102_1_Lecture11

102_1_Lecture11 - UCLA Spring 2009-2010 Systmes and Signals...

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UCLA Spring 2009-2010 Systmes and Signals Lecture 11: Frequency Response of LTI Systems April 28, 2010 EE102: Systems and Signals; Spr 09-10, Lee 1
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Linear Time-Invariant Systems, Revisited A linear time-invariant system is completely characterized by its impulse response h ( t ) . For a linear system with an input signal x ( t ) , the output is given by the convolution y ( t ) = ( x * h )( t ) = -∞ x ( τ ) h ( t - τ ) d τ y ( t ) x ( t ) * h ( t ) EE102: Systems and Signals; Spr 09-10, Lee 2
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The Fourier transform of the convolution is Y ( j ω ) = H ( j ω ) X ( j ω ) where X ( j ω ) is the input spectrum, Y ( j ω ) is the output spectrum, and H ( j ω ) is the Fourier transform of the impulse response h ( t ) . H ( j ω ) is called the frequency response or transfer function of the system. Each frequency in the input spectrum X ( j ω ) is Scaled by the system amplitude response | H ( j ω ) | , | Y ( j ω ) | = | H ( j ω ) || X ( j ω ) | Phase shifted by the system phase response H ( j ω ) , Y ( j ω ) = H ( j ω ) + X ( j ω ) EE102: Systems and Signals; Spr 09-10, Lee 3
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to produce the output spectrum Y ( j ω ) . If the input is to a system is a complex exponential e j ω 0 t , the input spectrum is X ( j ω ) = F e j ω 0 t = 2 πδ ( ω - ω 0 ) . The output spectrum is Y ( j ω ) = H ( j ω )(2 πδ ( ω - ω 0 )) = H ( j ω 0 )(2 πδ ( ω - ω 0 )) . The ouput signal is y ( t ) = F - 1 [ Y ( j ω )] = F - 1 [ H ( j ω 0 )(2 πδ ( ω - ω 0 ))] EE102: Systems and Signals; Spr 09-10, Lee 4
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= H ( j ω 0 ) e j ω 0 t = | H ( j ω 0 ) | e j ( ω t + H ( j ω 0 )) A sinusoidal input e j ω 0 t to an LTI system produces a sinusoidal output at the Same frequency, Scaled in amplitude, and Phase shifted. This corresponds to multiplication by a complex number H ( j ω 0 ) . EE102: Systems and Signals; Spr 09-10, Lee 5
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Frequency Response Example An input signal x ( t ) = 2 cos( t ) + 3 cos (3 t/ 2) + cos(2 t ) is applied to a system with an impulse response h ( t ) h ( t ) = 2 π sinc 2 ( t/ π ) Find the output signal ( x * h )( t ) . First, the frequency response or transfer function of the system is F 2 π sinc 2 ( t/ π ) = 2 π π Δ ( πω / 2 π ) = 2 Δ ( ω / 2) EE102: Systems and Signals; Spr 09-10, Lee 6
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so H ( j ω ) = 2 Δ ( ω / 2) The input spectrum is X ( j ω ) = 2 π [ δ ( ω - 1) + δ ( ω + 1)] + 3 π [ δ ( ω - 3 / 2) + δ ( ω + 3 / 2)] + π [ δ ( ω - 2) + δ ( ω + 2)] The output spectrum is the product of the input spectrum, and the transfer function, as shown on the next page: EE102: Systems and Signals; Spr 09-10, Lee 7
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! - 2 0 - 1 1 2 ! - 2 0 - 1 1 2 ! - 2 0 - 1 1 2 1 2 = × 3 ! 4 ! ! 2 ! ! 2 ! 3 ! 2 ! 2 ! 2 ! 3 ! / 2 3 ! / 2 4 ! 2 ! X ( j ! ) H ( j ! ) = 2 " ( ! / 2 ) Y ( j ! ) = H ( j ! ) X ( j ! ) EE102: Systems and Signals; Spr 09-10, Lee 8
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The output signal spectrum is then Y ( j ω ) = 2 π [ δ ( ω - 1) + δ ( ω + 1)] + 3 π 2 [ δ ( ω - 3 / 2) + δ ( ω + 3 / 2)] and the output signal is y ( t ) = 2 cos( t ) + 3 2 cos(3 t/ 2) .
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