class 06 “time domain analysis” part i 1st order...

Class 06

“Time domain

analysis”

Part I

1st order systems

outputinput

First order systems

of type

cbs

a)s(G

+=

S

Time domain analysis - 1st order systems______________________________________________________________________________________________________________________________________________________________________________________

1st order

systems

cbs

a

)s(R

)s(Y

+=

cbs

a

+

Ko

,

1sc

bc

a

c

)cbs(c

a

)s(R

)s(Y

+=+=

T

that is:

outputinput

1st order

systems


cbs

a

+

that is:

outputinput

1Ts

K

)s(R

)s(Y o

+=

1Ts

K o

+

the transfer function can be

rewritten as:

1st order

systems


1Ts

K

)s(R

)s(Y o

+=

1Ts

K o

+

the transfer function:

Ko = gain of the system

T = time constant of the system

outputinput

1st order

systems


Example 1:

4s5

2

)s(R

)s(Y

+=

Ko = 3

Example 2:

4s

12

)s(R

)s(Y

+=

Ko = 2/4 = 0,5 T = 5/4 = 1,25

pole:

s = – 0,8

pole:

s = – 4

T = ¼ = 0,25


What is the output?

(step response)

unit step input

1Ts

K o

+outputinput

r(t) =

1st order

systems


)s(R1Ts

K)s(Y o ⋅

+=

in order to calculate:

)1Ts(

TK

s

K

s

1

)1Ts(

K)s(Y ooo

+−=⋅

+=

1Ts

K o

+outputinput

1st order

systems


0t,)e1(K)t(y T/t

o >−= −

[ ])s(Y)t(y 1−= L

hence, the unit step response is:

1Ts

K o

+outputinput

1st order

systems


the unit step response is:

0t,)e1(K)t(y T/t

o >−= −

1Ts

K o

+outputinput

r(t) =

1st order

systems


unit step input

0t,)e1(K)t(y T/t

o >−= −

the unit step response is:


0t,)e1(K)t(y T/t

o >−= −

oK632,0)e1(K)T(yTtIf 1

o =−=⇒= −

oK865,0)e1(K)T2(yT2tIf 2

o =−=⇒= −


o =−=⇒= −


o =−=⇒= −


o =−=⇒= −


Observe that, for the unit step response:

outputinput

What is the output?

(impulse response)unit impulse input

1Ts

K o

+

r(t) =

1st order

systems


)s(R1Ts

K)s(Y o ⋅

+=


)1Ts(

K1

)1Ts(

K)s(Y oo

+=⋅

+=

outputinput 1Ts

K o

+1st order

systems


0t,eT

K)t(y T/to >⋅= −

[ ])s(Y)t(y 1−= L

hence, the unit impulse response is:

outputinput 1Ts

K o

+1st order

systems


the unit impulse response is:

0t,eT

K)t(y T/to >⋅= −

outputinput 1Ts

K o

+

r(t) =

1st order

systems


unit impulse input

the unit impulse response is:

0t,eT

K)t(y T/to >⋅= −


0t,eT

K)t(y T/to >⋅= −

)T/K(368,0eK)T(yTtIf oo1 ⋅=⋅=⇒= −

)T/K(135,0eK)T2(yT2tIf oo2 ⋅=⋅=⇒= −

)T/K(05,0eK)T3(yT3tIf oo3 ⋅=⋅=⇒= −

)T/K(02,0eK)T4(yT4tIf oo4 ⋅=⋅=⇒= −

)T/K(007,0eK)T5(yT5tIf oo5 ⋅=⋅=⇒= −


Observe that, for the impulse response:

What is the output?

(ramp response)unit ramp input

outputinput 1Ts

K o

+

r(t) =

1st order

systems


)s(R1Ts

K)s(Y o ⋅

+=


)1Ts(

TK

s

TK

s

K

s

1

)1Ts(

K)s(Y

2

22

oooo

++−=⋅

+=

outputinput 1Ts

K o

+1st order

systems


0t,)eTTt(K)t(y T/t

o >⋅+−= −

[ ])s(Y)t(y 1−= L

hence, the unit ramp response is:

outputinput 1Ts

K o

+1st order

systems


0t,eTTt)t(y T/t >⋅+−= −

If Ko = 1, the unit ramp response is:

outputinput 1Ts

K o

+1st order

systems


the unit ramp response for Ko = 1:

0t,eTTt)t(y T/t >⋅+−= −

outputinput 1Ts

K o

+

r(t) =

1st order

systems


unit ramp input

the unit ramp response is:

0t,eTTt T/t >⋅+−= −

1K o =


Error for the unit ramp input


Error for the unit ramp input, Ko = 1:

( )1Ts

Ts

s

1

1Ts

11Ts

s

1

1Ts

11

s

1

s

1

1Ts

1

s

1)s(E

22

222

+⋅=

+−+⋅=

=

+−⋅=⋅

+−=

outputinput 1Ts

K o

+1st order

systems


Steady state error:

T1Ts

Tlim

1Ts

Ts

s

1slim)s(Eslime

0s

20s0sss

=+

=

=+

⋅⋅=⋅=

→

→→

Tess =

input

1st order

systems


Error for the unit ramp input


Thank you!

Felippe de Souza

[email protected]

Departamento de Engenharia Eletromecânica

class 06 “time domain analysis” part i 1st order...

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