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Unformatted text preview: 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 Time Domain Analysis of Continuous Time Systems Today’s topics • Zeroinput and zerostate 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 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 + 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 Solutions for the System Equation Solving the system equation tells us the output for a given input. The output consists of two components: • The zeroinput 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 y ( t ) x ( t ) = EE102: Systems and Signals; Fall 2011, Jin Hyung Lee 4 • The zerostate response, which is the output of the system with all initial conditions zero. t H y ( t ) x ( t ) t If H is a linear system, its zeroinput 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 x ( t ) = y ( t ) = t EE102: Systems and Signals; Fall 2011, Jin Hyung Lee 5 Example: Solve for the voltage across the capacitor y ( t ) for an arbitrary input voltage x ( t ) , given an initial value y (0) = Y . + R C y ( t ) + x ( t ) i ( t ) From Kirchhoff’s voltage law x ( t ) = Ri ( t ) + y ( t ) Using i ( t ) = Cy ( t ) RCy ( 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 ( t ) + y ( t ) = 0 . EE102: Systems and Signals; Fall 2011, Jin Hyung Lee 6 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 ( t ) e t/RC 1 RC A ( t ) e t/RC + A ( t ) e t/RC = x ( t ) Simplying A ( t ) = x ( t ) 1 RC e t/RC which can be integrated from t = 0 to get A ( t ) = Z t x ( τ ) 1 RC e τ/RC dτ + A (0) EE102: Systems and Signals; Fall 2011, Jin Hyung Lee 7 Then y ( t ) = A ( t ) e t/RC = e t/RC Z t x ( τ ) 1 RC e τ/RC dτ + A (0) e t/RC = Z t x ( τ ) 1 RC e ( t τ ) /RC dτ + A (0) e t/RC At t = 0 , y (0) = Y , so this gives A (0) = Y y ( t ) = Z t x ( τ ) 1 RC e ( t τ ) /RC dτ  {z } zero state response + Y e t/RC  {z } zero input response ....
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This note was uploaded on 01/30/2012 for the course ELEC ENGR 102 taught by Professor Jinhyunglee during the Fall '11 term at UCLA.
 Fall '11
 JinHyungLee

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