4. System Response

• This module is concern with the response of LTI system.

• L.T. is used to investigate the response of first and second order systems. Higher order systems can be considered to be the sum of the response of first and second order system.

• Unit step, ramp, and sinusoidal signal play important role in control system analysis. It is therefore we will investigate this signals.

4. System Response

(1)

2. ConvolutionOutput of LTI system is the convolution of its input and its impulse response:

h(t)r(t) c(t)

tdthrthtrtc )()()(*)()(

tdhtr )()(

Taking the L.T. of (1) yields C(s) = R(s)H(s) (2)Where C(s) is the L.T. of c(t) R(s) is the L.T. of r(t) H(s) is the L.T. of h(t) H(s) is called the transfer function (T.F)3. Derivative If the input is r’(t) then the output is c’(t) where r’(t) denotes the derivative of r(t)

Review of some LTI propertiesWe will express system as in figure below, system input is r(t), output is c(t), and impulse response is h(t)

4. Integral

If the input is r(t)dt then the output is c(t)dt

1. Impulse responseImpulse response, denoted by h(t), is the output of the system when its input is impulse (t). h(t) is called the impulse response of the system or the weighting function

5. Poles and ZeroT.F. is usually rational and therefore can be expressed as N(s)/D(s). Poles is the values of s resulting T.F to be infinite. Zeroes is the values of s resulting T.F to be zero

4.1. Time Response of the First Order Systems.

We can found the differential equation first we write (1) as

The diff. Eq. is the inverse L.T. of (2)

Now we take the L.T of (3) and include the initial condition term to get

1τ)(

)()(

s

K

sR

sCsG (1)

)(τ

)(τ

1sR

KsCs

(2)

)()(1

)( trτ

Ktc

τtc (3)

)(τ

)(τ

1)0()( sR

KsCcssC (4)

Solving for C(s) yields

)τ/1(

τ)R(s)/(

)τ/1(

)0()(

s

K

s

csC (5)

Note that the initial condition as an input has a Laplace transform of c(0), which is constant.

The inverse L.T of a constant is impulse (t). Hence the initial condition appears as the impulse function

Here we can see that the impulse function has a practical meaning, even though it is not a realizable signal

τ1

τ

s

K

τ1

1

s

R(s)

c(0)

+

+

C(s)

Here we will investigate the time response of the first order systems. The transfer function of a general first order system can be written as:

The eq. can be shown in the block diagram as shown in the figure bellow.

4.1. Time Response of the First Order Systems.

Unit step responseFor unit step input R(s)=1/s, then

Since we usually ignore the initial condition in block diagram, we use the system block diagram as shown bellow.

)()τ/1(

τ)/()( sR

s

KsC

(1)

Suppose that the initial condition is zero then

τ1

τ

s

KR(s) C(s)

τ/1s

1

τ)/1(

τ)/()(

s

K

s

K

s

KsC (2)

Taking the inverse L.T of (2) yields

)1()( τt

eKtc (3)

The first term originates in the pole of input R(s) andis called the forced response or steady state responseThe second term originates in the pole of the transferfunction G(s) and is called the natural responseFigure below plot c(t)

t

c(t)

Kτ

)1( τt

eK

The final value or the steady state value of c(t) is Kthat is

lim c(t)= Kt

c(t) is considered to reach final value after reaching 98% of its final value.The parameter is called the time constant. The smaller the time constant the faster the system reaches the final value.

4.1. Time Response of the First Order Systems.

A general procedure to find the steady state value is using final value theorem

lim c(t) = lim sC(s) = lim sG(s)R(s) t s0 s0

For Unit step input then the final value is

css(t) = lim G(s) s0

since R(s) = 1/s

System DC gain is the steady state gain to

a constant input for the case that the output

has a final value.

Ramp ResponseFor the input equal to unit ramp function

r(t) = t and R(s) = 1/s2, C(s) is

τ/1s

1

τ)/1(

τ)/()(

22

s

K

s

K

s

K

s

KsC (1)

therefore

))( τt

eKKKttc (2)

This ramp response is composed of three term 1. a ramp2. a constant3. an exponential.

t

c(t)

r(t)=tc(t)

4.2. Time Response of Second Order System

22

2

2)(

)()(

nn

n

sssR

sCsG

(1)

The standard form second order system is

Where = damping ration = natural frequency, or undamped frequency.

Consider the unit step response of this system

sssRsGsC

nn

n

)2()()()(

22

2

(2)

Case 1: 0<<1 (complex poles), c(t) is

)sin(1)( ttc n(3)

).(tanand1 define 1

For this system we haveTime constant = = 1/n ; frequency = n

n t

c(t)=0

0.7

2

0.2

The poles of the TF is s = n jω(12)

This system is said to be underdamped

Case 2: >1 (real unequal poles), c(t) is

21 /2

/11)( tt ekektc (4)

This system is said to be overdamped

Case 3: =1 (real equal poles), c(t) is

/2

/11)( tt tekektc (5)

This system is said to be critically damped

Case 4: =0 (imaginary poles), c(t) is

)sin(1)( 1 tetc n

tn (3)

This system is said to be undamped