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

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Unformatted text preview: f f (t) = / l(t) + h (t - 6 + 5 [ e-(t-2) - e- 2(t-2)] u (t - 2) E xercise E6_3 Find the Laplace transform of the signal illustrated in Fig. 6.6. Answer: - ; (1 - 3e- 2s s 6 f (t) &lt;=&gt; F (8) 2) = ( 2e- t - e- 2t ) u (t) + 2 e- 3 ,) 3 e- 2s F (s) = (s _ 1)(s + 2) 2- t hen d f &lt;=&gt; 8 F(8) - f(O-) dt (6.34a) R epeated a pplication o f t his p roperty y ields 'V E xercise E 6A Find the inverse Laplace transform of Answer: [e t - • e- 2(t-2)] u (t - 2) 'V (6.34b) t The dual of the time-differentiation property is the frequency-differentiation property, which states t hat 6 Continuous-Time System Analysis Using the Laplace ' Ifansform 386 (6.34c) where fer) ( 0-) is dr f / dt r a t t = 0 -. Proof: 6.2 Some P roperties of t he Laplace ' Ifansform • E xample 6 .7 F ind t he L aplace t ransform o f t he s ignal f (t) i n Fig. 6.7a using Table 6.1 a nd t he t ime-differentiation a nd t ime-shifting properties o f t he L aplace transform. Figures 6.7b a nd 6.8c show t he first two derivatives o f f (t). Recall t hat t he d erivative a t a p oint o f j ump d iscontinuity is a n i mpulse o f s trength e qual t o t he a mount o f j ump [see Eq. (1.27)]. Therefore 21 d 2 = 6(t) _ 36(t - 2) dt I ntegrating by p arts, we o btain OO f I:. [d ] = f (t)e-stI 0dt +s 387 + 26(t - 3) T he L aplace t ransform o f t his e quation yields roo f(t)e- st dt Jo- £. ( For the Laplace integral t o converge ( that is for F (8) to exist), i t is necessary t hat f (t)e- st - ; 0 a s t - ; 0 0 for t he values of s in the region of convergence for F (s). T hus, ~:t) = c. [6(t) - 36(t - 2) + 26(t - 3)] Using t he t ime-differentiation p roperty (6.34b), t he t ime-shifting p roperty (6.29a) a nd t he f acts t hat 1 (0-) = j (O-) = 0, a nd 6(t) &lt; ==} 1, we o btain T herefore which confirms t he e arlier result in Exercise E6.3. j{&lt;)2~ o 2 3 4. t- • The Time-Integration Propertyt T his property states t hat if (a) f (t) ~ F (s) df then dt 1 t o -2 (b) 2 3 f(r)dr~ 0- t~ F (s) s (6.35) and LJ 1 t f (r)dr ~ - 00 F (s) -- J &quot;'-__ + -~~f(r)dr S S (6.36) Proof: We define 2 so t hat o -3 d (e) 2 I 3 - 9(t) = f (t) dt t~ F ig. 6 .7 F inding t he Laplace transform o f a piecewise-linear function using t he timedifferentiation p roperty. a nd N ow, if 9 (t) ~ G (s) t The dual of the time-integration property is t he frequency-integration property, which states t hat 1~t) &lt; ==} ['0 F (z)dz 6 388 Continuous-Time S ystem Analysis Using t he Laplace Transform 6.2 Some P roperties of t he Laplace Transform 389 T able 6 .2 t hen F (s) =£ [ ftg(t)] = sG(s) - g(O-) = sG(s) T he L aplace T ransform P roperties T herefore G(s) = F (s) Operation s or i {=}- s 0- A ddition h (t) S calar m ultiplication f (r)dr = l~ f (r)dr + i~ f (r)dr + f2(t) k f(t) To prove Eq. (6.36), observe t hat loo F (s) F (s) t f (r)dr f (t) T ime differentiation Note t hat t he first term on t he r ight-hand side is a constant. Taking the Laplace transform of t he above equation a nd using Eq. (6.35), we o btain FI(S) + F2(S) k F(s) df dt d2 f dt2 d3 f dt 3 Scaling T ime i ntegration T his p roperty s tates t hat if f (t) f rat) {=} 0 -;F(s) ~F (~) Time shift Frequency shift F (s)e- sto FI(S) a nd f2(t) {=} * f2(t) f (t) 1 f...
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