Unit 2

Time domain analysis

2.1 Transient Response and Steady State Response

For a second order control system, the total time response is analysed 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.

Fig:1 Transient Response

The above fig: a clearly shows steady state response is that part of response when the transient has 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 –

Fig 2 Open loop system block diagram

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

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

Fig 3 Closed loop system block diagram

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)

Key takeaway

The error signal is the difference between the unity feedback signal and input signal.

2.2 Standard Test Inputs of Time Domain Analysis

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 has zero amplitude everywhere except at the origin. Fig below shown the representation of Impulse signal.

Fig 4 Unit 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.

Fig 5 Unit Step Signal

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

Fig 6 Ramp Signal

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.

Fig 7 Parabolic Signal

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)

Key takeaway

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)

2.3 Order and Type of a System

The order of the system is defined as the maximum power of S in the denominator. The type of system is defined as the number of poles at the origin.

For example, G(s)= K/S(S+1)

It is order 2 and type 1 system

G(S)= K(S+1)/S2(S+2)

It is order 3 and type 2 system

Key takeaway

The type and order of system is not dependent on the number of zeros.

2.4 Transient Analysis of First and Second Order Systems

2.4.1 Transient Analysis of First Order System:

OLTF G(s) = 1/TS

CLTF 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 Step:

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]

Fig 8 Output response for Unit Step input

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

Fig 9 Output response for ramp input

Key takeaway

1)     From both the Cases input unit step, ramp Unit step

Unit Step      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

2)     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

Fig 10 Location of poles

G(s) = 1/Ts

ATf = 1/1+TS

1st order system

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

2.4.2 Transient Analysis of Second Order System:

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 characteristic Equn (location of poles) is dependent only on (wn constant for a given system)

S1 = -wn + jwn 1-2

S2= - wn - jwn1-2

CASE 1: (=0)

S1=jwn, S2, = -jwn

Fig 11 Location of poles for =0

Undamped

CASE 2:(0<<1)

S1= -wn + jwn 0.75

=-wn/2 +jwn(0.26)

S2 = -wn/2 – jwn (0.86)

Fig 12 Location of poles for <1

CASE:3        (=1)

S1= S2 = -wn = -wn

Fig 13 Location of poles for =1

CASE4: (=-2)

S1 = -2wn +jwn -3

=-2wn – jwn (1.73)

S1  = -3.73wn

S2 = -2wn –jwn -3

= -2wn + jwn (1.73)

= -0.27 wn

Fig 14 Location of poles for >1

Overdamped

Key takeaway

All practical systems are 2 order to, if R(e) = (t) R(s) = 1 :. CLTF = 2nd order st 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

Fig 15 Undamped oscillations

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)

Fig 16 Underdamped oscillations

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

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)=

Fig 17 Critically damped oscillations

No damping obtained at so is called CRITICALLY DAMPED.

CONDITIONS 4:>1

C(s) = wn2/s22wnS+Wn2

S1, s2 = WN+-jwn

Fig 18 location of poles for over damped oscillations

b)    UNIT Step 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

Fig 19 C(t) = 1+cos wnt

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 20. 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% )

Key takeaway

Rise Time tr = -ø/wd

Peak Time tp = n/wd   for n=1

tp = wd

Peak Overshoot Value

Mp% = e-∏ξ / √1 –ξ2

Settling Time

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) Settling 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

2.5 Time Domain Specifications of Second Order Under Damped System from its Step Response

Please refer section 2.4 it is already covered.

2.6 Steady State Error and Static Error Constants

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

1)     R(s) = unit step

2)     R (s) = Ramp

3) R(s) = parabolic

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

Key takeaway:

ess	Type 0	Type 1	Type 2
Unit step	≠0	0	0
Ramp		≠0	0
Parabolic			≠0

References:

1. Benjamin C. Kuo, “Automatic control systems”, Prentice Hall of India, 7th Edition.

2. M. Gopal, “Control System – Principles and Design”, Tata McGraw Hill, 4th Edition.

3. N. J. Nagrath and M. Gopal, “Control System Engineering”, New Age International Publishers, 5th Edition.

4. K. Ogata, “Modern Control Engineering”, Prentice Hall India Learning Private Limited; 5th Edition.