[email protected] SYSTEM RESPONSEsigmedia/pmwiki/uploads/Teaching.3C1/handout2.… · HANDOUT 2. Dr Anil Kokaram Electronic and Electrical Engineering Dept. [email protected]

SIGNALS AND SYSTEMS: PAPER 3C1

HANDOUT 2.

Dr Anil Kokaram

Electronic and Electrical Engineering Dept.

[email protected] www.mee.tcd.ie∼ack

SYSTEM RESPONSE

• Ultimate goal is to control a process or analyze a signal To do this need

to be able to predict the response of a system to various inputs

• Given known system dynamics (i.e. we know enough about the system

to write down ODE’s describing its operation) we can use the Laplace

Transform to work out the response of the system to any input whose

Laplace Transform can be found.

• However, real inputs are not predictable; and it may not be possible to

write down a system’s ODE’s. The process used in Mobile Phones for

canceling echoes is a good example of a system which operates despite

the inability to analytically define the geometry/dynamics of the phone’s

immediate environment.

• Still can identify some basic system characteristics which allow handling

of system response. This is possible using purely time domain analysis.

• Time domain analysis relies on the fact that the response of a system to an

impulse or to a step tells you everything about the system (in principle).

These responses can be calculated if the ODE’s for the system can be

written, or they can be measured by experiment beforehand or on-line

(e.g. mobile phones, noise cancellation etc.)

• We will explore both techniques for analyzing systems. Start with Laplace

analysis as you met this in 2nd year.

3C1 Signals and Systems 1 www.mee.tcd.ie/∼sigmedia

1 REVIEW OF LAPLACE ANALYSIS

1 Review of Laplace Analysis

We’ll be using the Laplace Transform to solve differential equations.

Need to be confident in using it.

1. Laplace Transform of a signal f (t) is

L(

f (t)

)=

∫ ∞

0

f (t)e−stdt (1)

s is a COMPLEX number, e.g. σ + jω !! The result of taking

the Laplace Transform of a signal f (t) is a function in s. So we

go from a 1-dimensional function i.e. a function of time only, to

a function of 2 variables σ and ω i.e. a function of a complex

variable. We will write the laplace Transform of f (t) as F(s).

Sometimes in books you might see f (s) whic is useful because

some capital letters look like the common ones when written by

hand e.g. V and v.

2. Inverse laplace Transform of F(s) is

f (t) = L−1

(F(s)

)=

1

2πj

∫ σ+j∞

σ−j∞F(s)estds (2)

This is a nasty1 integral to do, need to know about Contour

integration. Happily, being Engineers we can use tables of Laplace

Transform pairs to do this.

3. Sometimes people panic about the convergence of the inverse

Laplace Transform, because you find yourself having to say, for

instance ∫ ∞

0

e−stdt =

[−e−st

s

]∞

0

=1

s(3)

1Not really, but it makes you feel better if you think this is so.


1 REVIEW OF LAPLACE ANALYSIS

even though you don’t know what value s is. Don’t panic, in this

course we are going to assume that the Laplace integral always

converges (i.e. Re(s) > 0) and so[−e−st

s

]∣∣∣∣s=∞

= 0 (4)

BUT YOU NEED TO REMEMBER THAT CONVERGENCE

IS AN ISSUE

4. In addition, assume that e−(s+a)t is 0 for s = ∞, (i.e. you can

always assume that s + a > 0 to make this happen.)

5. Remember to spot the ‘shift’ theorem since it gives a slick way of

using tables to get Laplace transforms of functions (we’ll derive

this later for the Fourier transform as well).

Given L(

f (t)

)= F(s)

Then L(

e−ktf (t)

)= F(s + k)

In other words, if you want to find the Laplace transform

of e−ktf (t), then just find the Laplace transform of f (t)

and replace all occurrences of s with s + k. A similar

statement can be made for the inverse transform.


1.1 Transform Tables 1 REVIEW OF LAPLACE ANALYSIS

1.1 TABLE OF LAPLACE TRANSFORM RELATIONS

Waveform: Laplace Transform:

g(t) (defined for t ≥ 0) G(s) = Lg(t) =∫∞

0−g(t)e−stdt

δ(t) impulse 1

u(t) unit step 1s

tn n!sn+1

e−at 1s+a

sin(ω0t)ω0

s2+ω20

cos(ωot)s

s2+ω20

sinh(ω0t)ω0

s2−ω20

cosh(ω0t)s

s2−ω20

e−at[A cos(ω0t) + B sin(ω0t)]A(s+a)+Bω0

(s+a)2+ω20

e−atg(t) G(s + a) shift in s

g(t− τ)u(t− τ) where τ ≥ 0 e−sτG(s) shift in t

tg(t) − ddsG(s)

dgdt differentiation sG(s)− g(0)

d2gdt2 2nd differential s2G(s)− sg(0)−

(dgdt

)∣∣∣∣0

dngdtn snG(s)− sn−1g(0)− sn−2

(dgdt

)∣∣∣∣0− . . .−

(dn−1gdtt−1

)∣∣∣∣0∫ t

0 g(τ)dτ integration G(s)s

g1(t) ∗ g2(t) convolution G1(s)G2(s)

=∫ t

0 g1(t− τ)g2(τ)dτ


1.2 Laplace Transform Examples 1 REVIEW OF LAPLACE ANALYSIS

1.2 Laplace Transform Examples

1. Find the Laplace Transform of e−at from first principles

Le−at =

∫ ∞

0

e−ate−stdt

=

∫ ∞

0

e−(s+a)tdt

=

[− 1

s + ae−(s+a)t

]∞

0

= − 1

s + a

[0− 1

]

=1

s + a

2. Find the Laplace Transform of δ(t− a) from first principles

Lδ(t) =

∫ ∞

0

δ(t)e−stdt

= 1

3. Find the Laplace Transform of δ(t− a) from first principles

Lδ(t− a) =

∫ ∞

0

δ(t− a)e−stdt

= e−at

4. Laplace transform of e−at + e−bt. Laplace transform obeys super-

position so..

Le−at + e−bt = Le−at + Le−bt=

1

s + a+

1

s + b



5. Find the Laplace transform of e−at sin(βt + φ)from first principles (i.e. the hard way) :

L

e−at sin(βt + φ)

=

∫ ∞

0e−at sin(βt + φ)e−stdt

Using sin(x) =1

2j

(ejx − e−jx

)

=

∫ ∞

0e−at 1

2j

(ej(βt+φ) − e−j(βt+φ)

)e−stdt

=1

2j

∫ ∞

0e−(s+a)t

(ej(βt+φ) − e−j(βt+φ)

)dt

=1

2j

∫ ∞

0

(ejφe−(s+a−jβ)t − e−jφe−(s+a+jβ)t

)dt

=1

2j

(ejφ

[e−(s+a−jβ)t

−(s + a− jβ)

]∞

0− e−jφ

[e−(s+a+jβ)t

−(s + a + jβ)

]∞

0

)

=1

2j

(ejφ

(s + a− jβ)− e−jφ

(s + a + jβ)

)

=1

2j

((s + a + jβ)ejφ − (s + a− jβ)e−jφ

(s + a)2 + β2

)

=1

2j

((s + a)(ejφ − e−jφ) + jβ(ejφ − e−jφ)

(s + a)2 + β2

)

=(s + a) sin(φ)

(s + a)2 + β2 +β cos(φ)

(s + a)2 + β2



But using tables,

sin(ω0t) ↔ ω0

s2 + ω20

cos(ω0t) ↔ s

s2 + ω20

e−at sin(ω0t) ↔ Using the shift theorem: Replace s with s + a

↔ ω0

(s + a)2 + ω20

we can do the same thing the slick way (i.e. from tables and

spotting the shift theorem) :

L

e−at sin(βt + φ)

=

= L

e−at[sin(βt) cos(φ) + cos(βt) sin(φ)]

= L

e−at sin(βt) cos(φ)

+ L

e−atcos(βt) sin(φ)

= cos(φ)L

e−at sin(βt)︸︷︷︸

+ sin(φ)L

e−at cos(βt)︸︷︷︸

=β cos(φ)

(s + a)2 + β2+

(s + a) sin(φ)

(s + a)2 + β2


1.3 Partial Fractions 1 REVIEW OF LAPLACE ANALYSIS

1.3 Review of partial fractions

You will need to find the inverse Laplace Transform of sometimes complicates

expressions. The method of partial fractions allows you to do this by express-

ing a complicated fraction in terms of a sum of simpler fractions. The idea

is that the inverse laplace transform of the simpler fractions is easier to spot.

Basic ideak3s + 1

s(s + k1)(s + k2)=

A

s+

B

s + k1+

C

s + k2(1)

Expand r.h.s. and equate coeffs

=A(s + k1)(s + k2) + Bs(s + k2) + Cs(s + k1)

s(s + k1)(s + k2)

Equating coeffs in s2 ⇒0 = A + B + C

Equating coeffs in s ⇒k3 = A(k1 + k2) + Bk2 + Ck1

Equating constants ⇒1 = Ak1k2

Then solve simultaneous equations for A, B, C. Urrgh, could be a

pain. There’s another, easier way (cover up rule)

Multiply eqn. 1 by s, set s = 0, This gives A straightaway

⇒ k3s + 1

(s + k1)(s + k2)=

As

s+

Bs

s + k1+

Cs

s + k2(2)

s = 0 ⇒ 1

k1k2= A

Rule is Set s to be value that makes a denominator factor equal

0. Then COVER UP that factor in denominator and subsitute

to get coefficient value. To get C then . . .

s = −k2 ⇒ −k3k2 + 1

−k2(k1 − k2)= C (3)


2 SYSTEM RESPONSE (AT LAST)

2 System Response (at last)

• What is the response of the system shown to a unit impulse input

at x(t), i.e. What happens to y(t) when x(t) = δ(t) Volts ?

C y(t)x(t)

R

• Steps are

1. Write down the differential equations for modeling the system.

2. Given the stated initial conditions (if any) get an expression for y(t)

in terms of x(t). In other words : solve the differential equations.

3. Do this by

(a) First principles using P.I. and Homogeneous solution. (A pain).

(b) Laplace Transforms (much easier).

4. Insert the numerical values or expression for the input x(t) as well as

initial conditions.

5. Solution falls out.


2.1 Example 1 : 2 SYSTEM RESPONSE (AT LAST)

2.1 Example 1 :

Impulse response of this system is the response y(t) when x(t) = δ(t).

C y(t)x(t)

R

Write down differential equations to model the system.

x(t) = Ri(t) + y(t) = RCdy

dt+ y(t)

Take Laplace Transforms, and solve for Y(s)

X(s) = RC(sY(s)− y(0)) + Y(s)

Y(s)

(1 + RCs

)= X(s)−RCy(0)

⇒ Y(s) =X(s)−RCy(0)

1 + RCs(4)

Assuming initially, output is zero i.e. for t < 0, y(t) = 0

=1

1 + RCsX(s) = G(s)X(s) (5)

Find Laplace Transform of input x(t): X(s) = Lδ(t) = 1

Hence the impulse response, (we’ll refer to it as h(t) ) is therefore

h(t) = L−1

[G(s)X(s)

]= L−1

[G(s)× 1

]

= L−1

[1

1 + RCs

]= L−1

[1/(RCs)

(1/(RCs) + s

]

= (6)


2.2 Step response 2 SYSTEM RESPONSE (AT LAST)

2.2 Step response

Impulse response of this system is the response y(t) when x(t) = u(t)

(i.e. x(t) is a step function.

C y(t)x(t)

R

⇒ Y(s) =1

1 + RCsX(s) (7)

Find Laplace Transform of input x(t) : Lu(t) = 1s

Hence the step response y(t) is

y(t) = L−1

[Y(s)

]

= L−1

[1

1 + RCsX(s)

]

= L−1

[(1

1 + RCs

)(1

s

)]

use partial fractions:1

s(1 + RCs)=

A

s+

B

1 + RCs

Use Cover up rule to find coefficients A,B

Hence y(t) = L−1

[1

s− RC

1 + RCs

]

=

= u(t)− e−t/RC

=


2.3 PLOTS 2 SYSTEM RESPONSE (AT LAST)

2.3 PLOTS OF THE STEP AND IMPULSE RESPONSE

(assuming zero initial conditions, R=10KΩ, C = 10µF. )

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 50

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Time (secs)

Am

plitu

de (

volts

)

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 50

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Time (secs)

Am

plitu

de (

volts

)


3 IMPULSE AND STEP RESPONSE

3 A time domain relationship

• The step function is the integral of the delta function.

u(t) =

∫ t

0

δ(τ )dτ (8)

• All systems we deal with are LTI so SUPERPOSITION APPLIES

So if input xi gives output yi then∑

i xi gives output®

©

ª.

• Hence

Input

∫ t

0

x(τ )dτ → Output

∫ t

0

y(τ )dτ (9)

• So input u(t) (which is the integral of δ(t)) gives output

• Therefore The step response of a system is the integral of the

impulse response.

• Now do the example 1 again, but this time integrate the impulse

response to get the step response.


4 TRANSFER FUNCTIONS

4 TRANSFER FUNCTIONS

Recall for the capacitor and resistor circuit we had

Y(s) =X(s)−RCy(0)

1 + RCs

=X(s)

1 + RCs− RCy(0)

1 + RCs

⇒ Y(s) = F (s)X(s) + G(s)y(0) (10)

F (s) is called the transfer function of the system. Its all we need

to work out everything about the system if the initial conditions were

0. The Transfer function generalizes the idea of ‘Gain’ to dynamic

attributes of the system.

+

LAPLACE TRANSFORM OF OUTPUT SIGNAL

LAPLACE TRANSFORM OF INPUT SIGNAL

SYSTEM TRANSFER FUNCTION

OTHER TERMS

(INITIAL CONDITIONS)

= X

So the system block diagram can be drawn as follows.


5 LINKING H(T ) AND H(S)

5 A link between Impulse response

and Transfer Function

Suppose Impulse response of system is h(t). What is its Transfer

Function?


6 TRANSFER FUNCTION EXAMPLES

6 Transfer Function examples

Transfer Function G(s) =Laplace Transform of Output Signal

Laplace Transform of Input signal

=Y(s)

X(s)

y(t)RCx(t)

A

+x(t) y(t)

R

R

1

2

-

C

+x(t) y(t)

RA

-


6.1 A cascade of electrical systems 6 TRANSFER FUNCTION EXAMPLES

6.1 A cascade of electrical systems

+w(t) z(t)

x(t) y(t)

+

--

Assuming that each stage does not load the preceeding one (i.e. each

input impedance >> output impedance of previous stage), Then

X(s)

w(s)= −k1;

Y(s)

X(s)=

1

1 + sT;

z(s)

Y(s)= −k2


7 BLOCK DIAGRAM ALGEBRA

7 Block Diagram Algebra


8 MORE BLOCK DIAGRAM ALGEBRA

8 More Block Diagram Algebra



EXAMPLES OF THE DISTINCTION BETWEEN

TRANSFORMS OF SIGNALS AND TRANSFER

FUNCTIONS OF SYSTEMS.

SIGNALS SYSTEMS

1 ↔ δ(t) Y(s) = 1 ·X(s) ↔ y(t) = x(t)

1s ↔ u(t) Y(s) = 1

s ·X(s) ↔ y(t) =∫ t

0 x(τ )dτ

1s+a ↔ e−at Y(s) = 1

s+a ·X(s) ↔ y(t) + ay(t) = x(t)

ωs2+ω2 ↔ sin(ωt) Y(s) = ω

s2+ω2 ·X(s) ↔ y(t) + ω2y(t) = ωx(t)

e−sT ↔ δ(t− T ) Y(s) = e−sT ·X(s) ↔ y(t) = x(t− T )

Transfer functions arise through Laplace Analysis of

Differential Equations which represent the dynamic behaviour

of systems. They summarise the input/output behaviour of a

system in the s domain (i.e. they generalize the concept of ‘gain’).

Finding the Laplace transform of a signal is a step one takes when

it is required to find out what a particular system (represented by its

transfer function) does to an input signal.

It just turns out that the Laplace transform of some signals have a

form which occurs also as a Transfer function of particular systems.



Terminology (will be used later in examining stability)

Suppose

G(s) =n(s)

d(s)(11)

Then the roots of n(s) are called the ZEROS of the system G(s)

And the roots of d(s) are called the POLES of the system G(s)

EXAMPLE:

G(s) =4s2 − 8s− 60

s3 + 2s2 + 2s

=4(s + 3)(s− 5)

s(s + 1 + j)(s + 1− j)

Zeros of G(s) are the values of s that make the numerator = 0 hence

Poles of G(s) are the values of s that make the denominator = 0

hence


8.1 DC Motor 8 MORE BLOCK DIAGRAM ALGEBRA

8.1 Another Example: FIELD CONTROLLED DC MOTOR

Left: Circuit Diagram, Right: The Motor

DC Motors found in Hard Drives, CD Players, Scalextric Model car sets

How does θ(t) (the rotational angle of the axle) depend on e(t) (the voltage

applied to the field winding)? The following system equations apply:

Linearised Motor Equation (k=some const.)

τ(t) = ki(t) (1)

Torque = Inertia × acceleration

τ(t)−Bθ(t) = Jθ(t) (2)

Kirchoff

e(t) = Ri(t) + L∂i

∂t(3)

1. What is the the system transfer function θ(s)E(s)?

2. What is the impulse response of the system?

3. What is the response of the system to a unit step impulse at t = 0?

Assume θ(0) = 0, θ(t) = 0, i(0) = 0; First have a think about what you



expect . . .

Now some analysis:

• To get the transfer function we need to express the Laplace transform of

the output signal τ(s) (torque) as a function of the laplace transform of

the input signal I(s) (current). So take Laplace xforms of the

differential equations representing the dynamic behaviour of each

system (remember tables!) . . .

From eqn 1

⇒ τ (s) = KI(s) (4)

From eqn. 2

⇒ τ(s)−B[sΘ(s)− θ(0)] = J [s2Θ(s)− sθ(0)− (θ)(0)]

But from initial conditions, θ(0) = 0, θ(t) = 0, i(0) = 0 so

τ(s) = Θ(s)[Bs + Js2] (5)

From eqn. 3

E(s) = RI(s) + L[sI(s)− i(0)] (6)

= I(s)[R + Ls]



Let Tf = L/R, Tm = J/B (to reduce the amount of writing to do).

Hence from eqn. 6

I(s) =E(s)

R(1 + sTf)(7)

Still have some work to do ... don’t panic ... all we want to do is to get

Θ(s) = somefunctionof(E(s))

Subst I from eqn 7 into 4

⇒ τ(s) =

(K

R(1 + sTf)

)E(s) (8)

Subst τ from 8 into 5

K

R(1 + sTf)E(s) = Θ(s)[Bs + BTms2]

= Θ(s)Bs[1 + Tms]

So

Θ(s)

E(s)=

K

R(1 + sTf)

1

Bs[1 + Tms]

Required Transfer Function is

Θ(s)

E(s)=

K/(RB)

s(1 + sTf)(1 + Tms)(9)

We shall denote this system transfer function as H(s).

• Impulse Response: Put an impulse as input into the system and

calculate the output. Therefore set e(t) = δ(t) ⇒ E(s) = 1 Hence, to

calculate output (which is the impulse response when an impulse is

input)

Θ(s) = H(s)E(s) = H(s)× 1

= H(s) (10)



Therefore, the time domain impulse response h(t) is just L−1(H(s))

Remember: System transfer function is Laplace transform of the

impulse response, or the impulse response is the inverse Laplace

Transform of the system transfer function. So

h(t) = L−1

K/(RB)

s(1 + sTf)(1 + Tms)

(11)

Need to use partial fractions to express the transfer function in a

simpler form to make taking the inverse easier (i.e. to use tables).

K/(RB)

s(1 + sTf)(1 + Tms)=

A

s+

B

(1 + sTf)+

C

(1 + sTm)

Using “Cover Up” rule

A =K/(RB)

(1 + 0× Tf)(1 + Tm × 0)=

K

RB

B =K/(RB)

s(1 + Tms)

∣∣∣∣s=− 1

Tf

=K/(RB)

− 1Tf

(1− TmTf

)

=−KTf/(RB)

(1− TmTf

)

=−KT 2

f /(RB)

(Tf − Tm)

C =K/(RB)

s(1 + sTf)

∣∣∣∣s=− 1

Tm

=K/(RB)

− 1Tm

(1− Tf

Tm)

=−KTm/(RB)

(1− Tf

Tm

=KT 2

m/(RB)

(Tf − Tm)

Hence (after some simplification)

K/(RB)

s(1 + sTf)(1 + Tms)=

K

RB

[1

s− Tf

(Tf − Tm)

(1

s + 1Tf

)+

Tm

Tf − Tm

(1

s + 1Tm

)]



So now we can use tables to find the inverse Laplace Transform, hence

H(s) =K

RB

[1

s− Tf

(Tf − Tm)

(1

s + 1Tf

)+

Tm

Tf − Tm

(1

s + 1Tm

)]

⇒ h(t) = L−1 H(s)

=K

RB

[u(t)− Tf

(Tf − Tm)e−

(1

Tft

)

+Tm

Tf − Tme−( 1

Tmt)

]

Consider:

– Tf generally smaller than Tm.

– Electronic cct transient FAST, Mechanical cct transient SLOW

• Step response g(t) (say) : can get this either i) integrate the impulse

response OR ii) set e(t) = u(t) and then calculate output. You can spot

that the only terms that involve t in h(t) have simple exponentials, so

integration seems straightforward

g(t) =

∫ t

0h(τ)dτ

=K

RB

[∫ t

0

(u(τ)− Tf

(Tf − Tm)e−

(1

Tfτ)

+Tm

Tf − Tme−( 1

Tmτ)

)dτ

]

remember∫

e−t/a = 1−(1/a)e

−t/a = −ae−t/a

=K

RB

[τ +

T 2f

(Tf − Tm)e−

(τ

Tf−1

)

− T 2m

Tf − Tme−( τ

Tm−1)

]t

0

=K

RB

[t +

T 2f

(Tf − Tm)e−

(t

Tf−1

)− T 2

m

Tf − Tme−( t

Tm−1)

]

Steady state response is g(t) as t →∞. So steady state response is

gt→∞(t) =Kt

RB(12)


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[email protected] SYSTEM RESPONSEsigmedia/pmwiki/uploads/Teaching.3C1/handout2.… · HANDOUT 2. Dr Anil Kokaram Electronic and Electrical Engineering Dept. [email protected]