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Chapter 2 - EEL4657 Dr Haniph Latchman Chapter 2 Solutions...

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EEL4657 - Dr. Haniph Latchman Chapter 2 Solutions 1. L d 2 y dt 2 -→ s 2 Y ( s ) 2. Derive the Laplace Transform pairs from Table 2.1, page 33 of the text **Note that the Laplace Equation is given as: X ( s ) = - 0 x ( t ) e - st dt (a) x ( t ) = δ ( t ) X ( s ) = - 0 δ ( t ) e - st (since - δ ( t ) dt = 1) X ( s ) = - δ ( t ) e - st dt X ( s ) = 1 · e - s · 0 = 1 L [ δ ( t )] = 1 (b) x ( t ) = u ( t ) X ( s ) = -∞ u ( t ) e - st dt X ( s ) = 0 1 · e - st dt X ( s ) = -∞ u ( t ) e - st dt X ( s ) = - 1 s e - st 0 X ( s ) = 1 s e - s ( ) + 1 s e - s (0) L [ u ( t )] = 1 s 1
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(c) x ( t ) = e - at u ( t ) X ( s ) = 0 e - at · e - st dt X ( s ) = 0 e - t ( s + a ) dt X ( s ) = 1 - ( s + a ) e - t ( s + a ) 0 X ( s ) = 1 - ( s + a ) e -∞ ( s + a ) + 1 - ( s + a ) e 0( s + a ) X ( s ) = 1 s + a L [ e - at u ( t )] = 1 s + a (d) x ( t ) = cos ( ωt ) u ( t ) X ( s ) = 0 1 2 e jωt e - st dt + 0 1 2 e - jωt e - st dt X ( s ) = 1 2 0 e - t ( s - ) dt + 1 2 0 e - t ( s + ) dt X ( s ) = - e - t ( s - ) - 2( s - ) - e t ( s + ) 2( s + ) 0 X ( s ) = 1 - 2( s - ) + 1 2( s + ) X ( s ) = s + + s - 2( s - )( s + ) X ( s ) = s s 2 + ω 2 L [ cos ( ωt ) u ( t )] = s s 2 + ω 2 2
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(e) x ( t ) = sin ( ωt ) u ( t ) X ( s ) = 0 1 2 j e jωt e - st dt - 0 1 2 j e - jωt e - st dt X ( s ) = 1 2 j 0 e - t ( s - ) dt - 1 2 j 0 e - t ( s + ) dt X ( s ) = e - t ( s - ) 2 j ( s - ) - e - t ( s + ) 2 j ( s + ) 0 X ( s ) = 1 2 j ( s - ) - 1 2 j ( s + ) X ( s ) = s +
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