1. System State and Differential Equations Linear Systems and Signals Lecture 3 Spring 2008

2. 3 - 2 System State Example: Reformulate Nth-order differential equation into N simultaneous first-order differential equations Define three state variables After substituting the state variables, we obtain a state-space description as Lathi, 2 nd ed, Section 1.10

3. 3 - 3 System State By putting the state variables in vector x, If A is not a function of time, then solution is –exp(M) for matrix M yields a matrix

4. 3 - 4 System State General form of state-space description State-space descriptions can –Describe time-varying and nonlinear systems –Be simulated by computer (e.g. Spice for circuit simulation) Knowledge of state variables allows one to determine every possible output of the system State-space descriptions covered in controls courses, e.g. Intro. to Automatic Control

5. 3 - 5 Continuous-Time Domain Analysis Example: Differential systems –There are N derivatives of y(t) and M derivatives of f(t) –Constants a 1, …, a N and b N-M, b N-M+1, …, b N –Linear constant-coefficient differential equation Using short-hand notation, above equation becomes Lathi, 2 nd ed, Section 2.1

6. 3 - 6 Continuous-Time Domain Analysis Polynomials Q(D) and P(D) Normalization: a N = 1 N derivatives of y(t) M derivatives of f(t) This differential system behaves as (M-N)th- order differentiator if M > N (see Lathi, p. 151) Noise occupies both low and high frequencies Differentiator amplifies high frequencies (as we’ll see later in Lathi, Section 4.3-2) To avoid amplification of noise in high frequencies, we assume that M  N

7. 3 - 7 Continuous-Time Domain Analysis Linearity: for any complex constants c 1 and c 2,

8. 3 - 8 Continuous-Time Domain Analysis For a linear system, The two components are independent of each other Each component can be computed independently of the other T[·] y(t)y(t)f(t)f(t)

9. 3 - 9 Continuous-Time Domain Analysis Zero-input response –Response when f(t) = 0 –Results from internal system conditions only –Independent of f(t) –For most filtering applications (e.g. your stereo system), we want no zero-input response. Zero-state response –Response to non-zero f(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.

10. 3 - 10 Zero-Input Response Simplest case Solution: For arbitrary constant C –How is C determined? –Could C be complex-valued?

11. 3 - 11 Zero-Input Response General case: The linear combination of y 0 (t) and its N successive derivatives are zero Assume that y 0 (t) = C e t

12. 3 - 12 Zero-Input Response Substituting into the differential equation y 0 (t) = C e  t is a solution provided that Q( ) = 0 Factor Q( ) to obtain N solutions: Assuming that no two i terms are equal

13. 3 - 13 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,

14. 3 - 14 Zero-Input Response For repeated roots, the solution changes Simplest case of a root repeated twice: With r repeated roots

15. 3 - 15 System Response Characteristic equation –Q(D)[y(t)] = 0 –Polynomial Q( ) Characteristic of system Independent of the input –Q( ) roots 1, 2, …, N Characteristic roots a.k.a. characteristic values, eigenvalues, natural frequencies Characteristic modes (or natural modes) are the time-domain responses corresponding to the characteristic roots –Determine zero-input response –Influence zero-state response

16. 3 - 16 RLC Circuit [Similar to Lathi, Ex. 2.2] Component values L = 1 H, R = 4 , C = 1/40 F Realistic breadboard components? Loop equations [Ex. 1.10] (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) f(t)f(t) Envelope y 0 (t) = 4 e -2t cos(6t -  /3) A