Signals and Systems EE235 Lecture 19 Leo Lam © 2010-2012

Stanford The Stanford Linear Accelerator Center was known as SLAC, until the big earthquake, when it became known as SPLAC. SPLAC? Stanford Piecewise Linear Accelerator. Leo Lam © 2010-2012

Today’s menu LCCDE! Leo Lam © 2010-2012

Zero input response (example) 4 steps to solving Differential Equations: Step 1. Find the zero-input response = natural response yn(t) Step 2. Find the Particular Solution yp(t) Step 3. Combine the two Step 4. Determine the unknown constants using initial conditions 4 Leo Lam © 2010-2012

From earlier 5 Solve Guess solution: Substitute: We found: Solution: Characteristic roots = natural frequencies/ eigenvalues Unknown constants: Need initial conditions 5 Leo Lam © 2010-2012

Zero-state output of LTI system Total response(t)=Zero-input response (t)+Zero-state output(t) Response to our input x(t) LTI system: characterize the zero-state with h(t) Initial conditions are zero (characterizing zero-state output) Zero-state output: T d(t) h(t) 6 Leo Lam © 2010-2012

Zero-state output of LTI system Total response(t)=Zero-input response (t)+Zero-state output(t) Zero-input response: Zero-state output: Total response: “Zero-state”: d(t) is an input only at t=0 Also called: Particular Solution (PS) or Forced Solution 7 Leo Lam © 2010-2012

Zero-state output of LTI system Finding zero-state output (Particular Solution) Solve: Or: Guess and check Guess based on x(t) 8 Leo Lam © 2010-2012

Trial solutions for Particular Solutions Guess based on x(t) Input signal for time t> 0 x(t) Guess for the particular function yP   9 Leo Lam © 2010-2012

Particular Solution (example) Find the PS (All initial conditions = 0): Looking at the table: Guess: Its derivatives: 10 Leo Lam © 2010-2012

Particular Solution (example) Substitute with its derivatives: Compare: 11 Leo Lam © 2010-2012

Particular Solution (example) From We get: And so: 12 Leo Lam © 2010-2012

Particular Solution (example) Note this PS does not satisfy the initial conditions! Not 0! 13 Leo Lam © 2010-2012

Natural Response (doing it backwards) Guess: Characteristic equation: Therefore: 14 Leo Lam © 2010-2012

Complete solution (example) We have Complete Soln: Derivative: 15 Leo Lam © 2010-2012

Complete solution (example) Last step: Find C1 and C2 Complete Soln: Derivative: Subtituting: Two equations Two unknowns 16 Leo Lam © 2010-2012

Complete solution (example) Last step: Find C1 and C2 Solving: Subtitute back: Then add u(t): y n ( t ) y ( t ) u(t) is there to enforce the fact that the forced signal was applied at t=0, remember the input was sin(t)u(t). y p ( t ) 17 Leo Lam © 2010-2012

Another example Solve: Homogeneous equation for natural response: Characteristic Equation: Therefore: Input x(t) 18 Leo Lam © 2010-2012

Another example Solve: Particular Solution for Table lookup: Subtituting: Solving: b=-1, a=-2 Input signal for time t> 0 x(t) Guess for the particular function yP   No change in frequency! 19 Leo Lam © 2010-2012

Another example Solve: Total response: Solve for C with given initial condition y(0)=3 Tada! 20 Leo Lam © 2010-2012

Stability for LCCDE Stable with all Re(lj)<0 Given: A negative l means decaying exponentials Characteristic modes 21 Leo Lam © 2010-2012

Stability for LCCDE Graphically Stable with all Re(lj)<0 “Marginally Stable” if Re(lj)=0 IAOW: BIBO Stable iff Re(eigenvalues)<0 Im Re Roots over here are stable 22 Leo Lam © 2010-2012

Summary Differential equation as LTI system Leo Lam © 2010-2012