EE2010_Gp2_Chap4_1011s2 - Impulse Response Step Response...

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Impulse Response Step Response Transient Spec Zero Def Convolution Example Impulse Response : Definition I The impulse response of a LTI system is the system’s output signal when the input signal, u ( t ), is the unit impulse function, δ ( t ), with zero initial conditions. I Suppose the transfer function of a LTI system is G ( s ) = Y ( s ) U ( s ) . When u ( t ) = δ ( t ), then U ( s ) = L{ δ ( t ) } = 1 and Y ( s ) = G ( s ) U ( s ) = G ( s ) U ( s ) = 1 y ( t ) = L - 1 { G ( s ) } = Impulse response, h ( t ) System transfer function, G(s) = L{ Impulse Response } WW Tan Chapter 4 Page 1
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Impulse Response Step Response Transient Spec Zero Def Convolution Example Impulse response of a system, G ( s ), can be expressed as h ( t ) = L - 1 { G ( s ) } = L - 1 ± K ( s - z 1 )( s - z 2 ) . . . ( s - z M ) ( s - p 1 )( s - p 2 ) . . . ( s - p N ) ² = L - 1 ± A 1 s - p 1 + A 2 s - p 2 + . . . + A N s - p N ² = A 1 e p 1 t + . . . + A N e p N t = N X n =1 A n e pnt where p n ( n = 1 , . . . , N ) are the system poles. I For a stable system, all system poles p n ( n = 1 , . . . , N ) lie on the open left half plane. Then, impulse response decays to zero as t → ∞ . I Impulse response has no steady-state term. As L{ δ ( t ) } = 1, the absence of a input pole also indicate that there is no steady-state term. I Recall that the transient response is y tr ( t ) = N X n =1 K n e pnt I Impulse response and the transient response share the same number and type of elementary signals WW Tan Chapter 4 Page 2
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Impulse Response Step Response Transient Spec Zero Def Convolution Example Convolution Let f ( t ) and g ( t ) be piecewise continuous functions defined on (0 , ). The convolution of f ( t ) and g ( t ), denoted by f ( t ) * g ( t ), is defined by either (equivalent) integral f ( t ) * g ( t ) = Z t 0 f ( t - τ ) g ( τ ) d τ = Z t 0 f ( τ ) g ( t - τ ) d τ ( t 0) Convolution is an integral that expresses the amount of overlap of one function f ( t ) as it is shifted over another function g ( t ). I Essentially, convolution “blends” one function with another. I Possible to obtain the output signal of a system by blending two signals ? Laplace Transform of Convolution rule : If L{ p ( t ) } = P ( s ) and L{ q ( t ) } = Q ( s ), then L{ p ( t ) * q ( t ) } = P ( s ) Q ( s ) or p ( t ) * q ( t ) = L - 1 { P ( s ) Q ( s ) } WW Tan Chapter 4 Page 3
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Impulse Response Step Response Transient Spec Zero Def Convolution Example From the definition of a system transfer function G ( s ), Y ( s ) = G ( s ) U ( s ) where u ( t ) = L - 1 { U ( s ) } and y ( t ) = L - 1 { Y ( s ) } are the input and output signal respectively. Hence, the output signal is y ( t ) = L - 1 { Y ( s ) } = L - 1 { G ( s ) U ( s ) } Using the Laplace Transform of Convolution rule, y ( t ) = L - 1 { G ( s ) } * L - 1 { U ( s ) } Since L - 1 { G ( s ) } = h ( t ), the impulse response, y ( t ) = h ( t ) * u ( t ) The output signal, y ( t ), of a system may be obtained by convolving the impulse response, h ( t ), with the input/forcing signal, u ( t ). I Impulse response, h ( t ), is the basic building block for constructing the output signal corresponding to a general input signal. WW Tan Chapter 4 Page 4
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Impulse Response Step Response Transient Spec Zero Def Convolution Example Example Given that G ( s ) = 2 s + 1 and the input signal is u ( t ) = 1 - U ( t - 1).
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EE2010_Gp2_Chap4_1011s2 - Impulse Response Step Response...

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