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Unformatted text preview: . Letting x = >. o n b oth
sides of this e quation, we o btain dx To find ai, we conceal the factor (x + 1)3 in F{x), take the derivative of the remaining
expression, and then let x =  1:
2
3
al = ..'!:. [4X + 16x + 23x + 1 3] 1
=1
dx
(x + 2)
x =l
Similarly,
2
3
2
=3
2 1 d [4X + 16x + 23x + 1 3] 1
a =2!dx 2 ~ x =l
Therefore, ( x  >.)"F(x)!x=>. = ao .:!. [(x  E xample B .lO B.54 + {x + 1)2 + X + 1 + x + 2 • A Hybrid Method: Mixture of the Heaviside "CoverUp" and
Clearing Fractions For multiple roots, especially of higher order, t he Heaviside expansion method,
which requires repeated differentiation, can become cumbersome. For a function
which contains several repeated a nd u nrepeated roots, a hybrid of t he two procedures proves t he b est. T he simpler coefficients are determined by t he Heaviside
method, a nd t he r emaining coefficients are found by clearing fractions or shortcuts,
thus incorporating t he b est of t he two methods. We d emonstrate this procedure by
solving Example B.lO once again by this method.
In Example B.1O, coefficients k a nd ao a re relatively simple t o d etermine by
t he Heaviside expansion method. These values were found t o b e k l = 1 a nd ao = 2.
Therefore,
4 x 3 + 16x 2 + 2 3x + 13
2
al
a2
1
 ,:;:::: =    (x+l)3(x+2) ( x+l)3 +  +  + (x+l)2 We now multiply b oth sides of t he above equation by (x
fractions. This yields = 2 (x x +l x +2 + 1)3(x + 2) t o clear t he + 2) + a l(x + 1 )(x + 2) + a2(x + 1)2(x + 2) + (x + 1)3 = (1 + a2)x 3 + (al + 4a2 + 3 )x 2 + (5 + 3 al + 5a2)X + (4 + 2al + 2a2 + 1) F{ )
ao
al
a2
k
x = { x+1)3 + ( x+1)2 + x +1 + x +2 E quating coefficients of t he t hird a nd second powers of x o n b oth sides, we o btain The coefficient k is obtained by concealing the factor ( x+2) in F{x) and then substituting
x =  2 in the remaining expression:
2
3
k = 4x + 16x + 23x + 13 1
=1 ~X=2 ~X=l 131 F{x) = {x + 1)3 4 x 3 + 16x 2 + 23x + 13 Expand F{x) into partial fractions if
2
3
F{x) = 4x + 16x + 23x + 13
(x + 1)3{x + 2)
The partial fractions are To find ao, we conceal the factor (x + 1)3 in F{x) and let x
expression:
2
3
ao = 4x + 16x + 23x + 13 1
=2 31 B.5 P artial F raction Expansion =  1 in the remaining We m ay s top here if we wish because t he two desired coefficients, a l a nd a2, a re
now determined. However, equating t he coefficients of t he two remaining powers of
x yields a convenient check on t he answer. Equating t he coefficients of t he x l a nd
x O t erms, we o btain
23 = 5 + 3 al + 5a2 13 = 4 + 2al + 2a2 + 1 Background 32
T hese e quations a re satisfied by t he values a l = 1 a nd a2
providing an a dditional check for o ur answers. Therefore, 3, found earlier, 2
1
3
1
F (x) =    +    +   +  (x + 1)3
( x + 1)2
X+1
x+2 B.5 P artial F raction E xpansion 33 t he o nly difference between t he p roper a nd i mproper case is t he a ppearance o f a n
e xtra c onstant bn i n t he l atter. O therwise t he p rocedure remains t he s ame. T he
p roof is left as a n exercise for t he r eader.
• E xampl...
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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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