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Understanding Causal LTI Systems: Differential & Difference

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University of Texas**We aren't endorsed by this school
Course
ECE 313
Subject
Electrical Engineering
Date
Dec 11, 2024
Pages
8
Uploaded by DoctorRhinoceros2814
The University of Texas at AustinDepartment of Electrical and Computer EngineeringECE313: Linear Systems and Signals — Fall 2024Lecture 8Aryan MokhtariThursday, September 19, 2024Goal:In this lecture, we talk about causal LTI systems described by differential equations (forcontinuous-time systems) and difference equations (for discrete-time systems).1Introduction and MotivationAs we saw in previous lectures, sometimes it is not possible to come up with anexplicit expressionfor the output of a systemsy(t) ory[n] in terms of the values of inputx(t) andx[n]. For instance,recall the following examples that we discussed in Lecture 4.RC Circuit.Consider the following RC Circuit wherevsis our input signal andvcis the outputsignal. In the case, we can think of this circuit as a system.We observed that the relationship between the outputvcand inputvscan be written as the followingdifferential equationdvc(t)dt+1RCvc(t) =1RCvs(t),Bank Account Balance.Considery[n] as the balance of your bank account at the end ofnthmonth andx[n] the net deposit during thenth month.Here,x[n] is the input andy[n] is theoutput. If we assume that the monthly interest rate is 1%, then we can show thaty[n]−1.01y[n−1] =x[n]which is a difference equation.The above expressions provide an implicit specification of the system and to find out an explicitspecification of the system we need to solve these equations.1
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2Linear Constant-Coefficient Differential EquationsLet us first focus on a simple example. Consider the followingfirst-orderdifferential equationd y(t)dt+ 2y(t) =x(t)As you have seen in your DE class, to explicitly find the expression fory(t) we also need someadditional information. For instance, the initial condition. A common initial condition that we willuse for Causal LTI systems is that if the input for some time is zero, then the output is also zero.Remark 1.For a causal LTI system, ifx(t) = 0fort < t0, then the condition of initial restimplies thaty(t) = 0fort < t0.Note that the solution of a DE can be decomposed into the sum of a particular solutionyp(t) (whichdepends on input) and a homogenous solutionyh(t) (for the case that input isx(t) = 0), i.e.,y(t) =yp(t) +yh(t).Consider the case that the input is given byx(t) =Ke3tu(t)For this input the particular part which depends on the input, i.e.,yp(t) can be often written asan exponential signal with the same form, i.e., fort >0yp(t) =Me3t,where the constantMshould be determined. To do so, plug inyp(t) into to the DE to obtain3Me3t+ 2Me3t=Ke3tt >0by simplifying this expression we obtainM=K5⇒yp(t) =K5e3tfort >0.Now to find the homogeneous solution which is the solution ofd y(t)dt+ 2y(t) = 0we use the general form ofAest, whereAandsmust be determined. By applying this expressioninto the above homogeneous DE, we obtainAsest=−2Aest⇒s=−2⇒yh(t) =Ae−2tfort >0Hence, we obtain thaty(t) =Ae−2t+K5e3tfort >0.Now to determineAwe use the initial condition thaty(0) = 0, as the system is causal and LTIand the input is zero fort <0. Hence,0 =y(0) =Ae−0+K5e0=A+K5⇒A=−K5,Considering this result and the fact thaty(t) = 0 fort <0, we havey(t) =−K5e−2t+K5e3tu(t).2
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2.1GeneralNth-order linear constant-coefficient differential equationIn the previous example, we focus on a first-order ODE, where only the first derivative of the outputappears. The generalNth-order linear constant-coefficient differential equation is given byNXk=0akdky(t)dtk=MXk=0bkdkx(t)dtkNote that the order refers to the highest derivative of the outputy(t) appearing in the equation.Remark 2.In the case thatN= 0, then we havey(t) =MXk=0bkdkx(t)dtkand the output can be written explicitly as a function of the input and its derivatives.ForN≥1 we need to solve the DE to find an explicit expression fory(t).The first step would be finding the homogeneous part of the solution by solvingNXk=0akdky(t)dtk= 0.The solution of the homogeneous DE is also known as thenatural responseof the system.Then, we need to find a particular solution based on the form of the input. To do so, we need someadditional information to specify constants. For that, we used the condition that ifx(t) = 0 fort < t0, theny(t) = 0 fort < t0. Therefore, we havey(t0) =d y(t0)dt=d2y(t0)dt2=· · ·=dN−1y(t0)dtN−1= 03Linear Constant-Coefficient Difference EquationsLet us again start with a simple example. Consider the followingfirst-orderdifference equation:y[n]−12y[n−1] =x[n]where the input isx[n] =Kδ[n]We can use the previous framework to solve this difference equation, but a simpler approach alsoholds in the discrete setting.First, we writey[n] in terms of the other terms which implies thaty[n] =12y[n−1] +x[n]Note thatx[n] = 0 forn≤ −1, and since the system is causal and LTI we havey[n] = 0 forn≤ −1.Now, forn= 0 we can writey[0] =12y[−1] +x[0] = 0 +K=K3
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For the next terms, we can also use this recursion to obtain.y[1] =12y[0] +x[1] =12K+ 0 =12Ky[2] =12y[1] +x[2] =1212K+ 0 =122K...y[n] =12y[n−1] +x[n] =12nKTherefore, we havey[n] =12nKu[n]3.1GeneralNth-order linear constant-coefficient difference equationThe generalNth-order linear constant-coefficient difference equation can be written asNXk=0aky[n−k] =MXk=0bkx[n−k]Then, we havey[n] =1a0−NXk=1aky[n−k] +MXk=0bkx[n−k]!Therefore, if we have access toy[n0−1], . . . , y[n0−N], we can obtain the expression for anyn≥n0.This information can be obtained from the initial condition that we have access to using the factthat the system is causal and LTI.4Block DiagramsIn this section we discuss how to illustrate differential equations and difference equations usingblock diagrams.4.1Difference equationsWe first introduce the following sub-blocks and then use them to illustrate the following first-orderdifference equation:y[n] +ay[n−1] =bx[n]Sum of two signals:Product with scalar:4
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Delay:Using these expression and the fact thaty[n] +ay[n−1] =bx[n]⇐⇒y[n] =−ay[n−1] +bx[n]we have:4.2Differential equationsFirst, consider the following sub-blocks:Sum of two signals:Product with scalar:Derivative:Now consider the following DE:dy(t)dt+ay(t) =bx(t)Note that we can write it asy(t) =−1ady(t)dt+bax(t)Hence, we have5
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5Additional Examples1. Consider a casual LTI System whose inputx[n] and outputy[n] are related by the differenceequationy[n] =13y[n−1] +12x[n].•(a) Draw block diagram representation for this LTI system.Solution:•(b) Determiney[n] ifx[n] =δ[n].Solution:Note thatx[n] = 0 forn≤ −1, and since the system is causal and LTI wehavey[n] = 0 forn≤ −1. Now, forn= 0 we can writey[0] =13y[−1] +12x[0] = 0 +12=12For the next terms, we can also use this recursion to obtain.y[1] =13y[0] +12x[1] =13×12+ 0 =16y[2] =13y[1] +12x[2] =1313×12+ 0 =13212=118...y[n] =13y[n−1] +12x[n] =13n12Therefore, we havey[n] =13n12u[n]6
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2. (10 pts) Consider the cascade of the following two causal LTI systemsS1andS2asS1:w[n] =12w[n−1] +x[n]andS2:y[n] =αy[n−1] +βw[n]Further, we know that the difference equation relatingx[n] andy[n] isy[n] =−18y[n−2] +34y[n−1] +x[n](a) Determineαandβ.Solution:Note that sincex[n] depends on the difference of two termsw[n] and 0.5w[n−1], we need to come up with two expressions forythat depend onw[n] and 0.5w[n−1].We can show thaty[n] =αy[n−1] +βw[n]y[n−1] =αy[n−2] +βw[n−1]Multiply the second expression by 0.5 and subtract it from the first equality to obtainy[n]−12y[n−1] =αy[n−1] +βw[n]−12(αy[n−2] +βw[n−1])Regroup the terms to obtainy[n] =12+αy[n−1]−α2y[n−2] +βw[n]−12w[n−1]now using the expressionw[n] =12w[n−1] +x[n] we can replacew[n]−12w[n−1] byx[n] to obtainy[n] =12+αy[n−1]−α2y[n−2] +βx[n]Hence, it follows thatα=14,β= 1(b) Find the impulse response of the cascade connection ofS1andS2.Solution:We can solve this problem using two different approaches.We first find the impulse responses of System 1 and System 2 and then since the sys-tems are connected in a cascade form, the impulse response of the whole system is theconvolution of the impulse responses of System 1 and System 2.7
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To find the impulse response of System 1 we setx[n] =δ[n] andh1[n] =w[n] and usethe fact thath1n] = 0 forn <0 since the system is causal and LTI. Hence,h1[0] = 0.5h1[−1] +x[0] = 1h1[1] = 0.5h1[0] +x[1] = 0.5h1[2] = 0.5h1[1] +x[2] = 0.25. . .h1[n] =12nHence, the impulse response of System 1 ish1[n] =12nu[n].Next we use a similar approach to compute the impulse response of System 2 by settingw[n] =δ[n] andy[n] =h2[n]. Hence, we haveh2[0] = 0.25h2[−1] +x[0] = 1h2[1] = 0.25h2[0] +x[1] = 0.25h2[2] = 0.25h2[1] +x[2] = 0.252. . .h2[n] =14nHence, the impulse response of System 1 ish2[n] =14nu[n].Now since the two systems are connected in a cascade manner the impulse response oftheir combination if the convolution of their impulse responses, i.e.,h[n] =h1[n]∗h2[n]Therefore, we haveh[n] =∞Xk=−∞h1[k]h2[n−k] =∞Xk=−∞h1[k]h2[n−k]=∞Xk=−∞12ku[k]14n−ku[n−k]Note that thatu[k] is zero fork≤ −1 andu[n−k] is zero fork≥n+ 1. Hence, theabove sum is nonzero only for 0≤k≤nand forn≥0 we haveh[n] =nXk=012k14n−k=14nnXk=012k14−k=14nnXk=02k=14n(2n−1)Therefore, we haveh[n] =14n(2n−1)u[n] =12n−14nu[n]8
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