lect7 - EL 625 Lecture 7 1 EL 625 Lecture 7 Frequency...

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Unformatted text preview: EL 625 Lecture 7 1 EL 625 Lecture 7 Frequency domain analysis of time-invariant systems Theorem : If S is a linear time-invariant continuous-time system, S { e st } = H ( s ) e st Proof : S { e st } = S { e st } e st | {z } H ( t,s ) .e st = H ( t, s ) e st S { e s ( t- τ ) } = H ( t- τ, s ) e s ( t- τ ) = e- sτ S { e st } = e- sτ H ( t, s ) e st = ⇒ H ( t, s ) = H ( t- τ, s ) for any τ Thus, H ( t, s ) is independent of t . H ( t, s ) = H ( s ) Thus, S { e st } = H ( s ) e st The response of a linear time-invariant continuous-time system to a complex exponential is the same complex exponential with a change of complex magnitude (magnitude and phase). EL 625 Lecture 7 2 Laplace transform( L transform) : U ( s ) = Z ∞ u ( t ) e- st dt Inverse Laplace transform : 1 2 πj Z c + j ∞ c- j ∞ U ( s ) e st ds Output, y ( t ) = S { u ( t ) } = 1 2 πj Z c + j ∞ c- j ∞ U ( s ) S { e st } ds = 1 2 πj Z c + j ∞ c- j ∞ U ( s ) H ( s ) e st ds Y ( s ) = H ( s ) U ( s ) Differentiation theorem of L transforms : L{ ˙ x ( t ) } = s X ( s )- x (0) Convolution theorem of L transforms : EL 625 Lecture 7 3 If y ( t ) = R ∞ x 1 ( t- τ ) x 2 ( τ ) dτ ,then, Y ( s ) = X 1 ( s ) X 2 ( s ) ˙ x ( t ) = A x ( t ) + B u ( t ) y ( t ) = C x ( t ) + D u ( t ) L{ ˙ x ( t ) } = A L{ x ( t ) } + B L{ u ( t ) } L{ y ( t ) } = C L{ x ( t ) } + D L{ u ( t ) } = ⇒ s X ( s )- x (0) = A X ( s ) + B U ( s ) Y ( s ) = C X ( s ) + D U ( s ) Thus, X ( s ) = [ sI- A ]- 1 x (0) + [ sI- A ]- 1 B U ( s ) Y ( s ) = C [ sI- A ]- 1 x (0) + C [ sI- A ]- 1 B + D U ( s ) Using, x ( t ) = φ ( t ) x (0) + Z t φ ( t- τ ) B u (...
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This note was uploaded on 12/06/2010 for the course EL 625 taught by Professor Khorami during the Spring '10 term at NYU Poly.

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lect7 - EL 625 Lecture 7 1 EL 625 Lecture 7 Frequency...

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