Control and Intr 2 Indika

8/3/2019 Control and Intr 2 Indika

1/7

Part a

)11.0)(1(

10

)(

)()( 0

++==

sssV

swsG

++==sT

sssE

sVsG

c

106.01

)(

)()(

As we know

++

++==

sTs

sssGsGsG

cOL

106.015

)11.0)(1(

10)()()(

++

++=

sT

TssT

ss

106.05

)11.0)(1(

10 2

sTss

sTTs

)11.0)(1(

)106.0(50 2

++

++=

sTss

sTTs

)11.11.0(

505032

2

++

++=

sTTsTs

sTTssG

OL ++++

=23

2

1.11.0

50503)(

As we know)(1

)()(

sG

sGsG

OL

OL

CL +=

sTTsTs

sTTs

sTTsTs

sTTs

sGCL

++++

+

++++

=

23

2

23

2

1.11.0

505031

1.11.0

50503

)(

505031.11.0

50503223

2

+++++++

=sTTssTTsTs

sTTs

50511.41.0

50503)( 23

2

+++ ++= sTTsTssTTs

sGCL

1


2/7

Characteristics Equation 0)()(1 =+= sHsGK

Characteristics Equation 050511.41.0 23 =+++= sTTsTs

Dividing all terms by sTTsTs 511.41.0 23 ++

0511.41.0

50

511.41.0

511.41.02323

23

=++

+++++

=sTTsTssTTsTs

sTTsTs

[ ]0

511.41.0

501

23=

+++=

sssT

0511.41.0

5011

23=

++

+=

sssT

TK

1=

Commands

>>sys=tf([50],[0.1 4.1 51 0])

>>rlocus(sys)

Graph 1

From Graph 1 it can be noted:

At damping of 0.7

2


3/7

Gain = 6.05 K = 6.05

Therefore sK

T 165.005.6

11===

Pole = -7.51+7.66i

Part b

Put T = 0.165 in GCL(s)

50)165.0(51)165.0(1.4)165.0(1.0

50)165.0(50)165.0(3)(

23

2

+++

++=

sss

sssG

CL

50415.86765.00165.05025.8495.0)(

23

2

+++ ++= ssssssG

CL

Commands

>>sys=tf([0.495 8.25 50],[0.0165 0.6765 8.415 50])

>>step(sys)

Graph 2

From Graph 2:

Rise Time Tr= 0.0506s

3


4/7

% Overshoot = 13.7

Settling Time Ts = 0.339s

Final Value = 1

From Graph 2: % Overshoot = 13.7

% Overshoot 1001

exp2

=

=

21

exp

100

7.13

21)137.0ln(

=

2

2

2

1)99.1(

=

2

22

195.3

=

22295.395.3 =

)5.9(95.3 22 +=

286.0)95.3(

95.32

2 =+

=

Damping ratio: 535.0=

n

st

= 4

n

st

=

535.0

4

Settling time: sts 339.0=

n

=535.0

4339.0

4


5/7

4535.0339.0 =n

Natural frequency:535.0339.0

4

=

n

sradn /1.22=

Frequency of oscillation: 21 =nd

2)535.0(11.22 =

845.01.22 =

sradd

/67.18=

As Final value = 1, so the ess must be zero.

Calculations

Plotting pole -7.51+7.66i on the graph

22266.751.7 i

n+=

222 )66.7()51.7( +=n

1.1152 =n

Natural frequency: sradn /72.10=

Poles can be written as dn wws =

5


6/7

nw=51.7

Damping ratio: 7.0=

% Overshoot 1001

exp2

=

100)7.0(1

7.0exp

2

=

% Overshoot 6.4=

n

st

=

4

72.107.0

4

=

st

Settling time: sts 533.0=

Frequency of oscillation: 21 =nd

2)7.0(172.10 =

71.072.10 =

sradd

/66.7=

( )[ ])(1)(lim0

sGsRseCL

s

s s=

As the system is step response R(s) =s

1

+++

++=

50415.86765.00165.0

5025.8495.01

123

2

0lim

sss

ss

sse

s

ss

50

501=

sse

Steady State Error 0=sse

6


7/7

The table below shows the results estimated in part b and results from the

calculations.

Discussion and Conclusion

Estimates from Part b Calculation Results

Damping ratio 0.535 0.7

% Overshoot 13.7% 4.6%

Settling Time st 0.339s 0.533s

Natural frequencyn

22.1 rad/s 10.72 rad/s

Frequency of oscillation

d

18.67 rad/s 7.66 rad/s

Steady State Error ess 0 0

From the table above it can be seen that as the damping ratio is decreased in

part b, this causes and increase in frequency of oscillation of the system causing it to

settle quicker. In calculation results, damping ratio used was restricted not to be any

lesser than 0.7. The reason for this restriction could be to prevent it to oscillate withhigher frequency causing the system to fail. This system shows an under-damped

response which is stable as the function is reaching a steady state value of 1. This is

not sustained oscillation so increase the gain until the point of instability is reached.

7

Documents

Control and Intr 2 Indika