Signal Processing and Linear Systems-B.P.Lathi copy

# Many mathematicians trying t o find solutions t o t

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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 right-hand 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 time-shifting property to t u(t) (t - I)u(t - 1) -= 2. e- Also u (t) s 82 -= ~s and and ¢ ==&gt; 1 /8 2 yields (t - 2)u(t - 2) u (t - 4) (6.30b) -= ~se-4S - = ~e-2S s (6.31) Therefore (6.32) • 3 84 6 C ontinuous-Time 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) &lt;=&gt; F (s) t hen o f (t)e sot &lt;=&gt; 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 time-delay. I n such a case we s hould separate F (s) into terms with and without delay factor, as &quot;-v--&quot; roo f(t)esote-st dt = roo f(t)e-(s-so)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 frequency-shifting property. T he P air Sa is 8 cos b tu(t) ~b2 8 + F rom t he frequency-shifting 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 ime-shifting 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 frequency-shifting 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 ime-differentiation a nd t ime-integration p roperties. 5 ( e- t - e- 2t ) u (t) 3. Also, because T he Time-Differentiation Propertyt T his p roperty s tates t hat i...
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