Transient Response Analysis

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TRANSIENT RESPONSE ANALYSIS

Test signals:

• Impulse • Step • Ramp • Sin and/or cos

Transient Response: for t between 0 and T

Steady-state Response: for t: \

System Characteristics:

• Stability Æ transient • Relative stability Æ transient • Steady-state error Æ steady-state B2

First order systems

R(s) E(s) C (s) 1 1 C(s) = + + Ts R (s) Ts 1 -

Unit step response:

1 1 1 T C(s) = ⋅ = − Ts +1 s s sT +1

− t T c(t) = 1− e t

− t T e(t) = r(t) − c(t) = e e( c(T) = 1 – e-1 = 0.632

− t T dc ( t ) 1 e 1 = = = = dt t 0 T t 0 T B3

Unit ramp response

1 1 1 T T 2 C(s) = ⋅ = − + Ts+1 s2 s2 s Ts+1

− t c(t) = t −T +Te T t ≥ 0

 − t  e t = r t − c t = T  − e T  ( ) ( ) ( ) 1  t _

e() =T

Unit-ramp response of the system B4

Impulse response:

R(s) = 1 r(t) = /(t) 1 C ( s ) = sT + 1

e − t / T c ( t ) = _ T t

Unit-impulse response of the system

Input Output

− t = − + T Ramp r(t) = t t c(t) t T Te t _

− t T Step r(t) = 1 t c (t ) = 1 − e t _

/ e − t / T Impulse r(t) = (t) c (t ) = t _ T B5

Observation:

Response to the derivative of an input equals to derivative of the response to the original signal.

Y(s) = G(s) U(s) U(s): input

U1(s) = s U(s) Y1(s) = s Y(s) Y(s): output

G(s) U1(s) = G(s) s U(s) = s Y(s) = Y1(s)

How can we recognize if a system is 1st order ?

r(t) System c(t)

r(t) = step

Plot log |c(t) – c( _ If the plot is linear, then the system is 1st order

Explanation: − t = − T c(t) 1 e c(\

log |c(t) – c( _ ORJ _H-t/T |= t T B6

Second Order Systems

Block Diagram

F + K J -

R(s) + K Position signal C (s) s ()Js + F -

Transfer function:

C(s) = K R(s) Js2 + Fs+ K K = J  2  2  F  F  K F  F  K s + +   − s + −   −      2J  2J  J  2J  2J  J  B7

Substitute in the transfer function: K ² - J n F 2 J = n ζ = F 2 JK

damping ratio n: undamped natural frequency stability ratio to obtain

ω 2 C(s) = n 2 + ζω + ω 2 R(s) s 2 n s n

• Underdamped FDVH

F² - 4 J K < 0 two complex conjugate poles

• Critically damped FDVH

F² - 4 J K = 0 two equal real poles

• Overdamped FDVH !

F² - 4 J K > 0 two real poles B8

Under damped case (0 < < 1):

ω 2 C(s) = n ()()+ ζω + ω + ζω − ω R(s) s n j d s n j d

jd n ω − ζ 2 n 1 Re n σ ζ = = cos β ω n

ω = ω −ζ 2 d n 1

ωn : undamped natural frequency ωd : damped natural frequency ζ : damping ratio B9

Unit step response:

R(s) = 1/s

1 s +ζω ζω C(s) = − n − n ()+ζω 2 +ω 2 ()+ζω 2 +ω 2 s s n d s n d   −ζω ζ = − nt  ω + ω  ≥ c(t) 1 e cos dt sin dt t 0  1−ζ 2  o r 1 − ζω t c(t ) = 1 − e n sin ()ω β t + θ t ≥ 0 β n β β = − ζ 2 θ = tan − 1 1 ζ   −ζω t ζ = = n  ω + ω  ≥ e(t) r(t) - c(t) e cos dt sin dt  t 0  1− ζ 2 

2.0

1.0

0 1 2 3 4 5 6 7 8 9 10 11 12 Unit step response curves of a second order system B10

8QGDPSHG FDVH

Unit step response:

= − ω ≥ c(t) 1 cos nt t 0

&ULWLFDOO\ GDPSHG FDVH

Unit step Response: R(s) = 1/s

ω 2 ω 2 C(s) = n = n 2 + ζω + ω 2 + ω 2 R(s) s 2 n n (s n ) 1 C(s) = ()+ ω 2 s s n

−ω ()= − nt + ω ≥ c t 1 e (1 nt) t 0 B11

2YHUGDPSHG FDVH !

Unit step Response: R(s) = 1/s

ω 2 1 C(s) = n ⋅ ()+ζω +ω ζ 2 − ()+ζω −ω ζ 2 − s s n n 1 s n n 1

−s t −s t ω  e 1 e 2  c(t) =1+ n  −  t ≥ 0 2   2 ζ −1  s1 s2 

= (ζ + ζ 2 − )ω with s 1 1 n = (ζ − ζ 2 − )ω s2 1 n

if |s2| << |s1|, the transfer function can be approximated by s C (s) = 2 + R (s) s s 2 and for R(s) = 1/s

− c(t) = 1− e s2t t ≥ 0 with = (ζ − ζ 2 − )ω s2 1 n B12

Unit step response curves of a critically damped system. B13

Transient Response Specifications

Unit step response of a 2nd order underdamped system:

Allowable tolerance

0.05 or

0.02 ~

0~5

td delay time: time to reach 50% of c(\ Ior the first time. tr rise time : time to rise from 0 to 100% of c( tp peak time : time required to reach the first peak. − ∞ c(t p ) c( ) Mp maximum overshoot : ⋅100% c(∞) ts settling time : time to reach and stay within a 2% (or 5%) tolerance of the final value c(

0.4 < ζ < 0.8

Gives a good step response for an underdamped system B14

Rise time tr time from 0 to 100% of c(

−ζω ζ = ⇒ − dtr ω + ω = c(tr ) 1 1 e (cos d tr sin d tr ) 1 1−ζ 2 ζ ω + ω = cos d tr sin d tr 0 1−ζ 2 1−ζ 2 ω tanω t = − = − d d r ζ σ

1 −  ω  t = tan 1  d  r ω σ d  

Peak time tp: time to reach the first peak of c(t)

dc t ω −ζω t ( ) = ⇒ ω n n p = t = t 0 (sin dt p ) e 0 dt p 1− ζ 2

ω = sin d t p 0

π t = p ω d B15

Maximum overshoot Mp: π t = t = p ω d

−ζω ()π ω ζ = = − n / d π + π Mp c(tp) 1 e (cos sin ) 1−ζ 2

ζω π −ζπ −σπ − n ω 1−ζ 2 ω =e d = e = e d

Settling time ts:

−ζω t 2 n  −ζ  = − e ω + −1 1  c(t) 1 sin dt tan 2  ζ  1−ζ   −ζω n t = ± e approximate ts using envelope curves: env (t) 1 1 − ζ 2

Pair of envelope curves for the unit-step response curve 4 4 3 3 t = = t = = 2% band: s σ ζω 5% band s σ ζω n n B16

Settling time ts versus ζ curves {T = 1/(ζωn) }

T 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 B17

Impulse response of second-order systems

ω 2 C(s) = n 2 + ζω +ω 2 R(s) = 1 s 2 n s n

underdamped case (0< ζ < 1):

ω −ζω = n nt ω −ζ 2 ≥ c(t) e sin n 1 t t 0 1−ζ 2

the first peak occurs at t = t0

1 − ζ 2 tan − 1 ζ = t 0 ω − ζ 2 n 1 and the maximum peak is

 ζ −ζ 2  =ω − −1 1  c(t0) n exp tan   1−ζ 2 ζ  B18 critically damped case ( ζ = 1):

2 −ω = ω nt ≥ c(t) n te t 0 overdamped case ( ζ >1):

ω ω − s t − s t = n 1 − n 2 ≥ c(t) e e t 0 2 ζ 2 − 1 2 ζ 2 − 1 where

= (ζ − ζ 2 − )ω s1 1 n = (ζ + ζ 2 − )ω s2 1 n

Unit-impulse response for 2nd order systems B19

Remark: Impulse Response = d/dt (Step Response)

Unit-impulse response

Relationship between tp, Mp and the unit-impulse response curve of a system

Unit ramp response of a second order system

ω 2 1 C(s) = n ⋅ 2 2 + ζω + ω 2 2 R(s) = 1/ s s 2 n n s for an underdamped system (0 < ζ < 1)

 2  2ζ −ζω t 2ζ 2ζ −1 = − + n  ω + ω  ≥ c(t) t e cos d t sin d t t 0 ω  ω ω − ζ 2  n  n n 1  and the error: e(t) = r(t) – c(t) = t – c(t) at steady-state: 2ζ e()∞ = e ()t = lim ω t → ∞ n B20

Examples: a. Proportional Control 1 C(s) R(s) E(s) s()Js + F + K

ω 2 C(s) = n 2 + ζω + ω 2 R(s) s 2 n n with K ² J n F J = n ζ = F 2 JK Choose K to obtain ‘good’ performance for the closed-loop system For good transient response: 0.4 < ζ <0.8 Æ acceptable overshoot

n sufficiently large Æ good settling time For small stead- state error in ramp response:

2ζ 2F ζ F e()∞ = e ()t = = ⋅ = Æ lim ω large K t →∞ n 2 Kζ K K

Large K reduces e( ∞ ) but also leads to small ζ and large Mp Æcompromise necessary B21 b. Proportional plus derivative control:

C(s) R(s) E(s) Kp+K/s 1 + s(Js + F ) -

() K + K s C s = p d () 2 +()+ + R s Js F Kd s K p with F + K K ς = d ω = p n 2 K p J J

The error for a ramp response is:

s2 J + sF E()s = ⋅ R()s 2 + ()+ + s J s F K d K p and at steady-state: F e()∞ = lim sE ()s = s→0 K p K using z = p K d () ω 2 + C s = n ⋅ s z () 2 + ζω + ω 2 R s z s 2 n s n

Choose Kp, Kd to obtain ‘good’ performance of the closed-loop system

For small steady-state error in ramp response Æ Kp large

For good transient response Æ Kd so that 0.4 < ζ <0.8 B22 c. Servo mechanism with velocity feedback

s Θ(s) Θ(s) R(s) + + K 1 Js + F s - -

Transfer function Θ (s) = K 2 + ()+ + R(s) Js F KK h s K where F + KK ς = h 2 KJ ω = K n (not affected by velocity feedback) J F e(∞ ) = K for a ramp

Choose K, Kh to obtain ‘good’ performance for the closed-loop system For small steady-state error in ramp response Æ K large

For good transient response Æ Kh so that 0.4 < ζ <0.8

Remark: The damping ratio ζ can be increased without affecting the natural frequency ωn in this case. B23

Effect of a zero in the step response of a 2nd order system

C ()s ω 2 s + z = n ⋅ ζ = 0.5 () 2 + ζω + ω 2 R s z s 2 n s n

Unit-step response curves of 2nd order systems B24 Unit step Response of 3rd order systems

C()s ω 2 p = n ζ () ()2 + ζω +ω 2 ()+ 0 < < 1 R(s)=1/s R s s 2 ns n s p

− −ξω e pt e nt c()t =1− − • βζ 2()β −2 +1 βζ 2()β −2 +1  βζ[]ζ 2()β − +  βζ 2()β − −ζ 2ω + 2 1 ( −ζ 2ω )  2 cos 1 nt sin 1 nt   1−ζ 2  β = p where ζω n

Unit-step response curves of the third-order system, = 0.5

The effect of the pole at s = -p is: • Reducing the maximum overshoot • Increasing settling time B25 Transient response of higher-order systems

C (s) b s m + ... + b s + b K ()()s + z ... s + z = 0 m −1 m = 1 m n > m n + + + ()()+ + R(s) s ... d n −1s an s p1 ... s pn

Unit step response

m ()+ K∑ s zi = i=1 ⋅ 1 C(s) q r ()+ ()2 + ζ ω +ω 2 s ∑∑s p j s 2 k k s k j==11k

0 < ζk <1 k=1,…,r and q + 2r = n

a q a r b ()s+ζ ω +c ω 1−ζ 2 C(s) = +∑ j +∑ k k k k k k + 2 + ζ ω +ω 2 s j=1 s pj k=1 s 2 k k k

q r − p t −ζ ω t  2  c t = a + a e j + b e n k ω −ζ t  ( ) ∑ j ∑ k cos k 1 k  j =1 k =1 r −ζ ω t  2  + c e k k ω −ζ t  t ≥ ∑ k sin k 1 k  0 k =1

Dominant poles: the poles closest to the imaginary axis. B26 STABILITY ANALYSIS

m m − i ∑ b i s = B (s ) = i = 0 G (s ) n A(s ) n − i ∑ a i s i = 0 Conditions for Stability: A. Necessary condition for stability:

All coefficients of A(s) have the same sign.

B. Necessary and sufficient condition for stability: A(s) ≠ 0 for Re[s] _ or, equivalently All poles of G(s) in the left-half-plane (LHP)

Relative stability: The system is stable and further, all the poles of the system are located in a sub-area of the left-half-plane (LHP).

RRee B27 Necessary condition for stability:

()= n + n −1 + + + A s a0 s a1s ... an −1s a n = ()()()+ + + a0 s p1 s p2 ... s pn = n + ()+ + n −1 a0s a0 p1 p2 ... pn s + ()+ + n − 2 a0 p1 p2 ... pn −1 pn s + () a0 p1 p2... pn

-p1 to -pn are the poles of the system.

If the system is stable Æ all poles have negative real parts Æ the coefficients of a stable polynomial have the same sign.

Examples:

3 2 A()s = s + s + s +1 can be stable or unstable

3 2 A()s =s −s +s+1 is unstable

Stability testing Test whether all poles of G(s) (roots of A(s)) have negative real parts.

Find all roots of A(s) Æ too many computations

Easier Stability test? B28 Routh-Hurwitz Stability Test

n n-1 A(s) = α0s + α1s +… + αn-1 s +αn n s α0 α2 α4 … n-1 s α1 α3 α5 … n-1 s b1 b2 b3 c1 c2 …… 2 s e1 e2 1 s f1 0 s g1 1 a a a a −a a b = 0 2 = 1 2 0 3 1 − a1 a1 a3 a1 1 a a a a − a a b = 0 4 = 1 4 0 5 2 − a1 a1 a5 a1 1 a a a b − a b c = 1 3 = 3 1 1 2 1 − etc b1 b1 b2 b1 . Properties of the Ruth-Hurwitz table: 1. Polynomial A(s) is stable (i.e. all roots of A(s) have negative real parts) if there is no sign change in the first column. 2. The number of sign changes in the first column is equal to the number of roots of A(s) with positive real parts. B29 Examples:

2 A(s) = a0s + α1 s + α2

2 s a 0 a 2 1 s a1 0 s a 2

α0 >0, α1 > 0 , α2 > 0 or α0 <0, α1 < 0 , α2 < 0

For 2nd order systems, the condition that all coefficients of A(s) have the same sign is necessary and sufficient for stability.

3 2 A(s) = α0 s + α1s + α2s +α3

3 s a0 a2 2 s a1 a3 a a − a a s1 1 2 0 3 a1 0 s a3 α0 >0, α1>0, α3>0, α1α2 − α0α3 > 0

(or all first column entries are negative) B30 Special cases:

1. The properties of the table do not change when all the coefficients of a row are multiplied by the same positive number.

2. If the first-column term becomes zero, replace 0 by and continue. • If the signs above and below are the same, then there is a pair of (complex) imaginary roots. • If there is a sign change, then there are roots with positive real parts.

Examples:

A(s) = s3 +2s2 + s +2

s3 1 1 s 2 2 2 1 → ε s 0 Æ pair of imaginary roots ( s= ±j ) s 0 2

A(s) = s3- 3s +2 = (s-1)2(s+2)

s3 1 − 3 s 2 0 ≈ ε 2 2 s1 − 3 − Æ two roots with positive real parts ε s0 2 B31 3. If all coefficients in a line become 0, then A(s) has roots of equal magnitude radially opposed on the real or imaginary axis. Such roots can be obtained from the roots of the auxiliary polynomial.

Example:

A(s)= s5 + 2s4 +24 s3 + 48s2 -25s -50 s5 1 24 − 25 s4 2 48 − 50 3 s 0 0 Æ auxiliary polynomial p(s)

p(s) = 2s4 + 48s2 - 50 dp(s) = 8s3 + 96s ds s3 8 96 s2 24 − 50 s1 112.7 0 s0 − 50

• A(s) has two radially opposed root pairs (+1,-1) and (+5j,-5j) which can be obtained from the roots of p(s). • One sign change indicates A(s) has one root with positive real part.

Note: A(s) = (s+1) (s-1)(s+5j)(s-5j)(s+2) p(s) = 2(s2-1) (s2 +25) B32 Relative stability

Question: Have all the roots of A(s) a distance of at least IURP WKH LPDJLQDU\ D[LV"

Im Substitute s with s = z - in A(s) and apply the Routh-Hurwitz Re test to A(z)

Closed-loop System Stability Analysis

R(s) C(s) K G(s) + -

Question: For what value of K is the closed-loop system stable?

Apply the Routh-Hurwitz test to the denominator polynomial KG ( s ) of the closed-loop transfer function 1 + KG ( s ) . B33 Steady-State Error Analysis

E(s) C(s) R(s) + G(s) -

H(s)

Evaluate the steady-state performance of the closed-loop system using the steady-state error ess = = ess lim e(t) lim sE(s) t→∞ s→0

1 E(s) = ⋅ R(s) 1 + G(s)H (s) for the following input signals: Unit step input Unit ramp input Unit parabolic input

Assumption: the closed-loop system is stable

Question: How can we obtain the steady-state error ess of the closed-loop system from the open-loop transfer function G(s)H(s) ? B34 Classification of systems:

For an open-loop transfer function K(T s +1)(T s +1) G(s)H (s) = a b N ()()+ + s T1s 1 T2s 1

Type of system: Number of poles at the origin, i.e., N

Static Error Constants: Kp, Kv, Ka

Open-loop transfer function: G(s)H(s)

= G(s) Closed-loop transfer function: Gtot (s) 1 + G(s)H (s)

Static Position Error Constant: Kp

Unit step input to the closed-loop system shown in fig, p. B33.

= = 1 ess lim sE(s) R(s) = 1/s s→0 1+ G(0)H (0)

= = Define: K p lim G(s)H (s) G(0)H (0) s→0 1 e = Type 0 system Kp = K ss + 1 K p

Type 1 and higher Kp= \ ess = 0 B35

Static Velocity Error Constant: Kv

Unit ramp input to the closed-loop system shown if fig, p. B33.

2 = s ⋅ 1 = 1 R(s) = 1/s ess lim lim s→0 1+ G(s)H (s) s 2 s→0 sG(s)H (s) = Define: K v lim sG (s)H (s) s→0

Type 0 system Kv = 0 ess = \

Type 1 system Kv = K ess = 1/Kv

Type 2 and higher Kv = \ ess = 0

Static Acceleration Error Constant: Ka

Unit parabolic input to the closed-loop system shown in fig, p. B33

= s ⋅ 1 = 1 3 ess lim lim R(s) = 1/s s→0 1+ G(s)H(s) s3 s→0 s2G(s)H(s)

= 2 K a lim s H ( s)G ( s ) Define: s → 0

Type 0 system Ka = 0 ess = \

Type 1 system Ka = 0 ess = \

Type 2 system Ka = K ess = 1/ Ka

Type 3 and higher Ka = \ ess = 0 B36 Summary: Consider a closed-loop system:

E(s) C(s) R(s) + G(s) -

H(s) with an open-loop transfer function: K ()()T s + 1 ⋅ T s + 1 ... G ()s H ()s = a b N ()()+ ⋅ + s T1 s 1 T2 s 1 ... and static error constants defined as: = = K p lim G(s)H (s) G(0)H (0) s→0 = K v lim sG (s)H (s) s→0 = 2 K a lim s H ( s)G ( s ) s → 0

The steady-state error ess is given by: Unit step Unit ramp Unit r(t) = 1 r(t) = t parabolic r(t) = t2/2 1 1 Type 0 e = (= ) ess= \ ess= \ ss + + 1 K p 1 K = 1 = 1 \ Type 1 ess= 0 ess ( ) ess= K v K = 1 = 1 Type 2 ess= 0 ess= 0 ess ( ) K a K B37 Correlation between the Integral of error in step response and Steady-state error in ramp response

R(s) E(s) C(s) + G(s) -

∞ − E(s) = L[e(t)] = ∫ e st e(t)dt 0 ∞ ∞ − lim E(s) = lim ∫ e st e(t)dt = ∫ e(t)dt s→0 s→ 0 0 0 R(s) E(s) = substitute 1+ G(s) in the above eq. ∞ R(s) 1 lim = ∫ e(t) dt step : R(s) = → + G s s 0 1 ( ) 0 s ∞  1 1  1 1 e(t)dt =  ⋅  = = ∫ 1 + G(s) s s ⋅ G(s) lim→   lim→ 0 s 0 s 0 K v

1 = Steady-state error in unit-ramp input = essr KV ∞ = essr ∫ e(t)dt 0 B38 ∞ ∫ et()dt c(t) 0

t c(t) r(t) r(t) essr c(t)

0 t