Unit-3

Time domain analysis of Control System

Q1). 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.

Sol: 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?

Sol: 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?

Sol:

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

Q5) For the CLTF G(s) = . Find Kp, Kv and Ka?

Sol: Kp = G(s)  

             =  

             = 1

Kv = SG(s)      

     = S

      = 0

Ka =   s2 G(s)   

     =   s2

     = 0

Q6) For a unity feedback system G(s) = an input t3u(t) is applied. Find the steady-state error?

Sol: ess=

r(t) = t3u(t)

R(s) = 6/s4

H(s) =1

ess=

     =

     =

     = 5/3

Q7) For the OLTF with unity feedback is G(s)= . Determine the damping ratio, maximum overshoot, rise time?

Sol: The CLTF will be T(s) =

                                       T(s) =

For second order system,

S2 + 2ξWnS + Wn2

wn = = 5.1

2ξWn = 5

i)                   Damping Ratio     = 0.49

ii)                 Maximum overshoot Mp = e-∏ξ / √1 –ξ2 x 100

                                             = e-∏x0.49 / √1 –(0.49)2

                                             = 17.1%

iii)              Rise Time tr = ∏ - φ/Wd

             Wd = Wn√1 - ξ2

               = 5.1 √1 – (0.49)2

         = 4.45 sec

         φ= tan-1√1 – ξ2 / ξ = 1.059 rad

        tr = ∏ - φ/Wd

    = ∏ - 1.059/4.45

    =468.53msec

Q8) Define and derive the time domain specifications for a second-order system?

Sol:

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

 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% )

Q9) Determine the type and order of the system G(s)= K/S(S+1)

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

Sol: 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

Q10) Derive position error coefficient?

Sol: 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)