3.1 Standard test signals | unit 3 time response of feedback control systems

UNIT - 3

Time Response of 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 everywhere expect at the origin. fig below shown the representation of Impulse signal.

Fig 1 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

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 2 Unit Step Signal

r(t)= A t >=0

=0 t<0

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

Fig 3 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.

Fig 4 Parabolic Signal

r(t) At2 t>=0

=0 t<0

Then Laplace Transform is given as

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

3.2 Unit step response of First and second order systems

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 5 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 6 Output response for ramp input

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

1) 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 7 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.

Key takeaways

From both the Cases input unit step, ramp Unit step

3.3 Time response specifications

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 takeaways

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% )

3.4 Time response specifications of second order systems

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 8 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 9 Location of poles for <1

CASE:3 (=1)

S1= S2 = -wn = -wn

Fig 10 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 11 Location of poles for >1

Overdamped

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 12 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

S2+2wns +wn2 =0

S2, S1 = - wn ±jwn 1-2

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

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

Fig 13 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 14 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 15 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) = 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 16 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 17 Transient Response of second order system

Key takeaways

All practical systems are 2 order to, if R(e) = (t) R(s) = 1:. CLTF = 2nd order standard eqn and hence already the system is possible.

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 2nd 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

3.5 Steady – state errors and 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)

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

3.6 Introduction to PID Controllers (excluding design)

Effects of controller are viewed on time response and stability.

Fig 18 Controller with unity feedback system

Gc(S) = TF of the controller

G(S) = OLTF without controller

G’(S) = Gc(S).G(S)

= OLTF with controller

1. Proportional (P) 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 (I) controller –

The transfer function of this controller is

Gc(S) = Ki/S

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

Fig 19 Integral Controller

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

Fig 20 Pole Locations For given system

G’(S) = Ki/S(S + 5)

Fig 21 Pole Location after Integral controller

Fig(20), is more stable than (21) 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)

Fig 22 Root Locus Without controller

G’(S) = Ki/S(S + 5)(S + 10)

Fig 23 Root Locus Without controller

So, Fig 23 is less stable at root locus lies on the R.H.S,

3. Derivative (D) Controller -

Gc(S) = Kd(S)

They are used to improve the stability.

Fig 24 Derivative controller system

G’(S) = KdS/S2(S + 10)

Fig 25 Root Locus for system without controller

Fig 26 Root locus with controller

Fig 26 is stable i.e. more stable than the Fig 25.

Disadvantage –

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

4. Proportional Plus Integral [PI] controller -

Fig 27 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 (PD) Controller -

Fig 28 PD controller

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

6. PID controller -

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

Fig 29 PID Controller

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

Key takeaway

Que) The block diagram of a system using PI controllers is shown in the figure 30 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?

Fig 30 Block diagram with PI controller

(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?

Fig 31 Block diagram with PD 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)

References:

1. “Modern Control Engineering “, K. Ogata, Pearson Education Asia/

PHI, 4th Edition, 2002.

2. “Automatic Control Systems”, Benjamin C. Kuo, John Wiley India

Pvt. Ltd., 8th Edition, 2008.

3. “Feedback and Control System”, Joseph J Distefano III et al.,

Schaum’s Outlines, TMH, 2nd Edition 2007.

4. J. Nagarath and M.Gopal, “Control Systems Engineering”, New Age

International (P) Limited, Publishers, Fourth edition – 2005

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