Unit 2

Time Response Analysis

2.1 Time Response of Second Order System:

Introduction

Time Response Analysis:-

For a second order control system, the total time response is analyzed by its transient as well as ready state response.

Any system which contains energy along elements (L1 c etc) if suffer any disturbance in the energy state effect at the input end as at this output  or at both the ends takes source time to change  in  thus  form one state other . This change its then is called as Transient times and the values of reverent and voltage during this period is Transient Response.

The above fig: a clearly shows steady star response is that part of response when the transient have died. If the steady star response of the output does not match with input then the system has steady state errors.

Consider the following open loop system

L-1[c(s)]= L-1[ G(s) R (S)]

C(t) = L-1[ G(S) Q (s)]

E(s) = R(s) –c(s) [ error signal]

C(S)= [ R(s)- c(Q)] G(s)

C(s) = R(s) G(s)- c(s) G(s)

C(s) [ 1+G(s)] = R(s) G(s)

C(s)/R(s) = G(s)/1+G(s)

C(s)/R(s) = G(s)/1+G(s)

E(s) = R(s)- c(s)

=R(s)-R(s) G(s)/1+G(s0

E(s) = R(s) [1/1+G(s)

If no error then E(s) = 0, f(t) = 0 Hence , r(t) = c(t)

2.1 Time response analysis of second order system

1)     ORDER – 2

OLTF  G(s) = k/s(1+TS)

CLTFC(s)/R(s) = k/k+s(1+Ts)

= k/s2+ s/T +k/T

Comparing above equation with standard 2nd order eqn

Standard 2nd order equation

CEs2+2 wn s+wn2=  0

S= -2wn±/2

=2wn±2wn/2

S= wn±wn

S= -wn±wn

S= -wn ±wn

S= -wn ±gwn

Standard eqn:-

T(s) = wn2/s2+2wns+wn2

Our eqn T(s) = K/T/s2+1/T s+ K/T

Wn2 = k/T

2 wn = 1/T

2k/T = 1/T

k/T = 1/2T

= 1/2T T/K

= 12KT

Graphically showing the position of loops s1,s2 for offered of

As the charaEqun (location of polls ) is depend ent only on r (wn constant for a given  system)

S1 = -wn + jwn 1-2

S2= - wn - jwn1-2

CASE 1: (=0)

S1=jwn,S2, = -jwn

Undamped

CASE 2:(0<<1)

S1= -wn + jwn0.75

=-wn/2 +jwn(0.26)

S2 = -wn/2 – jwn (0.86)

CASE:3        (-1)

S1= S2 = -wn = -wn

CASE4:- (-1)

S1 = -2wn +jwn-3

=-2wn – jwn (1.73)

S1  = -3.73wn

S2 = -2wn –jwn-3

= -2wn + jwn (1.73)

= -0.27 wn

Overdamped

All prac sys are 2 order to, if R(e) = (t) R(s) = 1 :. CLTF = 2nd order steqn and hence already the system is possible.

CLTF = e(s)/ R(s) = wn2/s2+2wns+wn2

Now calculating c(s) , c(t) for different values of input

1)     Impulse I/p

R(t) = (t)

R(s) = 1

C(s) = R(s) wn2/s2+2wns+wn2

C(s) = wn2/s2 +2wn+wn2

Under this i/p (R(t) = (t)) the output varies with different values of . So,

CONDITION 1:    ( =0)

C(s) =wn2/s2 +wn2

Os2 +wn2  =0

S=+-jwn

C(t) = wnsinwnt

Sinwnt

As there in no damping i.e. oscillations at t= 0 are some at t so, called UNDAMPED

CONDITION 2:  0<<1

c(s) = R(s) wn2/s22wn+wn2 R(t) = (t)

R(s) =1

C(s) = wn2/s2+2wns+wn2 

CE

S2+2wns +wn2 =0

S2, S1 = - wn ±jwn 1-2

C(t) = e-wnt   sin(wn)t

C(t)=e-wnt sin (wdt)

The oscillations are present but at t- infinity the Oscillations are 0 so, it is UNDERDENPLZ

CONDITION 3 :=1

C(s) = R(s) wn2/s2+2wns+wn2

C(s)=wn2/S2 +2wns+wn2

=wn2/(s +wn)2

CE    S= -Wn

Diagram

C(t)=w2n /()2

C(t)=

No damping obtained at so is called CRITICALLY DAMPED.

CONDITIONS  4 :->1

C(s) =  wn2/s22wnS+Wn2

S1,s2 = WN+-jwn

b)UNITStep Input:-

R(s) = 1/s

C(s)/R(s) = wn2/s2+2wns+wn2

C(s) = R(s) wn2 /s2 +2wns+wn2

C(s) = R(s) wn2/s2+2wns+wn2

C(s) = R(s) wn2/s2+2wns+wn2

C(s) = wn2s(s2+2wns+wn2)

C(t) = 1- ewnt/1-es2 sin  (wdt + ø)

Wd = wn1-2

Ø=

Where

Wd = Damping frequency of oscillations

Wn = natural frequency of  oscillations

wn = damping coefficient.

T= Time constant

Condition 1= 0

C(s) = wn2 /s(s2+wn2)

C(t) = 1- e° sin wdt +ø

C(t)= 1- sin( wn +90)

C(t) = 1+cos wnt

Constant

C(t) = 1+constant

Condition 2:-  0<<1

C(s) = /s2+wns +wn2

C(s) =1/s –    s+ wn/s2+ wns +wn2

=1/s –    s+wn/(s+wn)2+wd2- wn/(s+wn2) +wd2

Wd = wn1-2

Taking Laplace inverse of above equation

L --1 s+wn/(s+wn) +wd2= e-wnt  cos wdt

L-1 s+wn/(s+wn)2+wd2 = e-wnt sinWdt

C(t) = 1-e-wnt  [coswdt +/1-2sinwdt]

= 1-ewnt /1-2 sin [wdt + 1-2/]    t>=0

C(t) = 1-ewnt/1-2 sin(wdt+ø)

Ø = 1+2/

Fig 6. Transient Response of second order system

Specifications:

1)     Rise Time (tp):- The time taken by the output to reach the already status value for the first time is known as Rise time.

C(t) = 1-e-wnt/1-2 sin (wdt+ø)

Sin (wd +ø) = 0

Wdt +ø = n

tr  =n-ø/wd

For first time so ,n=1.

2)     Peak Time (tp)

The peak value attained by the output is called peak time .The time required by the output to reach this value is lp.

d(cct) /dt = 0 (maxima)

d(t)/dt = peak value

tp = n/wd   for n=1

tp = wd

(3) Peak Overshoot Value :-

Maximum deviation of output from steady state value is called peak overshoot value(Mp ).

( ltp ) = 1 = Mp

(Sin(Wat + φ )

(Sin( Wd∏/Wd + φ)

Mp = e-∏ξ / √1 –ξ2

Condition 3 ξ = 1

C( S ) = R( S ) Wn2 / S2 + 2ξWnS + Wn2

C( S ) = Wn2 / S(S2 + 2WnS + Wn2)        [ R(S) = 1/S ]

C( S ) = Wn2 / S( S2 + Wn2 )

C( t ) = 1 – e-Wnt + tWne-Wnt

The response is critically damped.

(4). Settling Time (ts) :

ts = 3 / ξWn  ( 5% )

ts = 4 / ξWn  ( 2% )

Q.1. The open loop transfer function of a system with unity feedback gain G( S ) = 20 / S2 + 5S + 4. Determine the ξ, Mp, tr, tp.

Soln : Finding closed loop transfer function,

C( S ) / R( S ) = G( S ) / 1 + G( S ) + H( S )

As it is unity feedback so, H(S) = 1

C(S)/R(S) = G(S)/1 + G(S)

= 20/S2 + 5S + 4/1 + 20/S2 + 5S + 4

C(S)/R(S) = 20/S2 + 5S + 24

Standard equation for second order system,

S2 + 2ξWnS + Wn2 = 0

We have,

S2 + 5S + 24 = 0

Wn2 = 24

Wn = 4.89 rad/sec

2ξWn = 5

(a). ξ = 5/2 x 4.89 = 0.511

(b). Mp% = e-∏ξ / √1 –ξ2 x 100

= e-∏ x 0.511 / √1 – (0.511)2 x 100

Mp% = 15.4%

(c). tr = ∏ - φ / Wd

φ = tan-1√1 – ξ2 / ξ

φ=  tan-1√1 – (0.511)2 / (0.511)

φ = 1.03 rad.

tr = ∏ - 1.03/Wd

Wd = Wn√1 – ξ2

= 4.89 √1 – (0.511)2 

Wd = 4.20 rad/sec

tr = ∏ - 1.03/4.20

tr = 502.34 msec

(d). tp = ∏/4.20 = 747.9 msec

Q.2. A second order system has Wn = 5 rad/sec and is ξ = 0.7 subjected to unit step input. Find (i) closed loop transfer function. (ii) Peak time (iii) Rise time (iv) Setteling time (v) Peak overshoot.

Soln: The closed loop transfer function is

C(S)/R(S) = Wn2 / S2 + 2ξWnS + Wn2

= (5)2 / S2 + 2 x 0.7 x S + (5)2

C(S)/R(S) = 25 / S2 + 7s + 25

(ii). tp = ∏ / Wd

Wd = Wn√1 - ξ2

= 5√1 – (0.7)2

= 3.571 sec

(iii). tr = ∏ - φ/Wd

φ= tan-1√1 – ξ2 / ξ = 0.795 rad

tr = ∏ - 0.795 / 3.571

tr = 0.657 sec

(iv). For 2% setteling time

ts = 4 / ξWn = 4 / 0.7 x 5

ts = 1.143 sec

(v). Mp = e-∏ξ / √1 –ξ2 x 100

Mp = 4.59%

Q.3. The open loop transfer function of a unity feedback control system is given by

G(S) = K/S(1 + ST)

Calculate the value by which k should be multiplied so that damping ratio is increased from 0.2 to 0.4?

Soln: C(S)/R(S) = G(S) / 1 + G(S)H(S)          H(S) = 1

C(S)/R(S) = K/S(1 + ST) / 1 + K/S(1 + ST)

C(S)/R(S) = K/S(1 + ST) + K

C(S)/R(S) = K/T / S2 + S/T + K/T

For second order system,

S2 + 2ξWnS + Wn2

2ξWn = 1/T

ξ = 1/2WnT

Wn2 = K/T

Wn =√K/T

ξ = 1 / 2√K/T T

ξ = 1 / 2 √KT

Forξ1 = 0.2, for ξ2 = 0.4

ξ1 = 1 / 2 √K1T

ξ2 = 1 / 2 √K2T

ξ1/ ξ2 = √K2/K1

K2/K1 = (0.2/0.4)2

K2/K1 = 1 / 4

K1 = 4K2

Q.4. Consider the transfer function C(S)/R(S) = Wn2 / S2 + 2ξWnS + Wn2

Find ξ, Wn so that the system responds to a step input with 5% overshoot and settling time of 4 sec?

Soln:

Mp = 5% = 0.05

Mp = e-∏ξ / √1 –ξ2

0.05 = e-∏ξ / √1 –ξ2

Cn 0.05 = - ∏ξ / √1 –ξ2

-2.99 = - ∏ξ / √1 –ξ2

8.97(1 – ξ2) = ξ2∏2

0.91 – 0.91 ξ2 = ξ2

0.91 = 1.91 ξ2

ξ2 = 0.69

(ii). ts = 4/ ξWn

4 = 4/ ξWn

Wn = 1/ ξ = 1/ 0.69

Wn = 1.45 rad/sec

2.2 Design consideration of Higher Order System :

The time domain analysis for second order system can be extended to third order system. Let transfer function for third order system be,

C(S)/R(S) = Wn2/(1 + ST)(S2 + 2ξWnS + Wn2)

Let 1/T = P

C(S)/R(S) = Wn2P/(S + P) (S2 + 2ξWnS + Wn2)

Taking Laplace inverse of above equation (for unit step response will be)

Cp(t) = 1 – K1e-pt + K2e-ξWntSin[ Wn√(1-ξ2)t - φ ]

Where K1 = Wn2 / Wn2 - 2ξWnp + p2

K2 = 1 / √1 – ξ2(1 - 2ξWn/p + Wn2/p2)

φ = tan-1√(1 – ξ2) / -ξ + tan-1Wn√1- ξ2/p – ξWn

Fromabove equations we can conclude that :

(1). The system has three poles S = -p and S = - ξWn+-√(1 – ξ2). So the position of these poles effect the system. As the pole S = -p moves towards the origin, the output Cp(t) is dominated by only this pole.

(2). As the pole S = -p moves away from the origin, the response Cp(t) is dominated by the complex poles at S = -ξWn+-(√1-ξ2)jWn.

The response of third order system is shown in figure ||(a).

The relative location of dominant poles of the third order system is shown below in fig ||(b).

2.3 Stability and Relative Stability

Concept of Stability:-

A stable system always gives bounded output for bounded input and the system is known as BIBO stable

A linear time invariant(LTI) system is stable if,

The system is BIBO stable

In absence of the input the output tends towards zero

For system;

1)          R(S) 

For R(s) = 1

C(S) =

R (t) = (t)          C (t) = t

R(t) (t)

T    Bounded                                    Unbounded

So, system unstable.

Eg.2             R(s)       

[R(s) = 1]

= 

C(t) = e-10tC(t)

        t=00       

e-10t                t=0

t

e-10tC(t)                   1

0                                                         t

BIBO stable fig. 2

Effect of location of Poles on stability

A)  Real and negative     G(s)=

+

C(t)

-a                            

Stable                                                      fig2(a)

B)    Poles on right half of S- plane

G(s) = 

Iw                                    C(t)

                                        a                  

0                                                               t

Unstable:-                                                            Unbounded Output

Fig 2(b)

(C)Poles at origin

G(s)=

Jw C(t)

k

Fig 2.(c)

Marginally stable

4) Complex poles on left half of S-plane

S1,S2=  -α ± iw

=

C(t) = k e- αt  Cos wt

Fig 2(d)

At  t - ∞        C(t) – 0   so, system stable

5) Complex poles on right half of S-plane

S1,S2 = α ± iw.

C(t) = k eαt COS wt.

Fig 2(e)

C(t) increases exponentially with t             ∞

So, unstable.

6) Poles on iw axis:

                  S1,S2  = ± iw.

C(t) = k coswt.

       0                    

Fig 2 (f)

The system is marginally stable than sustained oscillations.

Consider a system with characteristic equation

aoSm+ a1 Sm-1……………..am=0.

1)     All the coefficients of above equation should have same sign.

2)     There should be no missing term.

The above two condition are necessary but not sufficient condition for stability

Relative Stability:-

Routh stability criterion deals about absolute stability of any closed loop system. For relative stability we need to shift the S-plane and the apply the Routh criterion.

The above fig 10.shows the characteristic equation is modified by shifting the origin of S-plane to S1= -.

Reference Books :

1. Modern Control Engineering by K.Ogata, Pearson Education.

2.            Control Engineering by Kuo.

Unit 2

Time Response Analysis

2.1 Time Response of Second Order System:

Introduction

Time Response Analysis:-

For a second order control system, the total time response is analyzed by its transient as well as ready state response.

Any system which contains energy along elements (L1 c etc) if suffer any disturbance in the energy state effect at the input end as at this output  or at both the ends takes source time to change  in  thus  form one state other . This change its then is called as Transient times and the values of reverent and voltage during this period is Transient Response.

The above fig: a clearly shows steady star response is that part of response when the transient have died. If the steady star response of the output does not match with input then the system has steady state errors.

Consider the following open loop system

L-1[c(s)]= L-1[ G(s) R (S)]

C(t) = L-1[ G(S) Q (s)]

E(s) = R(s) –c(s) [ error signal]

C(S)= [ R(s)- c(Q)] G(s)

C(s) = R(s) G(s)- c(s) G(s)

C(s) [ 1+G(s)] = R(s) G(s)

C(s)/R(s) = G(s)/1+G(s)

C(s)/R(s) = G(s)/1+G(s)

E(s) = R(s)- c(s)

=R(s)-R(s) G(s)/1+G(s0

E(s) = R(s) [1/1+G(s)

If no error then E(s) = 0, f(t) = 0 Hence , r(t) = c(t)

2.1 Time response analysis of second order system

1)     ORDER – 2

OLTF  G(s) = k/s(1+TS)

CLTFC(s)/R(s) = k/k+s(1+Ts)

= k/s2+ s/T +k/T

Comparing above equation with standard 2nd order eqn

Standard 2nd order equation

CEs2+2 wn s+wn2=  0

S= -2wn±/2

=2wn±2wn/2

S= wn±wn

S= -wn±wn

S= -wn ±wn

S= -wn ±gwn

Standard eqn:-

T(s) = wn2/s2+2wns+wn2

Our eqn T(s) = K/T/s2+1/T s+ K/T

Wn2 = k/T

2 wn = 1/T

2k/T = 1/T

k/T = 1/2T

= 1/2T T/K

= 12KT

Graphically showing the position of loops s1,s2 for offered of

As the charaEqun (location of polls ) is depend ent only on r (wn constant for a given  system)

S1 = -wn + jwn 1-2

S2= - wn - jwn1-2

CASE 1: (=0)

S1=jwn,S2, = -jwn

Undamped

CASE 2:(0<<1)

S1= -wn + jwn0.75

=-wn/2 +jwn(0.26)

S2 = -wn/2 – jwn (0.86)

CASE:3        (-1)

S1= S2 = -wn = -wn

CASE4:- (-1)

S1 = -2wn +jwn-3

=-2wn – jwn (1.73)

S1  = -3.73wn

S2 = -2wn –jwn-3

= -2wn + jwn (1.73)

= -0.27 wn

Overdamped

All prac sys are 2 order to, if R(e) = (t) R(s) = 1 :. CLTF = 2nd order steqn and hence already the system is possible.

CLTF = e(s)/ R(s) = wn2/s2+2wns+wn2

Now calculating c(s) , c(t) for different values of input

1)     Impulse I/p

R(t) = (t)

R(s) = 1

C(s) = R(s) wn2/s2+2wns+wn2

C(s) = wn2/s2 +2wn+wn2

Under this i/p (R(t) = (t)) the output varies with different values of . So,

CONDITION 1:    ( =0)

C(s) =wn2/s2 +wn2

Os2 +wn2  =0

S=+-jwn

C(t) = wnsinwnt

Sinwnt

As there in no damping i.e. oscillations at t= 0 are some at t so, called UNDAMPED

CONDITION 2:  0<<1

c(s) = R(s) wn2/s22wn+wn2 R(t) = (t)

R(s) =1

C(s) = wn2/s2+2wns+wn2 

CE

S2+2wns +wn2 =0

S2, S1 = - wn ±jwn 1-2

C(t) = e-wnt   sin(wn)t

C(t)=e-wnt sin (wdt)

The oscillations are present but at t- infinity the Oscillations are 0 so, it is UNDERDENPLZ

CONDITION 3 :=1

C(s) = R(s) wn2/s2+2wns+wn2

C(s)=wn2/S2 +2wns+wn2

=wn2/(s +wn)2

CE    S= -Wn

Diagram

C(t)=w2n /()2

C(t)=

No damping obtained at so is called CRITICALLY DAMPED.

CONDITIONS  4 :->1

C(s) =  wn2/s22wnS+Wn2

S1,s2 = WN+-jwn

b)UNITStep Input:-

R(s) = 1/s

C(s)/R(s) = wn2/s2+2wns+wn2

C(s) = R(s) wn2 /s2 +2wns+wn2

C(s) = R(s) wn2/s2+2wns+wn2

C(s) = R(s) wn2/s2+2wns+wn2

C(s) = wn2s(s2+2wns+wn2)

C(t) = 1- ewnt/1-es2 sin  (wdt + ø)

Wd = wn1-2

Ø=

Where

Wd = Damping frequency of oscillations

Wn = natural frequency of  oscillations

wn = damping coefficient.

T= Time constant

Condition 1= 0

C(s) = wn2 /s(s2+wn2)

C(t) = 1- e° sin wdt +ø

C(t)= 1- sin( wn +90)

C(t) = 1+cos wnt

Constant

C(t) = 1+constant

Condition 2:-  0<<1

C(s) = /s2+wns +wn2

C(s) =1/s –    s+ wn/s2+ wns +wn2

=1/s –    s+wn/(s+wn)2+wd2- wn/(s+wn2) +wd2

Wd = wn1-2

Taking Laplace inverse of above equation

L --1 s+wn/(s+wn) +wd2= e-wnt  cos wdt

L-1 s+wn/(s+wn)2+wd2 = e-wnt sinWdt

C(t) = 1-e-wnt  [coswdt +/1-2sinwdt]

= 1-ewnt /1-2 sin [wdt + 1-2/]    t>=0

C(t) = 1-ewnt/1-2 sin(wdt+ø)

Ø = 1+2/

Fig 6. Transient Response of second order system

Specifications:

1)     Rise Time (tp):- The time taken by the output to reach the already status value for the first time is known as Rise time.

C(t) = 1-e-wnt/1-2 sin (wdt+ø)

Sin (wd +ø) = 0

Wdt +ø = n

tr  =n-ø/wd

For first time so ,n=1.

2)     Peak Time (tp)

The peak value attained by the output is called peak time .The time required by the output to reach this value is lp.

d(cct) /dt = 0 (maxima)

d(t)/dt = peak value

tp = n/wd   for n=1

tp = wd

(3) Peak Overshoot Value :-

Maximum deviation of output from steady state value is called peak overshoot value(Mp ).

( ltp ) = 1 = Mp

(Sin(Wat + φ )

(Sin( Wd∏/Wd + φ)

Mp = e-∏ξ / √1 –ξ2

Condition 3 ξ = 1

C( S ) = R( S ) Wn2 / S2 + 2ξWnS + Wn2

C( S ) = Wn2 / S(S2 + 2WnS + Wn2)        [ R(S) = 1/S ]

C( S ) = Wn2 / S( S2 + Wn2 )

C( t ) = 1 – e-Wnt + tWne-Wnt

The response is critically damped.

(4). Settling Time (ts) :

ts = 3 / ξWn  ( 5% )

ts = 4 / ξWn  ( 2% )

Q.1. The open loop transfer function of a system with unity feedback gain G( S ) = 20 / S2 + 5S + 4. Determine the ξ, Mp, tr, tp.

Soln : Finding closed loop transfer function,

C( S ) / R( S ) = G( S ) / 1 + G( S ) + H( S )

As it is unity feedback so, H(S) = 1

C(S)/R(S) = G(S)/1 + G(S)

= 20/S2 + 5S + 4/1 + 20/S2 + 5S + 4

C(S)/R(S) = 20/S2 + 5S + 24

Standard equation for second order system,

S2 + 2ξWnS + Wn2 = 0

We have,

S2 + 5S + 24 = 0

Wn2 = 24

Wn = 4.89 rad/sec

2ξWn = 5

(a). ξ = 5/2 x 4.89 = 0.511

(b). Mp% = e-∏ξ / √1 –ξ2 x 100

= e-∏ x 0.511 / √1 – (0.511)2 x 100

Mp% = 15.4%

(c). tr = ∏ - φ / Wd

φ = tan-1√1 – ξ2 / ξ

φ=  tan-1√1 – (0.511)2 / (0.511)

φ = 1.03 rad.

tr = ∏ - 1.03/Wd

Wd = Wn√1 – ξ2

= 4.89 √1 – (0.511)2 

Wd = 4.20 rad/sec

tr = ∏ - 1.03/4.20

tr = 502.34 msec

(d). tp = ∏/4.20 = 747.9 msec

Q.2. A second order system has Wn = 5 rad/sec and is ξ = 0.7 subjected to unit step input. Find (i) closed loop transfer function. (ii) Peak time (iii) Rise time (iv) Setteling time (v) Peak overshoot.

Soln: The closed loop transfer function is

C(S)/R(S) = Wn2 / S2 + 2ξWnS + Wn2

= (5)2 / S2 + 2 x 0.7 x S + (5)2

C(S)/R(S) = 25 / S2 + 7s + 25

(ii). tp = ∏ / Wd

Wd = Wn√1 - ξ2

= 5√1 – (0.7)2

= 3.571 sec

(iii). tr = ∏ - φ/Wd

φ= tan-1√1 – ξ2 / ξ = 0.795 rad

tr = ∏ - 0.795 / 3.571

tr = 0.657 sec

(iv). For 2% setteling time

ts = 4 / ξWn = 4 / 0.7 x 5

ts = 1.143 sec

(v). Mp = e-∏ξ / √1 –ξ2 x 100

Mp = 4.59%

Q.3. The open loop transfer function of a unity feedback control system is given by

G(S) = K/S(1 + ST)

Calculate the value by which k should be multiplied so that damping ratio is increased from 0.2 to 0.4?

Soln: C(S)/R(S) = G(S) / 1 + G(S)H(S)          H(S) = 1

C(S)/R(S) = K/S(1 + ST) / 1 + K/S(1 + ST)

C(S)/R(S) = K/S(1 + ST) + K

C(S)/R(S) = K/T / S2 + S/T + K/T

For second order system,

S2 + 2ξWnS + Wn2

2ξWn = 1/T

ξ = 1/2WnT

Wn2 = K/T

Wn =√K/T

ξ = 1 / 2√K/T T

ξ = 1 / 2 √KT

Forξ1 = 0.2, for ξ2 = 0.4

ξ1 = 1 / 2 √K1T

ξ2 = 1 / 2 √K2T

ξ1/ ξ2 = √K2/K1

K2/K1 = (0.2/0.4)2

K2/K1 = 1 / 4

K1 = 4K2

Q.4. Consider the transfer function C(S)/R(S) = Wn2 / S2 + 2ξWnS + Wn2

Find ξ, Wn so that the system responds to a step input with 5% overshoot and settling time of 4 sec?

Soln:

Mp = 5% = 0.05

Mp = e-∏ξ / √1 –ξ2

0.05 = e-∏ξ / √1 –ξ2

Cn 0.05 = - ∏ξ / √1 –ξ2

-2.99 = - ∏ξ / √1 –ξ2

8.97(1 – ξ2) = ξ2∏2

0.91 – 0.91 ξ2 = ξ2

0.91 = 1.91 ξ2

ξ2 = 0.69

(ii). ts = 4/ ξWn

4 = 4/ ξWn

Wn = 1/ ξ = 1/ 0.69

Wn = 1.45 rad/sec

2.2 Design consideration of Higher Order System :

The time domain analysis for second order system can be extended to third order system. Let transfer function for third order system be,

C(S)/R(S) = Wn2/(1 + ST)(S2 + 2ξWnS + Wn2)

Let 1/T = P

C(S)/R(S) = Wn2P/(S + P) (S2 + 2ξWnS + Wn2)

Taking Laplace inverse of above equation (for unit step response will be)

Cp(t) = 1 – K1e-pt + K2e-ξWntSin[ Wn√(1-ξ2)t - φ ]

Where K1 = Wn2 / Wn2 - 2ξWnp + p2

K2 = 1 / √1 – ξ2(1 - 2ξWn/p + Wn2/p2)

φ = tan-1√(1 – ξ2) / -ξ + tan-1Wn√1- ξ2/p – ξWn

Fromabove equations we can conclude that :

(1). The system has three poles S = -p and S = - ξWn+-√(1 – ξ2). So the position of these poles effect the system. As the pole S = -p moves towards the origin, the output Cp(t) is dominated by only this pole.

(2). As the pole S = -p moves away from the origin, the response Cp(t) is dominated by the complex poles at S = -ξWn+-(√1-ξ2)jWn.

The response of third order system is shown in figure ||(a).

The relative location of dominant poles of the third order system is shown below in fig ||(b).

2.3 Stability and Relative Stability

Concept of Stability:-

A stable system always gives bounded output for bounded input and the system is known as BIBO stable

A linear time invariant(LTI) system is stable if,

The system is BIBO stable

In absence of the input the output tends towards zero

For system;

1)          R(S) 

For R(s) = 1

C(S) =

R (t) = (t)          C (t) = t

R(t) (t)

T    Bounded                                    Unbounded

So, system unstable.

Eg.2             R(s)       

[R(s) = 1]

= 

C(t) = e-10tC(t)

        t=00       

e-10t                t=0

t

e-10tC(t)                   1

0                                                         t

BIBO stable fig. 2

Effect of location of Poles on stability

A)  Real and negative     G(s)=

+

C(t)

-a                            

Stable                                                      fig2(a)

B)    Poles on right half of S- plane

G(s) = 

Iw                                    C(t)

                                        a                  

0                                                               t

Unstable:-                                                            Unbounded Output

Fig 2(b)

(C)Poles at origin

G(s)=

Jw C(t)

k

Fig 2.(c)

Marginally stable

4) Complex poles on left half of S-plane

S1,S2=  -α ± iw

=

C(t) = k e- αt  Cos wt

Fig 2(d)

At  t - ∞        C(t) – 0   so, system stable

5) Complex poles on right half of S-plane

S1,S2 = α ± iw.

C(t) = k eαt COS wt.

Fig 2(e)

C(t) increases exponentially with t             ∞

So, unstable.

6) Poles on iw axis:

                  S1,S2  = ± iw.

C(t) = k coswt.

       0                    

Fig 2 (f)

The system is marginally stable than sustained oscillations.

Consider a system with characteristic equation

aoSm+ a1 Sm-1……………..am=0.

1)     All the coefficients of above equation should have same sign.

2)     There should be no missing term.

The above two condition are necessary but not sufficient condition for stability

Relative Stability:-

Routh stability criterion deals about absolute stability of any closed loop system. For relative stability we need to shift the S-plane and the apply the Routh criterion.

The above fig 10.shows the characteristic equation is modified by shifting the origin of S-plane to S1= -.

Reference Books :

1. Modern Control Engineering by K.Ogata, Pearson Education.

2.            Control Engineering by Kuo.