04_SEMD3_Ch3_LaplaceText

04_SEMD3_Ch3_LaplaceText - 1. Standard notation in dynamics...

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Unformatted text preview: 1. Standard notation in dynamics and control (shorthand notation) = how you communicate with other engineers. other engineers 2. Converts differential calculus mathematics to algebraic operations, very useful for: Initial Value Problems (IVPs). Chapter 3 Chapter 3 Laplace Transforms For “simple” simple Y(s) -- maybe a short cut ! 3. Advantageous for block diagram analysis = a “systems” way to analyze more complex systems. (study subsystems, then combine to large system) subsystems then combine to large system) Other Transforms Laplace Transform Definition ⎡d2 f L⎢ 2 ⎣ dt ∞ L [f (t )]= ∫ f (t )e- st dt = F ( s ) 0 = s [ sF ( s ) - f (0) ] − f ′(0) ∞ a ⎡ a⎤ a L [a]= ∫ ae dt = − ⎡e −st ⎤ = 0 − ⎢ − ⎥ = ⎣ ⎦0 s ⎣ s⎦ s 0 -st ∞ ∞ 0 0 L [e-bt ]= ∫ e-bt e-st dt = ∫ e-(b+s)t dt = ∞ 1 1 ⎡ − (b +s)t ⎦ = ⎤ ⎣-e 0 b+s s+b ∞ df ⎡ df ⎤ and a theorem ... L [f ′ ] = L ⎢ ⎥ = ∫ e-st dt = sF ( s ) − f (t = 0) ⎣ dt ⎦ 0 dt Very convenient for zero initial condition f(0) = 0 An example is when f = the “error” in a process control IVP, example is when the error process control IVP i.e at t = 0 the process is at a “nominal design” steady-state. Chapter 3 Chapte 3 Two derivations (“a” is a constant, f ’ is derivative wrt time): (“a” ∞ ⎤ df ⎡ dg ⎤ where we assign ⎥ = L ⎢ dt ⎥ where we assign g = dt = f ′(t ) ⎣⎦ ⎦ = sG ( s ) - g (0) = s 2 F ( s ) − sf (0) − f ′(0) n etc. for higher orders d f n dt ⎡ e- jωt + e + jωt ⎤ L [cos (ωt )] = L ⎢ ⎥ 2 ⎣ ⎦ 1⎛ 1 1⎞ =⎜ + ⎟ 2 ⎝ s + jω s − jω ⎠ 1 ⎛ s − jω s + jω ⎞ = ⎜2 + ⎟ 2 ⎝ s + ω 2 s2 + ω 2 ⎠ s =2 s + ω2 ⎡ e + jωt - e − jωt ⎤ L [sin (ωt )] = L ⎢ ⎥ 2j ⎣ ⎦ = Note: ω s + ω2 2 e jωt = cos ωt + j sin ωt e − jwt = cos ωt − j sin ωt j = −1 2. Examples 2 Numerator( s ) = 3s + 4s + 1 Quadratic Denominator( s ) b2 − 4ac = 16 − 12 = 4 > 0 , with a > 0, -b < 0 so this is an overdamped system = 2 exponential decays negative real roots s = −b ± b − 4ac . 2a 2 3s 2 + 4s + 1 = (3s + 1)( s + 1) = 3( s + 1 )( s + 1) 3 -t/3, e-t (real roots) Transforms to “pieces”: e 3 3 j )( s + 0.5 − j) 2 2 3 3 e −0.5t sin t , e −0.5t cos t 2 2 s 2 + s + 1 = ( s + 0.5 + 2 Chapter 3 Chapter 3 1. Y ( s ) = 2+s b2 − 4ac = 1 − 4 = −3 < 0 , an underdamped system. s + s +1 2 Transforms to (oscillation, complex roots to Denom(s) = 0) From Table 3.1, line 17 L e − bt sin ω t ←⎯ → 2 = s + s +1 2 (no oscillation, neg. real roots to Denom.(s) = 0) ω (s + b)2 + ω 2 2 ⎛ 3⎞ ( s + 0.5) + ⎜ ⎟ ⎝2⎠ L e − bt cos ωt ←⎯ → 2 2 s ( s + b) 2 + ω 2 ω = omega Difference of two step inputs S(t) – S(t-1) ( S(t-1) is step starting at t = h = 1 ) Let h→0, f(t) = δ(t) (Dirac delta) f(t) delta) F (s) = ∫ h 0 1 − st 1 e dt = (1 − e − hs ) h hs If h = 1, rectangular pulse input L(δ) = 1 L( Chapter 3 Chapter 3 By Laplace transform F ( s) = 1 1 −s −e ss Can be generalized to steps of different magnitudes be generalized to steps of different magnitudes (a1, a2). Please see Table 3.1 pg 42-43 in SEMD3 for examples of “pairs” of functions : f(t) and F(s) f( Laplace transforms can be used in process control for: 2. Analysis of linear control systems (frequency response) (frequency response) 3. Prediction of transient response “pieces” for different inputs, e.g. stability analysis. ff Chapter 3 Chapter 3 1. Solution of differential equations (linear) Please see Table 3.1 pg 42-43 in SEM for examples of “pairs” of functions : f(t) and F(s) f( Chapter 3 Chapter 3 Please see Table 3.1 pg 42-43 in SEMD3 for examples of “pairs” of functions : f(t) and F(s) f( One other useful feature of the Laplace transform is that one can analyze the denominator of the transform to determine its dynamic behavior. For example, if Example 3.1 dy 5 + 4y = 2 y (0) = 1 (3-26) dt First, take Laplace of both sides of (3-26), 2 5 ( sY ( s ) − 1) + 4Y ( s ) = s Rearrange, 5s + 2 (3-34) Y (s) = s ( 5s + 4 ) Find Standard form, ⎡ 5s + 2 ⎤ ⎤ s + 0.4 −1 ⎡ y ( t ) = L−1 ⎢ ⎥=L ⎢ ⎥ ⎣ s ( 5s + 4 ) ⎦ ⎣ ( s + 0) ( s + 0.8 ) ⎦ Using Table 3.1 #11 form, y ( t ) = 0.5 + 0.5e−0.8t Y(s)= Chapter 3 Chapter 3 Solve the ODE, the ODE (3-37) E (s) = sE (s) = a τs + 1 lim s→ 0 a = a = E( ∞ ) τs + 1 The “offset” = steady state error = a. B. Time-shift theorem y( y(t)=0 t < to Ly The step response of the process will have exponential terms step response of the process will have exponential terms e-2t and e-t, which indicates y(t) approaches zero. However, if 1 1 = s − s − 2 (s + 1 )(s − 2 ) 2 lim y(t)= lim sY(s) t →0 Useful in evaluating steady-state in evaluating steady error E(t) as time goes to infinity. 1 a τs + 1 s w h a t is th e f in a l v a lu e o f E a t th e n e w s te a d y -s ta te ? If α1 α +2 s + 2 s +1 C. Initial value theorem Initial theorem ( (t - θ )) = e θ -s Y(s) Y(s) Chapter 3 Chapter 3 s →0 Y(s)= We know that the system is unstable and has a transient response involving response involving e2t and e-t. e2t is unbounded for unbounded large time. We shall use this concept later in the analysis of feedback system stability. Final value theorem y(∞)= lim sY(s) the denominator can be factored into (s+2)(s+1). Using the partial fraction technique Y(s) = Other applications for y(t) = E(t) = the process control error A. 1 s + 3s + 2 2 s →∞ For Y(s)= y( 0 )=0 y(∞)= 1 3 4 s+2 s(s+1 )(s+2 )(s+3 ) by initial value theorem by final value theorem, Since roots of denominator of sY(s)=0 are -1, -2 and -3, i.e. all have negative real parts. For Y(s)=Num(s)/Denom(s) inverse transformed to yield y(t), the roots of Denom(s)=0 determine the kinds of “pieces” in y(t). Denom determine the kinds of pieces in If the roots are: 1) Individual real = the corresponding piece decays if root<0, it grows if real th if it if Chapter 3 root >0. A constant “piece” arises when only one root is at 0. 2) Repeated real = polynomial in t times exponential decay or growth Repeated real polynomial in exponential decay or growth corresponding to the real part of root <0 or real part of root >0 respectively. If a repeated root is at s = 0, response is not convergent. 3) Complex (conjugate) = Re + Im j and Re – Im j yields growing oscillation (Re>0) or decaying oscillation (Re<0). Always in “pairs”. Purely imaginary pair gives response that does not converge due to continuous oscillation as sine or cosine. ill 4) Repeated complex = polynomial in t times form from 3). Repeated imaginary pairs response is unbounded sine/cosine times polynomial imaginary pairs, response is unbounded, a sine/cosine times a polynomial in time, thus does not converge. ...
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This note was uploaded on 03/01/2012 for the course CHE 361 taught by Professor Staff during the Winter '08 term at Oregon State.

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