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Unformatted text preview: ECE220 Problem Lab #5 Solutions Differential equations and transfer functions Date: Week of February 25, 2008 The goal of this lab is to (a) summarize and reinforce all the methods we presented in chapter 7, about solving first order differential equations, (b) introduce the concept of transfer functions through a “filter design” problem you’ll see in ECE301. ————————————————————————————– A. First-order DEs Run the Matlab GUI first order DE . You will find it in the problem labs directory in the course locker, MATLAB GUIs subdirectory, and then FIRST ORDER DE subdirectory. This GUI depicts the complementary, particular and total solutions of the 1st order DE (see equation 7.46, page 348 in the textbook): dv dt + av = v s ( t ) with initial condition v (0). The driving force v s ( t ) is a unit-step, an exponential or a sinu- soidal function. The GUI also shows the effects of the constant a , and the initial condition. In summary, the GUI depicts graphically • the complementary solution v c ( t ) in equation 7.35, page 341, v c ( t ) = Ce- at u ( t ) • the particular solution v p ( t ) in theorem 7.4, page 348, v p ( t ) = A a u ( t ) , when v s ( t ) = Au ( t ), A √ 1+ a 2 sin( ω t + arctan( a/ω )- π/ 2) u ( t ) , when v s ( t ) = A sin( ω t ) u ( t ), A a + b e bt u ( t ) , when v s ( t ) = Ae bt u ( t ). 1 • the total solution v ( t ) = v c ( t ) + v p ( t ) In order to interpret the results of this lab in a “circuits” context, we will suppose that this DE represents the simple RC circuit in series (see equation 7.10, page 332). In this case, a = 1 /RC , A = 1 /RC and v s ( t ) ,v ( t ) are measured in Volts. The top graph always shows the driving force. The bottom graph shows the driving force and up to three solutions. You can choose which solution(s) to depict by checking the appropriate checkboxes, marked Complementary solution, Particular solution and Total solution. You can change a and v (0) via a slider and/or an edit box. To make sure graphs are “nice”, the smallest/largest values for a are limited to 0.01/10; the smallest/largest values for v (0) are limited to -10/10. I. Unit-step driving force 1. When you bring up the GUI, the default settings are v s ( t ) = u ( t ) ,a = 1 ,v (0) = 0 and the GUI depicts the Total solution. In other words, the bottom graph depicts the total solution to the DE dv dt + 1 · v = u ( t ) , v (0) = 0 . Question 1. Using equations from the textbook (which ones?) verify that what you see is correct. Solution: To solve a differential equation, we must find the complementary solution and the particular solution to get the total solution. We are given: dv dt + 1 · v = u ( t ) , v (0) = 0 Find the complementary solution (see equation 7.35, page 283)....
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This note was uploaded on 03/30/2009 for the course ECE 220 taught by Professor Nilson during the Spring '08 term at N.C. State.
- Spring '08