Unformatted text preview: gration in t he complex plane, a s ubject b eyond t he scope of this book. 2 For our
purpose, we c an find t he inverse transforms from t he t ransform table 6.1. All we
need is t o express F (s) as a s um o f simpler functions o f t he form listed in t he t able.
Most of t he t ransforms F (8) o f practical interest a re r ational f unctions; t hat is,
ratios o f polynomials in s . Such functions can be expressed as a s um o f s impler
functions by using p artial f raction expansion (see Sec. B.5). Values of 8 for which
F (s) = 0 a re called t he z eros o f F (8)j t he values of s for which F (s)  + 00 a re
called t he p oles o f F (s). I f F (s) is a r ational function of t he form P (s)IQ(s), t he
r oots of P (8) a re t he zeros a nd t he r oots of Q (8) a re t he poles of F (8).
• E xample 6 .3 Find the inverse Laplace transforms of
78  6 ( a) 8 2 86 28 2 + 5
6(8 + 34)
8 s + 10
s 2+38+2 ( e) s (s2+108+34) ( d) ( 8+1)(s+2)3 ( a)
7s  6 F (s) E xercise E 6.1
B y direct integration, find the Laplace transform F (s) and the region of convergence of F (s)
for the signals shown in Fig. 6.3.
Answer: ( a) ~(l e 2s ) for all s. ( b) } (l e 2s )e 2s for ails. ' 7 T he d efinition of t he L aplace transform is identical to t hat o f t he Fourier transform w ith j w r eplaced by 8 . I t is r easonable t o e xpect F (8), t he Laplace transform
of f (t), t o b e t he s ame as F (w ), t he Fourier transform of f (t) w ith j w replaced by
s . For example, we found t he F ourier transform of e atu(t) t o b e 1 /(jw + a ). Replacing j w w ith s in t he Fourier transform results in 1 /(s + a ), which is t he Laplace
transform as s een from Eq. (6.16b). Unfortunately this procedure is n ot valid for
all f (t). We m ay use it only if t he region of convergence for F (s) includes the
imaginary ( jw) axis. For instance, t he Fourier transform of t he u nit s tep function
is IT.5 (w) + (1 / j w ). T he c orresponding Laplace transform is 1I 8, a nd its region of
convergence, which is R e 8 > 0, does n ot include the imaginary axis. I n t his case the
connection between t he Fourier a nd Laplace transforms is n ot so simple. T he reason
for this complication is r elated t o t he convergence of the Fourier integral, where the ) The inverse transform of none of the above functions is directly available in Table
6.1. We need to expand these functions into partial fractions discussed in Sec. B.5. /::; Connection t o t he Fourier Transform b
( = (s + 2)(s  3) kl
k2
= s+2+s3 To determine k l, corresponding to the term (s + 2), we cover up (conceal) the term (s + 2)
in F (s) and substitute s =  2 (the value of s t hat makes s + 2 = 0) in the remaining
expression (see Sec. B. 5 2)
7 s6 (8  3) I =  146 = 4 8 =2 2  3 Similarly, to determine k2 corresponding to the term (s  3), we cover up the term (s  3)
in F (s) and substitute s = 3 in the remaining expression Therefore
7s  6 F (s) = (s + 2)(s _ 4 3) = s +2 + 8 3
 3 (6.25a) 372 6 C ontinuousTime S ystem Analysis Using t he Laplace T ransform
T ab...
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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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