ME121 Week1 2 Review

8/18/2019 ME121 Week1 2 Review

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University of California at Riverside, Department of Mechanical Engineering

ME 121, Spring 2016, Elisa Franco

Week 1

Review

Contents

I Modeling physical systems 1

I.1 A cart system . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

I.2 Mass-spring-damper systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

I.3 Electrical circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

I.4 Using Matlab to numerically solve ordinary differential equations . . . . . . . . . . . . . . . . . . 8

II The Laplace transform method to solve ordinary differential equations 10

II.1 Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

II.2 Common Laplace transforms (LT) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

II.3 Properties of Laplace transforms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

II.4 Inverse Laplace Transform . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

II.5 Using Laplace Transforms to solve ODEs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

II.6 Poles and zeros . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

III Block diagrams 19

III.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

I Modeling physical systems

I.1 Dynamics of a moving cart

M

u

Friction ~ bv

Figure 1: Sketch of a cart moving on a surface with viscous friction

Figure 1 shows a scheme for one of the simplest mechanical systems: a cart having mass M , moving onwheels on a surface with dynamic friction coefficient b. The cart could represent a vehicle or a dolly used tomove loads in a factory. We want do describe the dynamics of its speed. Suppose the cart has an engine,

which produces a force u in the direction shown in the figure.We start from the fundamental laws:

M a = u − bv


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Now remember that a = dvdt

. Therefore, we instantly get a simple ordinary differential equation (ODE) model:

v̇ = − bM

v + u

M

What is the solution?

This differential equation is LINEAR. Therefore, we can use the superposition principle to find the solution.

When u = 0, and v(0) = v0 we find the natural response of the system:

vN (t) = v0e− b

M t

When initial conditions are zero, but we have an input u = 0, we find the forced response using the convolu-tion1 integral:

vF (t) =

t0

e− bM (t−τ ) 1

M u(τ )dτ

Note: If you never heard of convolution - that is how “forced” response solutions to ODEs are computed. Do not

worry too much about it, since we will hardly use convolution throughout this course. The overall solution is:

v(t) = vN (t) + vF (t) = v0e− b

M

t

+ t

0 e− b

M

(t−τ ) 1

M u(τ )dτ

Let us take u = 1 for simplicity:

v(t) = v0e− b

M t +

1

M

t0

e− bM (t−τ )dτ

= v0e− b

M t +

1

M e−

bM t

t0

e bM τ dτ

= v0e− b

M t +

1

M e−

bM tM

b

e bM τ τ =tτ =0

= v0e− b

M t + e−

bM t 1

b e bM t

−1

= v0e− b

M t

→0 as t→∞

+ 1

b

1− e− bM t

→ 1

b as t→∞

As you probably remember from your Linear Systems course, there are better ways to find solutions to linear

ODEs. The Laplace transform method will be our standard choice in the following weeks.

We call this system a FIRST ORDER system, since there is only ONE dynamic variable, an the natural response

dynamics are ALWAYS of exponential type, regardless of b, M and u.The only dynamic variable is called the STATE of the system. We were able to find analytically the state for this

simple case. As you can see, v(t) only depends on INITIAL CONDITIONS, INPUT and PARAMETERS. Wesay that the system is CAUSAL, as it does not depend on future values of the external inputs. In addition, we

say that the system is LINEAR TIME INVARIANT, because the dynamics are linear, and because the solution

does not depend on the specific times at which we begin or finish observing it, but only on their difference.

For first order system, whose natural response is of type e pt we can define a characteristic timescale τ = 1 p .

At t = τ , the steady state of the system has reached a value (1− e−1) = %63 of its steady state value. Finally,1Convolution integral : given f (t), g(t), with f (t) = g(t) = 0 for t ≤ 0: h(t) f (t) g(t) =

t0

f (τ )g(t− τ )dτ


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engineers like to represent dynamical systems with simple block diagrams: see figure 2.

vu

1

M

Z

b

M

+

-

Figure 2: Block diagram representing the cart dynamics

I.2 Mass-spring-damper systems

1. A mass-spring system (Figure 3). Suppose a mass M hangs from support with a spring having springconstant k. For example, M could represent a cable cart or a load on a crane. We assume that theresting length of the spring is zero. We are interested in describing mathematically the dynamics of the

mass. A static description would not give us a lot of useful information: for example, suppose M changessuddenly. Does the system oscillate? If it does, how long? What happens if M changes very slowly overtime or very fast?

0

k

Mg

r(t

Figure 3: Mass-spring system

We use the fundamental laws of physics, and simply write the resultant force on the mass. We choose

our axes as in the figure, and define r as the position of the mass.

M a = M g − kr ⇒⇒ M v̇ = M g − kr, v = ṙ⇒ M ̈r = M g − kr

We want to reduce this equation to a first order differential equation.

We define two variables x1 = r and x2 = ṙ, and we find:

ẋ1 = x2

ẋ2 = g − kM

r = g − kM

x1

We can rewrite the equations above in matrix form:ẋ1ẋ2

=

0 1

− kM

0

x1x2

+

01

g


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Variables x1 and x2 are the STATES of our system. Because we have two states, this is a SECONDORDER system. Second order systems can oscillate, in contrast with first order systems which we just

saw in the first example. We will learn more about this in the next lectures.

By solving the differential equation we can find x1(t) and x2(t), which depend only on the initial conditions,the parameters of the system, and on the external inputs (in this case only gravity).

In general, we will write linear dynamical system in state space form as:

ẋ = Ax + Bu,

where:

• x ∈ Rn is the state vector• u ∈ Rm is a vector of inputs• A is a matrix in Rn×n (also called the state matrix, describes how state affect the dynamics of the

states)

• B is a matrix in Rn×m (describes how external inputs affect the dynamics of the states)2. A mass-damper system (Figure 4). In this case, we assume that viscous friction “slows” the system

proportionally to the speed. This is a standard assumption when we model hydraulic dampers commonlyused in industry.

0

Mg

r(t

b

Figure 4: Mass-damper system

M a = M g − bvv = ṙ, a = v̇ = r̈ ⇒

⇒ M ̈r = M g − bṙ

Now as done before, define two variables x1 = r and x2 = ṙ, and we find:

ẋ1 = x2

ẋ2 = g − bM

b = g − bM

x2

In matrix form: ẋ1ẋ2

=

0 1

0 − bM

x1x2

+

01

g ⇒ ẋ = Ax + Bu


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0

Mg

r

(t

)

bk

Figure 5: Mass-spring-damper system

3. A mass-spring-damper system (Figure 5). This example combines the ones we just saw.

M a = M g − kr − bvM ̈r = M g

−kr

−bṙ

We define two states as before

ẋ1 = x2

ẋ2 = g − bM

b = g − kM

x1 − bM

x2


=

0 1

− kM − b

M

x1x2

+

01

g ⇒ ẋ = Ax + Bu

4. A mass-spring-damper system with an active element (Figure 6). The active element can be, for example,a motor that we can control. Thus, it is a new external input. (This is useful for your homework Matlab

project!)

0

Mg

r(t)

bk F

F

Figure 6: Mass-spring-damper system with an active element

Newton’s law gives:

M v̇ = M g − kr − bv + F


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In matrix form:ẋ1ẋ2

=

0 1

− kM − b

M

x1x2

+

01

g +

01M

F =

0 1

− kM − b

M

x1x2

+

0 01 1

M

g F

Implementing an AUTOMATIC CONTROL action on the system means determining F as a function ofx1 = r and x2 = v. Sensors will measure the actual x1 and x2, provide that information to a computer

that will compare those values to the desired ones, and feed-back automatically a value of F that willdecrease the error. Just to give a simple example, take F = K (xd1−x1) = −Ke, where xd1 is the desiredposition of the mass M and e stands for error, e = x1 − xd1. Note: if x1 > xd1, then e > 0 but F < 0;if instead x1 < x

d1, then e 0. Error and control action have the opposite sign: this is called

negative feedback. The system model will change:

ẋ1ẋ2

=

0 1

− kM − b

M

x1x2

+

01

g +

0K M

(xd1 − x1)

= 0 1

−k+K M

− bM

New matrix!

x1x2 +

0 01 1

M g xd1

5. Finally, we can represent with simple schemes all the dynamic relationships we just described with ODEs:

Z

b

M

+-

Z

xd

1

g

x1

x2

-

-+

+

Z +

-

Z

d

1

g

x1

x2

-

-+

+

K

b

k

1

M

M

K

M

Equivalent scheme

k

M

Figure 7: Mass-spring-damper system: block diagram. Note: the common factor 1/M can be moved in thediagram as long as the overall physical interconnections are consistent.

I.3 Electrical circuits

Electronic devices are present in almost every technological application! Control theory developed around

analysis of electrical circuits.

1. Resistances, capacities and inductors. These are the fundamental elements of any circuit, and are char-

acterized by specific dynamic relationships between their currents and voltages.

Resistance Inductor Capacity

V = Ri V = Ldi

dt i = C

dV

dt


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+

i(t)

+

+

_

i(t)

V (t)

__

V (t) V (t)

i(t)

Figure 8: Fundamental elements of electrical circuits

R

Lu(t) i(t)

Figure 9: A series RL circuit

2. A series RL circuit.

We use Kirchhoff’s voltage law: u = V R + V L = Ri + Ldidt . We immediately have a dynamic relationship

describing the only state of the system:di

dt =

1

Lu− R

Li

This is a first order system and has exponential dynamics.

3. An RC circuit: derive its state space representation as homework.

R

u(t) i(t) C

Figure 10: A series RC circuit

4. An LC circuit.

Now Kirchhoff’s voltage law tells us:

u = V L + V C = Ldi

dt + V C

Now, note that i = iL = iC , and i = C dV dt

. This yields:

u = LC d2V C

dt2 + V C

We have now a second order differential equation; as done for the mechanical system, we define x1 = V C ,x1 =

dV cdt

:

ẋ1 = x2

ẋ2 = − 1LC

x1 + 1

LC u


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u(t) i(t) C

L

Figure 11: A series LC circuit


=

0 1− 1

LC 0

x1x2

+

01LC

u = Ax + Bu

Notice the similarity of the A matrix structure with that of the mass-spring example.

5. An RLC circuit: derive its state space representation as homework.

R L

Cu(t) i(t)

Figure 12: A series RLC circuit

I.4 Using Matlab to numerically solve ordinary differential equations

Matlab is extremely useful to solve differential equations numerically.

Matlab offers several ordinary differential equation solvers. The syntax is the same for all of them. The most

important feature: the dynamics of the system have to be defined in a separate function file (see below). (In

some cases, functions can be nested in the main file.) As an example, here is the code that will simulate themass-spring-damper system. Try to run this code, change the damping and spring constants and see what

happens!

Main file:

%MAIN FILE

clear all

close all

% DEFINE PARAMETERS:

par.k= 14.8e3; % N/m

par.b= 800; % Ns/m

par.M= 400; % Kg

% NOTE: WE ARE PUTTING ALL THE PARAMETERS IN A SINGLE struct ELEMENT, SO

% IT IS EASIER TO PASS IT TO THE SOLVER!

% WE COULD ALSO DEFINE THEM AS

% SEPARATE VARIALBES AND PASS EACH OF THEM TO THE SOLVER


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% INITIAL CONDITION

x0=[10 1];

% TIME DURATION OF THE SIMULATION

tspan=[0:.1:10];

%OPTIONS FOR THE SOLVER

options = odeset(’RelTol’,10^-11,’AbsTol’,10^-13);

[t,x]=ode45(@MassSpringDamper,tspan,x0,options,par);

plot(t,x(:,1),’b’,’LineWidth’,3)

hold on

plot(t,x(:,2),’r’,’LineWidth’,3)

hold on

% PLOT, PRINT AND SAVE %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

legend(’Position’, ’Speed’)legend boxoff

Fsize=16;

FontForPlot=’Arial’;

set(gca,’Fontsize’,Fsize, ’FontName’,FontForPlot) % SETS FONT AND FONTSIZE FOR AXES

xlabel(’Time, [s]’,’Fontsize’,Fsize, ’FontName’,FontForPlot)

ylabel(’Response’,’Fontsize’,Fsize, ’FontName’,FontForPlot)

set(gcf, ’PaperPositionMode’, ’manual’);set(gcf, ’PaperUnits’, ’centimeters’);

set(gcf, ’PaperPosition’, [0 0 15 10])

print(gcf,’-dpdf’, [’MassSpringDamper’])

Function fileMust be saved with the same name as the function! In this case, save as MassSpringDamper.m

function dx=MassSpringDamper(t,x,par)

% F(1) AND F(2) ARE THE DYNAMICS OF THE SYSTEM

F(1)= x(2);

F(2)= -(par.k/par.M)*x(1) - (par.b/par.M)*x(2);

dx=F’; % MUST RETURN A COLUMN VECTOR


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II The Laplace transform method to solve ordinary differential equations

Laplace transforms are useful to solve linear time invariant differential equations. Differential equations are

turned into algebraic equations; the solution in the time domain can be found by decomposing the algebraic

solution in the Laplace domain into terms that are easy to reverse-transform.

II.1 DefinitionLaplace Transform (LT): bijective (invertible) transformation that maps functions of time f (t) with functions ofcomplex argument F (s), s = σ + jω ∈ C.

t ∈ RFunctions of time

Differential equations

L

L−1

Functions of complex variables

Algebraic equationss ∈ C

THE SCHEME ABOVE IS THE MOST IMPORTANT THING YOU SHOULD REMEMBER ABOUT LAPLACE

TRANSFORMS!

Given f (t), piecewise continuous and bounded, f (t) = 0 for t ≥ 0

F (s) = L[f (t)] +∞0−

f (t)e−stdt

For those values of s ∈ C such that the integral converges.If the integral converges for s0, then it does ∀ s such that Re{s} > Re{s0} = σ0.The value σ0 limits the region of convergence , or ROC, of the transform (Figure 13).

σ

j ω

σ0

Figure 13: Example of region of convergence of Laplace Transforms

II.2 Common Laplace transforms (LT)

Notation: I will indicate as 1(t) a step function, which is zero for negative time, and equal to one for t ≥ 0.When we multiply a function by 1(t), for instance et 1(t), it means we consider the function only for t ≥ 0.

1. LT of Dirac’s δ function:


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Dirac’s delta function, can be defined in many ways... Intuitively, it is a very large signal with a very

short duration. Also called impulse.

δ σ(x− x0) = limσ→0 1σ√ πe−(x−x0)

2

σ2

x x 0

In the limit for σ → 0:

t

0f (τ )δ σ(τ − τ 0)dτ = f (τ 0)

L [δ (t)] = +∞0−

δ (t)e−stdt =

+∞−∞

δ (t)e−stdt = 1

2. LT of the step function:

L [1(t)] = +∞0−

1(t)e−stdt = −1s

[e−st]∞0 = 1

s

3. LT of the exponential function:

L ekt1(t) = +∞

0− ekt

e−st

dt = +∞

0− e(k−s)t

dt =

1

k − s [e(k−s)t

]+∞

0 =

1

s− k

4. LT of time to the power n: tn

L [tn1(t)] = (−1)n{(−1)n n!sn+1

} = n!sn+1

5. LT of the sine function sin(ωt).Use Euler formulas e jφ = cos φ + j sin φ , e− jφ = cos φ− j sin φ.

L e jωt − e− jωt

2 j · 1(t) =

1

2 j 1

s

− jω − 1

s + jω = 1

2 j

2 jω

s2 + ω2 =

ω

s2 + ω2

6. LT of the cosine function cos(ωt).

L

e jωt + e− jωt

2 · 1(t)

=

1

2

1

s − jω + 1

s + jω

=

1

2

2s

s2 + ω2 =

s

s2 + ω2


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II.3 Properties of Laplace transforms

1. Linearity The integral operation is linear, therefore the LT inherits such property:

L[c1f 1(t) + c2f 2(t)] = c1L[f 1(t)] + c2L[f 2(t)]

2. Transformation of the integral Given f (t) and g(t) =

t

0− f (τ )dτ , where the LT of f (t) is F (s):

L[g(t)] = 1sL[f (t)]

3. Trasformation of the derivative Given f (t) and its derivative f (t), where the LT of f (t) is F (s):

L f (t) = sL [f (t)]− f (0−)This can be extended to the derivative of order n:

L

dn

dtnf (t)

= snL [f (t)] −

n−1k=0

sn−k−1f k(0−)

4. Frequency shift Given F (s), the LT of f (t), and k

∈C. Then, the LT of function ektf (t) is:

L

ektf (t)

= F (s− k)In fact: +∞

0−ektf (t)e−stdt =

+∞0−

f (t)e−(s−k)tdt

5. Multiplication by tn Assume n = 1, and develop the derivative of F (s):

d

dsF (s) =

d

ds

+∞0−

f (t)e−stdt =

+∞0−

f (t) d

dse−stdt

=

− +∞

0−

f (t)te−stdt =

− +∞

0−

tf (t)e−stdt =

−L[tf (t)]

This yields the general formula:

L [tnf (t)] = (−1)n dn

dsnF (s)

6. Convolution VERY IMPORTANT property of the LT.

Given f (t), g(t), f (t) = g(t) = 0 for t ≤ 0,

z(t) f (t) g(t) =

t0

f (τ )g(t− τ )dτ

Then the Laplace transform of z(t) isZ (s) = F (s)G(s)

LAPLACE TRANSFORMS CONVERT CONVOLUTION INTEGRALS IN SIMPLE PRODUCTS!

Application: remember the general expression for the forced response of an linear time invariant system?

xF (t) =

t0−

eA(t−τ )Bu(τ )dτ =

t0−

h(t− τ )w(τ )dτ,

where h(t) := eA(t−τ ) is the state transition matrix and w(t) := Bu(t). We find:

X (s) = H (s)W (s)

We will see that this product is very easy to handle!


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7. Time shift If f (t) admits a LT, the introduction of a time shift t0 > 0 still yields a LT with the same ROCand :

L [f (t− t0) · 1(t− t0)] = e−st0F (s)This can be shown with a change of variable:

+∞

0−

f (t

−t0)e

−stdt = +∞

−t0

f (τ )e−sτ e−st0dt = e−st0F (s)

8. Time scaling (Not too frequently used)

L [f (at)] = 1a

F s

a

, a ∈ R+

Again this can be proved with a change of variables: +∞0−

f (at)e−stdt =

+∞0−

f (t)e−sτ a dτ

a

=

1

aF s

a

II.4 Inverse Laplace Transform

Given F (s) = L [f (t)], its Inverse Laplace Transform (ILT) is denoted as f (t) = L−1 [F (s)]; the operatorL−1 is linear. The general formula is:

f (t) = 1

2π j

σ̄+ j∞σ̄− j∞

F (s) estds

where σ̄ > σ0 delimits the ROC and the integration is along the vertical line σ = σ̄.There is a lot of theory on the ILTs. This is outside of the scope of this course.

Our approach to inverting LTs will be to try to reduce a general complex function to a sum of

elementary, known LTs which we can straightforwardly invert.Most LTs we will encounter are rational polynomial functions which can be expanded in partial fractions.

PARTIAL FRACTION EXPANSION

(a) Example 1 Antitransform

G(s) = 1

s2 − 3s + 2We want to reduce this to a sum of simple monomial fractions.

i. Find the roots of the denominator of G(s)

s1,2 = 3±√ 9 − 8

2 =

21

ii. Expand in partial fractions

G(s) = 1

s2 − 3s + 2 = A

s− 2 + B

s − 1 = A(s − 1) + B(s − 2)

(s− 2)(s− 1)

s(A + B)− (A + 2B) = 1 ⇔

A + B = 0

A + 2B = −1


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So we find A = 1, B = −1. Therefore:

G(s) = 1

s − 2 − 1

s − 1 ⇒ L−1 [G(s)] = g(t) = e2t − et

Which was easy to find using the Laplace transform of the exponential function.

(b) Example 2 Antitransform:

G(s) =

1

s(s2 + 1)

Using partial fractions expansion:

G(s) = 1

s

1

s2 + 1 =

−ss2 + 1

+ 1

s

The time function is g(t) = 1(t)− cos(t)(c) Example 3 Antitransform:

G(s) = 1

s2 + 2s + 2

G(s) = 1

s2

+ 2s + 2

= 1

(s + 1 − j)(s + 1 + j)=

1

(s + 1)2 + 1

Now we apply the frequency shift property, and we can find:

g(t) = e−t sin t

II.5 Using Laplace Transforms to solve ODEs

We will start with an example. Consider this differential equation:d2ydt

+ 3 dydt

+ 2y(t) = (1 + 3t)u(t)

y(0−) = 1, dydt |

t=0− = 0

If u(t) = 1(t), let’s apply the property of LT for the derivatives:s2 Y (s)− s y(0−) − ẏ(0−) + 3 [s Y (s)− y(0−)] + 2 Y (s) = 1

s +

3

s2

s2 Y (s)− s

+ {3 [s Y (s)− 1]}+ 2 Y (s) = 1

s +

3

s2

(s2 + 3 s + 2)Y (s)− s − 3 = 1

s + 3

s2

This is an algebraic equation in s we can easily solve for Y (s) :

Y (s) = s + 3

s2 + 3 s + 2 +

1

s2 + 3 s + 2

1

s +

3

s2

= Y N (s) + Y F (s)

Y N (s): Natural response , depends on initial conditions!Y F (s): Forced response , depends on the input u(t)!Now we need to go back to the time domain to find y(t)!


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• Natural response Y N .The roots of the denominator are: s1 = −1, s2 = −2. Then:

Y N = C 1(s + 1)

+ C 2(s + 2)

Find C 1 and C 2:

Y N = C 1(s + 2) + C 2(s + 1)(s2 + 3s + 2)

= s(C 1 + C 2) + 2 C 1 + C 2(s2 + 3s + 2)

= s + 3s2 + 3 s + 2

Equating the numerators: C 1 + C 2 = 1

2 C 1 + C 2 = 3

We get C 1 = 2, C 2 = −1.

Y N = 2

(s + 1) − 1

(s + 2)

These two terms are in our list of common LTs...

=⇒ L−1 [Y N ] =⇒ yN (t) = 2 e−t − e−2t

• Forced response Y F .Roots of the denominator: s1 = −1, s2 = −2, s3 = 0, where zero is a root that is repeated twice.Expansion in simple fractions:

Y F = C 1(s + 1)

+ C 2(s + 2)

+ C 3,1

s2 +

C 3,2s

The final inverse transform is:

yF (t) = 3

2t− 7

41(t) + 2 e−t − 1

4e−2t

9. Let’s recap:

(a) We have a method to write down systems in state space form and solve the ODEs.

(b) We have seen how to find the solution to an ODEs applying directly the Laplace transform method

to the differential equation.

Now let’s use this method to analyze a model of a physical system and find the solution to the ODEs.

Consider a modification of our mass-spring system, operating on a horizontal plane with dynamic friction

- Figure 14.

0

kl0 u(t)

l(t)

Figure 14: Another mass-spring system with dynamic friction

We know how to obtain the differential equation that describes it:

M d2(l − l0)

dt2 + k(l − l0) + b d(l − l0)

dt = u(t)


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b ddt

(l − l0) is a term describing viscous friction.Defining y = (l − l0):

M d2y

dt2 + k y + b

dy

dt = u(t)

Applying directly the Laplace Transform Method:

M [s2Y (s)− sy(0) − ẏ(0)] + kY (s) + b [sY (s)− y(0)] = U (s)Solve for Y (s):

Y (s)(M s2 + bs + k) = sM y(0) + M ẏ(0) + by(0) + U (s)

Y (s) = (sM + b)y(0) + M ẏ(0)

M s2 + bs + k natural response

+ U (s)

M s2 + bs + k forced response

II.6 Poles and zeros

The most common type of Laplace transforms we will encounter is:

F (s) = N (s)

D(s), N (s), D(s) polynomials in s = σ + jω

deg(D) = nd, deg(N ) = nd, nn < nd

This can be rewritten as:

F (s) = N (s)

D(s) = K

(s− z1)r1(s− z2)r2 · · · (s − zr)rq(s−

p1)m1(s−

p2)m2

· · ·(s−

pn)mn

IMPORTANT DEFINITIONS:

• z j zeros, j = 1,...,q . The scalar r j is called multiplicity of each zero.• pi poles of the function, i = 1,...,n. The scalar mi is called multiplicity or each pole.

Expand in partial fractions:

F (s) = C 1,1(s− p1) + ... +

C 1,m1(s− p1)m1 +

+ C 2,1(s−

p2) + ... +

C 2,m2(s−

p2)m2+

+ C n,1(s− pn) + ... +

C n,mn(s− pn)mn

Linearity:

L−1 [F (s)] =ni=1

mik=1

L−1

C i,k(s − pi)k

The ILT of each simple fraction has the form:

L−1

C i,k(s − pi)k

=

C i,k(k − 1)! t

(k−1) e pit 1(t)


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NOTE: Repeated poles are associated to exponential functions multiplied by a power of time factor. What drives

the convergence of each term is the sign of the real part of the pole. (Exponential functions converge or diverge

faster than any power.)

Going back to the use of Laplace transforms to solve linear time invariant systems.: recall our definition of

eigenvalues of the A matrix in state space form. The poles are a subset of the eigenvalues of A.

Real poles:

L−1

1

(k − 1)! t(k−1) e pt

For p > 0: time function diverges!!

0 5 100

0.2

0.4

0.6

0.8

1

p0

0 5 100

0.1

0.2

0.3

0.4

k = 2

0 5 100

2

4

6

8

10

0 5 100

0.5

1

1.5

2

2.5x 10

5

0 5 100

0.1

0.2

0.3

0.4

k = 3

Time0 5 10

0

10

20

30

40

50

Time0 5 10

0

5

10

15x 10

5

Time


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Complex poles

L−1

1

(k − 1)! t(k−1) e(σ+ jω)t

For σ > 0: time function diverges!! ω =5rad/s

0 5 10−1

−0.5

0

0.5

1

σ0

0 5 10−0.4

−0.2

0

0.2

0.4

k = 2

0 5 10−10

−5

0

5

10

0 5 10−2

−1

0

1

2

3x 10

5

0 5 10−0.4

−0.2

0

0.2

0.4

k = 3

Time

0 5 10−50

0

50

Time

0 5 10−1

−0.5

0

0.5

1

1.5x 10

6

Time


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III Block diagrams

As a consequence of the convolution property of Laplace transforms (convolution integrals become products!)

we can represent dynamical systems in a very simple and convenient way.

As an example, consider again the mass-spring-damper system in Fig. 15:

0

Mg

r(t)

bk

Figure 15: Mass-spring-damper system

M ̈r = M g − kr − bṙAssuming zero initial conditions, and applying the derivative rule:

M s2R(s) = M g − kR(s)− sbR(s)R(s) =

M

M s2 + sb + kg

See Figure 16. This is an input-output (I/O) representation.

We do not include information about initial conditions.

g R(s)Ms2 + sb+ k

Figure 16: Mass-spring-damper system: block diagram in the Laplace complex domain

With Laplace transforms we can represent interconnected systems of differential equations with blocks. And

(because of the convolution property) there are some useful simplification rules. These block diagrams are

much simpler than the diagrams in the time domain - see, for instance, Fig. 7.1) Series, Figure 17

H 1(s) H 2(s)U (s) Y (s)

H 1(s) · H 2(s)U (s) Y (s)

Figure 17: Representation of blocks in series the Laplace domain

2) Parallel, Figure 18. 3) Negative feedback: In this case, we need to do a few steps to derive the mathematical

expression of the equivalent block in Figure 19.


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U (s) Y (s) U (s) Y (s)H 1(s) + H 2(s)

H 1(s)

H 2(s)

Figure 18: Representation of blocks in parallel the Laplace domain

If our error is E (s) = Y d(s)−H (s)Y (s), where Y d(s) is the desired output in the Laplace domain, we have thatY (s) = G(s) [Y d(s)−H (s)Y (s)]. Therefore, we collect Y (s) and derive:

Y (s)(1 + G(s)H (s)) = G(s)Y d(s) =⇒ Y (s) = G(s)1 + G(s)H (s)

Y d(s)

Y s Y (s)+ _

H (s)

G(s)E (s) G(s)

1 + G(s)H (s)

Y d(s) Y d(s)

Figure 19: Representation of blocks in a negative feedback interconnection, in the Laplace domain


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III.1 Exercises

Antitransform the following complex functions:

• Exercise 1:F (s) =

s + 3

s + 5

Solution: f (t) = δ (t)− 2e−5t


s + 1

s2 + 2s + 5

Solution: f (t) = e−t cos(2t)


12

(s + 2)2 + 9

Solution: f (t) = 4e−2t sin(3t).


(s − 1)(s2 − 6s + 25)

Solution: f (t) = e3t(cos(4t) + 12 sin(4t))

ME121 Week1 2 Review

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