Unformatted text preview: gnal Comparison: Correlation 1
5 T he Best Friends. Worst Enemies. and Complete Strangers E f6 We c an r eadily verify t hat if I (t) = K x(t), t hen en = 1 w hen K is a ny positive
constant, a nd e n =  1 when K is any negative constant. Also en = 0 if I (t) a nd
x (t) are orthogonal. Thus, t he m aximum similarity [when I (t) = K x(t)] is i ndicated
by en = 1, t he m aximum dissimilarity [when I (t) =  Kx(t)] is indicated by en =
 1. W hen t he two signals are orthogonal, t he similarity is zero. Qualitatively
speaking, we m ay view orthogonal signals as unrelated signals. Note t hat m aximum
dissimilarity is d ifferent from unrelated ness qualitatively. For example, we have t he
b est friends (en = 1), t he worst enemies (en =  1), a nd complete strangers, who
do n ot c are w hether we e xist or n ot (en = 0). T he enemies are n ot s trangers, b ut
in m any ways, people who t hink like us, only in opposite ways.
We can r eadily e xtend t his discussion t o complex signal comparison. We generalize t he d efinition of en t o include complex signals as 1 00 ~
1 en = V EjEx (3.27) I (t)x*(t) dt  00 x (t) (d) 0 .51, o o 5
(c) (b) (a) e o 5 o 5 1 f       '  t15
(e) ( I) e r o t Fig. 3 .4 Signals for Example 3.2. • E xample 3 .2
Find the correlation coefficient en between the pulse x (t) and the pulses Ii (t), i =
1, 2, 3, 4, 5, and 6 illustrated in Fig. 3.4.
We shall compute en using Eq. (3.25) for each of the 6 cases. Let us first compute
the energies of a ll the signals. 1
5 Ex = 2
x (t) dt = 1
5 dt = 5 (3.28) In the same way we find E ll = 5, E h = 1.25, and E!3 = 5. Also, to determine Ef4 and
Efso we determine the energy E of eatu(t) over the interval t = 0 to T : E = T r io (e at )2 dt = T r e 2at dt = ~a(1 _ e 2aT )
2 ~ For 14(t), a = 1 /5 and T = 5. Therefore E f4 = 2.1617. For 15(t), a = 1 and T =
Therefore E /5 = 0.5. The energy of Ef6 is given by 0 0. = 2 sin 27rt dt = 2.5 Using Eq. (3.25), the correlation coefficients for the six cases are found as
(1) _1_ (3)
( 5) [ d t= 1 [ 0.5 dt =1 ( 2)
1
V ( 1.25)(5) ~([(l)dt=l
5)(5) ( 4) 1
V (2.1617)(5) ~[
( 0.5)(5) ( 6) ~ 105 sin 27rtdt = 0
( 2.5)(5) V (5)(5) 0 0 0 e  t dt = 0.628 0( [ 0 ) e  t / 5 dt = 0.961 • Comments on the results: Because h (t) = x (t), t he two signals have t he m aximum
possible similarity a nd en = 1. However, t he signal 12 (t) also shows maximum
possible similarity with en = 1. T he r eason is t hat we have defined en t o m easure
t he s imilarity of t he waveshapes; it is i ndependent of t he a mplitude (strength) of t he
signals compared. T he s ignal h (t) is i dentical t o x (t) in shape; only t he a mplitude
(strength) is different. Hence en = 1. T he signal h (t), o n t he o ther h and, has
t he maximum possible dissimilarity w ith x (t) b ecause i t is e qual to  x(t). F or
14(t), e n = 0.961, implying a high degree of similarity w ith x (t). T his is r easonable
because 14(t) is very similar to x (t) over t he d uration o f x (t) (for 0 :S t :S 5). J ust
by inspection, we n otice t hat t he v ariations or changes in b oth x (t) a nd 14(t) a re
a t similar rate...
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This note was uploaded on 04/14/2013 for the course ENG 350 taught by Professor Bayliss during the Spring '13 term at Northwestern.
 Spring '13
 Bayliss
 Signal Processing, The Land

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