sm10_23

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Problem 10.23. For M-ary PAM, 
(a) Show that the formula for probability of error, namely, 
12e
M AP Q
M σ
−⎛ ⎞ ⎛= ⎜ ⎟ ⎜⎝ ⎠ ⎝
⎞⎟⎠ 
holds for M = 2, 3, and 4. By mathematical induction, show that it holds for all M. 
 
(b) Show the formula for average power, namely, 
2 2( 1)
3
M AP −= 
holds for M = 2, and 3. Show it holds for all M. 
 
Solution 
(a) M-ary PAM with the separation between nearest neighbours as 2A. Assume that all M 
symbols are equally transmitted. 
 
(i) For M=2, we have the result given in the text for binary PAM 
 
2
12
PAM
e
AP Q
M AQ
M
σ
σ
⎛ ⎞= ⎜ ⎟⎝ ⎠
− ⎛ ⎞= ⎜ ⎟⎝ ⎠
 
 
for M = 2. 
 
(ii) For M = 3, the constellation is: 
 
[ ] [ ]
[ ]
2 2
2 2
2 2
2 2
1 1| ( 2 ) is transmitted or | 0 is transmitted
3 3
1 | (2 ) is transmitted
3
1 1 ( 2 ) 1 1exp exp
3 2 3 22 2
1 1 1 1 ( 2 )exp exp
3 2 3 22 2
e
A A
A
P P y A A P y A y A
P y A A
y A ydy dy
y ydy
σ σπσ πσ
σ σπσ πσ
∞ ∞
−
−
−∞ −∞
= > − − + > < −
+ <
⎛ ⎞ ⎛ ⎞+= − + −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
⎛ ⎞ ⎛ ⎞−+ − + −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
∫ ∫
∫ A
4
3
A
dy
AQ σ
⎛ ⎞= ⎜ ⎟⎝ ⎠
∫
 
 
Continued on next slide 
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Problem 10.23 continued 
 
From the formula ⎟⎠
⎞⎜⎝
⎛−= σ
AQ
M
MPe
)1(2 , when M=3, ⎟⎠
⎞⎜⎝
⎛= σ
AQPe 3
4 . Thus the formula 
⎟⎠
⎞⎜⎝
⎛−= σ
AQ
M
MPe
)1(2 holds for M = 3. 
 
(iii) For M = 4, the constellation is: 
 
- 3A + A-A + 3A 
[ ] [ ]
[ ] [ ]
( ) ( )
2
22
2 2
2 20
1 12 | ( 3 ) is transmitted 2 or 0 | is transmitted
4 4
1 10 or 2 | is transmitted 2 | (3 ) is transmitted
4 4
1 1 ( 3 )exp
4 22
1 1 1 12 exp exp
4 2 4 22 2
e
A
P P y A A P y A y A
P y y A A P y A A
y A dy
y A y A
dy
σπσ
σ σπσ πσ
∞
−
∞
= > − − + < − > −
+ < > + + <
⎛ ⎞+= −⎜ ⎟⎝ ⎠
⎛ ⎞ ⎛+ ++ − + −⎜ ⎟⎜ ⎟⎝ ⎠
∫
∫ 2
2
2
1 1 ( 3 )exp
4 22
6
4
A
A
dy
y A dy
AQ
σπσ
σ
−
−∞
−∞
⎡ ⎤⎞⎢ ⎥⎜ ⎟⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦
⎛ ⎞−+ −⎜ ⎟⎝ ⎠
⎛ ⎞= ⎜ ⎟⎝ ⎠
∫
∫
 
 
where the factor 2 in the third last line, comes from the symmetry of the second and third 
terms of the first equation. From the formula ⎟⎠
⎞⎜⎝
⎛−= σ
AQ
M
MPe
)1(2 , when M=4, 
6
4e
AP Q σ
⎛ ⎞= ⎜ ⎟⎝ ⎠ . Thus the formula ⎟⎠
⎞⎜⎝
⎛−= σ
AQ
M
MPe
)1(2 holds for M = 4. 
 
 
(iv) Assume that the formula of Pe holds for (M-1)-ary PAM. By mathematical induction, 
we need to show it also holds for M-ary PAM. The (M-1)-ary PAM constellation may be 
illustrated as shown: 
K
 
 
Continued on next slide 
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only 
to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 
Problem 10.23 continued 
 
By adding one point P2 on the (M-1)-ary PAM constellation, which has the distance 2A 
from point P1, we obtain M-ary PAM constellation as follows (in practice, the average or 
dc level may be adjusted as well but this has no effect on the symbol error rate): 
K
 
 
Since error probabilities of P1 symbol on the (M-1)-ary PAM is the same as that of P2 
point on the M-ary PAM, the error probability of M-ary PAM is 
 
( 1)1 1 symbol error prob. of P1symbol on -aryM ary M arye e
MP P M
M M
− − −−= + ⋅ (1) 
 
where 1/M is the probability that P1 is transmitted and (M-1)/M is the probability that one 
of the other constellation points is transmitted. The probability of error formula for 
(M-1)-ary PAM is given by 
⎟⎠
⎞⎜⎝
⎛
−
−=−− σ
AQ
M
MP aryMe )1(
)2(2)1( . (2) 
 
The symbol error rate of P1 symbol on M-ary PAM is 
 
[ ]1 1 ( ),or ( ) | P1 is transmittedPP P y A y AM µ µ= < − > + 
 
 where µ is the signal level of P1 symbol. 
 
2 2
1 2 2
1 ( ) 1 ( )exp exp
2 2
2
A
P A
y yP dy
M M
AQ
M
µ
µ
µ µ
σ σ
σ
− +∞
−∞ +
⎛ ⎞ ⎛ ⎞− −= − + −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝
⎛ ⎞= ⎜ ⎟⎝ ⎠
∫ ∫ dy⎠ (3) 
 
 
Substituting Eqs. (2) and (3) into (1), we obtain the symbol error probability of M-ary 
PAM 
 
1 2 22
1
12
M PAM
e
M M AP Q
M M M
M AQ
M
AQσ σ
σ
− − − ⎛ ⎞ ⎛ ⎞= ⋅ ⋅ +⎜ ⎟ ⎜ ⎟− ⎝ ⎠ ⎝ ⎠
− ⎛ ⎞= ⎜ ⎟⎝ ⎠
. 
 
Continued on next slide 
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to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 
Problem 10.23 continued 
 
The formula holds for M-ary PAM. Therefore, by mathematical induction, the formula 
holds for all M. 
 
 
(b) To compute the average symbol power we note: 
i) For M = 2, the average symbol power is A2 and the formula 
2 2( 1)
3
M AP −= holds for 
M=2. 
 
ii) For M = 3, the average symbol energy is 
 
( )2 2 21 8(2 ) 0 (2 )3 3P A A= + + = 2A . 
The formula 
2 2( 1)
3
M AP −= holds for M=3. 
 
iii) For general even M, the M-ary PAM constellation points are 
 
{ }( 1) , , 3 , , ,3 , , ( 1)M A A A A A M− − − − −L L A . 
 
The average symbol energy is 
 
2 2 2
2
2 / 2
2
1
2 / 2 / 2 / 2
2 2
1 1 1
2
2 2
2 ( 1) ( 3) 3 1
2 (2 1)
2 2 4 1
2 ( / 2 1)( 1) ( / 2 1)4 4
2 6 2 2 2
( 1)
3
M
k
M M M
k k k
M M
P A
M
A k
M
A k k
M
A M M M M M M
M
M A
=
= = =
⎡ ⎤− + − + + +⎣ ⎦=
= −
⎡ ⎤= − +⎢ ⎥⎣ ⎦
+ + +⎡ ⎤= − +⎢ ⎥⎣ ⎦
−=
∑
∑ ∑ ∑
L
� �
. 
 
where we have used the summation formulas of Appendix 6. 
 
iv) For general odd M, the M-ary PAM constellation points are 
 
{ }( 1) , , 2 ,0,2 , ( 1)M A A A M− − − −L L A . 
 
 
Continued on next slide 
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to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 
Problem 10.23 continued 
 
 The average symbol energy is 
 
2 2 2
2
2 22
2 2
2 ( 1) / 2
2
1
2
2 2
2 ( 1) ( 3) 2
2 1 32 ... 1
2 2
8
8 ( 1)( 1)( )
2 2 6
( 1)
3
M
k
M M
P A
M
A M M
M
A k
M
A M M M
M
M A
−
=
⎡ ⎤− + − + +⎣ ⎦=
⎡ ⎤− −⎛ ⎞ ⎛ ⎞= + + +⎢ ⎥⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎢ ⎥⎣ ⎦
=
− +=
−=
∑
L
� �
 
 
where the fourth line uses the summation formula found in Appendix 6. 
 
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only 
to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.