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

Size: px
Start display at page:

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

Transcription

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, non-zero 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 right-handhand 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 zero-input response and y ( t) is the zi zero-state 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) steady-state 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 state-space 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 higher-order s-domain expressions in terms of a sum of lower-order 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 steady-state 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 left-half s-plane + 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?

### Laplace Transform Part 1: Introduction (I&N Chap 13)

Laplace Transform Part 1: Introduction (I&N Chap 13) Definition of the L.T. L.T. of Singularity Functions L.T. Pairs Properties of the L.T. Inverse L.T. Convolution IVT(initial value theorem) & FVT (final

### Lecture 7: Laplace Transform and Its Applications Dr.-Ing. Sudchai Boonto

Dr-Ing Sudchai Boonto Department of Control System and Instrumentation Engineering King Mongkut s Unniversity of Technology Thonburi Thailand Outline Motivation The Laplace Transform The Laplace Transform

### ECEN 420 LINEAR CONTROL SYSTEMS. Lecture 2 Laplace Transform I 1/52

1/52 ECEN 420 LINEAR CONTROL SYSTEMS Lecture 2 Laplace Transform I Linear Time Invariant Systems A general LTI system may be described by the linear constant coefficient differential equation: a n d n

### Laplace Transforms and use in Automatic Control

Laplace Transforms and use in Automatic Control P.S. Gandhi Mechanical Engineering IIT Bombay Acknowledgements: P.Santosh Krishna, SYSCON Recap Fourier series Fourier transform: aperiodic Convolution integral

### Unit 2: Modeling in the Frequency Domain Part 2: The Laplace Transform. The Laplace Transform. The need for Laplace

Unit : Modeling in the Frequency Domain Part : Engineering 81: Control Systems I Faculty of Engineering & Applied Science Memorial University of Newfoundland January 1, 010 1 Pair Table Unit, Part : Unit,

### Laplace Transforms Chapter 3

Laplace Transforms Important analytical method for solving linear ordinary differential equations. - Application to nonlinear ODEs? Must linearize first. Laplace transforms play a key role in important

### Introduction & Laplace Transforms Lectures 1 & 2

Introduction & Lectures 1 & 2, Professor Department of Electrical and Computer Engineering Colorado State University Fall 2016 Control System Definition of a Control System Group of components that collectively

### GATE EE Topic wise Questions SIGNALS & SYSTEMS

www.gatehelp.com GATE EE Topic wise Questions YEAR 010 ONE MARK Question. 1 For the system /( s + 1), the approximate time taken for a step response to reach 98% of the final value is (A) 1 s (B) s (C)

### School of Mechanical Engineering Purdue University. ME375 Dynamic Response - 1

Dynamic Response of Linear Systems Linear System Response Superposition Principle Responses to Specific Inputs Dynamic Response of f1 1st to Order Systems Characteristic Equation - Free Response Stable

### Advanced Analog Building Blocks. Prof. Dr. Peter Fischer, Dr. Wei Shen, Dr. Albert Comerma, Dr. Johannes Schemmel, etc

Advanced Analog Building Blocks Prof. Dr. Peter Fischer, Dr. Wei Shen, Dr. Albert Comerma, Dr. Johannes Schemmel, etc 1 Topics 1. S domain and Laplace Transform Zeros and Poles 2. Basic and Advanced current

### Chap. 3 Laplace Transforms and Applications

Chap 3 Laplace Transforms and Applications LS 1 Basic Concepts Bilateral Laplace Transform: where is a complex variable Region of Convergence (ROC): The region of s for which the integral converges Transform

### 2.161 Signal Processing: Continuous and Discrete Fall 2008

MIT OpenCourseWare http://ocw.mit.edu 2.6 Signal Processing: Continuous and Discrete Fall 2008 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. MASSACHUSETTS

### 9.5 The Transfer Function

Lecture Notes on Control Systems/D. Ghose/2012 0 9.5 The Transfer Function Consider the n-th order linear, time-invariant dynamical system. dy a 0 y + a 1 dt + a d 2 y 2 dt + + a d n y 2 n dt b du 0u +

### Time Response Analysis (Part II)

Time Response Analysis (Part II). A critically damped, continuous-time, second order system, when sampled, will have (in Z domain) (a) A simple pole (b) Double pole on real axis (c) Double pole on imaginary

### 8 sin 3 V. For the circuit given, determine the voltage v for all time t. Assume that no energy is stored in the circuit before t = 0.

For the circuit given, determine the voltage v for all time t. Assume that no energy is stored in the circuit before t = 0. Spring 2015, Exam #5, Problem #1 4t Answer: e tut 8 sin 3 V 1 For the circuit

### EE102 Homework 2, 3, and 4 Solutions

EE12 Prof. S. Boyd EE12 Homework 2, 3, and 4 Solutions 7. Some convolution systems. Consider a convolution system, y(t) = + u(t τ)h(τ) dτ, where h is a function called the kernel or impulse response of

### Laplace Transforms. Chapter 3. Pierre Simon Laplace Born: 23 March 1749 in Beaumont-en-Auge, Normandy, France Died: 5 March 1827 in Paris, France

Pierre Simon Laplace Born: 23 March 1749 in Beaumont-en-Auge, Normandy, France Died: 5 March 1827 in Paris, France Laplace Transforms Dr. M. A. A. Shoukat Choudhury 1 Laplace Transforms Important analytical

### Chapter 10: Sinusoids and Phasors

Chapter 10: Sinusoids and Phasors 1. Motivation 2. Sinusoid Features 3. Phasors 4. Phasor Relationships for Circuit Elements 5. Impedance and Admittance 6. Kirchhoff s Laws in the Frequency Domain 7. Impedance

### STABILITY. Have looked at modeling dynamic systems using differential equations. and used the Laplace transform to help find step and impulse

SIGNALS AND SYSTEMS: PAPER 3C1 HANDOUT 4. Dr David Corrigan 1. Electronic and Electrical Engineering Dept. corrigad@tcd.ie www.sigmedia.tv STABILITY Have looked at modeling dynamic systems using differential

### Transform Solutions to LTI Systems Part 3

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

### Dr. Ian R. Manchester

Dr Ian R. Manchester Week Content Notes 1 Introduction 2 Frequency Domain Modelling 3 Transient Performance and the s-plane 4 Block Diagrams 5 Feedback System Characteristics Assign 1 Due 6 Root Locus

### Dynamic Response. Assoc. Prof. Enver Tatlicioglu. Department of Electrical & Electronics Engineering Izmir Institute of Technology.

Dynamic Response Assoc. Prof. Enver Tatlicioglu Department of Electrical & Electronics Engineering Izmir Institute of Technology Chapter 3 Assoc. Prof. Enver Tatlicioglu (EEE@IYTE) EE362 Feedback Control

### Chapter 3 : Linear Differential Eqn. Chapter 3 : Linear Differential Eqn.

1.0 Introduction Linear differential equations is all about to find the total solution y(t), where : y(t) = homogeneous solution [ y h (t) ] + particular solution y p (t) General form of differential equation

### Source-Free RC Circuit

First Order Circuits Source-Free RC Circuit Initial charge on capacitor q = Cv(0) so that voltage at time 0 is v(0). What is v(t)? Prof Carruthers (ECE @ BU) EK307 Notes Summer 2018 150 / 264 First Order

### Chapter 6: The Laplace Transform. Chih-Wei Liu

Chapter 6: The Laplace Transform Chih-Wei Liu Outline Introduction The Laplace Transform The Unilateral Laplace Transform Properties of the Unilateral Laplace Transform Inversion of the Unilateral Laplace

### Linear Systems Theory

ME 3253 Linear Systems Theory Review Class Overview and Introduction 1. How to build dynamic system model for physical system? 2. How to analyze the dynamic system? -- Time domain -- Frequency domain (Laplace

### e st f (t) dt = e st tf(t) dt = L {t f(t)} s

Additional operational properties How to find the Laplace transform of a function f (t) that is multiplied by a monomial t n, the transform of a special type of integral, and the transform of a periodic

### EE/ME/AE324: Dynamical Systems. Chapter 8: Transfer Function Analysis

EE/ME/AE34: Dynamical Sytem Chapter 8: Tranfer Function Analyi The Sytem Tranfer Function Conider the ytem decribed by the nth-order I/O eqn.: ( n) ( n 1) ( m) y + a y + + a y = b u + + bu n 1 0 m 0 Taking

### LAPLACE TRANSFORMATION AND APPLICATIONS. Laplace transformation It s a transformation method used for solving differential equation.

LAPLACE TRANSFORMATION AND APPLICATIONS Laplace transformation It s a transformation method used for solving differential equation. Advantages The solution of differential equation using LT, progresses

### Review of Linear Time-Invariant Network Analysis

D1 APPENDIX D Review of Linear Time-Invariant Network Analysis Consider a network with input x(t) and output y(t) as shown in Figure D-1. If an input x 1 (t) produces an output y 1 (t), and an input x

### Basic Procedures for Common Problems

Basic Procedures for Common Problems ECHE 550, Fall 2002 Steady State Multivariable Modeling and Control 1 Determine what variables are available to manipulate (inputs, u) and what variables are available

### ENGIN 211, Engineering Math. Laplace Transforms

ENGIN 211, Engineering Math Laplace Transforms 1 Why Laplace Transform? Laplace transform converts a function in the time domain to its frequency domain. It is a powerful, systematic method in solving

### The Laplace Transform

The Laplace Transform Generalizing the Fourier Transform The CTFT expresses a time-domain signal as a linear combination of complex sinusoids of the form e jωt. In the generalization of the CTFT to the

### Physics 116A Notes Fall 2004

Physics 116A Notes Fall 2004 David E. Pellett Draft v.0.9 Notes Copyright 2004 David E. Pellett unless stated otherwise. References: Text for course: Fundamentals of Electrical Engineering, second edition,

### To find the step response of an RC circuit

To find the step response of an RC circuit v( t) v( ) [ v( t) v( )] e tt The time constant = RC The final capacitor voltage v() The initial capacitor voltage v(t ) To find the step response of an RL circuit

### Control Systems. Laplace domain analysis

Control Systems Laplace domain analysis L. Lanari outline introduce the Laplace unilateral transform define its properties show its advantages in turning ODEs to algebraic equations define an Input/Output

### Time Response of Systems

Chapter 0 Time Response of Systems 0. Some Standard Time Responses Let us try to get some impulse time responses just by inspection: Poles F (s) f(t) s-plane Time response p =0 s p =0,p 2 =0 s 2 t p =

### 09/29/2009 Reading: Hambley Chapter 5 and Appendix A

EE40 Lec 10 Complex Numbers and Phasors Prof. Nathan Cheung 09/29/2009 Reading: Hambley Chapter 5 and Appendix A Slide 1 OUTLINE Phasors as notation for Sinusoids Arithmetic with Complex Numbers Complex

### First and Second Order Circuits. Claudio Talarico, Gonzaga University Spring 2015

First and Second Order Circuits Claudio Talarico, Gonzaga University Spring 2015 Capacitors and Inductors intuition: bucket of charge q = Cv i = C dv dt Resist change of voltage DC open circuit Store voltage

### Course roadmap. ME451: Control Systems. Example of Laplace transform. Lecture 2 Laplace transform. Laplace transform

ME45: Control Systems Lecture 2 Prof. Jongeun Choi Department of Mechanical Engineering Michigan State University Modeling Transfer function Models for systems electrical mechanical electromechanical Block

### MODELING OF CONTROL SYSTEMS

1 MODELING OF CONTROL SYSTEMS Feb-15 Dr. Mohammed Morsy Outline Introduction Differential equations and Linearization of nonlinear mathematical models Transfer function and impulse response function Laplace

### Control Systems I. Lecture 6: Poles and Zeros. Readings: Emilio Frazzoli. Institute for Dynamic Systems and Control D-MAVT ETH Zürich

Control Systems I Lecture 6: Poles and Zeros Readings: Emilio Frazzoli Institute for Dynamic Systems and Control D-MAVT ETH Zürich October 27, 2017 E. Frazzoli (ETH) Lecture 6: Control Systems I 27/10/2017

### 7. Find the Fourier transform of f (t)=2 cos(2π t)[u (t) u(t 1)]. 8. (a) Show that a periodic signal with exponential Fourier series f (t)= δ (ω nω 0

Fourier Transform Problems 1. Find the Fourier transform of the following signals: a) f 1 (t )=e 3 t sin(10 t)u (t) b) f 1 (t )=e 4 t cos(10 t)u (t) 2. Find the Fourier transform of the following signals:

### ELG 3150 Introduction to Control Systems. TA: Fouad Khalil, P.Eng., Ph.D. Student

ELG 350 Introduction to Control Systems TA: Fouad Khalil, P.Eng., Ph.D. Student fkhalil@site.uottawa.ca My agenda for this tutorial session I will introduce the Laplace Transforms as a useful tool for

### Control Systems Engineering (Chapter 2. Modeling in the Frequency Domain) Prof. Kwang-Chun Ho Tel: Fax:

Control Systems Engineering (Chapter 2. Modeling in the Frequency Domain) Prof. Kwang-Chun Ho kwangho@hansung.ac.kr Tel: 02-760-4253 Fax:02-760-4435 Overview Review on Laplace transform Learn about transfer

### Ordinary Differential Equations. Session 7

Ordinary Differential Equations. Session 7 Dr. Marco A Roque Sol 11/16/2018 Laplace Transform Among the tools that are very useful for solving linear differential equations are integral transforms. An

### Math 307 Lecture 19. Laplace Transforms of Discontinuous Functions. W.R. Casper. Department of Mathematics University of Washington.

Math 307 Lecture 19 Laplace Transforms of Discontinuous Functions W.R. Casper Department of Mathematics University of Washington November 26, 2014 Today! Last time: Step Functions This time: Laplace Transforms

### Problem Set 3: Solution Due on Mon. 7 th Oct. in class. Fall 2013

EE 56: Digital Control Systems Problem Set 3: Solution Due on Mon 7 th Oct in class Fall 23 Problem For the causal LTI system described by the difference equation y k + 2 y k = x k, () (a) By first finding

### Electrical Circuits (2)

Electrical Circuits (2) Lecture 7 Transient Analysis Dr.Eng. Basem ElHalawany Extra Reference for this Lecture Chapter 16 Schaum's Outline Of Theory And Problems Of Electric Circuits https://archive.org/details/theoryandproblemsofelectriccircuits

### INC 341 Feedback Control Systems: Lecture 2 Transfer Function of Dynamic Systems I Asst. Prof. Dr.-Ing. Sudchai Boonto

INC 341 Feedback Control Systems: Lecture 2 Transfer Function of Dynamic Systems I Asst. Prof. Dr.-Ing. Sudchai Boonto Department of Control Systems and Instrumentation Engineering King Mongkut s University

### Control Systems. System response. L. Lanari

Control Systems m i l e r p r a in r e v y n is o System response L. Lanari Outline What we are going to see: how to compute in the s-domain the forced response (zero-state response) using the transfer

### Dynamic circuits: Frequency domain analysis

Electronic Circuits 1 Dynamic circuits: Contents Free oscillation and natural frequency Transfer functions Frequency response Bode plots 1 System behaviour: overview 2 System behaviour : review solution

### Module 4. Related web links and videos. 1. FT and ZT

Module 4 Laplace transforms, ROC, rational systems, Z transform, properties of LT and ZT, rational functions, system properties from ROC, inverse transforms Related web links and videos Sl no Web link

### DESIGN OF CMOS ANALOG INTEGRATED CIRCUITS

DESIGN OF CMOS ANALOG INEGRAED CIRCUIS Franco Maloberti Integrated Microsistems Laboratory University of Pavia Discrete ime Signal Processing F. Maloberti: Design of CMOS Analog Integrated Circuits Discrete

### The Laplace Transform

The Laplace Transform Syllabus ECE 316, Spring 2015 Final Grades Homework (6 problems per week): 25% Exams (midterm and final): 50% (25:25) Random Quiz: 25% Textbook M. Roberts, Signals and Systems, 2nd

### Figure Circuit for Question 1. Figure Circuit for Question 2

Exercises 10.7 Exercises Multiple Choice 1. For the circuit of Figure 10.44 the time constant is A. 0.5 ms 71.43 µs 2, 000 s D. 0.2 ms 4 Ω 2 Ω 12 Ω 1 mh 12u 0 () t V Figure 10.44. Circuit for Question

### Introduction to Modern Control MT 2016

CDT Autonomous and Intelligent Machines & Systems Introduction to Modern Control MT 2016 Alessandro Abate Lecture 2 First-order ordinary differential equations (ODE) Solution of a linear ODE Hints to nonlinear

### Control Systems. Frequency domain analysis. L. Lanari

Control Systems m i l e r p r a in r e v y n is o Frequency domain analysis L. Lanari outline introduce the Laplace unilateral transform define its properties show its advantages in turning ODEs to algebraic

### Linear System Theory

Linear System Theory - Laplace Transform Prof. Robert X. Gao Department of Mechanical Engineering University of Connecticut Storrs, CT 06269 Outline What we ve learned so far: Setting up Modeling Equations

### Outline. Classical Control. Lecture 2

Outline Outline Outline Review of Material from Lecture 2 New Stuff - Outline Review of Lecture System Performance Effect of Poles Review of Material from Lecture System Performance Effect of Poles 2 New

### C(s) R(s) 1 C(s) C(s) C(s) = s - T. Ts + 1 = 1 s - 1. s + (1 T) Taking the inverse Laplace transform of Equation (5 2), we obtain

analyses of the step response, ramp response, and impulse response of the second-order systems are presented. Section 5 4 discusses the transient-response analysis of higherorder systems. Section 5 5 gives

### EEE105 Teori Litar I Chapter 7 Lecture #3. Dr. Shahrel Azmin Suandi Emel:

EEE105 Teori Litar I Chapter 7 Lecture #3 Dr. Shahrel Azmin Suandi Emel: shahrel@eng.usm.my What we have learnt so far? Chapter 7 introduced us to first-order circuit From the last lecture, we have learnt

### Circuit Analysis Using Fourier and Laplace Transforms

EE2015: Electrical Circuits and Networks Nagendra Krishnapura https://wwweeiitmacin/ nagendra/ Department of Electrical Engineering Indian Institute of Technology, Madras Chennai, 600036, India July-November

### MATHEMATICAL MODELING OF CONTROL SYSTEMS

1 MATHEMATICAL MODELING OF CONTROL SYSTEMS Sep-14 Dr. Mohammed Morsy Outline Introduction Transfer function and impulse response function Laplace Transform Review Automatic control systems Signal Flow

### Chapter 1 Fundamental Concepts

Chapter 1 Fundamental Concepts 1 Signals A signal is a pattern of variation of a physical quantity, often as a function of time (but also space, distance, position, etc). These quantities are usually the

### ECE Circuit Theory. Final Examination. December 5, 2008

ECE 212 H1F Pg 1 of 12 ECE 212 - Circuit Theory Final Examination December 5, 2008 1. Policy: closed book, calculators allowed. Show all work. 2. Work in the provided space. 3. The exam has 3 problems

### CHEE 319 Tutorial 3 Solutions. 1. Using partial fraction expansions, find the causal function f whose Laplace transform. F (s) F (s) = C 1 s + C 2

CHEE 39 Tutorial 3 Solutions. Using partial fraction expansions, find the causal function f whose Laplace transform is given by: F (s) 0 f(t)e st dt (.) F (s) = s(s+) ; Solution: Note that the polynomial

### Electric Circuits. Overview. Hani Mehrpouyan,

Electric Circuits Hani Mehrpouyan, Department of Electrical and Computer Engineering, Lecture 15 (First Order Circuits) Nov 16 th, 2015 Hani Mehrpouyan (hani.mehr@ieee.org) Boise State c 2015 1 1 Overview

### Control System. Contents

Contents Chapter Topic Page Chapter- Chapter- Chapter-3 Chapter-4 Introduction Transfer Function, Block Diagrams and Signal Flow Graphs Mathematical Modeling Control System 35 Time Response Analysis of

### Definition of the Laplace transform. 0 x(t)e st dt

Definition of the Laplace transform Bilateral Laplace Transform: X(s) = x(t)e st dt Unilateral (or one-sided) Laplace Transform: X(s) = 0 x(t)e st dt ECE352 1 Definition of the Laplace transform (cont.)

### Review of 1 st Order Circuit Analysis

ECEN 60 Circuits/Electronics Spring 007-7-07 P. Mathys Review of st Order Circuit Analysis First Order Differential Equation Consider the following circuit with input voltage v S (t) and output voltage

### School of Mechanical Engineering Purdue University

Case Study ME375 Frequency Response - 1 Case Study SUPPORT POWER WIRE DROPPERS Electric train derives power through a pantograph, which contacts the power wire, which is suspended from a catenary. During

### ME 375 Final Examination Thursday, May 7, 2015 SOLUTION

ME 375 Final Examination Thursday, May 7, 2015 SOLUTION POBLEM 1 (25%) negligible mass wheels negligible mass wheels v motor no slip ω r r F D O no slip e in Motor% Cart%with%motor%a,ached% The coupled

### Solving a RLC Circuit using Convolution with DERIVE for Windows

Solving a RLC Circuit using Convolution with DERIVE for Windows Michel Beaudin École de technologie supérieure, rue Notre-Dame Ouest Montréal (Québec) Canada, H3C K3 mbeaudin@seg.etsmtl.ca - Introduction

### A system that is both linear and time-invariant is called linear time-invariant (LTI).

The Cooper Union Department of Electrical Engineering ECE111 Signal Processing & Systems Analysis Lecture Notes: Time, Frequency & Transform Domains February 28, 2012 Signals & Systems Signals are mapped

### ANALOG AND DIGITAL SIGNAL PROCESSING CHAPTER 3 : LINEAR SYSTEM RESPONSE (GENERAL CASE)

3. Linear System Response (general case) 3. INTRODUCTION In chapter 2, we determined that : a) If the system is linear (or operate in a linear domain) b) If the input signal can be assumed as periodic

### Math 308 Exam II Practice Problems

Math 38 Exam II Practice Problems This review should not be used as your sole source for preparation for the exam. You should also re-work all examples given in lecture and all suggested homework problems..

### ECE-202 FINAL April 30, 2018 CIRCLE YOUR DIVISION

ECE 202 Final, Spring 8 ECE-202 FINAL April 30, 208 Name: (Please print clearly.) Student Email: CIRCLE YOUR DIVISION DeCarlo- 7:30-8:30 DeCarlo-:30-2:45 2025 202 INSTRUCTIONS There are 34 multiple choice

### Dr Ian R. Manchester Dr Ian R. Manchester AMME 3500 : Review

Week Date Content Notes 1 6 Mar Introduction 2 13 Mar Frequency Domain Modelling 3 20 Mar Transient Performance and the s-plane 4 27 Mar Block Diagrams Assign 1 Due 5 3 Apr Feedback System Characteristics

### Raktim Bhattacharya. . AERO 422: Active Controls for Aerospace Vehicles. Dynamic Response

.. AERO 422: Active Controls for Aerospace Vehicles Dynamic Response Raktim Bhattacharya Laboratory For Uncertainty Quantification Aerospace Engineering, Texas A&M University. . Previous Class...........

### Stability. X(s) Y(s) = (s + 2) 2 (s 2) System has 2 poles: points where Y(s) -> at s = +2 and s = -2. Y(s) 8X(s) G 1 G 2

Stability 8X(s) X(s) Y(s) = (s 2) 2 (s 2) System has 2 poles: points where Y(s) -> at s = 2 and s = -2 If all poles are in region where s < 0, system is stable in Fourier language s = jω G 0 - x3 x7 Y(s)

### EE100Su08 Lecture #11 (July 21 st 2008)

EE100Su08 Lecture #11 (July 21 st 2008) Bureaucratic Stuff Lecture videos should be up by tonight HW #2: Pick up from office hours today, will leave them in lab. REGRADE DEADLINE: Monday, July 28 th 2008,

### Problem Weight Score Total 100

EE 350 EXAM IV 15 December 2010 Last Name (Print): First Name (Print): ID number (Last 4 digits): Section: DO NOT TURN THIS PAGE UNTIL YOU ARE TOLD TO DO SO Problem Weight Score 1 25 2 25 3 25 4 25 Total

### Introduction to Controls

EE 474 Review Exam 1 Name Answer each of the questions. Show your work. Note were essay-type answers are requested. Answer with complete sentences. Incomplete sentences will count heavily against the grade.

### EE 3054: Signals, Systems, and Transforms Summer It is observed of some continuous-time LTI system that the input signal.

EE 34: Signals, Systems, and Transforms Summer 7 Test No notes, closed book. Show your work. Simplify your answers. 3. It is observed of some continuous-time LTI system that the input signal = 3 u(t) produces

### Explanations and Discussion of Some Laplace Methods: Transfer Functions and Frequency Response. Y(s) = b 1

Engs 22 p. 1 Explanations Discussion of Some Laplace Methods: Transfer Functions Frequency Response I. Anatomy of Differential Equations in the S-Domain Parts of the s-domain solution. We will consider

### EE 3CL4: Introduction to Control Systems Lab 4: Lead Compensation

EE 3CL4: Introduction to Control Systems Lab 4: Lead Compensation Tim Davidson Ext. 27352 davidson@mcmaster.ca Objective To use the root locus technique to design a lead compensator for a marginally-stable

### Chapter 7. Digital Control Systems

Chapter 7 Digital Control Systems 1 1 Introduction In this chapter, we introduce analysis and design of stability, steady-state error, and transient response for computer-controlled systems. Transfer functions,

### EE Experiment 11 The Laplace Transform and Control System Characteristics

EE216:11 1 EE 216 - Experiment 11 The Laplace Transform and Control System Characteristics Objectives: To illustrate computer usage in determining inverse Laplace transforms. Also to determine useful signal

### Passive RL and RC Filters

NDSU Passive RL and RC Filters ECE 3 Passive RL and RC Filters A filter is a system whose gain changes with frequency. Essentially, all dynamic systems are filters. -Stage Low-Pass Filter For the following

### ENGR 2405 Chapter 8. Second Order Circuits

ENGR 2405 Chapter 8 Second Order Circuits Overview The previous chapter introduced the concept of first order circuits. This chapter will expand on that with second order circuits: those that need a second

### Once again a practical exposition, not fully mathematically rigorous Definition F(s) =

Laplace transforms Once again a practical exposition, not fully mathematically rigorous Definition F(s) = 0 f(t).e -st.dt NB lower limit of integral = 0 unilateral LT more rigorously F(s) = 0 f(t).e -st.dt

### Control System Design

ELEC ENG 4CL4: Control System Design Notes for Lecture #4 Monday, January 13, 2003 Dr. Ian C. Bruce Room: CRL-229 Phone ext.: 26984 Email: ibruce@mail.ece.mcmaster.ca Impulse and Step Responses of Continuous-Time

### MAS107 Control Theory Exam Solutions 2008

MAS07 CONTROL THEORY. HOVLAND: EXAM SOLUTION 2008 MAS07 Control Theory Exam Solutions 2008 Geir Hovland, Mechatronics Group, Grimstad, Norway June 30, 2008 C. Repeat question B, but plot the phase curve

### Electric Circuits Fall 2015 Solution #5

RULES: Please try to work on your own. Discussion is permissible, but identical submissions are unacceptable! Please show all intermeate steps: a correct solution without an explanation will get zero cret.

### Some of the different forms of a signal, obtained by transformations, are shown in the figure. jwt e z. jwt z e

Transform methods Some of the different forms of a signal, obtained by transformations, are shown in the figure. X(s) X(t) L - L F - F jw s s jw X(jw) X*(t) F - F X*(jw) jwt e z jwt z e X(nT) Z - Z X(z)

### 4.1. If the input of the system consists of the superposition of M functions, M

4. The Zero-State Response: The system state refers to all information required at a point in time in order that a unique solution for the future output can be compute from the input. In the case of LTIC

### A sufficient condition for the existence of the Fourier transform of f : R C is. f(t) dt <. f(t) = 0 otherwise. dt =

Fourier transform Definition.. Let f : R C. F [ft)] = ˆf : R C defined by The Fourier transform of f is the function F [ft)]ω) = ˆfω) := ft)e iωt dt. The inverse Fourier transform of f is the function