4. System Response

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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.

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Presentation Transcript
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.

h(t)

r(t)

c(t)

4. System Response

Review of some LTI properties

We will express system as in figure below, system input is r(t), output is c(t), and impulse response is h(t)

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)

1. Impulse response

Impulse 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

4. Integral

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

5. Poles and Zero

T.F.isusually 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

2. Convolution

Output of LTI system is the convolution of its input and its impulse response:

(1)

(4)

(1)

(5)

(2)

+

+

R(s)

C(s)

(3)

c(0)

4.1. Time Response of the First Order Systems.

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:

Solving for C(s) yields

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

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

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

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

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

c(t)

τ

K

(1)

t

(2)

R(s)

C(s)

(3)

4.1. Time Response of the First Order Systems.

Since we usually ignore the initial condition

in block diagram, we use the system block

diagram as shown bellow.

The first term originates in the pole of input R(s) and

is called the forced response or steady state response

The second term originates in the pole of the transfer

function G(s) and is called the natural response

Figure below plot c(t)

Suppose that the initial condition is zero then

Unit step response

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

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

that is

lim c(t)= K

t

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.

Taking the inverse L.T of (2) yields

(2)

c(t)

t

r(t)=t

c(t)

(1)

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 s0 s0

For Unit step input then the final value is

css(t)= lim G(s)

s0

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 Response

For the input equal to unit ramp function

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

therefore

• This ramp response is composed of three term
• a ramp
• a constant
• an exponential.

(1)

(5)

2

(3)

c(t)

(2)

=0

0.2

0.7

(3)

nt

(4)

4.2. Time Response of Second Order System

The standard form second order system is

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

This system is said to be critically damped

The poles of the TF is s = n jω(12)

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

Where  = damping ratio

n = natural frequency, or undamped frequency.

Consider the unit step response of this system

This system is said to be undamped

For this system we have

Time constant =  = 1/n ; frequency = n

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

This system is said to be underdamped

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

This system is said to be overdamped