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)