EE/ME/AE324: Dynamical Systems. Chapter 7: Transform Solutions of Linear Models


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1 EE/ME/AE324: Dynamical Systems Chapter 7: Transform Solutions of Linear Models
2 The Laplace Transform Converts systems or signals from the real time domain, e.g., functions of the real variable t, to the complex frequency domain, e.g., functions of the complex variable s, where s = σ + j ω Can be used to convert linear differential eqns. into algebraic eqns. that are easier to solve Given a time domain signal f(t), the Laplace transform is defined as st F s f t e dt L[ f () t ] () () f t = 0
3 Transform Pairs: Step Function ( ) Suppose f t = A for t > 0, then by definition : t A st st F() s = Ae dt = e s 0 t= 0 A A 0 [ 0 1] A = e e s = = s s These signals are known as transform pairs: A F () s = L [ A ] = f () t = A, for t > 0 s A = 1 is a special case known as the Unit Step fcn., denoted U(t) = 1 for all t > 0
4 Transform Pairs: Exp. Function ( ) at Suppose f t = Ae for t > 0, then by definition : at at st L 0 0 ( ) s+ a t F() s = Ae = Ae e dt = Ae dt t + A A ( s a) t 0 = e = e e s + a s t= 0 A A = [ 0 1 ] = s s+ a
5 Transform Pairs: Exp. Function Plots below illustrate t the exp. fcn. for three cases of a : The exp. fcn. is equivalent to U(t) when A= 1, a= 0
6 Transform Pairs: Ramp Function Suppose f t = At for t > 0, then by definition i i : ( ) st F() s L At Ate dt [ ] = = 0 t st Ate 1 st = + Ae dt s s t= 0 0 A s = 2 = 0 1 Step Fcn. = s
7 Transform Pairs: Trig. Functions Suppose f t = Asin( ω t ) for t > 0, then by def.: ( ) L[ ] st F () s A sin( i( ωt) A sin( i( ωte ) dt = = st Ae = = 0 t [ ssin( ωt) ωcos( ωt) ] ( 2 2 s) + ω t= 0 s Aω + ω 2 2 L[ A ω t ] = 2 2 Similarly, cos( ) s As + ω
8 Transform Pairs: Rectangular Pulse Suppose f t is a pulse of width L and height A: ( ) L A A Fs () = Ae dt= e = 1 e s s ( s ) L ( ) st st sl 0 0 L, A= 1 1 L = 0 0 A Height of arrow is area under curve L A= 1
9 Transform Pairs: Unit Impulse The Laplace transform of the unit impulse δ () t can be derived from the pulse transform using the assumption that L = 1 0: A ( sl 1 e ) 0 δ () s = lim =, an indeterminant idt i tform L 0 sl 0 Applying LHospital L'Hospital'sRule: s δ () s = lim ( sl 1 ) d e sl dl se = lim = 1 δ ( t) d sl L s ( ) L 0 0 dl
10 This definition will be used in EE321 (not EE32) 4! Transform Pairs: Unit Impulse Assuming L = 1 0, the definition of the unit A impulse δ ( t) can be expressed heuristically as: δ ( t) = 0, for t 0 only exists at the origin ε δ ( tdt ) = 1, for ε > 0 unity area under curve ε Assuming a continuous function f( t), a formal definition of the unit impulse δ () t is given by: b a f( t) δ ( t) dt = f(0), for a> 0, b> 0
11 Transform Properties Tables E.1 and E.2 in Appendix E contain a list of helpful Laplace transform pairs and properties; key transform properties will now be discussed Multiplication by a Constant: L [ ] st st af() t af() t e dt a f() t e dt af() s = = = 0 0 Superposition: L af ( t) + bg( t) = af( s) + bg( s) [ ] This implies that t the Laplace transform is a linear operator, like integration and differentiation!
12 Multiplication Properties Generally, L [ f() t gt () ] L [ f() t ] L [ gt () ] Multiplication by an Exponential: ( ) at s+ a t L e f() t = f() t e dt = F s+ a Common cases: 0 L te at = 1 ( ) ( s+ a ) at L e cos( ω t ) =, ( ) 2 2 s+ a + ω at ω L e sin( ωt) = s+ a + ω s ( ) a ( ) 2,
13 Multiplication by Time Signal multiplication by time occurs commonly: L d ds [ t f () t ] = { Fs ( )} Special Case: n n!! L t = n L 1 s L = L = L = = s s L s s [] [] 2 3 1, t, t, t 2 3 4
14 Differentiation by Time 2 2 For example, if f( t) = t F( s) =, 3 s then gt ( ) = f ( t) = 2t df () t G () s = L dt = sf () s f (0) 2 2 = s = s 3 s 2 This can be verified from Table E.1!
15 Differentiation by Time 1 Given U( t) = 1, for t > 0 U ( s ) =, s du () t if f () t =, then dt du () t Fs () = L dt = su ( s) U (0) 1 = s = 1 =δ ( s ) s du () t This implies that δ ( t) =! dt
16 Integration by Time Signal integration by time occurs commonly: t 1 L f ( λ ) dλ = Fs ( ) 0 s t Specific Cases: L λ d λ = 2 = s s s 0 3 A a ( 1 ) t at aλ A e = Ae dλ 0 s s + a ( ) t 0 g (0) Fs () g() t = g( 0) + f( λ) dλ G( s) = + s s
17 Application to Systems Analysis You will learn, e.g., in EE321, that [ f() t g() t ] = [ f() t ] [ g() t ] L L L where is the convolution operator defined as: ht ( )* ut ( ) h( λ) ut ( λ) dλ 0 In general, multiplication in one domain implies convolution in the other domain [ ) ] L [ h( ) ] yt () = u () t h () t L u () t h( t) = Y () s U( s) H( s)
18 Application to Systems Analysis This fact is useful when analyzing a system's response where h(t) is the system impulse response H(s) the system transfer fcn., u(t) is the system input and y(t) is the system output: ht () = y () t ut () = δ ()ICs (), t = 0 ut () y() t = h() t u() t System U() s Y() s = H() s U() s H () s = Y () s U ( s ) = 1, ICs= 0
19 Finding System Responses Using Laplace Transforms A system response (regardless of its order) can be solved using the Laplace transform in three steps: 1. Write and immediately transform the intergraldifferential eqns. describing the system for t>0, evaluating all ICs and inputs, e.g., unit step fcns. 2. Solve the resulting algebraic eqns., which are now fcns. of the complex frequency domain variable s, for the transform of the desired outputs. 3. Evaluate the inverse transform to obtain the output as a fcn. of time. This is done using Tables E.1 and E.2 along with some techniques to be discussed in Section 7.6 of the text.
20 Unit Step and Impulse Responses If the procedure on the prior slide is applied using a unit step fcn., U(t), (), as the system input and zero ICs, then the system output is called the Step Response and is denoted in the text as y U (t) () If the procedure is applied using a unit impluse δ ( t) as the system input and zero ICs, then the system output is called the Impulse Respons e and is denoted ht ()
21 Unit Step and Impulse Responses Knowing a system's s Impulse Response ht ( ) H( s) is important since this can be used to calculate the system's response to any other input u() t U ( s ), since: yt ( ) Y( s) = H( s) U( s) ht () = yt () ut () = δ (), t ICs= 0 ut () yt () = ht () ut () System U() s Y() s = H() s U() s H () s
22 Unit Step and Impulse Responses Based on the Laplace operator properties, it can be shown that: dyu () t ht ( ) = H( s) dt A system's impulse response can be obtained indirectly by differentiating its step response!
23 General First Order Models Consider a general first order system of the form: 1 y + y = f( t) τ where τ is a real, nonzero constant called the system time constant If the fcn. f( t) is a constant A for all t > 0, the Laplace transform of the eqn. is given by 1 A sy () s y(0) () + Y () s = τ s y(0) A Y () s = s + s s τ + τ
24 General First Order Models The inverse transforms of the terms on the righthandhand side of the eqn. can be found directly using Tables E.1 and E.2 in Appendix E, as shown: Y y(0) A () s = s + s s τ + τ t τ τ yt () = y(0) e + τa τae y zi () t y () t where y ( t) is the zeroinput response and y ( t) is the zi zerostate response zs t zs
25 General First Order Models If the input is zero, i.e., y () t = 0, then the stability of the system can be determined from the limit of zs y zi ( t) as t ; this is determined d by the value of τ as shown: τ < 0 unstable τ > 0 stable τ = 0 marginally stable
26 General First Order Models The system response can also be expressedinterms of its transient y ( t) and y ( t) steadystate responses: tr ss t t t τ τ τ yt () = y(0) e + τ A τae = τa + y(0) τae y () t yzs() t yss() t y tr () t 0 as t zi [ ] tr The response settles to within 2% of y in 4τ y ss τ is the time for the response to reach 63% of its total transition; alternately, the time for the initial slope to reach y ss as shown
27 Ex. 7.2: Rotational Mass Damper Find the system'ss step response assuming the system is initially at rest with no stored energy and the input is ω () t = A for t > a 0, ie i.e., ascaledstepfcn: step fcn.: The system EoM is: ( B + B ) B J ω B B ω B ω t ω ω ω t J J ( ) = 1 a () = a ()
28 Ex. 7.2: Rotational Mass Damper Substituting the input ω ( t a ) = A yields: ( B ) 1+ B2 BA 1 ω1+ ω1 = J J ( B ) 1+ B2 BA 1 sω1() s ω1(0) + ω1() s = J Js ( B ) 1 B2 BA 1 s ω + 1() s + 1() s 1(0) ω = + J Js ω 1 BA 1 J s+ ω 1( s) =, where τ = τ Js B + B ( ) 1 2 BA ( t ) τ ω BA 1 1 ω1 () s = 1 () t = 1 e τ 1 B1+ B2 Js s + τ
29 Ex. 7.3: RL Circuit Find for > 0 assuming the system input ( ) e t > e t = A o for t > 0, with zero ICs: i KCL yields: 1 1 ( e ) i eo + i(0) + eodλλ = 0 R L t 0
30 Ex. 7.3: RL Circuit Substituting i (0) = 0, e = A, taking the Laplace transform and solving system eqn. for i e o ( s) yields: 1 A 1 A eo() s = eo() s eo() s = R s Ls R s + L Taking the inverse Laplace transform of e o ()y s yields: A Rt eo () s = e ( ) L o t = Ae for t > 0 R s + L
31 Ex. 7.4: Switched Circuit Assumetheswitchisopenwithnoenergystored switch is with no energy stored in the capacitor for t < 0 and the switch closes at t = 0. Find e () t for t > 0: o KCL at nodes A and O for t > 0: 1 e e = 2 e e = 6e + e ( ) 3 o A A o A A e + ( e e ) = ( 24 e ) e = 6 + e o o A o o A
32 Ex. 7.4: Switched Circuit Transforming the eqns. term by term yields: e = 6e + e o A A e () 6 () (0) o s = sea s e ( ) A + ea() s = 6s+ 1 ea () s eo = 6 + ea eo ( s ) = + ea ( s ) 2 s 2 Combining the eqns. to solve for e o () s e () s yields: 6s = = s s+ s+ s s o ( ) t t t e () t = 6e e = 12 6e o
33 Ex. 7.4: Switched Circuit t 12 Given eo () t = 12 6e e + e (0 ) = 6 V and e = lim e ( t) = 12 V o oss o t The system has a time constant τ = 12 sec system settles to SS within 5 τ = 60 sec The system can be written in statespace form as shown: 1 1 e A = ea + 1 and eo = ea
34 Ex. 7.4: Simulink Model and Plot
35 Exam 2 Review Electrical circuits (Chapter 6): How to model resistors, capacitors, inductors, open/closed switches, op amps, etc. Model using Node Voltage Method to obtain: Input Output models State space t models First order system responses (through Section 7.5): Application of the Laplace transform to both mechanical and electrical systems, use of Tables E.1 2 How to handle common inputs, e.g., unit step and unit impulse, and initial conditions Understand the concepts of zero state and zero input responses, transient and steady state responses, system time constant and stability
36 Section 7.6: Transform Inversion We now discuss a method for computing an inverse Laplace transform that allows us to compute system output yt ( ) from Y () s assuming that it can be written as a strictly yproperp rational function, i.e., m< n, where: m Ns () bs m + + b0 Y( s) = = n n 1 Ds () s + a s + + a n 1 0 In addition, assume that Ds () can be factored into nroots: D( s) = ( s s )( s s ) ( s s ) 1 2 where the s terms are called pol es of Y () s and their locations i determine the stability and shape of the system response Note, the poles are generally complex valued s = σ + jω i i i n
37 Partial Fraction Expansion Method Partial Fraction Exapnsion (PFE) allows us to write higherorder sdomain expressions in terms of a sum of lowerorder order expressions having known inverses There are two basic cases when computing PFEs: Dist inct Poles (Real or Complex) and Repeated Poles
38 Distinct Poles (Real or Complex) Let's examine the case of Distinct Poles, ie i.e., s s s ; assume Y( s) can be written as: 1 2 n A1 A2 An Ai Y() s = = ( s s1) ( s s2) ( s sn) i= 1 ( s si) Known as a residue, the term A is computed as follows A i A = ( s s ) Y( s), for i = 1,, n i i s= s i The inverse Laplace transform can then be computed as n 1 1 A i st i yt ( ) = L [ Y( s)] = L = Ae i, for t> 0 i= 1 ( s si ) i= 1 n n
39 Distinct Poles (Real or Complex) Complex poles are treated the same as real poles other than the need to combine the resulting complex conjugate terms so that yt () is a real fcn. of time For example, assume Y( s) contains complex poles at s = σ ± jω so that: 1,2 Y () s 1 A A ( s+ σ jω) ( s+ σ + jω) = where A = ( s + σ j ω) Y( s) = Ae jφ, s= σ + jω * 2 1 j A = ( s+ σ + jω) Y( s) = A = Ae φ with A = σ + ω and φ = tan s= σ jω ( ω ) σ
40 Distinct Poles (Real or Complex) Transforming the complex poles back into the time domain using the same transform pair as in the real poles case and applying Euler's formula yields: Y( s) yt () = jφ jφ Ae Ae = + + ( s + σ j ω ) ( s + σ + j ω ) jφ ( j Ae e σ + ω ) t jφ ( j ) t Ae e σ ω + + ( ω + φ) ( ω + φ) j t j t σ t e + e = 2Ae 2 σ t = 2Ae cos ωt + φ, for t > 0 ( ) where y, t, σ, ω, A and φ are all real
41 Ex. 7.9: PFE w/district Real Poles Find the inverse transform of s+ 5 s+ 5 A A s s + 4 s + 1 s + 4 ( s + 1) ( s + 4) 1 2 ( ) = = = +, Y s 1 2 ( )( ) where s = 1, s = 4 are the system poles A = ( s s ) Y ( s ) = ( s+ 1) Y ( s ) 1 1 s= s = 1 s 1 s = = = = 2 ( s + 4) s= 1 ( 1+ 4) 3 A = 2 ( s s2) Y( s ) s = s2 = ( s + 4) Y( s) s= 4 s = = = = 3 ( s + 1) s= 4 ( 4+ 1) 3
42 Ex. 7.9: PFE w/district Real Poles Given the PFE for Y (), s we can now write: 2 3 t 4t Y () s = y () t = 2e 3 e, for t > 0 ( s+ 1) ( s+ 4) Plotting this fcn. over 5τ using Matlab: Do the initial and final values of the figure make sense, e.g., y(0) and y SS?
43 Ex PFE: w/district Complex Poles Find yt () given: 4s+ 8 4s+ 8 Y() s = = 2 s + 2 s + 5 s + 1 j 2 s+ 1+ j 2 ( )( ) A1 A2 = +, ( s + 1 j 2) ( s j 2) where s = 1+ j2, s = 1 j2 = s are the system poles * A = ( s s ) Y() s = ( s + 1 j2) Y() s 1 1 = s= s1 s= 1+ j j j 8 j = = 2 j = 5e ( 1+ j2+ 1+ j2) j4 ( ) A 2 = ( s s 2 ) Y () s s= s = ( s j 2) Y () s 2 s= j * = 2 + j = 5e j = A 1 1 2
44 Ex. 7.11: PFE w/district Complex Poles Given the PFE for Y(): s j0.464 j e 5e Y () s = + s+ 1 j2 s+ 1+ j2 t yt ( ) = 2 5 e cos 2 t 0.464, for t> > 0 ( ) Plotting this fcn. over 5τ using Matlab yields: Do the initial and final values of the figure make sense, e.g., y(0) and y SS?
45 Repeated Poles (Real or Complex) Poles are said to be repeated if several roots of Ds ( ) = 0 are the same; assuming r repeated poles s1 = s2 = = s r, we represent the PFE of Y () s as: A1, r A1, r 1 A11 Y( s) = r ( s s ) ( s s ) ( s s ) yt = + () A 1,rr ( ) r 1 st 1 A 1, r 1t+ + A t e + ( ) 11 r 1! ( p 1) 1 d r where A 1, p = ( s s ( ) ( 1) 1) Y ( s) p p 1! ds = r d r A11 = ( s s1 ) Y() s, A12 = ( s s1 ) Y() s, s= s 1 ds s s 1 s= s 1
46 Ex. 7.10: Repeated Real Poles Find the inverse Laplace transform of: 5s+ 16 5s+ 16 Y( s) = = 3 2 s + 9s + 24s+ 20 s+ 2 s+ 5 A12 A11 A2 = ( s+ 2) ( s+ 2) ( s+ 5) t yt () = A + At e + Ae 2 t 5 ( ) ( ) ( ) 2 5s + 16 where A 11 = ( s+ 2) Y( s) = = 2 s = 2 s + 5 s= 2 Matlab plot d A = s + 2) 2 d s+ Y s 12 = = = 1 2 ( ds () ds s 5 + s + 5 ( ) 2 s= 2 s= 2 s= 2 >>eval(limit(simple(diff('(5*s+16)/(s+5)',s)),s,  2) ) A = 2 ( s + 5) Y( s ) s= 5 = 1
47 Long Division When m= n, PFE can't be applied directly to Y (); s performing long division fixes the problem as shown: N () s 1 N () s Y() s = A+ y() t = Aδ () t + L Ds () Ds () where N ( s) has order p< n PFE can be applied N () s to Y ( s) = as shown previously Ds ()
48 Ex. 7.13: Long Division 2 2 s + 7 s + 8 s + 4 For example, if Y( s) = = s + 3s+ 2 s + 3s+ 2 yt ( ) = 2 δ ( t) + L 1 s + 2 s s s + 3s+ 2 2s + 7s s + 6 s + 4 s + 4
49 Additional Transform Properties There are a few additional Laplace transform properties that are helpful to know when analyzing system responses, including: modeling Time Delays, the Initial and Final Value Theorems To model ltime delay, consider the unit step fcn., 0, for t 0 Ut () = 1, for t > 0 and a unit step delayed by T seconds 0, for t T Ut ( T) = 1, 1 for t > T
50 If f () t = f () t U () t Time Delay [ st Fs ( ) L[ f () tu () t] = f () tu () te dt, = then f( t T) = f( t T) U( t T) L [ f( t T) U( t T) ] = ( ) ( ) 0 = = a 0 st f t T U t T e dt ( ) st st st T f ( t T ) e dt e f ( t T ) e dt a st s λ st st = e f ( λ ) e d λ = e F () s f ( t T ) e F ( s ) a
51 Example of Time Delay Find the Laplace transform of ft ( ) = Ut ( ) Ut ( 2) st ( st ) 1 e Fs () = U( s) e U() s = 1 e Us () = s st
52 Initial and Final Value Theorems The Initial Value Theorem (IVT) and Final Value Theorem (FVT) allow computation of y( t) at zero and steadystate from Y( s) as shown: + y(0 ) = lim y( t) = lim sy( s), when the limit in s exists y SS t t 0 = lim y( t) = lim s 0 s sy ( s), if all poles in lefthalf splane + For example, find y(0 ) and y for Y( s) when 4s + 8 Y () s =, where s1,2 1 j 2 stable tbl poles 2 s + 2s+ 5 = ± SS
53 Initial and Final Value Theorems s + 8 s From IVT, y(0 ) = lim sy( s) = lim = 4 s s s + 2s s + 8s From FVT, yss = lim sy( s) = lim = 0 s 0 s 0 2 s + 2s+ 5 These results agree with those of Ex (see graph)!
54 Questions?
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