《信号与系统 Signals and Systems》课程教学资料（英文版）lecture19 CT System Function Properties

H AA Signals and systems Fall 2003 Lecture #19 1 8 November 2003 1. CT System Function Properties 2. System Function Algebra and Block diagrams 3. Unilateral Laplace Transform and plications

Signals and Systems Fall 2003 Lecture #19 18 November 2003 1. CT System Function Properties 2. System Function Algebra and Block Diagrams 3. Unilateral Laplace Transform and Applications

CT System Function Properties H(8) H(S)=system function Y(S=H(SX( 1) System is stabl/d-1→b() right-sided T C +1 t→t+T t→t+T e-(+Tu(t+r)+0 at t<0Non-causal

CT System Function Properties 2) Causality ⇒ h ( t) right-sided signal ⇒ ROC of H( s) is a right-half plane Question: If the ROC of H( s) is a right-half plane, is the system causal? |h ( t) | dt < ∞ −∞ ∞ ∫ 1) System is stable ⇔ ⇔ ROC of H( s) includes j ω axis Ex. H(s) = “system function” Non-causal

Properties of ct rational System Functions a) However, if H(s)is rational, then The system is causal The roc of H(s)is to the right of the rightmost pole b) If H(s)is rational and is the system function of a causal system, then The system is stable jo-axis is in ROC A all poles are in lhp

Properties of CT Rational System Functions a) However, if H( s) is rational, then The system is causal ⇔ The ROC of H( s) is to the right of the rightmost pole j ω-axis is in ROC ⇔ all poles are in LHP b) If H( s) is rational and is the system function of a causal system, then The system is stable ⇔

Checking if all Poles are in the left-half plane Poles are the roots of d(s)=sm+an-1Sn- +..+a1s+a0 Method#1: Calculate all the roots and see Method #2: Routh-Hurwitz- Without having to solve for roots Polynomial Condition so that all roots are in the lhp First -order S+ ao a0>0 econd-order s2+ars+ao a1>0,a0>0 Third-order a18+a0a2>0,a1>0,a0>0 and ao alas

Checking if All Poles Are In the Left-Half Plane Method #1: Calculate all the roots and see! Method #2: Routh-Hurwitz – Without having to solve for roots

Initial-and final-Value Theorems If x(t)=0 for to, then Im SA(S Final value S→

Initial- and Final-Value Theorems If x(t) = 0 for t < 0 and there are no impulses or higher order discontinuities at the origin, then Initial value If x(t) = 0 for t < 0 and x(t) has a finite limit as t → ∞, then Final value

Applications of the lnitial-and Final-Value Theorem F X D(s) n-order of polynomial N(s), d-order of polynomial D(s) d>m+1 C X Im SAS )= finite≠0d=m+1 d<m+1 E. g. X(s) S+1 Final value Ifx(∞x)= lim sX(s)=0→lmX(s)< → No poles at s=0

Applications of the Initial- and Final-Value Theorem • Initial value: • Fin a l v a l u e For

LTI Systems Described by lccdes ∑am0D=∑k dt Repeated use of differentiation property. dk R k dt M ∑aksY(s)=∑bksX Y(S)=H(SX(s roots of numerator zero where roots of denominator= poles h-o ak s Rational ROC= Depends on: 1) 1)Locations of all poles 2) Boundary conditions, i.e right-, left-, two-sided signals

LTI Systems Described by LCCDEs ROC =? Depends on: 1) Locations of all poles. 2) Boundary conditions, i.e. right-, left-, two-sided signals. roots of numerator ⇒ zeros roots of denominator ⇒ poles

System Function Algebra Example: a basic feedback system consisting of causal blocks (t)->(+e(t)h,( H y(t) 1(S z()h2(t) H2(S) E(s=Xs-Z(s=X(s-H2(sr(s) Y(S)=HIsE(s)=H1(sX(s-H2(s)Y(s) H(s=Y(s) H1( More on this later X(s)1+H1(s)H2( feedback ROC: Determined by the roots of 1+H(H2(s), instead of H,(s)

System Function Algebra Example: A basic feedback system consisting of causal blocks ROC: Determined by the roots of 1+H1( s ) H2 ( s), instead of H1( s ) More on this later in feedback

Block Diagram for Causal LtI Systems with rational system Functions Example: Y(S=H(SX(S) 2s2+4s-6 82+3s+2 2+3s+2 (28+48-6)-Can be viewed as cascade of two systems +3s+2 -0 du dt2 +s dt+ 2w(t)=c(t), initially at rest Or dt r()-3 dt 2(t) Similarly (s)=(2s2+4s-6)W(s) dw(t) dw( +4 6u(t) dt

Block Diagram for Causal LTI Systems with Rational System Functions — Can be viewed as cascade of two systems. Example:

Example(continued H(S Instead of 2s2+4s-6 3s+2 We can construct H(s)using do(t) 200 dt dt dw(t),do(t) y(t) w(t) w(t) 3 Notation: 1/s-an integrator

Example (continued) Instead of 1 s 2 + 3s + 2 2s2 + 4s − 6 H(s) Notation: 1/s — an integrator We can construct H(s) using: x(t) y(t)