performance parameters - module ii reading iii-2

8/20/2019 Performance Parameters - Module II Reading III-2

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ECM2105 - Control Engineering Dr Mustafa M Aziz (2013)

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1

SYSTEM RESPONSE

1. Introduction

2. Response Analysis of First-Order Systems

3. Second-Order Systems

4. Sinusoidal Response of the System

5. Bode Diagrams

6. Basic Facts About Engineering Systems

1. Introduction

The order of a system is defined as being the highest power of derivative in the differential

equation, or being the highest power of s in the denominator of the transfer function. A first-ordersystem only has s to the power one in the denominator, while a second-order system has the highest

power of s in the denominator being two.

Types of the input functions (or test input signals) commonly used are:

• Impulse function: In the time domain, u(t) = cδ(t). In the s domain, U(s) = c.• Step function: In the time domain, u(t) = c. In the s domain, U(s) = c/s.

• Ramp function: In the time domain, u(t) = ct. In the s domain, U(s) = c/s2.• Sinusoidal function: In the time domain, u(t) = csin(ωt). In the s domain, U(s) = cω /(s2+ω2).where c is a constant in all the above.

With these test signals, mathematical and experimental analyses of control systems can be carried

out easily since the signals are very simple functions of time.

Which of these typical signals to use for analysing system characteristics may be determined by the

form of the input that the system will be subjected to most frequently under normal operation. If

the inputs to a control system are gradually changing functions of time, then a ramp function of time

may be a good test signal. Similarly, if a system is subjected to sudden disturbances, a step function

of time may be a good test signal, and for a system subjected to a shock input, a pulse or an impulse

function may be best.

Exercise: What are the orders of the systems described by the following transfer functions:

a)k bsms

1)s(G

2 ++=

b)1RCs

1)s(G

+=

c)1RCsLCs

1)s(G

2 ++=


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2

The time response of a control system consists of two parts: the transient response and the steady-

state response. The transient response is defined as the part of the time response which goes from

the initial state to the final state and reduces to zero as time becomes very large. The steady-state

response is defined as the behaviour of the system as t approaches infinity after the transients have

died out. Thus the system response y(t) may be written as:

y(t) = yt(t) + yss(t)

where yt(t) denotes the transient response, and yss(t) denotes the steady-state response.

2. Response Analysis of First-Order Systems

Many systems are approximately first-order. The important feature is that the storage of mass,

momentum and energy can be captured by one parameter. Examples of first-order systems are

velocity of a car on the road, control of the velocity of a rotating system, electric systems where

energy storage is essentially in one capacitor or one inductor, incompressible fluid flow in a pipe,

level control of a tank, pressure control in a gas tank, temperature in a body with essentiallyuniform temperature distribution (e.g. steam filled vessel). Next we will present several examples

to show how to obtain the dynamic equations of first-order systems.

Example 1: Mechanical system

m is the mass, u(t) is the external force, y(t) is the velocity

and b is the friction coefficient. By Newton’s law, we

have the following differential equation:

)t(u)t(bydt

)t(dy

m =+

Example 2: Electrical system

R is the resistance, C is the capacitance, u(t) is the input

voltage and y(t) is the output voltage. By Kirchhoff’s law:

u(t) = Ri(t) + y(t) and i(t) = Cdy(t)/dt

Thus

)t(u)t(ydt

)t(dyRC =+

A general form of a first-order system can be represented by the block diagram.

1/TsY(s)

+ _

R(s)

= 1Ts

1

+

Y(s)R(s)

mu(t)

y(t)

by(t)

u(t) y(t)

R

C


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Tasks: Write the system outputs or responses to inputs such as the unit-step, unit-ramp, and unit-

impulse functions, respectively. The initial conditions are assumed to be zero. Draw the response

curves. T is the time constant of the system.

2.1. Unit-step response of first-order systems

R(s) = 1/s, and therefore the unit-step response is:)1Ts(s

1)s(Y

+=

Expanding Y(s) into partial fractions:T / 1s

1

s

1

1Ts

T

s

1)s(Y

+−=

+−=

Take the inverse Laplace transform: y(t) = 1 - e-t/T

, t ≥ 0.

The solution has two parts: a steady-state response: yss(t) = 1, and a transient response:T / t

t e)t(y −= , which decays to zero as t → ∞.

y(t)

Unit-step response, T = 1

0 1 2 3 4 5 60

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Slope = 1/T

0.632

Tr

Ts

t / T

The slope of the tangent line at t = 0 is 1/T.

Pole location in the s plane: s = -1/T.

At t = T, y(T) = 1 – e-1

= 0.632. T is called the time constant , and it is the time it takes for the

step response to rise to 63.2% of its final value.

y(2T) = 0.865; y(3T) = 0.95; y(4T) = 0.982; y(5T) = 0.993 ... It can be seen that for t ≥ 4T, theresponse y(t) remains within 2% of the final value; this time is known as the settling time, Ts.


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The rise time, Tr, is defined as the time for the waveform to go from 10% to 90% of its final

value.

The steady-state error is the error after the transient response has decayed leaving only the

continuous response. The error signal:

e(t) = r(t) - y(t) = 1 - 1 + e-t/T

= e-t/T

As t approaches infinity, e-t/T

approaches zero and the steady-state error is:

[ ])t(y)t(rlim)(eet

ss −=∞=∞→

= 0

The larger the time constant T is, the slower the system response is.

It is noted that the transient response dominates the response of the system at times

‘immediately’ after the input is applied and can make significant contribution to the systemresponse when the time constant is large.

Exercise: A RC circuit has the following transfer function:4s10

2

)s(R

)s(Y

+=

For a step input r(t) = 2V, what is the time taken for the output of the RC circuit to reach 95% of its

steady-state response?

Exercise: A system has transfer function:50s

50

)s(R

)s(Y

+=

Find the time constant, T, the settling time, Ts, and the rise time, Tr for a unit-step input.


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2.2. Unit-ramp response of first-order systems

The input, r(t) = t for t ≥ 0.

Laplace transform: R(s) = 1/s2.

The output transform:1Ts

1

s

1)s(Y

2 +=

Expanding Y(s) into partial fractions:

1Ts

T

s

T

s

1)s(Y

2

2 ++−=

Taking the inverse Laplace transform:

y(t) = t - T + Te-t/T

, t ≥ 0.

Steady-state error:

[ ])t(y)t(rlim)(eet

ss −=∞=∞→

= T

2.3. Unit-impulse response of first-order systems

The unit-impulse input, r(t) = δ(t), t ≥ 0

=∞

≠=δ

0 t

0 t0)t(

Laplace transform: R(s) = 1.

The output transform:1Ts

1)s(Y

+=

Taking the inverse Laplace transform:

y(t) = e-t/T /T, t ≥ 0.

Note that the impulse input yields the

transfer function of the system as output.

t / T

y(t)

Unit-ramp response, T = 1

0 1 2 3 4 5 60

1

2

3

4

5

6

r(t) = t

Steady-state error

t / T

y(t)

Unit-impulse response, T = 1

0 1 2 3 4 5 60

0.2

0.4

0.6

0.8

1

1.2

1/T


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3. Second-Order Systems

Example 1: Mechanical system

For the mechanical system shown in the figure, m is the

mass, k is the spring constant, b is the friction

coefficient, u(t) is the external force and y(t) is the

displacement. From Newton’s second law ∑force = ma:

)t(u)t(kydt

)t(dyb

dt

)t(ydm

2

2

=++

Example 2: Electrical system: RLC circuit

Using Kirchhoff’s law:

)t(ydt

)t(diL)t(Ri)t(u ++=

wheredt

)t(dyL)t(i =

Hence:

)t(u)t(ydt

)t(dyRC

dt

)t(ydLC

2

2

=++

• A general form of a second order system is:

)t(uk )t(ydt

)t(dy2dt

)t(yd 2n

2

nn2

2

ω=ω+ζω+

• Transfer function:

2

nn

2

2

n

s2s

k

)s(R

)s(Y

ω+ζω+

ω=

k: the gain of the system

ζ: the damping ratio of the systemωn: the (undamped) natural frequency of the system

• Solutions (roots, or poles of the system) of the characteristic equation are:

1s 2nn1 −ζω−ζω−= and 1s2

nn2 −ζω+ζω−=

• Three cases:

ζ = 1, critically damped case

ζ > 1, overdamped case0 < ζ < 1, underdamped case

We will study the above cases when k = 1 for simplicity.

mk

b

y(t)

u(t)

u(t) y(t)

R L

C


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3.1. Step Response of Second-Order Systems

3.1.1. Critically damped case ( ζ ζζ ζ = 1)

Two equal poles: s1 = s2 = -ζωn

For a unit-step input R(s) = 1/s, the output is:2

n

2

n

)s(s)s(Y

ω+ω=

Expanding Y(s) into partial fractions:2

n

n

n )s(s

1

s

1)s(Y

ω+

ω−

ω+−=

Taking the inverse Laplace transform: t)

(1e1y(t) nt

n +−= −

Steady-state error: e(∞) = 0

t (sec)

y(t)

Unit-step responses of 2nd-order system (critically damped case)

0 1 2 3 4 50

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

ωn=2

ωn=3

σ

jω

-ζωn

s-plane

Exercise: A system has the following transfer function:

16s8s

1

)s(R

)s(Y

2 ++=

What is the state of damping of the system when it is subjected to a unit-step input? Determine the

natural frequency of the system.


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3.1.2. Overdamped case ( ζ ζζ ζ > 1)

We can write the transfer function of a second-order system by factoring the denominator as:

)ss)(ss()1s)(1s()s(R

)s(Y

21

2

n

2

nn

2

nn

2

n

−−

ω=

−ζω−ζω+−ζω+ζω+

ω=

Two poles in the s-plane: 1s 2nn1 −ζω−ζω−= and 1s2

nn2 −ζω+ζω−=

For a unit-step input, the output is:)ss)(ss(s

)s(Y21

2

n

−−

ω=

Taking the inverse Laplace transform yields the time response (prove this time response as an

exercise):

−

−−=

2

ts

1

ts

2

n

s

e

s

e

1

2

1y(t)21

When ζ is much greater than unity, i.e. ζ >> 1, then |s1| >> |s2| and the term involving s1 in thetime response will decay faster than the term involving s2. The term involving s1 can therefore

be neglected and the system becomes first-order decided mainly by the pole s2:

2

1

2

n|s||s|

21

1

2

n

21

2

n

ss

s /

)ss)(1s / s(

s /

)ss)(ss()s(R

)s(Y 21

−

ω−≈

−−

ω=

−−

ω=

>>

Since 2n21ss ω= , the transfer function becomes:2

2

ss

s

)s(R

)s(Y

−

−=

The unit-step time response is:)t1

(

2ne1y(t)

−−−−= , t ≥ 0

t (sec)

y(t)

Unit-step responses of 2nd order system (overdamped case), ζ=2, ωn=1

0 5 10 15 20 250

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

−

−−=

2

ts

1

ts

2

n

s

e

s

e

1

2

1y(t)21

ts2e1y(t) −=

σ

jωs-plane

)1( 2n −ζ+ζω−

)1( 2n −ζ−ζω−

ζ>>1, s1 can be neglected.


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t (sec)

y(t)

Unit-step responses of 2nd order system (overdamped case), ζ=1.1, ωn=1

0 1 2 3 4 5 6 7 8 90

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

−

−−=

2

ts

1

ts

2

n

s

e

s

e

121y(t)

21

ts2e1y(t) −=

ζ≈1, s1 cannot be neglected.

3.1.3. Underdamped case (0 < ζ ζζ ζ < 1)

Transfer function:

) js)( js()s(R

)s(Y

dndn

2

n

ω−ζω+ω+ζω+

ω=

where 2nd 1 ζ−ω=ω is called the damped natural frequency.

The two poles are:

2

nn1 1 js ζ−ω−ζω−= and2

nn2 1 js ζ−ω+ζω−=

ζ

ζ−

=θ

−2

1 1

tan ζ = cos(θ)

For unit-step input R(s) = 1/s, the output is:

2

d

2

n

n

2

d

2

n

n

2

nn

2

n

)s()s(

s

s

1

s2s

2s

s

1)s(Y

ω+ζω+

ζω−

ω+ζω+

ζω+−=

ω+ζω+

ζω+−=

Taking the inverse Laplace transform using the table of Laplace transforms yields:

-1 )tcos(e)s(

sd

t

2

d

2

n

n n ω=

ω+ζω+ζω+ ζω−

-1 )tsin(e)s(

dt

2

d

2

n

n n ω=

ω+ζω+ζω ζω−

σ

jωs-plane

ωd

-ζωn

-ωd

θ

-ζωn+jωd

-ζωn-jωd

ωn


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The time response is:

)tsin(1

e

1

)tsin(1

)tcos(e1)t(y

d2

t

d2

d

t

n

n

θ+ωζ−−=

ω

ζ−

ζ+ω−=

ζω−

ζω−

where

ζ

ζ−=θ −

2

1 1tan .

When ζ = 0, the response becomes undamped and oscillations continue indefinitely at frequency

ωn. The time response in this case becomes:y(t) = 1 – cos(ωnt)

The natural undamped frequency, ωn, is the frequency of oscillation of the system withoutdamping.

t (sec)

y(t)

Unit-step responses of 2nd order system (underdamped case), ωn=3

0 1 2 3 4 5 60

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

ζ=0

ζ=0.1

ζ=0.5

ζ=0.9


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3.2. Transient system specifications

t (sec)

y(t)

0 1 2 3 4 5 6 70

0.2

0.4

0.6

0.8

1

1.2

1.4

Tr

Tp

Ts

Maximum overshoot

Unit-step response of a 2nd order system, ζ=0.9, ωn=3

0.02

3.2.1. Maximum overshoot

The maximum amount by which the system output response proceeds beyond the desired response.

Let ymax denotes the maximum value of y(t), and yss = y(∞) the steady-state value of y(t), then themaximum overshoot of y(t) is defined as:

maximum overshoot = ymax - yss

The maximum overshoot is often represented by a percentage of the final value of the step response:

ζ−

ζπ−=×

−=

2ss

ssmax

1exp100%100

y

yy)(PO overshoot percent

3.2.2. Peak time, T p

The time required for the response to reach the first peak of the overshoot:

2

n

p

1T

ζ−ω

π=

3.2.3. Rise time, T r

The time required for the step response to rise from 10% to 90% of its final value for critical and

overdamped cases, and from 0% to 100% for underdamped cases. For the underdamped case:

2

n

21

d

r

1) / 1(tanT

ζ−ωζζ−−π=

ωθ−π=

−


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3.2.4. Settling time, T s

The time required for the step response to settle within a certain percentage of its final value. A

frequently used figure is 2% in which case the settling time is approximately:

n

s

4T

ζω≈

For varying within 5% of the final value, the setting time is:n

s

3T

ζω≈

Note that NOT all these specifications necessarily apply to any given case. For example, for an

overdamped system, the terms peak time and maximum overshoot do not apply.

The transient behaviour of a second-order system can be described by:

− the swiftness of the response, as represented by Tr and Tp − the closeness of the response to the desired response, as represented by PO and Ts.

From the design requirement, the swifter and closer, the better. However, when ωn is fixed,small Tr and Tp require a small ζ, while small Ts and PO require a large ζ. Note that

21 / exp100PO ζ−ζπ−= , ]1 /[T 2np ζ−ωπ= , ]1 /[)] / 1(tan[T

2

n

21

r ζ−ωζζ−−π= −

.

These lead to conflicting requirements. A compromise must be obtained sometime.

Exercise: A second-order system is underdamped with a damping ratio of 0.4 and a natural

frequency of 10Hz. Find:

a) the transfer function

b) the time response when it is subjected to a unit-step input

c) the percentage overshoot with such an input

d) the rise time


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Exercise: Given the transfer function:

100s15s

100)s(G

2 ++=

find Tp, PO, Ts and Tr.

3.3. Ramp response of a second-order system

The Laplace transform of a unit-ramp input is R(s) = 1/s2

The output is:)s2s(s

k )s(Y

2

nn

22

2

n

ω+ζω+

ω=

Again, there are three cases:

ζ = 1, critically damped caseζ > 1, overdamped case

0 < ζ < 1, underdamped case


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3.4. Impulse response of a second-order system

The Laplace transform of a unit-impulse input is R(s) = 1.

The output transform is therefore equal to the transfer function of the system, i.e.:

2

nn

2

2

n

s2s

k )s(Y

ω+ζω+

ω=

Fact: the unit-impulse function is the time derivative of the unit-step function.

Therefore, the impulse response of a LTI system can be found from the time derivative of the

step response for a given damping.

Taking the example of a critically damped system where ζ=1, the unit-step response is given by:

t)

(1e1y(t) nt n +−= −

The unit-impulse response is therefore:

t 2

nnte

dt

dy(t) −ω=

Note, again, that the impulse input gives the transfer function of the system.


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4. Sinusoidal Response of the System

Although step responses are commonly used in both simulation and experimental tests, it is also

common to undertake frequency response tests on the system.

The frequency response of a system is defined as the steady-state response of the system to asinusoidal input signal.

Uo

u(t) = Uosin(ωt) yss(t) = Uo|G(jω)|sin(ωt+φ)

φ

Uo|G(jω)|

t

The linear, time-invariant system G(s) subjected to a sinusoidal input of amplitude Uo and

frequency ω describe by:

u(t) = Uosin(ωωωωt)

will, at steady state, have a sinusoidal output of the same frequency as the input but, generally, with

different amplitude and phase given by:

yss(t) = Uo|G(jωωωω)|sin(ωωωωt + φφφφ(ωωωω))

where Uo|G(jω)| is the amplitude of the output sine wave:22 )]} j(G{Im[)]} j(G{Re[|) j(G| ω+ω=ω

and φ(ω) is the phase shift in radians or degrees given by:

) j(G)] j(GRe[

)] j(GIm[tan)( 1 ω∠=

ω

ω=ωφ −

The sinusoidal transfer function of any linear system is obtained by substituting jωωωω for s in thetransfer function of the system.

Proof:

Consider a system described by:

)s(G)s(U

)s(Y=

The input u(t) is a sine wave with and amplitude Uo and frequency ω:u(t) = Uosin(ωt)

The Laplace transform of u(t) is:22

o

sU)s(U

ω+ω=

G(s)y(s)u(t)

U(s) Y(s)


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With zero initial conditions, the Laplace transform of the output is:

22

o

s

U)s(G)s(Y

ω+

ω=

A partial fraction expansion of a general system (assuming the poles of G(s) are distinct) yields:

ω−+

ω++

+++

++

+=

js

c

js

c

ps

c

ps

c

ps

c)s(Y

*00

n

n

2

2

1

1

G(s)fromermsfraction tPartial4 4 4 4 4 34 4 4 4 4 21

L

where cn are constants and c0 and*

0c are a complex conjugate pair that can be obtained using the

cover-up rule:

j2

U) j(Gc o0

ω−−=

j2

U) j(Gc o

*

0

ω=

Since G(jω) is a complex quantity, it can be written in the form:G(jω) = |G(jω)|e jφ

where |G(jω)| is the magnitude and φ is the phase given respectively by:

22)]} j(G{Im[)]} j(G{Re[|) j(G| ω+ω=ω ) j(G

)] j(GRe[

)] j(GIm[tan)( 1 ω∠=

ω

ω=ωφ −

Similarly: G(-jω) = |G(-jω)|e-jφ = |G(jω)|e-jφ

Therefore: j2

Ue|) j(G|c

o

j

0

φ−ω−= j2

Ue|) j(G|c

o

j*

0

φω=

The time response that corresponds to Y(s) is:t j*

0

t j

0

tp

n

tp

2

tp

1 ececececec)t(yn21 ωω−−−− +++++= L

If all the poles of the system represent a stable behaviour, the natural unforced response ( tpnnec

−

decays to zero at t→∞) will die out eventually and therefore the steady-state response of the systemwill be due solely to the sinusoidal term which is caused by the sinusoidal excitation, i.e.

t j*

0

t j

0t

ss ecec)t(ylim)t(y ωω−

∞→+==

Substituting for c0 and*

0c and noting that sin(x) = (e jx

– e-jx

)/(2j), gives the steady-state output:

yss(t) = Uo|G(jω)|sin(ωt + φ)

Advantages of frequency domain analysis:

− The transfer functions of complicated components can be determined experimentally by

frequency response tests (without deriving their mathematical models) using available signal

generators and precise measurement equipment (e.g. spectrum analysers).


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− The amplitude and phase of the frequency response can be used to predict both time-domain

transient and steady-state system performances.

− Systems may be designed to achieve transient and steady-state requirements using frequency

response analysis, and such analysis and design may be extended to certain nonlinear control

systems.

Exercise: For the sinusoidal input u(t) = sin(10t) applied to the system:2s

1)s(G

+= ,

determine the steady-state output of the system.

Time (sec.)

A m p l i t u d e

Linear Simulation Results

0 2 4 6 8 10-0.1

0

0.1

0.2


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There are two commonly used representations of sinusoidal transfer functions:

1) Nyquist or polar plot, and

2) Bode diagram

We shall focus on the more popular Bode analysis and show how we can use MATLAB to produce

these plots.

5. Bode Diagrams (Asymptotic Approximation)

The frequency response of a system G(s) can be described as:

)( jX)(R)] j(GIm[)] j(GRe[) j(G)s(G js

ω+ω=ω+ω=ω=ω=

A Bode diagram consists of two graphs:

- plot of the logarithm of the magnitude of the a sinusoidal transfer function, |G(jω)|

- plot of the phase angle, φ(ω)

both are plotted against the frequency ω (rad/s) on a logarithmic (base 10) scale.

Logarithmic magnitude (also called gain) of G(jω), M = 20log10|G(jωωωω)| (unit in decibels, dB)

Phase:

ω

ω=ωφ −

)(R

)(Xtan)( 1

Advantages of Bode diagrams:

− Bode plots of systems in series simply add, which is quite convenient. For example,

consider the transfer function:

)ps()ps)(ps(

)zs()zs)(zs(b)s(G

n21

m21m

−−−

−−−=

L

L

The magnitude of the frequency response of the system is given by:

ω→−−−

−−−=ω

jsn21

m21m

)ps()ps()ps(

)zs()zs()zs(b) j(G

L

L

Taking the logarithm yields:

20log10|G(jω)| = 20log10bm + 20log10|(s - z1) + 20log10|(s – z2)| + ⋅⋅⋅ - 20log10|(s - p1) - 20log10|(s – p2)| - ⋅⋅⋅|s→ jω

Knowing the respone of each term, the algebraic sum would give the total response in dB.

− Bode's important phase-gain relationship is plotted using asymptotic approximations on a

logarithmic scale, which means a wider range of system behaviour, from low to high

frequencies, can be displayed on a single plot.

− Dynamic compensator (controller) design can be based entirely on Bode plots.


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Example: Find the Bode plots for the following RC filter

with:1T j

1

1) j(RC

1) j(G

+ω=

+ω=ω

Solution:

Magnitude:2222 T / 1

T / 1

T1

1) j(G

ω+=

ω+=ω

Logarithmic Gain (dB) 2 / 122101010 )T / 1(log20)T / 1(log20) j(Glog20 ω+−=ω=

At low frequencies (ω → 0): gain (dB) 0)T / 1(log20)T / 1(log20 1010 ≈−≈ dB

At high frequencies (ω → ∞): gain (dB) )(log20)T / 1(log20 1010 ω−≈

If we plot the logarithmic gain (dB) against )(log10 ω , then the above equation becomes the

straight line:

y = c + mx

where y = ) j(Glog20 10 ω , c = intercept, m = slope, and )(logx 10 ω= . Thus the logarithmic

gain reduces by 20dB (negative slope) per factor of 10 (decade) increase in frequency ω (20dB/decade).

The transition between the high and low frequency asymptotes is found by equating the low and

high frequency limits – this is known as the corner or cut-off frequency:

T

1c =ω

Since )( jX)(RT1

T j

T1

1

T j1

1) j(G

2222 ω+ω=

ω+

ω−

ω+=

ω+=ω

Phase: ( )TtanRX

tan)(

11

ω−=

=ωφ

−−

(rad)

At low frequencies (ω → 0): ( ) 00tan)0( 1 =−≈φ − rad

At high frequencies (ω → ∞): ( ) 2 / tan)( 1 π−=∞−≈∞φ − rad

At corner frequency (ω → 1/T): ( ) 4 / 1tan)T / 1( 1 π−=−≈φ − rad

u(t) y(t)

R

C


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20

-20

-15

-10

-5

0

5

1/T0.1/T0.01/T 10/T

Corner frequencyAsymptote

Asymptote

ω

dB

Exact curve

Asymptotic curve of |G(jω)|

1/T0.1/T0.01/T 10/T

ω

-π /2

-π /4

0

Asymptotic curve of ∠G(jω)

Asymptote

Exact curve

φ

Exercise: Sketch the Bode plot for the following transfer function:1s4

20)s(G

+=


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21

ω (rad/sec)

P h a s e ( d

e g )

M a g n i t u d e ( d

B )

Bode Diagrams

10

15

20

25

30

10-2

10-1

100

-80

-60

-40

-20

0

5.1. Bode plots with MATLAB

The MATLAB command bode(SYS) computes the logarithmic gain and phase angles of the

frequency response of the LTI SYS=tf(num,den), where num and den are the numerator and

denominator coefficients of the system, respectively. For example, to plot the Bode diagrams

shown for the transfer function of the previous exercise, we enter on the MATLAB command line:

num = [0 20];

den = [4 1];

SYS = tf(num,den);

bode(SYS) or bode(num,den)


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22

Frequency (rad/sec)

P h a s e ( d e g ) ; M a g n i t u d e ( d B )

Bode Diagrams

10

15

20

25

30

10-2

10-1

100

-80

-60

-40

-20

0

6. Basic Facts About Engineering Systems

ENGINEERING SYSTEMS OFTEN ARE REQUIRED TO OPERATE IN A STEADY OUTPUT

CONDITION – with the system designed on the basis of achieving the best output from the given

raw material or power available. The design of the system to produce the desired steady-state is a

problem in its own right but not the subject matter of this course.

OPERATING CONDITIONS ARE OFTEN CHANGED BY OPERATOR OR COMPUTER

INTERVENTION – such as changes in power supplied by a power station to the national grid due

to changes in the overall national power consumption.

ENGINEERING SYSTMES HAVE DYNAMICAL BEHAVIOURS – they do not just produce the

desired output temperature, pressure, concentration, voltage, current, frequency, position, velocity,

acceleration, force, torque, flow, level, concentration, reaction rate, .. etc even if that output variable

is required to be constant. This can be due to the effect of disturbances (known or unknown),

physical effects within the system or human intervention/interference such as changes required in

operation condition.

ENGINEERING SYSTEMS REQUIRE CONTROL – to counteract the dynamic behaviours

“preferred” by the system and replace them by acceptable dynamic responses, the process must be

augmented by a control system incorporating features of output measurement (sensors), input

variable changes (actuators) and a (dynamical) data processing device that processes both sensor

data and desired output specifications to generate a desired plant input signal to the actuator. This

system almost always has a FEEDBACK structure reflecting the fact that its output is used to create

the desired input in real time.

CONTROL SYSTEMS REQUIRE DESIGNING – notwithstanding the impression left by the

popular TV science programme “Tomorrow’s World”, control systems are designed rather thanbeing available (without thought required from the user) in a form that simply needs to be hooked


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23

up to the plant. Even those controllers that are available commercially require an element of design

either off-line or on-line and hence an effective engineer requires an appreciation of the design

process and design tools currently available.

CONTROLLERS DEPEND ON THE DETAILED DYNAMICAL CHARACTERISTICS OF THE

PROCESS TO BE CONTROLLED – If that were not the case then it would only be necessary tohave one control system on sale. Practical experience has shown that this is not feasible – it is

found that it is necessary to have appropriate data and some understanding of the process to be

controlled. This typically takes the form of EITHER

1) a mathematical model expressed in differential equation, transfer function or state space

form, AND/OR

2) data on the behaviour of the plant output(s) in response to known inputs (such as steps or

sinusoids) from which

(a) the desired parameters for the model (obtained in (1)) can be estimated or

(b) if a physical model is not available a model can be constructed from a curve

fitting or a, so-called, “identification” procedure.

THE DEGREE OF CONTROL OBTAINED DEPENDES ON THE AVAILABILITY OF

SUITABLE MEASUREMENTS OF SYSTEM BEHAVIOUR – and the more accurate and

extensive these measurements are, the better control will be.

YOU CANNOT ALWAYS MEASURE WHAT YOU WANT TO MEASURE – if you cannot

measure the output variable of interest (due to extreme physical conditions of speed, temperature or

pressure .. etc, it is necessary to create an “intelligent” device which observes the (available)

measurement and uses them to create a useful estimate of the (unavailable) output. As an example,

how is it possible to control the temperature in the centre of a furnace when the temperature sensors

are placed on the external wall?

CONTROLLERS CAN DO AMAZING THINS – As control requirements are as varied as the

applications and needs for new products, even the virtually impossible has been asked for in the

search for the “intelligent controller”.

This requires the development of an abstract way of thinking but has amazing consequences e.g.

(a) the development of control elements capable of observing and accurately estimating

variables that cannot be measured

(b) the development of control systems capable of adapting to new situations and learning

from experience.


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TUTORIAL PROBLEM SHEET 4

1. Identify the gain and time constant of the following first-order transfer function:

10s

3)s(G

+

=

2. A RC circuit has the following transfer function:

4s10

2

)s(R

)s(Y)s(G

+==

For a step input u(t) = 2V for t ≥ 0:a) What is the steady-state response of the circuit?

b) What is the time taken for the output of the RC circuit to reach 95% of its steady-state response?

c) Check your result with MATLAB.

3. The general form of a first-order system is described by:

)t(ku)t(ydt

)t(dyT =+

where T is the time constant, k is the gain, u(t) and y(t) are the input and output of the system

respectively. The unit-step response of a first-order system is shown in the following figure.

Determine the parameters k and T from this figure.

Time (sec.)

A m p l i t u d e

Step Response

0 2 4 6 8 10 120

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45

0.5


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4. Consider the first-order system,

1Ts

1

)s(R

)s(Y

+=

Obtain the unit-step response curves for T = 0.1, 0.5, 1.0, 5.0 and 10.0 respectively, with

MATLAB.

5. Consider the first-order system,

1s

k

)s(R

)s(Y

+=

Obtain the unit-step response curves for k = 0.1, 0.5, 1.0, 5.0, and 10.0 respectively, with

MATLAB.

6. A general second-order system has the form:

2

nn

2

2

n

s2s

k

)s(R

)s(Y

ω+ζω+

ω=

What are the values of k, ζ, and ωn for the following system:

9s2s

3

)s(R

)s(Y2 ++

=

7. A second-order system is described by the differential equation:

)t(u25)t(y25dt

)t(dy5

dt

)t(yd2

2

=++

a) Write down the transfer function Y(s)/U(s) of the system, where U(s) and Y(s) are the Laplacetransforms of u(t) and y(t), respectively.

b) Obtain the damping ratio ζ and the natural frequency ωn of the system.c) Calculate the rise time and percent overshoot of the system.

d) Evaluate y(t) for a unit-step input u(t).

e) Check your answers of the above with MATLAB.

8. For the control system shown by the block diagram, the numerical value of J = 1 kg-m2 and B =

1 N-m/(rad/sec).

K1

+_

R(s)

K2

Y(s)1/s

+ BJs

1

+_

a) Find the transfer function Y(s)/R(s).

b) Determine the values of the gain K1 and velocity feedback constant K2 so that the maximum

overshoot in the unit-step response is 0.2 and peak time is 1 sec.

c) With these values of K1 and K2, obtain the rise time and settling time.

d) Obtain the response y(t) for the unit-step input r(t).e) Check the above calculations with MATLAB.


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9. When the second-order system

KsTs

K

)s(R

)s(Y2 ++

=

is subjected to a unit-step input, the system output responds as shown in the following figure.

Determine K and T from the response curve.

Time (sec.)

A m p l i t u d e

Step Response

0 0.5 1 1.5 2 2.5 30

0.2

0.4

0.6

0.8

1

1.2

1.4

1.31

0.55

10. A sinusoidal input u(t) = 2sin(2t) is applied to a system with transfer function:

)2s(s

2

)s(U

)s(Y

+=

Determine the steady-state output, yss(t), of the system.

11. The figure below shows a block diagram of a space vehicle attitude control system where R and

Y are the Laplace transforms of the reference (or desired) and actual attitude angles

respectively. Determine the values of KP and KD to yield a settling time of 0.5 second and 20%

overshoot in the close-loop system for a unit-step input.

Y(s)

2s

1KP

R(s)

−+

KDs

−+


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12. Consider the second-order system

1s2s

1

)s(R

)s(Y2 +ζ+

=

Obtain the unit-impulse response curves for ζ = 0.1, 0.3, 0.5, 0.7, 1.0, and 4.0 respectively, withMATLAB.

13. Consider the second-order system

2

nn

2

2

n

s2s

k

)s(R

)s(Y

ω+ζω+

ω=

Assuming that ωn = 2, k = 2, obtain the unit-impulse response curves for ζ = 0.1, 0.3, 0.5, 0.7, 1.0,and 4.0 respectively, with MATLAB.

performance parameters - module ii reading iii-2

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