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<∞分 RoC of h(s) includes jo axis 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? T Ex. H(S) 8+’Re{s}>-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 t<0 and there are no impulses or higher order discontinuities at the origin then (0= lim sX(s) Initial value S→0 If x(t)=0 for t<0 and x(t has a finite limit as t>o, 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