s  5 and  b 
log 2 M 4
SNR per bit in 16-QAM is half SNR per bit for 4-QAM.
a) For sync pulse BT s 
2. Problem 6.2
 d

 d

 d

a) Pb  p(mˆ  0 | m  1) p(m  1)  p(mˆ  1 | m  0) p(m  0)  0.7Q min   0.3Q min   Q min 
 2N 
 2N 
 2N 
0 
0 
0 



 2 A2
Pb  Q
 N0





b) We use the following decoding scheme
ˆ  0 if r  s  n  a
m
ˆ  1 if r  s  n  a
m
Therefore
 Aa 

p(mˆ  0 | m  1) p(m  1)  0.7 p( A  n  a)  0.7 p(n  a  A)  0.7Q
 N /2 
0


 A a 

p(mˆ  1 | m  0) p(m  0)  0.3 p( A  n  a)  0.3 p(n  a  A)  0.3Q
 N /2 
0


Given A and N 0 the value of the threshold a can be found from the following
equation.
 Aa 


  0.3Q A  a 
0.7Q
 N /2
 N /2
0
0




Clearly a  0 .
c)
p(mˆ  0 | m  1) p(m  1)  p(mˆ  1 | m  0) p(m  0)
 A 


  0.3Q B 
0.7Q
 N /2 
 N /2 
0
0




The above equation can be used to find B when A and N 0 are given. Clearly A  B .
d)
Eb
 10  Eb  1  A 2  A  1
N0
2 

In (a), Pb  Q 2 A   3.87  10 6


N 0 
In (b), a  0.0203
In (c), B  0.9587
3. Problem 6.4
Tb
T
 Ac b
 Ac 
sin( 4f c t )   AcTb
0 s(t ) cos(2f ct )dt  0  Ac cos (2f ct )dt  2 0 (1  cos(4f ct ))dt  2 Tb  4f c   2  1
0 

Tb
Tb
2
Note that the second term above is zero because f c  1 .
x  1  n(t )  1  1.1e j  (1  1.1cos )  j sin 
Now the probability of error will be as follows.
Pe  p0 P(1  1.1 cos   0)  p1 P(1  1.1 cos   0)
P(1  1.1cos  0)  P(cos   
1
1
)  P(   ) 
1.1
6
P(1  1.1cos  0)  P(cos  
1
1
)  P(  0) 
1.1
6
1
1
Pe  p0 P(1  1.1cos  0)  p1 P(1  1.1cos  0)  ( p0  p1 ) 
6
6
4. Problem 6.5
We use the nearest neighbor approximation.
 d

Ps  N d min Q min 
 2N 
0 

a) V.29: d min  2a
We note that 4 outer points each have three neighbors and 12 inner points each have five
neighbors (Note, however, that not all neighbors are at minimum distance, so this merely
establishes a conservative approximation).
N d min 
 2a 

Ps  4.5Q
 2N 
0 

3  4  5  12
 4.5
16
 2a 
4 2  8 3  4 4

 3 , Ps  3Q
 2N 
16
0 

 2a 
2  3  3 2


 2.4 , Ps  2.4Q
 2N 
5
0 

 3a
4  2  4  3  1 4

 2.67 , Ps  2.67Q
 2N
9
0

b) 16-QAM: d min  2a , N d min 
c) 5-QAM: d min  2a , N d min
d) 9-QAM: d min  3a , N d min




5. Problem 6.10
Assuming a binary transmission Tb  Ts 
Tc 
1
1
 c

 
BD v /  v fv
1
 16  sec
64 10 3
3 108
 0.75 sec  Ts So outage probability is a good
At 1mph: Tc 
900 10 6 16 / 36
measure.
At 10mph: Tc 
3 108
 75m sec  Ts So outage probability is a good
900 10 6 10 16 / 36
measure.
3 108
 7.5m sec  Ts So outage or outage combined
900 10 6 100 16 / 36
with average probability of error can be a good measure.
At 100mph: Tc 
6. Problem 6.13
d min


 E s sin
 d min  2 sin
E s  0.39 E s
a)
2
16
16
b) Using nearest neighbor approximation
 d

 0.39 Es
Ps  2Q min   2Q
 2N 
 2N
0 
0




  2Q 0.076 Es


N0






  2Q 0.076 s   M Q  m s


Where  M  2 and  M  0.076
c) We have seen in the course that using Moment Generating Function we can simplify
the average error probability calculation.






Ps   Ps ( s ) f  ( s )d s    M Q  m s f  ( s )d s  M

0
0
M
s
where g 
2
 0.038 . Using the identity given in the hint we have.
M 
g s
Ps 
1 
2 
1  g s
d) Pb 
s
1

g s 
0 1  sin 2   d
2

  
1
  1  1 
   0.038 s

1
 
 2
1
  
  0.076 s

Ps
4
e) For BPSK Pb 
1
 10 3 leads to  b  250
4 b
For 16PSK, since  s  4 b and Pb 
thus Pb  10 3 implies  b  822 .
Ps
1
1

, (c) leads to Pb 
and
0.076  16 b 1.216 b
4

So the power penalty in going from BPSK to 16PSK is 10 log
822
 5.2 dB.
250
7. Problem 6.17

3
d 
 1 
7
a) Pr  Pt K  0   10 1 
  10 mW
 100 
d 
1
 10 4   b  2500  34 dB
b) Pb 
4 b
c) Assuming a rectangular pulse shape BTs  1 and thus
SNR   b 
Pr
10 7
 14
 25.2 dB
N 0 B 10  3 10 4
Given  d B  4 dB, the percentage of location in the cell where Pb  10 4 can be found
as follows.
 b 
  Q 34  25.2   Q(2.2)  0.0139  1.39%
Q b
  
4


dB


8. [Final exam, Spring 2012] In a slow fading wireless channel, determine the required
 s for 16-QAM such that for 98% of the time (or in 98% of locations), Ps ( s )  10 6 .
 3 0 
 where  0 is the SNR per symbol.
Hint: In MQAM, Ps  4Q

M

1


Solution:
From Q-table we find out that Ps  10 6 requires
3 0
 5 and  0  125  21.0 dB.
M 1
We will then have
s 
0
 Log (1  Pout )

125
 6187  37.9 dB
 Log (0.98)
9. [Final exam, Spring 2012] Find dmin for an 8-PSK constellation having the same
energy as an 8-PAM constellation with d min  1 .
Solution:
For 8-PSK: d min

 d

2
 2 R sin , E s  R   min

8

 2 sin
8

For 8-PSK: d min  2d , E s 







2

1
21 2
21
M 2  1 d 2  21d 2  d min

3
4
4
Therefore dmin can be found as follows.
2
d min
 21sin 2

8
 d min  1.8
10. [Final exam, Spring 2015] In a slow fading wireless channel, the average received
SNR  s is measured to be 42.9dB. Find the maximum size MPAM (M is assumed to
be power of 2) modulation which can provide Ps  10 4 for 95% of the times.
 6 0
Hint: For large size MPAM constellations, Ps  2Q 
2
 M
symbol.

 where  0 is the SNR per


Solution:
s 
0
 Log (1  Pout )
 10
42.9
10

0
 Log (0.95)
  0  1000  30dB
From Q-table we find out that Ps  10 4 requires
M
6 0
 3.9 . Thus
M2
6 0
6000

 19.8
3.9
3.9
We therefore use 16QAM.
11. [Final exam, Spring 2016] In the following modulation schemes transmitting in an
AWGN environment, find out how much transmit power has to increase (by how
many dB’s) in order for the number of bits per symbol to go up by one with no
change in the symbol error rate ( Ps ).
a) MPAM
b) MQAM
Hint:
 6 s 

Ps  2Q
 M2 


 3 s 

Ps  4Q
 M 


MPAM
MQAM
Solution:
a) For the Ps to remain the same, we need
6 s ,1
M 12

6 s , 2
M 22
. Increasing number of bits by
one means that M 2  2M 1 . Therefore
 s,2 
M 22
 s ,1  4 s ,1   s , 2,dB  10 log 4   s ,1,dB  6   s ,1,dB
M 12
For the fixed noise, SNR goes up by 6dB when the transmit power goes up by 6dB.
b) For the Ps to remain the same, we need
3 s ,1
M1

3 s , 2
M2
. Increasing number of bits by
one means that M 2  2M 1 . Therefore
 s,2 
M2
 s ,1  2 s ,1   s , 2,dB  10 log 2   s ,1,dB  3   s ,1,dB
M1
For the fixed noise, SNR goes up by 3dB when the transmit power goes up by 3dB.