102_1_lecture5

102_1_lecture5 - UCLA Fall 2011 Systems and Signals Lecture...

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UCLA Fall 2011 Systems and Signals Lecture 5: Time Domain System Analysis October 10, 2011 EE102: Systems and Signals; Fall 2011, Jin Hyung Lee 1
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Time Domain Analysis of Continuous Time Systems Today’s topics Zero-input and zero-state responses of a system Impulse response LTI System response to arbitrary inputs Convolution Properties of convolution EE102: Systems and Signals; Fall 2011, Jin Hyung Lee 2
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System Equation The System Equation relates the outputs of a system to its inputs. Example from last time: the system described by the block diagram + + - Z a x y has a system equation y 0 + ay = x. In addition, the initial conditions must be given to uniquely specifiy a solution. EE102: Systems and Signals; Fall 2011, Jin Hyung Lee 3
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Solutions for the System Equation Solving the system equation tells us the output for a given input. The output consists of two components: The zero-input response, which is what the system does with no input at all. This is due to initial conditions, such as energy stored in capacitors and inductors. t H t 0 0 y ( t ) x ( t ) = 0 EE102: Systems and Signals; Fall 2011, Jin Hyung Lee 4
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The zero-state response, which is the output of the system with all initial conditions zero. t H 0 0 y ( t ) x ( t ) t If H is a linear system, its zero-input response is zero. Homogeneity states if y = F ( ax ) , then y = aF ( x ) . If a = 0 then a zero input requires a zero output. t H 0 0 x ( t ) = 0 y ( t ) = 0 t EE102: Systems and Signals; Fall 2011, Jin Hyung Lee 5
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Example: Solve for the voltage across the capacitor y ( t ) for an arbitrary input voltage x ( t ) , given an initial value y (0) = Y 0 . + - R C y ( t ) + - x ( t ) i ( t ) From Kirchhoff’s voltage law x ( t ) = Ri ( t ) + y ( t ) Using i ( t ) = Cy 0 ( t ) RCy 0 ( t ) + y ( t ) = x ( t ) . This is a first order LCCODE, which is linear with zero initial conditions. First we solve for the homogeneous solution by setting the right side (the input) to zero RCy 0 ( t ) + y ( t ) = 0 . EE102: Systems and Signals; Fall 2011, Jin Hyung Lee 6
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The solution to this is y ( t ) = Ae - t/RC which can be verified by direct substitution. To solve for the total response, we let the undetermined coefficient be a function of time y ( t ) = A ( t ) e - t/RC . Substituting this into the differential equation RC ± A 0 ( t ) e - t/RC - 1 RC A ( t ) e - t/RC ² + A ( t ) e - t/RC = x ( t ) Simplying A 0 ( t ) = x ( t ) ± 1 RC e t/RC ² which can be integrated from t = 0 to get A ( t ) = Z t 0 x ( τ ) ± 1 RC e τ/RC ² + A (0) EE102: Systems and Signals; Fall 2011, Jin Hyung Lee 7
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y ( t ) = A ( t ) e - t/RC = e - t/RC Z t 0 x ( τ ) ± 1 RC e τ/RC ² + A (0) e - t/RC = Z t 0 x ( τ ) ± 1 RC e - ( t - τ ) /RC ² + A (0) e - t/RC At t = 0 , y (0) = Y 0 , so this gives A (0) = Y 0 y ( t ) = Z t 0 x ( τ ) ± 1 RC e - ( t - τ ) /RC ² | {z } zero - state response + Y 0 e - t/RC | {z } zero - input response . EE102: Systems and Signals; Fall 2011, Jin Hyung Lee
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This note was uploaded on 12/13/2011 for the course EE 102 taught by Professor Levan during the Fall '08 term at UCLA.

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102_1_lecture5 - UCLA Fall 2011 Systems and Signals Lecture...

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