Differential Equations

1. EE 313 Linear Systems and Signals Fall 2010 Differential Equations Prof. Brian L. Evans Dept. of Electrical and Computer Engineering The University of Texas at Austin Initial conversion of content to PowerPointby Dr. Wade C. Schwartzkopf

2. y(t) x(t) System Time-Domain Analysis • For a system governed by alinear constant coefficientdifferential equation, Each component can be computed independently of other System satisfies linearity property if zero-input response is zero (i.e. all initial conditions are zero) Zero-state response is convolution of impulse response and input signal

3. Zero-input response Response when x(t) = 0 Results from internal system conditions only Independent of x(t) For most filtering applications (e.g. your stereo system), we want a zero-valued zero-input response Zero-state response Response to non-zero x(t) when system is relaxed A system in zero state cannot generate any response for zero input Zero state corresponds to initial conditions being zero Time-Domain Analysis

4. Zero-Input Response • Simplest case (first-order equation) • Solution: • For arbitrary constant C How is C determined? Could C be complex-valued? • How about the following Nth-order equation?

5. Zero-Input Response • For the Nth-order equation Guess solution has form y0(t) = C e l t Substitute form into differential equation Factor common terms to obtain y0(t) = C el t is a solution provided that Q(l) = 0 Factor Q(l) to obtain N characteristic roots

6. Zero-Input Response • Assuming that no two literms are equal • For repeated roots, solution changes Simplest case of root l repeated twice: With r repeated roots • Characteristic modes e l t Determine zero-input response Influence zero-state response

7. Zero-Input Response • Could libe complex? If complex, we can write it in Cartesian form Exponential solution el t becomes product of two terms • For conjugate symmetric roots, and conjugate symmetric constants,

8. L R  x(t) C y(t) Envelope Zero-Input Solution Example • Component values L = 1 H, R = 4 W, C = 1/40 F Realistic breadboard components? • Loop equations (D2 + 4 D + 40) [y0(t)] = 0 • Characteristic polynomial l2 + 4 l + 40 =(l + 2 - j 6)(l + 2 + j 6) • Initial conditions y(0) = 2 A ý(0) = 16.78 A/s y0(t) = 4 e-2t cos(6t - p/3) A

9. y(t) x(t) System Impulse Response • Response to unit impulse Set x(t) = d(t) and solve for y(t) • Linear constant coefficient differential equation • With zero initial conditions, impulse response is b0 is coefficient of dNx(t)/dtN term and could be 0

10. Impulse Response • Where did b0 come from? • In solving these differential equations for t 0, • Funny things happen to y’(t) and y”(t) • In differential equations class, solved for m(t) Likely ignored d(t) and d’(t) terms Solution for m(t) is really valid for t 0+