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

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5 - 2 Time-Domain Analysis For a system governed by a linear constant coefficient differential 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 x(t)x(t) y(t)y(t)

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5 - 3 Time-Domain Analysis 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

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5 - 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?

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

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5 - 6 Zero-Input Response Assuming that no two i terms are equal For repeated roots, solution changes Simplest case of root repeated twice: With r repeated roots Characteristic modes e t Determine zero-input response Influence zero-state response

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5 - 7 Zero-Input Response Could i be complex? If complex, we can write it in Cartesian form Exponential solution e t becomes product of two terms For conjugate symmetric roots, and conjugate symmetric constants,

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5 - 8 Zero-Input Solution Example Component values L = 1 H, R = 4 , C = 1/40 F Realistic breadboard components? Loop equations (D 2 + 4 D + 40) [y 0 (t)] = 0 Characteristic polynomial 2 + 4 + 40 = ( + 2 - j 6)( + 2 + j 6) Initial conditions y(0) = 2 A ý(0) = 16.78 A/s LR C y(t)y(t) x(t)x(t) Envelope y 0 (t) = 4 e -2t cos(6t - /3) A

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5 - 9 Impulse Response Response to unit impulse Set x(t) = (t) and solve for y(t) Linear constant coefficient differential equation With zero initial conditions, impulse response is b 0 is coefficient of d N x(t)/dt N term and could be 0 x(t)x(t) y(t)y(t)

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5 - 10 Impulse Response Where did b 0 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 (t) and ’(t) terms Solution for m(t) is really valid for t 0 +

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