Transform Solutions to LTI Systems Part 3

Transcription

1 Transform Solutions to LTI Systems Part 3 Example of second order system solution: Same example with increased damping: k=5 N/m, b=6 Ns/m, F=2 N, m=1 Kg Given x(0) = 0, x (0) = 0, find x(t). The revised equation for X(s) with b=6 Ns/m is The solutions to s 2 + 6s + 5 = 0 are: s = 6 ± 36 4(5) 2 Hence s 2 + 6s + 5 = (s + 5)(s + 1) 5 s(s 2 + 6s + 5) = 3 ± 1 16 = 3 ± 2 = 5, s(s 2 + 6s + 5) = 5 s(s + 5)(s + 1) = A s + B s = A(s + 5)(s + 1) + Bs(s + 1) + Cs(s + 5) s = 0 5 = A(5)(1) A = 1 C s + 1 s = 1 5 = C( 1)( 1 + 5) C = 5 4 s = 5 5 = B( 5)( 5 + 1) B = 1 4

2 Hence = 1 s ( 1 s + 5 ) 5 4 ( 1 s + 1 ) x(t) = e 5t 5 4 e t x(t) t Third Order System Solution: We know how to solve 1 st order odes and 2 nd order odes. What about higher order odes? Consider a 3 rd order input-output relationship: x + a 2 x + a 1 x + a 0 x = u 1 Input: u Output: x Assume initial conditions are zero. Take the Laplace transform of 1 s 3 X(s) s 2 x(0) sx (0) x (0) + a 2 *s 2 X(s) sx(0) x (0)+ + a 1 *sx(s) x(0)+ + a 0 U(s)

3 Since the initial conditions are zero, *s 3 + a 2 s 2 + a 1 s + a 0 + U(s) X(s) U(s) = 1 s 3 + a 2 s 2 + a 1 s + a 0. 2 In the Laplace domain, the ratio of output to input is a rational function containing polynomials of s in the numerators and denominators. This rational function is called a transfer function. The characteristic equation is the equation obtained by setting the denominator of the transfer function to zero. e.g. the characteristic equation for 2 is s 3 + a 2 s 2 + a 1 s + a 0 = Equation 3 can be factored into either (s + a)(s + b)(s + c) = 0 Or (s + a)(s 2 + bs + c) = 0 For the coefficients of the characteristic equation to be all real, complex roots always must occur in complex-conjugate pairs: If σ + jω is a root, σ jω must also be a root. j = 1 (s σ jω)(s σ + jω) = s 2 σs + jωs σs + σ 2 jσω jωs + σjω j 2 ω 2 = s 2 2σs + σ 2 j 2 ω 2 = s 2 2σs + σ 2 + ω 2 (all real coefficients) Hence for a 3 rd order ode, the characteristic polynomial will always have at least 1 real root.

4 Higher Order Systems Likewise, the characteristic equation for any higher ode is factorizable into 1 st and 2 nd order polynomials: X(s) U(s) = 1 (s + a)(s + b). (s + d)(s 2 + e 1 s + f 1 )(s 2 + e 2 s + f 2 ).. (s n + e n s + f n ) Partial fraction expansion can be used to obtain terms with first order and second order polynomials in the denominator. Thus, you can find the analytical solution to any high order ode using Laplace transforms. Characteristic Equation, Poles and Stability Consider a linear time invariant (LTI) ode written in input-output form. For example: mx + bx + kx = F(t).. 1 Input : F(t) Output: x(t) On taking Laplace transforms, the output response of the system is related to the initial conditions and the input through transfer functions. A transfer function is a rational function with the numerator and denominator being polynomials in s. For example, for the system 1, m,s 2 X(s) sx(0) x (0)- + b,sx(s) x(0)- + k F(s) Hence

5 ,ms 2 + bs + k- F(s) + x(0),ms + b- + mx (0) 1 ms 2 + bs + k x(0),ms + b- + mx (0) F(s) + ms bs + k The transfer functions relating the output to the input and to the initial conditions have the same denominator. Hence the responses to the input and to the initial conditions are going to be the same type of functions. For all real-world systems, the order of the denominator order of the numerator. If order of denominator order of numerator the transfer function is said to be proper. If order of denominator > order of numerator the transfer function is said to be strictly proper. Setting the denominator to zero yields a polynomial equation in s called the characteristic equation. For example, the characteristic equation for the transfer function in 3 is : ms 2 + bs + k = The roots of the characteristic equation are called the poles of the system. Since the denominator can be factored using the poles, the behavior of the system is governed by the nature of the poles. For example, if the transfer function is:

6 The pole is s + a = 0 s = a And the response is: x(t) = Ae at If a > 0, pole is negative and real, A s + a, x(t) 0as t The system is said to be stable. If a < 0, pole is positive and real, x(t) as t. The system is said to be unstable. Second example: Consider As + B ms 2 + bs + k The poles of this system are given by the roots of ms 2 + bs + k = 0 If the roots(poles) are real, then the transfer function can be written as where s = a, b are poles. As + B (s + a)(s + b) After partial fraction expansion, it can be found that the response of the system is: x(t) = Ee at + Fe bt 5 Hence if a>0, b>0( both poles are real and ve), Then the response x(t) 0 as t. On the other hand, if either a < 0, or b < 0, then x(t) as t. The system in this case is said to be unstable.

7 Consider the case where the poles of the second order system in the example are complex. As + B ms 2 + bs + k In this case, the system function can be written as: As + B (s + a) 2 + ω 2 The poles are a + jω and a jω Complex poles always occur in conjugate pairs. Check: (s + a) 2 + ω 2 = 0 (s + a) 2 = ω 2 (s + a) = ± ω 2 = ±jω, j = 1 s = a ± jω Hence the response of the system will be of the type: x(t) = Ee at sinωt + Be at cosωt 6 If a>0, (real-part of the pole is ve) Then x(t) 0as t If a<0, (real-part of the pole is +ve) Then x(t) as t The imaginary part of the pole (ω) determines the frequency of oscillations, but does not determine stability. For any higher order system, the denominator of the transfer function can always be factored into 1 st order and 2 nd order terms. Hence, the solution of any LTI system is always going to be of the type

8 x(t) = A 1 e at + A 2 e bt + B 1 e α1t sinω 1 t + B 2 e α1t cosω 1 t + C 1 e α2t sinω 2 t + C 2 e α2t cosω 2 t + The real parts of all poles are ve The system is said to be stable. The real parts of one or more of the poles are +ve The system is said to be unstable. The real part of one or more of the poles is zero The system is said to be marginally stable. In this case, the response of the system contains terms of the type Ae 0t, e 0t sinωt, e 0t cosωt, etc or A, sinωt, cosωt, etc. Initial Value Theorem Given F(s), what is f(0)? Without having to find the Laplace inverse of F(s), the initial value can be determined using the following Initial Value Theorem: f(0 + ) = lim s sf(s) Conditions under which the initial value theorem (IVT) can be applied: Degree of denominator > degree of numerator in F(s) Example: a) F(s) = A s f(0 + ) = lim s s A s = lim s A = A

9 b) F(s) = A s 2 f(0 + ) = lim s A s s 2 = lim A s s = 0 c) ω F(s) = s 2 + ω 2 s. ω lim s s 2 + ω 2 = 0

10 d) s F(s) = s 2 + ω 2 f(0 + ) = lim s s. s s 2 + ω 2 = lim s s 2 + ω 2 = lim s Note: f(t) = cosωt, f(0 + ) = 1 s ω2 s 2 = 1 Final Value Theorem Given F(s), how can we find lim t f(t)? FVT: When is the FVT applicable? limf(t) = lim sf(s) t s 0 It is applicable when f(t) has a well-define final value Details in next lecture.