Chapter 9_Laplace Transforms - The Analysis and Design of...

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The Analysis and Design of Linear Circuits Seventh Edition 9 Laplace Transforms 9.1 Exercise Solutions Exercise 9–1. Find the Laplace transform of v ( t ) = 7 u ( t ) V. Apply the definition of a Laplace transform. V ( s ) = integraldisplay 0 - 7 u ( t ) e st dt = integraldisplay 0 - 7 e st dt = 7 e st s vextendsingle vextendsingle vextendsingle 0 - = 7 s (0 1) = 7 s V-s Exercise 9–2. Find the Laplace transform of v ( t ) = 8 e 5 t u ( t ) V. Apply the results of Example 9–2. V ( s ) = integraldisplay 0 - 8 e 5 t u ( t ) e st dt = 8 integraldisplay 0 - e 5 t e st dt = (8) 1 s + 5 = 8 s + 5 V-s Exercise 9–3. Find the Laplace transform of i ( t ) = 0 . 5 δ ( t ) A. Apply the definition of a Laplace transform. I ( s ) = integraldisplay 0 - 0 . 5 δ ( t ) e st dt = 0 . 5 integraldisplay 0 + 0 - δ ( t ) e st dt = 0 . 5 integraldisplay 0 + 0 - δ ( t ) dt = 0 . 5 A-s Exercise 9–4. Transform the response v ( t ) = bracketleftbig 10 e 1000 t 5 bracketrightbig u ( t ) V of a particular RC circuit into the Laplace domain. Apply the linearity property to transform each component. V ( s ) = 10 s + 1000 5 s = 10 s 5 s 5000 s ( s + 1000) = 5( s 1000) s ( s + 1000) V-s Exercise 9–5. Transform the sinusoid i ( t ) = 100 [sin(200 t )] u ( t ) mA into the Laplace domain. Apply the results from Example 9–5. I ( s ) = (100)(200) s 2 + 200 2 = 20000 s 2 + 40000 mA-s Exercise 9–6. Let v 1 ( t ) = V A e αt u ( t ) V. Show that the Laplace transform of v 2 ( t ) = integraltext t 0 V A e αx dx V is equal to V 1 ( s ) /s . Solution Manual Chapter 9 Page 9-1
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The Analysis and Design of Linear Circuits Seventh Edition The calculations are shown below. integraldisplay t 0 V A e αx dx = V A α e αx vextendsingle vextendsingle vextendsingle t 0 = V A α ( e αt 1 ) = V A α ( 1 e αt ) L braceleftbigg V A α ( 1 e αt ) bracerightbigg = V A α parenleftbigg 1 s 1 s + α parenrightbigg = V A α bracketleftbigg s + α s s ( s + α ) bracketrightbigg = V A s ( s + α ) V 1 ( s ) = V A s + α V 1 ( s ) s = V A s ( s + α ) Exercise 9–7. If i ( t ) = 6 e 1000 t u ( t ) mA, find the Laplace transform of v ( t ) = 1 10 6 integraltext t 0 i ( x ) dx V. Apply the integration property and linearity. I ( s ) = 6 s + 1000 mA-s V ( s ) = 10 6 I ( s ) s = 10 6 0 . 006 s ( s + 1000) = 6000 s ( s + 1000) V-s Exercise 9–8. Let v 1 ( t ) = V A r ( t ) V. Show that the Laplace transform of v 2 ( t ) = dV A r ( t ) /dt V is equal to sV 1 ( s ) v 1 (0 ). The calculations are shown below. dV A r ( t ) dt = dV A tu ( t ) dt = V A ( t ) + V A u ( t ) = V A u ( t ) L { V A u ( t ) } = V A s sV 1 ( s ) v 1 (0 ) = s V A s 2 0 = V A s Exercise 9–9. If i ( t ) = 30 e 1200 t u ( t ) mA, find the Laplace transform of v ( t ) = 0 . 1 di ( t ) dt V. We can solve this problem in two ways. First, find I ( s ) and apply the differentiation property. I ( s ) = L { i ( t ) } = L braceleftbig 0 . 03 e 1200 t u ( t ) bracerightbig = 0 . 03 L braceleftbig e 1200 t u ( t ) bracerightbig = 0 . 03 s + 1200 V ( s ) = L braceleftbigg 0 . 1 di ( t ) dt bracerightbigg = 0 . 1 L braceleftbigg di ( t ) dt bracerightbigg = 0 . 1 bracketleftbig sI ( s ) i (0 ) bracketrightbig = 0 . 1 bracketleftbigg 0 . 03 s s + 1200 0 bracketrightbigg V ( s ) = 0 . 003 s s + 1200 V-s = 3 s s + 1200 mV-s Second, confirm this answer by finding an expression for v ( t ) in the time domain and taking the Laplace Solution Manual Chapter 9 Page 9-2
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