Unit-3

Time domain Analysis of Control Systems

3.1 Time response of continuous data Systems

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 suffers 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 this form one state other. This change is then is called Transient times and the values of reverent and voltage during this period is Transient Response.

Fig:1 Transient Response

The above fig: clearly shows steady-state response is that part of the 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 the input signal.

3.2 Typical test signals for the time response of 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 has zero amplitude everywhere except at the origin. fig below showing the representation of the Impulse signal.

Fig 4 Unit Impulse Signal

The mathematical representations

A(t) = 0 for t ≠0

dt = A                e

Where A represents the energy or area of the Laplace Transform of the 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 changes from zero levels 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 a 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 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)

ess=    S[R(s)/1+G/(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)

3.3 Unit step response and time-domain specifications of a 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

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 a 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)

= 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 a 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.

3.4 Transient response of a prototype 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 + jwn0.75

=-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

CASE: 4(=-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 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

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

Fig 15 Location of Poles and Undamped oscillation

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

Wd=wn)

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

Fig 16 Location of Poles and 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 For Critically Damped

No damping obtained at so is called CRITICALLY DAMPED.

CONDITION 4 :->1

C(s) =  wn2/s22wnS+Wn2

S1,s2 = WN+-jwn

Fig 18 Location of poles for over damped oscillations

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 = 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 20. Transient Response of the 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 Rising time.

C(t) = 1-e-wnt/1-2 sin (wdt+ø)

Sin (wd +ø) = 0

Wdt +ø = n

tr  =n-ø/wd

for the 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:

The 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.

Sol: 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

3.5 Steady-State error

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)

2. 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/kvkv =   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)

Kp :

Kp =

 = 1/(s+a)(s+b)

Kp= 1/ab

ess=1/1+np = 1/1+(1/ab)   

o/p not locking the input

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

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)

Kp:-

Kp =   G(s)

= 1/s(s2+as+ab)

Kp = 1/0   =

ess = 1/1+kp = 0

o/p is tracking the input

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

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)

Kp:-

Kp = G(s)

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

Kp =

ess = 1/1+kp = 0

o/p tracking the i/p

kv:-

Kv =s  G(s)  

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

kv =

ess =1/ = 0

o/p tracking the o/p

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:

3.7 Static and dynamic error coefficients

There are two types of error coefficient.

1. Static error coefficient

2. Dynamic error coefficient

Static error coefficient

The ability of the system to reduce or eliminate the steady state error is static error coefficient. They are of 3 different types.

1. position error coefficient:- related to the output of the system. it is denoted by kp.

2. velocity error coefficient:- related to the rate of change of output of the system. it is denoted by kv.

3.acceleration error coefficient:- related to the rate of change of output. it denoted by ka.

Feature of static error are as follows

1.) Higher error coefficient increase the steady state performance of the system

2.) Cannot be used to calculate the error of unstable system

3.) Do not indicate the correct manner in which the error changes with time

4.) It is difficult to stabilized the system

Dynamic error coefficient

3.8 Error analysis for different types of systems

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)

2)     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/kvkv =   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

Reference:

1. I. J. Nagrath& M. Gopal, “Control System Engineering”, New Age International Publishers

2. A. Ambikapathy, Control Systems, Khanna Publishing House, Delhi.

2. Joseph J. Distefano III, Allen R. Stubberud, Ivan J. Williams, “Control Systems” Schaums Outlines Series, 3rdEdition, Tata McGraw Hill, Special Indian Edition 2010.

3. William A. Wolovich, “Automatic Control Systems”, Oxford University Press, 2010.