Section 6 Time Domain Analysis - College of Engineeringweb.engr.oregonstate.edu/~webbky/MAE3401_files/Section 6... · 2014-11-04 · output due solely to initial conditions ... Next,

MAE 3401 – Modeling and Simulation

SECTION 6: TIME‐DOMAIN ANALYSIS

K. Webb MAE 3401

Natural and Forced Responses2

This first sub‐section of notes continues where the previous section left off, and will explore the difference between the forced and natural responses of a dynamic system.

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Natural and Forced Responses

In the previous section we used Laplace transforms to determine the response of a system to a step input, given zero initial conditions The driven response

Now, consider the response of the same system to non‐zero initial conditions only The natural response

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Natural Response

0 (1)

Use the derivative property to Laplace transform (1) Allow for non‐zero initial‐conditions

0 0 0 0 (2)

Same spring/mass/damper system Set the input to zero Second‐order ODE for displacement

of the mass:

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Natural Response

Solving (2) for gives the Laplace transform of the output due solely to initial conditions

Laplace transform of the natural response

(3)

Consider the under‐damped system with the following initial conditions

1

16

4 ∙

0 0.15

0 0.1

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Natural Response

Substituting component parameters and initial conditions into (3)

. . (4)

Remember, it’s the roots of the denominator polynomial that dictate the form of the response Real roots – decaying exponentials Complex roots – decaying sinusoids

For the under‐damped case, roots are complex Complete the square Partial fraction expansion has the form

. . ..

(5)

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Natural Response

. . ..

(5)

Multiply both sides of (5) by the denominator of the left‐hand side

0.15 0.7 2 3.464

Equating coefficients and solving for and gives

0.15, 0.115

The Laplace transform of the natural response:.

.. .

.(6)

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Natural Response

Inverse Laplace transform is the natural response0.15 cos 3.464 ∙ 0.115 sin 3.464 ∙ (7)

Under‐damped response is the sum of decaying sine and cosine terms

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Now, Laplace transform, allowing for both non‐zero input and initial conditions

0 0 0

Solving for , gives the Laplace transform of the response to both the input and the initial conditions

(8)

Driven Response with Non‐Zero I.C.’s

1

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Laplace transform of the response has two components

Total response is a superposition of the initial condition response and the driven response

Both have the same denominator polynomial Same roots, same type of response

Over‐, under‐, critically‐damped


Natural response ‐ initial conditions Driven response ‐ input

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Laplace transform of the total response

0.15 0.7 10.15 0.7 1

Transform back to time domain via partial fraction expansion

22 3.464

3.4642 3.464

Solving for the residues gives

0.0625, 0.0875, 0.0794

1

16

4 ∙

0 0.15

0 0.1

1 ∙ 1

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Total response:0.0625 0.0875 cos 3.464 ∙ 0.0794 sin 3.464 ∙

Superposition of two components Natural responsedue to initial conditions

Driven response due to the input

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Next, we’ll apply the Laplace transform to the entire state‐space model in matrix form, just as we did for single differential equations.

Solving the State‐Space Model13

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Solving the State‐Space Model

We’ve seen how to use the Laplace transform to solve individual differential equations

Now, we’ll apply the Laplace transform to the full state‐space system model

First, we’ll look at the same simple example Later, we’ll take a more generalized approach

State‐space model is

10

10

(1)0 1

Note that, because this model was derived from a bond graph model, the state variables are now momentum and displacement

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Laplace Transform of the State‐Space Model

For now, focus on the state equation Output is a linear combination of states and inputs Determining the state trajectory is the important thing

Use the derivative property to Laplace transform the state equation

00 0

10

Rearranging to put all transformed state vectors on the left‐hand side

000

10 (2)

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000

10 (2)

Can factor out the transformed state vector from the left‐hand side Must multiply by a 2 2 identity matrix

000

10

00 0

00

10

00

10 (3)

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00

10 (3)

Note the form of (3) The LHS is , where is the system matrix Everything on the RHS reduces to a vector A known matrix times a vector of unknowns equals a known vector

If we can solve for and/or , we can inverse transform to get and/or Use Cramer’s Rule

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Cramer’s Rule

Given a matrix equation

We can solve for elements of as follows

The matrix is formed by replacing the column of with the vector

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According to Cramer’s Rule

(4)

According to the output equation from (1)

Equation (4) is identical to (8) from the previous subsection of notes, which we arrived at differently

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(5)

Next, sub in parameter values, I.C.’s and an input Use PFE to inverse transform to Again, consider the under‐damped system:

Let the input be a step:

1

16

4 ∙

0 0.15

0 0.1 ∙

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The Laplace transform of the output becomes. . .

. . (6)

Inverse transform via partial fraction expansion. . .

.(7)

Multiply both sides by left‐hand‐side denominator0.15 0.7 1 4 16 2 3.464

Equating coefficients and solving yields

0.0625, 0.0875, 0.0794

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The Laplace transform of the system response is. .

.. .

.(8)

The time‐domain response is0.0625 0.0875 cos 3.464 0.0794 sin 3.464 (9)

Transient portion Due to initial conditions

and input step Decays to zero

Steady‐State portion Due to constant input Does not decay

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Now, we’ll apply the Laplace transform to the solution of the state‐space model in general form

For now, focus on the state equation only Output is derived from states and inputs

Laplace transform of the state equation

0

Rearranging

0

Factoring out the transformed state from the left‐hand side

0 (10)

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0 (10)

Remember the dimensions of each term in (10) : : 1 : 1

: 1 0 : 1

Apply Cramer’s rule to solve for the Laplace transform of the state variable

(11)

The matrix is formed by replacing the column of with 0

An 1 vector of known values

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(11)

Denominator of (11) is the determinant of is an matrix

Each diagonal term is a first‐order polynomial in One term in the determinant is the trace of the matrix, the product terms along

the diagonal is an ‐order polynomial in

The characteristic polynomial:

Δ (12)

Roots of Δ are values of that satisfy the characteristic equation

Δ 0 (13) Poles of (11) Eigenvalues of system matrix,

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Denominator of every state variable’s Laplace transform contains the characteristic polynomial

(14)

A characteristic of the system Remember, denominator roots (i.e. poles) determine the nature of

the response Real roots – decaying exponentials Complex roots – decaying sinusoids

Responses of all state variables have same components Numerators of transforms determine the differences

Output transform has the same denominator, Δ Linear combination of states and input Response includes the same sinusoidal and/or exponential components

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Assume: zero initial conditions: 0 SISO system: single input – is a scalar transform

Can factor out the input from the numerator of (14)

∗

where ∗ is the matrix formed by replacing the column of with the vector

appears in every term of one column of appears in every term of the determinant

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Can now write the Laplace transform of the state variable response as

∗(15)

is, in general, different for each state variable At most, an 1 order polynomial in

Components of every state variable (and output) response determined by The characteristic polynomial, Δ The input,

Numerator, , determines exact response Weighting of each sinusoidal and/or exponential component

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(15)

Laplace transform of each state variable response, , is the Laplace transform of the input scaled by

Δ

In the next sub‐section, we’ll explore a related concept –transfer functions

Δ

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Transfer Functions30

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Transfer Functions

Now, come back to the full state‐space model, including the output equation – (SISO case assumed here ‐ and are scalars)

D

Assume zero initial conditions and Laplace transform the whole model

(1)

D (2)

Simplify the state equation as before

Solving for the state vector

(3)

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Transfer Functions

Substituting (3) into (2) gives the Laplace transform of the output

D

Factoring out the inputD (4)

Transform of the output is the input scaled by the stuff in the square brackets

Dividing through by the input gives the transfer function

D (5)

Ratio of system’s output to input in the Laplace domain, assuming zero initial conditions

An alternative to the state‐space (time‐domain) model for mathematically representing a system

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Transfer Matrix – MIMO Systems

For MIMO systems

inputs: is 1, is outputs: is 1, is

Transfer function becomes a matrix

Transfer function relates the output to the input

We’ll continue to assume SISO systems in this course

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Transfer Functions

System output in the Laplace domain is the input multiplied by the transfer function

We saw earlier that state variables are given by

where is the characteristic polynomial

Output is linear combination of states and input, so we’d expect the denominator of to be as well Is it? What is the denominator of ?

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Transfer Functions

D (5)

The matrix inverse term in (5) is given by

where the numerator is the adjoint of Equation (5) can be rewritten as

(6)

Transfer function denominator is the characteristic polynomial Poles of the transfer function are roots of Δ

System poles or eigenvalues Eigenvalues of the system matrix, Along with the input, system poles determine the nature of the time‐domain

response

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This sub‐section of notes takes a bit of a tangent to explain the use of the term eigenvalues when referring to system poles.

Eigenvalues36

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Eigenvalues

We’ve been using the term eigenvalue when referring to system poles – why? Recall from linear algebra, the eigenvalue problem

(1)

where: is an matrixis an 1 vector – an eigenvectoris a scalar – an eigenvalue

Eigenvalue problem involves finding both the eigenvalues and the eigenvectorsthat satisfy (1)

Eigenvalues and eigenvectors are specific to (characteristics of) the matrix An matrix will have, at most, eigenvalues and corresponding eigenvectors

Equation (1) says: An 1 eigenvector, , left‐multiplied by an matrix, , results in an 1 vector The resulting vector is the eigenvector scaled by the eigenvalue, Result is in the same direction as – i.e., not rotated

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Eigenvalues and Eigenvectors

Geometrically, multiplication of a vector by a matrix results in two things Scaling and rotation

Consider the matrix

1 34 2

And the vectors

11 , 1

0 Compute the product

In both cases, results have different magnitudes and different directions

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Multiplication of a matrix and one of its eigenvectors results in scaling only No rotation

The 2 2matrix1 34 2

has two eigenvectors (normalized)

0.7070.707 and 0.6

0.8and two corresponding eigenvalues

λ 2 and λ 5

such that

and

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A full‐rank, matrix will have pairs of eigenvalues and eigenvectors

To find all eigenvalues and eigenvectors that satisfy (1)

(1)rearrange

and factor out the eigenvector term

0 (2)

If exists, then , which is the trivial solution and of no interest

We’re interested in values of and that satisfy (2) when is not invertible – when it is singular

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Want to find values of for which is singular A matrix is singular if its determinant is zero

0 (3)

Equation (3) is the characteristic equation for is the characteristic polynomial, Δ An ‐order polynomial in

Eigenvalues of matrix are all values of that satisfy (3) Roots of the characteristic polynomial Find the corresponding eigenvectors by substituting into (2) and solving for

Letting , (3) becomes the denominator of the system transfer function,

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Using the Transfer Function to Determine System Response42

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Using to determine System Response

System output in the Laplace domain can be expressed in terms of the transfer function as

(1)

Laplace‐domain output is the product of the Laplace‐domain input and the transfer function

Response to two specific types of inputs often used to characterize dynamic systems Impulse response Step response

We’ll use the approach of (1) to determine these responses

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Impulse response

Impulse function

0, 0

1

Laplace transform of the impulse function is

1

Impulse response in the Laplace domain is

1 ∙

The transfer function is the Laplace transform of the impulse response Impulse response in the time domain is the inverse transform of the

transfer function

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Step Response

Step function: 1 0 01 0

Laplace transform of the step function

1

Laplace‐domain step response

∙

Time‐domain step response

∙

Recall the integral property of the Laplace transform

∙ , and ∙

The step response is the integral of the impulse response

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First‐ and Second‐Order Systems

All transfer functions can be decomposed into 1st‐ and 2nd‐order terms by factoring Δ Real poles – 1st‐order terms Complex‐conjugate poles – 2nd‐order terms

These terms and, therefore, the poles determine the nature of the time‐domain response Real poles – decaying exponentials Complex‐conjugate poles ‐ decaying sinusoids

All time‐domain responses will be a superposition of decaying exponentials and decaying sinusoids These are the natural modes or eigenmodes of the system

Instructive to examine the responses of 1st‐ and 2nd‐order systems Gain insight into relationships between pole location and response

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Response of First‐Order Systems47

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First‐Order System – Impulse Response

First‐order transfer function:

Single real pole at 1

where is the system time constant Impulse response:

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First‐Order System – Impulse Response

Initial slope is inversely proportional to time constant

Response completes 63% of transition after one time constant

Decays to zero as long as the pole is negative

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Impulse Response vs. Pole Location

Increasing corresponds to decreasing and a faster response

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First‐Order System – Step Response Step response in the Laplace domain

∙

Inverse transform back to time domain via partial fraction expansion

: →

: 0 →

/ /

Time‐domain step response

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First‐Order System – Step Response

Initial slope is inversely proportional to time constant

Response completes 63% of transition after one time constant

Almost completely settled after 7

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Step Response vs. Pole Location

Increasing corresponds to decreasing and a faster response

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Pole Location and Stability

First‐order transfer function

where is the system pole Impulse response is

If 0, decays to zero Pole in the left half‐plane System is stable

If 0, grows without bound Pole in the right half‐plane System is unstable

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Response of Second‐Order Systems55

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Second‐Order Systems

Second‐order transfer function

(1)

where is the damped natural frequency

Can also express the 2nd‐order transfer function as

(2)

where is the un‐damped natural frequency, and is the damping ratio

1

Two poles at

, 1

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Categories of Second‐Order Systems

The 2nd‐order system poles are

, 1

Value of determines the nature of the poles and, therefore, the response

: Over‐damped Two distinct, real poles – sum of decaying exponentials – treat as two first‐order terms ,

: Critically‐damped Two identical, real poles – time‐scaled decaying exponentials ,

: Under‐damped Complex‐conjugate pair of poles – sum of decaying sinusoids , 1

: Un‐damped Purely‐imaginary, conjugate pair of poles – sum of non‐decaying sinusoids ,

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2nd‐Order Pole Locations and Damping

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Second‐Order Poles ‐

Can relate , , , and to pole location geometry

is the magnitude of the poles is a measure of system

damping

sin

0 Two purely imaginary poles

1 Two identical real poles

1 Two distinct real poles

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Impulse Response – Critically‐Damped

For , the transfer function reduces to

2

Impulse response

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Impulse Response – Critically‐Damped

Speed of response is proportional to

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Impulse Response – Under‐Damped

For 0 1, the transfer function is

2

Complete the square on the denominator

1

Rewrite in the form of a damped sinusoid

Inverse Laplace transform for the time‐domain impulse response

sin

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Under‐Damped Impulse Response vs.

sin sin 1

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Under‐Damped Impulse Response vs.

sin sin 1

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Impulse Response – Un‐Damped

For 0, the transfer function reduces to

Putting into the form of a sinusoid

Inverse transform to get the time‐domain impulse response

An un‐damped sinusoid

sin

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Un‐Damped Impulse Response vs.

sin

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Second‐Order Step Response

The Laplace transform of the step response is1

The time‐domain step response for each damping case can be derived as the the inverse transform of

or as the integral of the corresponding impulse response

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Critically‐Damped Step Response vs.

11

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Under‐Damped Step Response vs.

11 cos sin

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Under‐Damped Step Response vs.

11 cos sin

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Un‐Damped Step Response vs.

11 cos

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2nd‐Order Step Response Characteristics72

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Step Response – Risetime

Risetime is the time it takes a signal to transition between two set levels Typically 10% to 90% of full transition

Sometimes 20% to 80%

A measure of the speed of a response

Very rough approximation:

.

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Step Response – Overshoot

Overshoot is the magnitude of a signal’s excursion beyond its final value Expressed as a percentage of full‐scale swing

Overshoot increases as decreases

%OS

0.45 20

0.5 16

0.6 10

0.7 5

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Step Response –Settling Time

Settling time is the time it takes a signal to settle, finally, to within some percentage of its final value Typically 1% or 5%

Inversely proportional to the real part of the poles,

For 1% settling:

. .

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In this sub‐section, we’ll see that the time‐domain output of a system is given by the convolution of its time‐domain input and its impulse response.

The Convolution Integral76

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Convolution Integral

Laplace transform of a system output is given by the product of the transform of the input signal and the transfer function

∙

Recall that multiplication in the Laplace domain corresponds to convolution in the time domain

∗

Time‐domain output given by the convolution of the input signal and the impulse response

∗

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Convolution

Time‐domain output is the input convolved with the impulse response

∗

Impulse response is flipped in time and shifted by

Multiply input and flipped/shifted

Integrate over 0…

Each time point of given by result of integral with

shifted by

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Convolution

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Convolution

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A few of MATLAB’s many built‐in functions that are useful for simulating linear systems are listed in the following sub‐section.

Time‐Domain Analysis in MATLAB81

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System Objects

MATLAB has data types dedicated to linear system models

Two primary system model objects: State‐space model Transfer function model

Objects created by calling MATLAB functions ss.m – creates a state‐space model tf.m – creates a transfer function model

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State‐Space Model – ss(…)

sys = ss(A,B,C,D)

A: system matrix ‐ B: input matrix ‐ C: output matrix ‐ D: feed‐through matrix ‐ sys: state‐space model object

State‐space model object will be used as an input to other MATLAB functions

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Transfer Function Model – tf(…)

sys = tf(Num,Den)

Num: vector of numerator polynomial coefficients Den: vector of denominator polynomial coefficients sys: transfer function model object

Transfer function is assumed to be of the form⋯⋯

Inputs to tf(…) are Num = [b1,b2,…,br+1]; Den = [a1,a2,…,an+1];

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Step Response Simulation – step(…)

[y,t] = step(sys,t)

sys: system model – state‐space or transfer function t: optional time vector or final time value y: output step response t: output time vector

If no outputs are specified, step response is automatically plotted

Time vector (or final value) input is optional If not specified, MATLAB will generate automatically

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Impulse Response Simulation – impulse(…)

[y,t] = impulse(sys,t)

sys: system model – state‐space or transfer function t: optional time vector or final time value y: output impulse response t: output time vector

If no outputs are specified, impulse response is automatically plotted


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Natural Response – initial(…)

[y,t,x] = initial(sys,x0,t)

sys: state‐space system model function x0: initial value of the state ‐ 1 vector t: optional time vector or final time value y: response to initial conditions ‐ length(t) 1 vector t: output time vector x: trajectory of all states ‐ length(t) matrix

If no outputs are specified, response to initial conditions is automatically plotted


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Linear System Simulation – lsim(…)

[y,t,x] = lsim(sys,u,t,x0)

sys: system model – state‐space or transfer function u: input signal vector t: time vector corresponding to the input signal x0: optional initial conditions – (for ss model only) y: output response t: output time vector x: optional trajectory of all states – (for ss model only)

If no outputs are specified, response is automatically plotted

Input can be any arbitrary signal

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More MATLAB Functions

A few more useful MATLAB functions

Pole/zero analysis: pzmap(…) pole(…) zero(…) eig(…)

Input signal generation: gensig(…)

Refer to MATLAB help documentation for more information

Documents

Section 6 Time Domain Analysis - College of Engineeringweb.engr.oregonstate.edu/~webbky/MAE3401_files/Section 6... · 2014-11-04 · output due solely to initial conditions ... Next,