Unit - 2

Time domain analysis

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.

A1)

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

Q2) 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.

A2)

(i) 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%

Q3) 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?

A3)

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

Q4) 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?

A4)

(i) 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?

A5)

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?

A6)

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?

A7)

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?

A8)

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.

tr = -ø/wd

T=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)

A9)

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?

A10)

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

Kp = G(s)  

(Position error coefficient)

ess = 1/1+kp