# Laplace Transform Math Review with Matlab:

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Laplace Transform Math Review with Matlab:
3/31/2017 Math Review with Matlab: Laplace Transform Application: Linear Time Invariant (LTI) Systems S. Awad, Ph.D. M. Corless, M.S.E.E. E.C.E. Department University of Michigan-Dearborn

Linear Time Invariant (LTI) Systems
Definition of a Linear Time Invariant System Impulse Response Transfer Function Simple Systems Simple System Example Pulse Response Example Transient and Steady State Example

System Definition A system can be thought of as a black box with an input and an output Output Signal y(t) Input Signal x(t) Excitation Response The signal connected to the input is called the Excitation The system performs a Transformation, T, (function) on the input Given an input excitation, the output signal is called the Response

Differential Equations
Output Signal y(t) Input Signal x(t) Time domain systems are often described using a Differential Equation Recall that time domain Differentiation corresponds to Laplace Transform domain Multiplication by s with subtraction of Initial Conditions

Linear Systems A system is Linear if it satisfies the Superposition Principle ( where a and b are constants ): This can be restated given the excitation and response relationships: Then an Excitation of: Results in a Response of:

Time Invariance A system is time-invariant if its input-output relationship does not change as time evolves t T t T

Impulse Response The Impulse Response signal, h(t), of a linear system is determined by applying an Impulse to the Input, x(t), and determining the output response, y(t) Due to the properties of a Linear Time Invariant System, the Impulse Response Completely Characterizes the relationship between x and y for all x such that: Where * denotes the Convolution operation

Laplace Transform Since Convolution may be Mathematically Intensive, the Laplace Transform is often used as an aid to analyze the Linear Time Invariant Systems. Recall the relationship between Convolution in the Time-Domain and Multiplication in the Laplace Transform-Domain LT

Transfer Function The Transfer Function, H(s), of a system is the Laplace Transform of the Impulse Response, h(t) The Transfer Function completely specifies the relationship between the excitation (input) and response (output) in the Laplace Transform-Domain

Simple Systems Most systems can be created by combining the following simple system building blocks: Linear Operations: Multiplication by a Constant Addition of Signals Time-Domain Differentiation Time-Domain Integration Time-Domain Delay

Multiplication by a Constant
Linear Operations Multiplication by a Constant Addition of Signals Time-Domain Time-Domain Laplace Transform-Domain Laplace Transform- Domain Linear operations have a direct correlation between the Time-Domain and Laplace Transform-Domain (s-domain) counterparts

Time-Domain Differentiation
Time-Domain Differentiation Operation Equivalent Laplace Transform-Domain Operation

Time-Domain Integration
Time-Domain Integration Operation (no initial conditions) Equivalent Laplace Transform-Domain Operation

Time-Domain Delay Time-Domain Delay Operation
Equivalent Laplace Transform-Domain Operation

Examples of LTI Systems
The building blocks described previously can be used to model and analyze real world systems such as: Audio Equalizers (band pass filters) Automatic Gain Controls for a radio Car Mufflers (mechanical filter) Suspension Systems (mechanical low pass filter) Cruise Control (motor speed control)

System Example Create a system to implement the differential equation:
1) Determine the Transfer Function directly from the Differential Equation 2) Draw the system in the Time-Domain 3) Draw the system in the Laplace Transform-Domain 4) Write the Transfer Function from the System Diagram 5) Determine the Impulse Response

Directly Determine H(s)
The Transfer Function H(s) can be directly determined by taking the Laplace Transform of the differential equation and manipulating terms LT By definition, H(s) = Y(s) / X(s)

Time-Domain-System + - Draw time-domain system representation for:
1) Reorder terms to create a Function for y(t) 2) Start by drawing Input and Output at far ends 3) Draw Differentiation Block connected to y(t) 4) Draw Summation Block and its connections + -

Laplace Transform-Domain
The Laplace Transform-Domain System can be drawn by leaving the linear summation block and replacing the differentiating block with a multiplication by s + - Time-Domain + - Laplace Transform Domain

Verify H(s) The Transfer Function H(s) can also be determined by writing an expression from the Laplace Transform-Domain System + - System directly yields: Reordering terms gives the same result as taking the Laplace Transform of the Differential Equation

Impulse Response The Impulse Response of the system, h(t), is simply the Inverse Laplace Transform of the Transfer Function, H(s)

Pulse Response Example
Given a system with an Impulse Response, h(t)=e-2tu(t) t T Input Pulse Impulse Response Pulse Response Output 1) Find the Transfer Function for the system, H(s) 2) Find the General Pulse Response,y(t) 3) Plot the Pulse Response for T=1 sec and T=2 sec

Transfer Function The transfer function of the system is simply the Laplace Transform of the Impulse Response: LT The Transfer Function can be used to find the Laplace Transform of the pulse response, Y(s), using:

Laplace Transform of Input
Given the equation for a General Pulse of period T LT The general Laplace Transform is thus: Combining Terms:

Determine Y(s) Y(s) is found using: Substituting for H(s) and X(s)
Distributing terms Rewrite in terms of a new Y1(t) where

Partial Fraction Expansion
The Matlab function residue can be used to perform Partial Fraction Expansion on Y1(s) [R,P,K] = RESIDUE(B,A) B = Numerator polynomial Coefficient Vector A = Denominator Polynomial Coefficient Vector R = Residues Vector P = Poles Vector K = Direct Term Constant

Expand Y1(s) Use residue to perform partial fraction expansion
3/31/2017 Expand Y1(s) Use residue to perform partial fraction expansion » B=[0 0 1];A=[1 2 0]; » [R,P,K]=residue(B,A) R = 0.5000 P = -2 K = []

General Solution y(t) Find y(t) by taking Inverse Laplace Transforms and substituting y1(t) back into y(t) LT-1 LT-1

Matlab Declarations The General Pulse Response can be verified using Matlab Variables must be carefully declared using proper syntax » syms h H t s » h=exp(-2*t) » T=sym('T','positive') » x=sym('Heaviside(t)-Heaviside(t-T)') Assuming the system to be causal, T must be explicitly declared as a positive number The Heaviside function is equivalent to the unit-step

Matlab Verification » H=laplace(h) H = 1/(s+2) » X=laplace(x) X =
1/s-exp(-T*s)/s » Y=H*X Y = 1/(s+2)*(1/s-exp(-T*s)/s) » y=ilaplace(Y) y = -1/2*exp(-2*t)+1/2+ 1/2*Heaviside(t-T)*exp(-2*t+2*T) -1/2*Heaviside(t-T)

Matlab Vector Code NOTE as of Matlab 6, ezplot cannot plot functions containing declarations of Heaviside or Dirac (Impulse) The following code recreates the Pulse Response as vectors for T=1 sec and T=2 sec t=[0:0.01:4]; % Time Vector tmax=size(t,2); % Index to last Time Value T1=find(t==1); % Index to 1 second T2=find(t==2); % Index to 2 seconds yexp=0.5*(1-exp(-2*t)); % Base exponential vector y1T=[zeros(1,T1),yexp(1:tmax-T1)]; y1=yexp-y1T; % Pulse Response T=1 y2T=[zeros(1,T2),yexp(1:tmax-T2)]; y2=yexp-y2T; % Pulse Response T=2

Matlab Plots The response for T=1 and T=2 is plotted
subplot(2,1,1);plot(t,y1); title('Pulse Response T=1'); grid on; subplot(2,1,2);plot(t,y2); title('Pulse Response T=2'); xlabel('Time in seconds');

Determine an equation for the output of a system, y(t), described by the transfer function H(s) and input x(t) From the output y(t): 1. Identify the Transient Response, ytrans(t), of the system (portion that goes to zero as t increases) 2. Identify the Steady State Response , yss(t), of the system (portion that repeats for all t)

Laplace Transform of Input
Recall the Laplace Transform of a general sine signal with an angular frequency w0 Find the Laplace Transform of the input signal x(t)

Purely Imaginary Roots
Roots of Y(s) Determine an expression for output signal Y(s) Determine general form for roots (poles) of denominator of Y(s) Purely Imaginary Roots Complex Roots

Verify Poles in Matlab » poles=roots( conv( [1 0 2], [1 2 2]) )

Partial Fraction Expansion
Note that since poles are complex conjugates, coefficients will also be complex conjugates

Find Coefficients in Matlab
» syms s t » p1=j*2^0.5; p1c=conj(p1); p2=(-1+j); p2c=conj(p2); » c1=(2*2^0.5)/(s-p1c)/(s-p2)/(s-p2c); » c1=subs(c1,'s',p1) c1 = i » c2=(2*2^0.5)/(s-p1)/(s-p1c)/(s-p2c); » c2=subs(c2,'s',p2) c2 = i

Inverse Laplace Take Inverse Laplace Transform of Y(s)
Reduce terms by combining complex conjugates

Substitute Values When substituting coefficients, it is useful to use the polar representation to simplify cosine conversions

The complex signal can be converted into a function of cosines Steady State Response (Repeats as t increases) Transient Response (Goes to 0 at t increases)