The complete solution to systems with inputs

Transcription

1 The complete solution to systems with inputs Slide

2 Learning Objectives Analyze linear time-invariant systems with inputs Solve for the homogeneous response of the system Natural response without inputs Solve for the particular solution Identify forced response for different input functions Obtain the complete solution using initial conditions Complete solution = homogeneous solution + particular solution Derive the transfer function Slide 2

3 Systems with input In general, systems have inputs Applied force in mechanical systems Voltage and current sources in circuits E.g., battery, power-supply, antenna, scope probe, etc. Systems also have outputs Displays, speakers, voltmeters, etc. We need to be able to analyze the system response to inputs Two methods: Solution to linear constant-coefficients differential equations Transfer function methods Slide 3

4 Linear constant coefficient differential equations E.g., dx(t) + 2x(t)=!u(t) Where x is the state variable and u is the input The complete solution is of the form: x(t)=!x p (t) + x h (t) where!x p!is!the!particular!solution!(when!input spefified) and x h!is!the!homogeneous!solution!to!the!de!when!u(t) = 0!!!!!!!!!!!!!!!!!!!!!!!!i.e.,! dx(t) + 2x(t)= 0 Thus far we have only considered homogeneous systems Slide 4

5 The particular solution dx(t) " + 2x(t)= u(t),!!!!u(t) = # 0 t < 0 $ e 3t t! 0 A common method for solving for the particular solution is to try a solution of the same form as the input This is called the forced response So try, x p (t) = ae 3t!! To solve for the constant a, we plug the solution to the original equation dx(t) + 2x(t)= e 3t! 3ae 3t + 2ae 3t = e 3t! a = / 5 Slide 5 Particular!solution :!!x p (t)= e3t 5,!!t > 0

6 The complete solution homogeneous!solution:!!!x h (t) = Be st!! Bse st + 2Be st = 0!!!!Bs + 2B = 0! B(s + 2) = 0! s = "2! x h (t) = Be "2t! x(t)!=! e3t 5 + Be"2t!!!,!!t > 0 In order to solve for B, must know initial conditions. E.g., x(0)!=!0! e0 5 + Be0 = 0! B = " 5 Slide 6 x(t) = 5 #$ e3t " e "2t % &,!!!t > 0

7 Key points Solution consists of homogeneous and particular solution Homogeneous solution is also called the natural response It is the response to zero input The particular solution often takes on the form of the input It is therefore referred to as the forced response The complete solution requires specification of initial conditions An n th order system would have n initial condition Apply initial conditions to the complete solution in order to obtain the constants The initial conditions are on the complete solution, not just the homogeneous part Slide 7

8 Example: RC circuit with inputs e R =!i - V c + dv = i C " dv =!e RC e (t) = v (t) + u(t) u(t) = e!5t + - C i R Y(t) " dv =!v (t) RC! u(t) RC C = F, R = ohm!c=,!r= " dv =!v (t)! u(t) Slide 8 Homogeneous!Solution:!!u(t) = 0 Guess!v (t) = ae st! ase st = "ae st! as = "a! s = "! v H = ae "t

9 The complete solution Forced!Response: v F (t) = Be!5t,!!v F =!5Be!5t dv =!v (t)! u(t) "!5Be!5t =!Be!5t! e!5t "!5B =!B! " B = / 4 " v F (t) = e!5t 4 v (t) = v H (t) + v F (t) = ae!t + e!5t 4 Initial!conditions:!v (0) = 0v " ae 0 + e0 4 " a + 4 = 0 " a =! 4 v (t) =!e!t + e!5t,!!y(t) = v (t) + u(t) = e!5t + e!5t! e!t 4 4 Slide 9

10 Example: RLC circuit with inputs R u(t) = V + - C i + v - + v 2 i 2 - L Y(t) Initial!conditions:!!V c (0) =!2V,! i L (0) = 2A Output = Y(t) = Voltage across inductor = v 2 (t) Node!equation!at!v : v (t)! u(t) R dv = i C " dv = v CR + u CR! i 2 C + i + i 2 = 0 " i = v R + u R! i 2 Slide 0 di 2 = L v = v 2 L

11 The homogeneous solution (aka: the natural response) d! v $!' / RC # " i & = ' / C $! v $ # 2 % " / L 0 & # % " i & 2 % +! / RC $ # " 0 & % u(t) homogeneous solution: take u(t)=0 d! # " v i 2 $!' / RC & = ' / C $! # % " / L 0 & #!# #" ### $ % " A v i 2 $ & %! -2 -$! s+2 $ Let!C= F,!R=/2 ",!L=2 H ( A= # " /2 0 & ( SI ' A = # % "-/2 s & % characteristic equation: s 2 + 2s + / 2 = 0 ( s = ' + 2, s = '' 2 Slide Natural!response:!!e ('+/ 2 )t (''/ 2 )t,!!e

12 The homogeneous solution, continued Finding the eigen-vectors: s =! + #! 2 & 2 " E s = % ( % 2 + (,!!s 2 =!! 2 " E s 2 = $ % '( # % % $ %! 2 2! & ( ( '( v n = a(! )e(!+ 2 )t + b(! 2 2! )e(!! 2 )t,!!i 2n = ae (!+ 2 )t + be (!! 2 )t Complete!solutions:!!v = v n + v f,!!!i 2 = i 2n + i 2 f Slide 2

13 u(t) = v The particular solution (aka: the forced response) The forced response would be a constant. I.e., v f = A,!i 2 f = B dv f = 0 =!v f RC + u RC! i 2 C =!A RC + RC! B C C = F, R = / 2", L = 2H # 0 = 2A + 2! B # 2A + B = 2 di 2 f = v L = A L = 0 # A = 0 # v f = 0V 2A + B = 2 # B!= 2 # i 2 f = 2A Slide 3 Does this solution make sense?

14 The complete solution Initial!conditions:v (0) = 2,!i 2 (0) = 2 Complete!solutions:!!v = v n + v f,!!!i 2 = i 2n + i 2 f Slide 4 v = a(! )e(!+ i 2 = ae (!+ 2 )t + be (!! 2 )t )t + b(! 2 2! )e(!! 2 )t + 0 v (0) = 2 " a(! ) + b(! 2 2! ) = 2 # % $ " a = i 2 (0) = 2 " a + b = 2 " a =!b % 2,!!b =! 2 & v (t) =! 2 + e(!+ i 2 (t) = 2 e(!+ 2 )t + 2! e(!! 2 )t! 2 e(!! 2 )t )t Forced response, v f = 0 Forced response, i 2f = 2