Introduction & Laplace Transforms Lectures 1 & 2


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1 Introduction & Lectures 1 & 2, Professor Department of Electrical and Computer Engineering Colorado State University Fall 2016
2 Control System Definition of a Control System Group of components that collectively perform certain desired tasks or maintain a desired result. Control system components can be electrical, mechanical, hydraulic, pneumatic, etc. or any hybrid combination. Basic Ingredients where c(t) = f(u(t))
3 Types of OpenLoop Output variable has no way of influencing the actuating signal. Errors cannot be reduced. ClosedLoop A feedback is added for more accurate and reliable control. Error signal, e(t) = r(t) f(t), drives the actuation via the feedback unit.
4 Open Loop Control System Effects of Feedback: 1 Gain: Reduces the overall gain (negative feedback). 2 Bandwidth (BW): Increases BW so that Gain BW = constant. 3 Stability: Improves stability with proper design. 4 Sensitivity: Def: Sensitivity is measured by variations of system parameters (e.g., gain, pole locations, etc.) wrt environmental changes and aging. Sensitivity for some parameters can be reduced via proper feedback. 5 Noise: Any real system is subject to noise (e.g., thermal noise, brush noise in motors, etc.). Effect of feedback on noise depends on where in the system the noise is introduced.
5  PierreSimon Laplace 1809 Why use the Laplace Transform (LT)? 1  Provides complete solution of differential equation (homogeneous and particular together). 2  Handles initial conditions in the solution. Def. 1: Let G(s) be a function of s = σ + jw. Singularities of G(s) correspond to points on the splane where G(s) and its derivative do NOT exist. Example: G(s) = 1 (s + 1)(s + 2) G(s) does not exist anywhere the denominator equals zero. In this case at s = 1 and s = 2 (i.e. poles of G(s)). (1)
6 Def. 2: G(s) is analytic in region R in splane if it does not have any singularities in R. (So in the example above, G(s) is analytic everywhere except at s = 1, s = 2). Theorem: Given a time function, f(t), Laplace Transform (LT) and its inverse exist if and only if: 1 For every interval t 1 t t 2, f(t) is bounded and has a finite number of max and min and finite number of discontinuities. 2 There exists a real constant α such that 0 e αt f(t) dt is convergent. (Abscissa of Convergence) In other words, e αt f(t) should not be more powerful that e αt (e.g., for e t2 LT doesn t exist).
7 Laplace Transform and its Inverse Then we have: Laplace Transform: L {f(t)} = F (s) = ˆ 0 f(t)e st dt (2) Note: The onesided LT should start from t = 0 to account for discontinuities or an impulse at t = 0. Inverse Laplace Transform: f(t) = L 1 {F (s)} = 1 2π ˆ c+j c j F (s)e st ds (3) where c is a real constant greater than the real parts of all the singularities of F (s). Thus, we have onetoone correspondence between f(t) and F (s) via LT i.e. f(t) F (s). Fortunately, the inverse LT can be obtained using the tables.
8 Properties of Laplace Transform The utilities of LT lies in the following properties. 1  Linearity Let f 1 (t) F 1 (s) and f 2 (t) F 2 (s) then a 1 f 1 (t) + a 2 f 2 (t) a 1 F 1 (s) + a 2 F 2 (s) 2  Differentiation Let f(t) F (s) then d k f(t) dt k s k F (s) s k 1 f(0 ) s k 2 f (1) (0 )... f (k 1) (0 ) where f (i) (t) = di f(t) dt i 3  Integration Let f(t) F (s) then t 0 f(τ)dτ 1 s F (s) In general: t a f(τ)dτ 1 s F (s) s a f(τ)dτ
9 Properties of Laplace Transform Cont. 4  Shift in timedomain Let f(t) F (s) then f(t T )u s (t T ) e T s F (s) where u s (t) is a unit step function. u s (t) = { 1 t 0 0 else 5  Shift in sdomain Let f(t) F (s) then e αt f(t) F (s ± α) 6  Initial Value Theorem Let f(t) F (s) then lim t 0 f(t) = lim s sf (s)
10 Properties of Laplace TransformCont. 7  Final Value Theorem (FVT) Let f(t) F (s) then lim t f(t) = lim s 0 sf (s). Condition: sf (s) is analytic on the jωaxis and in RHS (righthalf side) of splane. Very useful property for steadystate error analysis. Example: Consider F (s) = ω s 2 +ω. If we apply FVT, we get lim 2 s 0 sf (s) = 0. But we know that f(t) = sin ωt with no final value. The reason for this discrepancy is that F (s) has complex conjugate poles on the imaginary axis.
11 Properties of Laplace TransformCont. 8  Convolution in timedomain Suppose f 1 (t) F 1 (s) and f 2 (t) F 2 (s). Then: ˆ t f 1 (τ)f 2 (t τ)dτ = ˆ t 0 0 f 1 (t τ)f 2 (τ)dτ = f 1 (t) f 2 (t) F 1 (s)f 2 (s) where designates convolution. This is a useful property for system analysis. y(t) = u(t) h(t) Y (s) = U(s)H(s) and y(t) = L 1 {Y (s)}
12 Properties of 9  Convolution in sdomain This is dual of property of 8. Let f 1 (t) F 1 (s) and f 2 (t) F 2 (s). Then: f 1 (t)f 2 (t) 1 2πj F 1(s) F 2 (s) = 1 c+j 2πj c j F 1(s ξ)f 2 (ξ)dξ 10  Differentiation in sdomain Let f(t) F (s). Then: tf(t) F (s) = df (s) ds 11Integration in sdomain Let f(t) F (s). Then: f(t) t F (ς)dς s Condition: lim t 0 f(t) t = fixed
13 Inverse Laplace Transform Finding the inverse laplace transform Let X(s) = N(s) D(s), where N(s) is a polynomial of order m and D(s) is a polynomial of order n, n m. We want to find f(t) =. Remark: If n m function X(s) is said to be proper. If n > m it is strictly proper. For n < m, the system becomes noncausal. Case 1: D(s) has real and distinct roots. X(s) = a i s: real and distinct roots of D(s) Apply PFE (partial fraction expansion): N(s) (s + a 1 )... (s + a n ) X(s) = A 1 (s + a 1 ) + + A n (s + a n ) A i = (s + a i )X(s) s= ai
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