Unit 2

Time-Domain Analysis

2.1 Time domain performance specifications

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)

2.2 Transient response of first and second order system

Now calculating errors of when order is varied

1)     Order 1 :

OLIF  G(s) = 1/TS

CLIF c(s)/R(s) = 1/1+TS

CE :  HTS = 0

S= -1/T

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

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

So, Calculating value of c(s) , c(t) for different input

a)     Unit slip:-

R(t) = u(t)

R(s) = 1/s

C(s) = 1/s(1+ST)

1/s(1+ST) = A/s + o/1+TS

1/s(1+ST) = A/S + 0/1+TS

1= A(1+ST) + BS

AT +B =0

A = 1

B= -T

C(s) = 1/s + (-T)/1+TS

= 1+ (-T)/1-(-T)s  = 1-I/n/1+ass+1/T  =e-t/s

C(t) = [ 1-e-t/T]

Ramp:-

R(t) = t

R(s) = 1/s2

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

C(s) = 1/s2 1/(1+TS) + c

1/s2 (1+TS) = A/(s) + B/s2 + c(1+TS)

1        = A(s+s2. T) + B(1+TS) +cs2

TA +c = 0

A+TB =0

B =1

A = -T

C = T2

C(s) = -T/s + 1/s2 +T2/1+ST

=(-T) u(t) + t(u) + T e-t/T u(t)

= [ -T + t +Te-t/T  u(t)

= [ -T+ t+Te-t/T] u(t)

C(s) = t- T+ Te-t/T

ess =   R(t) – c(t)

= [ t- t+ 7-Te –t/T]

= T( 1-e-t/T]

ess = T  The less the value  of T the less in the errors.

Inference:-

From both  the Cases input unit step, ramp Unit slip

r(t) = 1      Ramp

c(t) = 1-e-t/T      r(t) = t

ess = e-t/T (r/t)-(t)    c(t) = t- T+te-t/T

ess= e-t/T (r/t) (t)    ess = T- Te-t/T

For :-    for :-

ess =0      ess  = T

In both the vases (values of T) must be  as small as possible ( so, that e-t/T) must be as small as possible which gives us

G(s) = 1/Ts

ATf = 1/1+TS

1st order system

Poles must be situated as for  as possible from origin i.e. deeper and deeper into the left half of s-place . Thus we get less errors.

Time response analysis of second order system

2)     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 chara Equn (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 + jwn 0.75

=-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 practical sys are 2 order to, if R(e) = (t) R(s) = 1 :. CLTF = 2nd order system eqn 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) = wn sinwnt

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 = wn 1-2

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 +/1-2 sinwdt]

= 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% settling 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

2.3   Steady state errors and Static error constants in unity feedback control systems

The Impulse signal, Ramp signal, unit step and parabolic signals are used as the standard test signals. All these signals are explained below.

Impulse Signal:-

This signal have zero amplitude every where expect at the origin. Fig below shown the representation  of Impulse signal.

The mathematical representations

A(t) = 0 for t ≠0

dt = A                e

Where A represent energy or area of the Laplace Transform of Impulse signal is

L [A (t) ]= A

UNIT IMPULSE SIGNAL :-

If A = 1

(t) = 0        for t ≠0

L [(t)] = 1

The transfer function of a linear time invariant

System is the Laplace transform of the impulse response of the system. If a unit impulse signal is applied to system  then Laplace  transform of the output c(s) is the transfer function G(s)

As we know  G(s)  = c(s)/R(S)

r(t) = (t)

R(s) = L [(t)] = 1

:. G(S) = C(s)

(b) Step signal:-

Step signal of size  A is a signal that change from zero level to A in zero time and stays there forever.

r(t)= A     t >=0

=0      t<0

L[r(t)]= R(s) = A/s

UNIT STEP SIGNAL :- If the magnitude of the slip  signal  is I then it is called unit step signal.

u(t) = 1  

t>=0

t<0

L[u(t)] = 1/s

(c) Ramp Signal:-

           The vamp signal increase linearly with time from initial value of zero at t= 0 as shown in fig is below

r(t) = At      t>=0

=0   t<0

A is the slope of the line The Laplace transform of ramp signal is

L[r(t)] = R(s) = A/s2

(d) Parabolic Signal:-

The instantaneous value of a parabolic signal varies as square of the time from an initial value of zero t=0. The signal representation in fig 14 below.

r(t) At2 t>=0

=0   t<0

Then Laplace Transform is given as

R(s) = L[At2] = 2A/s3

If no error then E(t) =0 , :. R(t) = c(t), output is tracking the input.

Steady state Errors signal (ess) :- ( t )

ess = t e(t)

Using final values the theorem

= ess = S.E (s)

Steady state error and error constant

So, for difficult input, ess will be different Input still be

1)     R(s) = unit step   (3) R(s) = parabolic

2)     R (s) = Ramp

1)     R(s)= Unit step

R(t) = u(t)

R(s) = 1/s

ess = s[ R(s)/1+G(s)]

= s[ys/1+G(s)]

ess= 1+1/1+G(s)

=1/1+lt G(s) s- 0

(Position error coefficient)

1)     R(s) = Ramp

R(t) = t

R(s) = 1/s2

ess= s[ R(s)/1+G(s)]

= s[ys2/1+G(s)]

= 1/s(1+G(S)

=  1/s+SG(s)

ess  = l/s-0  SG (s)

ess = t L/SG (s)

velocity errors coefficient

(3)Parabolic

R(t) = t2

R(s) =1/s3

Ess = s[ R(s)/1+G(s)]

= s(1/s3)/1+G(s)

=     1/s2+s2G(s)

Ess = 1/s2G(s)

Acceleration error coefficient

I/p                         error coefficient

(1)Unit step                                         ess= 1/1+kv        kv = G(S)

(2)Ramp      ess= 1/kv   kv =   sG(s)

(3)Parabolic    ess = 1/ka      ka = s2 G(s)

Now finding errors in  type 0,1,2 system

1)     Type  0:-

G(s) = 1/(s+a) (s+b)

2)     Kp :

Kp =

= 1/(s+a)(s+b)

Kp= 1/ab

Ess=1/1+np = 1/1+(1/ab)    

o/p not locking the input

(b) kv:-

Kv =   s G(s)

= s/(s+a) (s+b)

= s/(s+a) (s+b)

Kv = 0

ess = 1/kv =

o/p not locating the input

(3)ka:-

Ka = s2 G(s)

= s2 /(s+a) (s+b)

Ka =0

ess = 1/ka =

o/p not locking the input

(2)Type –I :-

G(s) = 1/s(s+a) (s+b)

(a)Kp:-

Kp =   G(s)

= 1/s(s2+as+ab)

Kp = 1/0   =

ess = 1/1+kp = 0

o/p is tracking the input

(b)kv:-

Kv = S G(s)

=1/(s+a) (s+b)

=1/ab

Kv = 1/ab

ess =1/ka = ab  

o/p is not tracking the input

3)ka :

Ka=  s2G(s)

= s2 /s(s+a)(s+b)

=   s/(s+a)(s+b)

Ka= 0

Ess = 1/ka  =

O/p is not tracking the input .

3) Type  2 :-

G(s) = 1+/(s2(s+a) (s+b)

a)     Kp:-

Kp = G(s)

= 1/s2 (s+a) (s+b)

Kp =

ess = 1/1+kp = 0

o/p tracking the i/p

b)    kv:-

Kv =s  G(s)  

= 1/s(s+a) (1+b)

kv =

ess =1/ = 0

o/p tracking the o/p

(b) ka:-

Ka= s2G(s)

=1/(s+a) (s+b)

Ka = 1/ab

ess = 1/(yab) ≠0

o/p not locking the i/p

ess                Type 0          Type 1        Type 2

Unit step     ≠0             0       0

Ramp                      ≠0        0

Parabolic       ≠0

2.4   Response with P, PI and PID controllers

Effects of controller are viewed on time response and stability.

Gc(S) = TF of the controller

G(S) = OLTF without controller

G’(S) = Gc(S).G(S)

= OLTF with controller

(1). Proportional Controller :-

Proportional means scalar Multiplier.

Gc(S) = Kp Stability can be controlled

Q. G(S) = 1/S(S + 8)

CLTF = 1/S2 + 8S + 1

w2n = 1

wn = 1

2ξWn = 8

ξ = 4

ξ>/1 so, overdamped.

Now introducing Gc(S) = K

G’(S) = Kp/S(S + 8)

CLTF = Kp/S2 + 8S + Kp

wn = √Kp

2 ξwn = 8

ξ = 4/√Kp

If,

Kp = 16; ξ = 1; critically damped

Kp> 16; ξ< 1; undamped

Kp< 16; ξ> 1; overdamped

(2). Integral controller:-

The transfer function of this controller is

Gc(S) = Ki/S

Integral controller are used to improve the steady state response or reduce the steady state error.

G’(S) = Gc(S).G(S)

G(S) = 1/S + 5, Gc(S) = Ki/S

G’(S) = Ki/S(S + 5). After applying the Gc(S) the type of system is increasing and hence, the steady state error is decreasing (Refer Time Response).

Disadvantage :-

By using integral controller the stability of closed loop system decreases.

G(S) = 1/S + 5

G’(S) = Ki/S(S + 5)

Fig(1), is more stable than (2) as more the away the pole from origin (imaginary axis) more is the stability.

Q. G(S) = 1/(S + 5)(S + 10)

G’(S) = Kp/S(S + 5)(S + 10)

G(S) = 1/(S + 5)(S + 10)

G’(S) = Ki/S(S + 5)(S + 10)

So, (b), is less stable at root locus lies on the R.H.S,

(3). Derivative Controller :-

Gc(S) = Kd(S)

They are used to improve the stability.

G’(S) = KdS/S2(S + 10)

(b) is stable i.e. more stable than the (a).

Disadvantage :-

It increases the steady state i.e. o/p will not track the input at steady state.

(4). Proportional Plus Integral [PI] controller :-

Gc(S) = KpS + Ki/S

The steady state error will decrease and the stability will depend on Kp i.e. if Kp is increased/decreased than according to it stability will change(Kpα stability).

Used to reduce ess without much affecting the stability.

(5). Proportional Plus Derivative Controller :-

Used to improve the stability without affecting much the steady state error.

(6). PID controller :-

Gc(S) = Kp + Ki/S + KdS

It improves the stability. It decreases the steady state error and both are proportional to Kp.

Q.The block diagram of a system using PI controllers is shown in the figure. Calculate :

(a). The steady state without and with controller for unit step input?

(b). Determine TF of newly constructed sys. With controller so, that a CL Poles is located at -5?

(a). Without :

C(S)/R(S) = 0.2/(S + 1)

C(S) = R(S) 0.2/(S + 1) + 0.2

For unit step

Ess = 1/1 + Kp

Kp = lt G(S)

S 0

= lt   0.2/(S + 1)

S 0

Kp = 0.2

Ess = 1/1.2 = 5/6 = 0.8

(b). With controller :

Gc(S) = Kp + Ki/S

G’(S) = Gc(S).G(S)

= ( Kp + Ki/S )0.2/(S + 1)

= (KpS + Ki)0.2/S(S + 1)

Ess = 1/1 + Kp = 0

So, the value of ess is decreased.

(b). Given

Kpi/Kp = 0.1

G’(S) = (Kp + Ki/S)(0.2/S + 1)

= (KpS + Ki)0.2/S(S + 1)

As a pole is to be added so, we have to examine the CE,

1 + G’(S) = 0

1 + (Kp + Ki)0.2/S(S + 1) = 0

S2 + S + 0.2KpS + 0.2Ki = 0

S2 + (0.2Kp + 1)S + 0.2Ki = 0

Given,

Kpi = 0.1Ki

Kp = 10Ki

S2 + (2Ki + 1)S + 0.2Ki = 0

Pole at S = -5

25 + (2Ki + 1)(-5) + 0.2Ki = 0

-10Ki – 5 + 25 + 0.2Ki = 0

-9.8Ki = -20

Ki = 2.05

Kp = 10Ki

= 20.5

Now,

G’(S) = (KpS + Ki)(0.2)/S(S + 1)

= (20.5S + 2.05)(0.2)/S(S + 1)

G’(S) = 4.1S + 0.41/S(S + 1)

Q.The block diagram of a system using Pd controller is shown, the PD is used to increase ξ to 0.8. Determine the T.F of controller?

(1). Kp = 1

Without controller:-

C(S)/R(S) = 16/S2 + 1.6S + 16

wn = 4

2 ξwn = 16

ξ = 1.6/2 x 4 = 0.2

(b). With derivative :-

ξS = 0.2 to 0.8

Undamped to critically damped,

G’(S) = (1 + KdS)(16)/(S2 + 1.6S)

CE:

S2 + 1.6S + 16(1 + KdS) = 0

S2 + (1.6 + Kd)S + 16 = 0

2 ξwn = 1.6 + Kd1.6

wn = 4

ξS = 0.8

2 x 4 x 0.8 = 1.6 + Kd1.6

6.4 – 1.6 = Kd1.6

4.8/16 = Kd

Kd = 0.3

TF = (1 + 0.3S)16/S(S + 16)

Ss2.5 Limitations of time domain analysis

The analysis in time domain is very difficult. The stability analysis in time domain is very tiresome. As the order of the system increases the calculation of stability of  system becomes very difficult.