EECE 301 Note Set 28 CT Use LT to Solve DiffEq

EECE 301 Note Set 28 CT Use LT to Solve DiffEq - 1/21 EECE...

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Unformatted text preview: 1/21 EECE 301 Signals & Systems Prof. Mark Fowler Note Set #28 • C-T Systems: Laplace Transform… Solving Differential Equations • Reading Assignment: Section 6.4 of Kamen and Heck 2/21 Ch. 1 Intro C-T Signal Model Functions on Real Line D-T Signal Model Functions on Integers System Properties LTI Causal Etc Ch. 2 Diff Eqs C-T System Model Differential Equations D-T Signal Model Difference Equations Zero-State Response Zero-Input Response Characteristic Eq. Ch. 2 Convolution C-T System Model Convolution Integral D-T System Model Convolution Sum Ch. 3: CT Fourier Signal Models Fourier Series Periodic Signals Fourier Transform (CTFT) Non-Periodic Signals New System Model New Signal Models Ch. 5: CT Fourier System Models Frequency Response Based on Fourier Transform New System Model Ch. 4: DT Fourier Signal Models DTFT (for “Hand” Analysis) DFT & FFT (for Computer Analysis) New Signal Model Powerful Analysis Tool Ch. 6 & 8: Laplace Models for CT Signals & Systems Transfer Function New System Model Ch. 7: Z Trans. Models for DT Signals & Systems Transfer Function New System Model Ch. 5: DT Fourier System Models Freq. Response for DT Based on DTFT New System Model Course Flow Diagram The arrows here show conceptual flow between ideas. Note the parallel structure between the pink blocks (C-T Freq. Analysis) and the blue blocks (D-T Freq. Analysis). 3/21 6.4 Using LT to solve Differential Equations In Ch. 2 we saw that the solution to a linear differential equation has two parts: ) ( ) ( ) ( t y t y t y zi zs total + = Here we’ll see how to get y total ( t ) using LT… … get both parts with one tool!!! We’ve seen how to find this using: “convolution w/ impulse response” or using “multiplication w/ frequency response” Ch. 2 Ch. 5 We’ve seen how to find this using the characteristic equation, its roots, and the so- called “characteristic modes” Ch. 2 4/21 Assume that for t <0 this has been switched on for “a long time” First-order case : Let’s see this for a 1 st-order Diff. Eq. with a causal input and a non-zero initial condition just before the causal input is applied. switch @ t = 0 The 1 st-order Diff. Eq. describes : a simple RC or RL circuit. The causal input means : we switch on some input at time t = 0. The initial condition means : just before we switch on the input the capacitor has a specified voltage on it (i.e., it holds some charge). Input : Time-Varying Voltage (e.g., guitar, microphone, etc.) Output : Time-Varying Voltage x ( t ) R C y ( t ) + + – – V IC + – Thus… the cap is fully charged to V IC volts Thus… the cap is fully charged to V IC volts 5/21 x ( t ) = 0, t < 0 With IC y (0- ) = V IC Cap voltage… just before x ( t ) “turns on” ) ( 1 ) ( 1 ) ( t x RC t y RC dt t dy = + This circuit is then described by this Diff. Eq.: ) ( ) ( ) ( t bx t ay dt t dy = + For this ex. we’ll solve the general 1 st-order Diff. Eq.: Now the key steps in using the LT are: • take the LT of both sides of the Differential Equation...
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This note was uploaded on 02/29/2012 for the course EECE 301 taught by Professor Fowler during the Fall '08 term at Binghamton.

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EECE 301 Note Set 28 CT Use LT to Solve DiffEq - 1/21 EECE...

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