first order system.pdf

8/13/2019 first order system.pdf

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ECM2105 - Control Engineering Dr Mustafa M Aziz (2013)________________________________________________________________________________

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 differentialequation, 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 highestpower 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) = c (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/s 2. Sinusoidal function: In the time domain, u(t) = csin( t). In the s domain, U(s) = c /(s 2+2).where c is a constant in all the above.

With these test signals, mathematical and experimental analyses of control systems can be carriedout 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 theform of the input that the system will be subjected to most frequently under normal operation. Ifthe inputs to a control system are gradually changing functions of time, then a ramp function of timemay be a good test signal. Similarly, if a system is subjected to sudden disturbances, a step functionof time may be a good test signal, and for a system subjected to a shock input, a pulse or an impulsefunction may be best.

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

k bsms1

)s(G 2 ++=

b)1RCs

1)s(G

+=

c)1RCsLCs

1)s(G 2 ++

=


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The time response of a control system consists of two parts: the transient response and the steady-state response. The t ransient response is defined as the part of the time response which goes fromthe initial state to the final state and reduces to zero as time becomes very large. The s teady-stateresponse is defined as the behaviour of the system as t approaches infinity after the transients havedied out. Thus the system response y(t) may be written as:

y(t) = y t(t) + y ss(t)

where y t(t) denotes the transient response, and y ss(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 arevelocity of a car on the road, control of the velocity of a rotating system, electric systems whereenergy 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 examplesto 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 velocityand b is the friction coefficient. By Newtons law, wehave 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 inputvoltage and y(t) is the output voltage. By Kirchhoffs 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 responsecurves. 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

1s1

1TsT

s1

)s(Y+

=+

=

Take the inverse Laplace transform: y(t) = 1 - e -t/T , t 0.

The solution has two parts: a steady-state response: y ss(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 thestep 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 , T s.


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The rise time , T r, is defined as the time for the waveform to go from 10% to 90% of its finalvalue.

The steady-state error is the error after the transient response has decayed leaving only thecontinuous 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)(eetss

==

= 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 itssteady-state response?

Exercise : A system has transfer function:50s

50)s(R)s(Y

+=

Find the time constant, T, the settling time, T s, and the rise time, T r 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/s 2.

The output transform:1Ts

1s1

)s(Y 2 +=

Expanding Y(s) into partial fractions:

1TsT

sT

s1

)s(Y2

2 ++=

Taking the inverse Laplace transform:y(t) = t - T + Te -t/T , t 0.

Steady-state error:[ ])t(y)t(rlim)(ee

tss ==

= T

2.3. Unit-impulse response of first-order systems

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

=

=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 thetransfer 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 themass, k is the spring constant, b is the frictioncoefficient, u(t) is the external force and y(t) is thedisplacement. From Newtons second law force = ma:

)t(u)t(kydt

)t(dyb

dt)t(yd

m 22

=++

Example 2: Electrical system: RLC circuit

Using Kirchhoffs law:

)t(ydt

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

wheredt

)t(dyL)t(i =

Hence:

)t(u)t(ydt

)t(dyRC

dt)t(yd

LC 22

=++

A general form of a second order system is:

)t(uk )t(ydt)t(dy

2dt)t(yd 2

n2nn2

2

=++

Transfer function:

2nn

2

2n

s2sk

)s(R)s(Y

++

=

k: the gain of the system : the damping ratio of the systemn: the (undamped) natural frequency of the system

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

1s 2nn1 = and 1s2

nn2 +=

Three cases:

= 1, critically damped case > 1, overdamped case0 < < 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 ( = 1)Two equal poles: s 1 = s 2 = - n

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

2

n

)s(s)s(Y

+=

Expanding Y(s) into partial fractions: 2n

n

n )s(s1

s1

)s(Y+

+=

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

n=2

n=3

j

- n

s-plane

Exercise : A system has the following transfer function:

16s8s

1

)s(R

)s(Y2 ++

=

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


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t (sec)

y(t)

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

se

se

121y(t)

21

ts2e1y(t) =

1, s 1 cannot be neglected.

3.1.3. Underdamped case (0 < < 1)Transfer function:

) js)( js()s(R

)s(Y

dndn

2n

+++

=

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

The two poles are:

2nn1 1 js = and

2nn2 1 js +=

=

21 1

tan = cos( )

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

2d

2n

n2d

2n

n2nn

2n

)s()s(s

s1

s2s2s

s1

)s(Y++

+++

=++

+=

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

-1 )tcos(e)s(

s dt

2d

2n

n n =++

+

-1 )tsin(e)s( d

t2d

2n

n n =++

js-plane

d

- n

-d

- n+jd

- n-jd

n


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The time response is:

)tsin(1

e1

)tsin(1

)tcos(e1)t(y

d2

t

d2dt

n

n

+=

+=

where

= 2

1 1tan .

When = 0, the response becomes undamped and oscillations continue indefinitely at frequencyn. The time response in this case becomes:

y(t) = 1 cos( nt)

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

t (sec)

y(t)

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

=0

=0.1

=0.5

=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

T r

TpTs

Maximum overshoot

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

0.02

3.2.1. Maximum overshootThe maximum amount by which the system output response proceeds beyond the desired response.Let y max denotes the maximum value of y(t), and y ss = y( ) the steady-state value of y(t), then themaximum overshoot of y(t) is defined as:

maximum overshoot = y max - y ss

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

=

=

2ss

ssmax

1exp100%100

yyy

)(PO overshoot percent

3.2.2. Peak time, T p The time required for the response to reach the first peak of the overshoot:

2n

p1

T

=

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

2n

21

dr

1) / 1(tanT

= =


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

ns

4T

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

s3

T

Note that NOT all these specifications necessarily apply to any given case. For example, for anoverdamped 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 T r and T p the closeness of the response to the desired response, as represented by PO and T s.

From the design requirement, the swifter and closer, the better. However, when n is fixed,small T r and T p require a small , while small T s and PO require a large . Note that

21 / exp100PO = , ]1 /[T 2np = , ]1 /[)] / 1(tan[T2

n21

r = .

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 naturalfrequency of 10Hz. Find:

a) the transfer functionb) the time response when it is subjected to a unit-step inputc) the percentage overshoot with such an inputd) the rise time


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Exercise : Given the transfer function:

100s15s100

)s(G 2 ++=

find T p, PO, T s and T r.

3.3. Ramp response of a second-order system

The Laplace transform of a unit-ramp input is R(s) = 1/s 2

The output is:)s2s(s

k )s(Y 2

nn22

2n

++

=

Again, there are three cases:

= 1, critically damped case > 1, overdamped case

0 < < 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.:

2nn

2

2

ns2s

k )s(Y ++

=

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 thestep response for a given damping.

Taking the example of a critically damped system where =1, the unit-step response is given by:t) (1e1y(t) n

t n +=

The unit-impulse response is therefore:t 2

nnte

dtdy(t) =

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 alsocommon 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) = U osin(t) y ss(t) = U o|G(j )|sin( t+)

Uo|G(j )|

t

The linear, time-invariant system G(s) subjected to a sinusoidal input of amplitude U o andfrequency describe by:

u(t) = U osin( t)

will, at steady state, have a sinusoidal output of the same frequency as the input but, generally, withdifferent amplitude and phase given by:

yss(t) = U o|G(j )|sin( t + ())

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

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

) j(G)] j(GRe[)] j(GIm[

tan)( 1 =

=

The sinusoidal transfer function of any linear system is obtained by substituting j 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 U o and frequency :u(t) = U osin( t)

The Laplace transform of u(t) is: 22o

sU)s(U

+=

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:

22o

sU

)s(G)s(Y+

=

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

+

++

+++

++

+=

jsc

jsc

psc

psc

psc

)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 c n are constants and c 0 and*0c are a complex conjugate pair that can be obtained using the

cover-up rule:

j2U) j(G

c o0

= j2

U) j(Gc o*0

=

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

where |G(j )| is the magnitude and is the phase given respectively by:22 )]} j(G{Im[)]} j(G{Re[|) j(G| += ) j(G

)] j(GRe[)] j(GIm[

tan)( 1 =

=

Similarly: G(-j ) = |G(-j )|e -j = |G(j )|e -j

Therefore: j2Ue|) j(G|

co

j

0

= j2

Ue|) j(G|c

o j

*0

=

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

0t j

0tp

ntp

2tp

1 ececececec)t(y n21 +++++= L

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

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 j0tss

ecec)t(ylim)t(y

+==

Substituting for c 0 and *0c and noting that sin(x) = (e jx e -jx)/(2j), gives the steady-state output:

yss(t) = U o|G(j )|sin( t + )

Advantages of frequency domain analysis:

The transfer functions of complicated components can be determined experimentally byfrequency response tests (without deriving their mathematical models) using available signalgenerators 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-domaintransient and steady-state system performances.

Systems may be designed to achieve transient and steady-state requirements using frequencyresponse 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, and2) Bode diagram

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

+=+===

A Bode diagram consists of two graphs:

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

- plot of the phase angle, ()

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

Logarithmic magnitude (also called gain) of G(j ), M = 20log 10 |G(j )| (unit in decibels, dB)

Phase:

=

)(R)(X

tan)( 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(Gn21

m21m

=L

L

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

=

jsn21

m21m

)ps()ps()ps(

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

L

L

Taking the logarithm yields:

20log 10 |G(j )| = 20log 10bm + 20log 10 |(s - z 1) + 20log 10 |(s z 2)| + - 20log 10 |(s - p 1) - 20log 10 |(s p 2)| - |s j

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 alogarithmic scale, which means a wider range of system behaviour, from low to highfrequencies, 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

11) j(RC

1) j(G

+=

+=

Solution:

Magnitude:2222 T / 1

T / 1

T1

1) j(G

+=

+=

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

At low frequencies ( 0): gain (dB) 0)T / 1(log20)T / 1(log20 1010 dB

At high frequencies ( ): gain (dB) )(log20)T / 1(log20 1010

If we plot the logarithmic gain (dB) against ) (log 10 , then the above equation becomes thestraight line:

y = c + mx

where y = ) j(Glog20 10 , c = intercept, m = slope, and )(logx 10 = . Thus the logarithmicgain reduces by 20dB (negative slope) per factor of 10 (decade) increase in frequency (20dB/decade).

The transition between the high and low frequency asymptotes is found by equating the low andhigh frequency limits this is known as the corner or cut-off frequency :

T1

c =

Since )( jX)(RT1

T jT1

1T j1

1) j(G 2222 +=+

+

=+

=

Phase: ( )TtanRX

tan)(11

=

=

(rad)

At low frequencies ( 0): ( ) 00tan)0( 1 = radAt high frequencies ( ): ( ) 2 / tan)( 1 = radAt corner frequency ( 1/T): ( ) 4 / 1tan)T / 1( 1 = rad

u(t) y(t)

R

C


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(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 thefrequency response of the LTI SYS=tf(num,den) , where num and den are the numerator anddenominator coefficients of the system, respectively. For example, to plot the Bode diagramsshown 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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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 OUTPUTCONDITION with the system designed on the basis of achieving the best output from the givenraw material or power available. The design of the system to produce the desired steady-state is aproblem in its own right but not the subject matter of this course.

OPERATING CONDITIONS ARE OFTEN CHANGED BY OPERATOR OR COMPUTERINTERVENTION such as changes in power supplied by a power station to the national grid dueto changes in the overall national power consumption.

ENGINEERING SYSTMES HAVE DYNAMICAL BEHAVIOURS they do not just produce thedesired output temperature, pressure, concentration, voltage, current, frequency, position, velocity,acceleration, force, torque, flow, level, concentration, reaction rate, .. etc even if that output variableis 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 behaviourspreferred by the system and replace them by acceptable dynamic responses, the process must beaugmented by a control system incorporating features of output measurement (sensors), inputvariable changes (actuators) and a (dynamical) data processing device that processes both sensordata and desired output specifications to generate a desired plant input signal to the actuator. Thissystem almost always has a FEEDBACK structure reflecting the fact that its output is used to createthe desired input in real time.

CONTROL SYSTEMS REQUIRE DESIGNING notwithstanding the impression left by the

popular TV science programme Tomorrows 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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up to the plant. Even those controllers that are available commercially require an element of designeither off-line or on-line and hence an effective engineer requires an appreciation of the designprocess 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 isfound that it is necessary to have appropriate data and some understanding of the process to becontrolled. This typically takes the form of EITHER

1) a mathematical model expressed in differential equation, transfer function or state spaceform, AND/OR

2) data on the behaviour of the plant output(s) in response to known inputs (such as steps orsinusoids) 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 OFSUITABLE MEASUREMENTS OF SYSTEM BEHAVIOUR and the more accurate andextensive these measurements are, the better control will be.

YOU CANNOT ALWAYS MEASURE WHAT YOU WANT TO MEASURE if you cannotmeasure the output variable of interest (due to extreme physical conditions of speed, temperature orpressure .. 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 sensorsare placed on the external wall?

CONTROLLERS CAN DO AMAZING THINS As control requirements are as varied as theapplications and needs for new products, even the virtually impossible has been asked for in thesearch 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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4. Consider the first-order system,

1Ts1

)s(R)s(Y

+=

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

5. Consider the first-order system,

1sk

)s(R)s(Y

+=

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

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

2nn

2

2n

s2sk

)s(R)s(Y

++

=

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

9s2s3

)s(R)s(Y

2 ++=

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

)t(u25)t(y25dt

)t(dy5

dt)t(yd

2

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 and the natural frequency n 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-m 2 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

KsTsK

)s(R)s(Y

2 ++=

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(s2

)s(U)s(Y

+=

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

11. The figure below shows a block diagram of a space vehicle attitude control system where R andY are the Laplace transforms of the reference (or desired) and actual attitude anglesrespectively. Determine the values of K P and K D 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

1s2s1

)s(R)s(Y

2 ++=

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

13. Consider the second-order system

2nn

2

2n

s2sk

)s(R)s(Y

++

=

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

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