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Unformatted text preview: f t he u nilateral Laplace transform holds only for positive to
b ecause if to w ere negative, t he s ignal j (t  to)u(t  to) would n ot b e causal.
We c an r eadily verify t his p roperty in Exercise E6.1. I f t he s ignal in Fig. 6.3a
is j (t)u(t), t hen t he s ignal in Fig. 6.3b is j (t  2)u(t  2). T he L aplace transform
for t he p ulse i n Fig. 6.3a is ~(l  e 28 ). T herefore, t he L aplace t ransform for t he
28
p ulse in Fig. 6 .3b is ~(l e 28 )e • (6.30a) 2)  u(t  4) T he first term on the righthand side is the signal t u(t) delayed by 1 second. Also, the
third and fourth terms are the signal u ( t) delayed by 2 and 4 seconds respectively. The
second term, however, cannot be interpreted as a delayed version of any entry in Table
6.1. For this reason, we rearrange it as
(t  I)u(t  2) = (t  2 + I )u(t  = (t  2) 2)u(t  2) + u (t  2) We have now expressed the second term in the desired form as t u(t) delayed by 2 seconds
plus u (t) delayed by 2 seconds. With this result, Eq. (6.30a) can be expressed as loco j (x)e 8X dx = F (8)e 8to u(t  4)] j (t) = (t  l)u(t  1)  (t  2)u(t  2)  u(t  4) Application of the timeshifting property to t u(t)
(t  I)u(t  1) = 2. e Also
u (t) s 82 = ~s and and ¢ ==> 1 /8 2 yields (t  2)u(t  2) u (t  4) (6.30b) = ~se4S  = ~e2S
s
(6.31) Therefore
(6.32) • 3 84 6 C ontinuousTime S ystem A nalysis U sing t he L aplace T ransform S ome P roperties o f t he L aplace T ransform 2. / (1) 6.2 Frequency Shifting 385 T his p roperty s tates t hat i f 2 f (t) <=> F (s)
t hen o f (t)e sot <=> F(8  1 P roof: E xample 6 .5
F ind t he inverse Laplace transform of = F (s) .c [J(t)esotj s +3+5e 2s
(s + 1 )(s + 2) Observe t he e xponential t erm e  2s in t he n umerator of F (s), i ndicating timedelay. I n
such a case we s hould separate F (s) into terms with and without delay factor, as "v" roo f(t)esotest dt = roo f(t)e(sso)t dt = F(8 _ ~ ~ 8 0) E xample 6 .6
Derive P air 9 a in Table 6.1 from P air Sa a nd t he frequencyshifting property.
T he P air Sa is
8
cos b tu(t)
~b2
8 + F rom t he frequencyshifting property (Eq. (6.33)J with 80 =  a we o btain F 2 (s)e 211 where s +3
( s + 1 )(s + 2 ) = F 2(s) = ( s + 1 )(s + 2) / l(t) = ( 2e t = h (t)  e 2 = s +1 5 Therefore = = '..' F l (8) F l(s) • 5 e 28 s +3 + 1)(s + 2) + (s + 1 )(s + 2) F (s) = (s (6.33) O bserve t he s ymmetry ( or d uality) b etween t his p roperty a nd t he t imeshifting
p roperty ( 6.29a). Fig. 6 .6 T he signal for Exercise E6.3.
• 80) s +1 s +2 COS ()
b tu t = 8 +a
( s+a)2+b 2 • +2 5 =s  at 6 5 e 2t ) u (t) E xercise E6.5
Derive Pair 6 in Table 6.1 from Pair 3 and the frequencyshifting property. 'V W e a re n ow r eady t o c onsider t he t wo o f t he m ost i mportant p roperties o f t he
L aplace t ransform: t he t imedifferentiation a nd t imeintegration p roperties. 5 ( e t  e 2t ) u (t) 3. Also, because T he TimeDifferentiation Propertyt
T his p roperty s tates t hat i...
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 Spring '13
 Bayliss
 Signal Processing, The Land

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