2nd order transfer function - Katedra výkonových …€¦ · · 2013-12-042nd Order transfer...

Second order transfer functions 1

© w t norris 2nd order transfer functions Monday 3 March 2008

2nd Order transfer function - Summary of results

The canonical 2nd order transfer function is expressed as

H(s) = ωn2

s2 + 2ζ ωn s + ωn2 …1

ωn is the natural frequency; ζ is the damping coefficient. …2

The Poles, s1, s2 are

s1,s2 = ωn ( ) – ζ ± √ζ2 – 1 …3

On the right is the root locus for fixed ωn and varying ζ.

s

σ

jω

ζ=0ζ=sinθ

ζ=0

ζ<1

ζ<1

ζ>1 ζ>1

ζ=1

jωn-ωn

jωn

-jωn

θ

The transfer function in the frequency domain is

H(ω) = ωn2

ωn2 – w2 + 2ζωnw =

ωn2

√(ωn2 – ω2)2 + 4 ζ2 ωn2 ω2 ejΦ

with tanΦ = – 2ζωn ω

ωn2 – ω2

The Bode plot is :–

For ζ = 1√2 we have the sharpest

corner without a maximum.

For ζ < 1√2 there is a peak in the plot of

magnitude

|H|max = 1

2ζ √1–ζ2 …5

This is above the level at ω = 0.

For the H(s) given at t= 0 then |H| = 1

The roll-off rate at high frequenciesis 40 dB per decade ( in frequency).

The overall phase change is -180˚.

For low ζ the swing in phase is allclose to the frequency at the maximum|H|.

The peak itself is at

ωp = ωn√ 1 – 2 ζ2

Beware; later we introduce

ωd = ω √1 – ζ2

-0.1 1 10-40

-30

-20

-10

0

10

H, dB

Frequency, rad s-1

ζ=2

ζ= 0.1

ζ = 1√2

0.01 0.1 1 10-180

-135

-90

-45

0

50

Frequency, rad s-1 ( log scale )

Phase, Φdegrees

ζ=2

ζ = 1√2

ζ = 0.1



Impulse Responses

ζ = 0 Undamped

y(t) = ωn sin( ωnt)

1/ωn

-1/ωn

y(t)

t

2π/ωn

ζ > 0 Damped

y(t) = 1

2ωn√ζ2 – 1 ( ) exp(s1t) – exp(s2t)

ζ = 1 Critically damped

y(t) = ωn2t exp( -ωn t)

Curve is very similar to damped response

2 3 4 5

-0.3

-0.2

-0.1

0.1

0.2

0.3

t

y(t)

exp( s2t)

exp( s1t)

ζ < 0 Underdamped

y(t) = ωn2

ωd exp(-ζωnt) sin(ωdt)

The logarithm of the ratio of successivepeaks is the logarithmic decrement or log-dec for short

δ = 2πζ

√1–ζ2 ≈ 2πζ if ζ << 1

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15 2025

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0.20.40.60.8

t

y(t)



Step responses

ζ = 0 Undamped

y(t) = ( ) 1 – cos(ωn t)

2 4 6 8 10

1

2

t

y(t)

ζ > 0 Damped

y(t) =

1 + 1

√ζ2 –1 [ s2 exp(s1t) – s1 exp(s2t) ]

10 15

0.2

0.4

0.6

0.8

1

t

y(t)

ζ = 1 Critically damped

y(t) = ( ) 1 – exp( – ωnt ) [1 + ωnt]

2 4 6 8 10

1

t

y(t)

ζ < 0 Underdamped y(t) =

1 – exp( – ζωn t)

cos( ωd t) + ζωnωd

sin(ωd t)

First peak* , ΜP = 1 + exp

– π ζ

√1–ζ2

Proportional overshoot:

ΟP = y –1 = exp

– π ζ

√1–ζ2

Time to be within δ , Ts = – ln(δ)ζωn

For δ = 0,02, i.e., 2% : Ts ≈ 4

ζωn

Root Total square error = √1+ 4ζ2

4 ζ ωn

* Some authors call MP the overshoot .

10 20

1

t

y(t)

MP

Tp Ts

Overshoot,OP

T = 2πωd

δ

Period,

Frequency of oscillation:

ωd = ωn √1 – ζ2

Period of oscillation:

T = 2πωd

= 2π

ωn√1-ζ2

For least root square total deviation ζ = 1 / 2.

Then OP ≈ 16%



Ramp response Now set f(t) = t asillustrated.

F = 1s2

f= t

t

Perhaps unkindly I leave derivations to the reader. The general response is: –

Y = H(s) × 1s2 =

ωn2

s2 (s2 + 2ζ ωn s + ωn2 )ζ = 0 Undamped

Y =

1s2 –

1s2 +ωn2

y =

t – sin(ωnt )

ωn

Plot is for ωn =1; y = t – sin(t)

Note: s1s2= ωn2 ; s1 + s2 = – 2ζ ωn

2 4 6 8 10

2

4

6

8

10

t

y(t)

Response

y=t

ζ > 0 Damped : s1, s2 = ωn ( ) -ζ ± √ζ2 – 1

Y =

1s2 +

s1+s2sωn2 +

1s1–s2

s2

s-s1 –

s1s–s2

y =

t –

2ζωn

+1

s1–s2( )s2es1t – s1es2t

Plot is with ωn= 1, ζ = 2

2 4 6 8 10

-4

-2

2

4

6

t

y(t)

Response

t - ωn 2ζ

ζ = 1 Critically Damped

Y = 1ωn

ωn

s2 – 2s +

2s+ωb

+ ωn

(s+ωn)2

y = 1ωn

( )ωn t – 2 + (2 + ωnt ) e–ωnt

The plot is for ωn = 1.

6 8 10-2

2

4

6

8

Y(t)

t

Response

2 4

t - ωn 2



ζ < 0 Underdamped: ωd = ωn √1 – ζ2

Y = 1ωn

ωn

s2 – 2ζs +

2ζ(s+ζωn) –ωnωd

(1–2ζ2)ωd

(s+ζωn)2s+ωn2(1–ζ2)

y = 1ωn

×

ωnt –2ζ+e–ωnζt[2ζcos(ωdt)–

ωnωd

(1–2ζ2) sin( ωdt)]

The plot is with ωn = 1 and ζ = 0·2

2 4 6 8 10

2

4

6

8

y(t)

t

Response

t - ωn 2ζ

Second order transfer function analysis

As an example consider the mass, damper and spring acted on by a force, f(t). We had forthe displacement, y :–

m y" + c y' + k y = f

Take the Laplace transform ( with no initial values ) :–

m s2Y + c s Y + k Y = F

or s2Y + cm s Y +

km Y = F/m

which with ωn = √ km and ζ =

c2√km

becomes If we redefine F ≡ F/ωn2m = F/k:–

s2 Y + 2ζ ωn s Y + ωn2 = Fωn2 : or (s2 + 2ζ ωn s + ωn2 ) Y = Fωn2

and so H(s) = FY =

ωn2 s2 + 2ζ ωn s + ωn2

…1

This equation 1 is a standard form for any second order transfer function. Whatever thephysical variables it helps to turn the expression to this format early in the analysis.

ωn is the natural frequency; ζ is the damping coefficient. …2

PolesPoles, s1, s2 are the values of s for which the denominator is zero.

s1,s2 = –2ζωn ± √4 ζ2ωn2 – 4 ωn2

2

= ωn ( ) – ζ ± √ζ2 – 1 …3



Cases;

(1) ζ = 0: poles at s = ± jωn,

(2) ζ < 1: two complex poles. These lie on a circle of radius ωn

(3) ζ = 1: two coincident poles at s = –ωn

(4) ζ > = 1: two poles on the negative real axis.

These are illustrated on the root locus diagram on the right forconstant ωn.

s

σ

jω

ζ=0

ζ=0

ζ<1

ζ<1

ζ>1 ζ>1

ζ=1

jωn-ωn

jωn

-jωn



Bode Plot

From eq. 1 H(s) = ωn2

s2 + 2ζ ωn s + ωn2

and putting s → jω H(ω) = ωn2

ωn2 – ω2 + j 2ζωnω…4a

= |H(ω)| × exp( jΦ) …4b

with |H(ω)| = ωn2

√(ωn2 – ω2)2 + 4 ζ2 ωn2 ω2…4c

and tanΦ = – 2ζωn ω

ωn2 – ω2 …4d

ω → 0 ω = ωn ω → ∞

| H | 1 12ζ →

ωn2

ω2

Φ 0 – π2 – π

For ζ < 1√2 there is a peak in the | H | curve. It is easier to find the peak of |H|2 which is at

the same value of ω as the peak of |H|. And easier still to find the minimum of 1

|H|2 whic

occurs at teh same value of ω as the peak in |H|

ωn4 d

dω

1

|H|2 =

ddω ( )(ωn2 – ω2)2 + 4 ζ2 ωn2 ω2 = 2(ωn2 – ω2)(-2ω) + 8ζ2ωn2ω

= 4ω ( ) ω2 – ωn2 ( 1–2ζ2)

so d

dω

1

|H|2 = 0 if ω = 0; or if ω = ωn√1 – 2 ζ2

In the latter case |H| = 1

√( 1– 1+ 2ζ2)2 + 4 ζ2( 1–2 ζ2)

= 1

√4 ζ2( 1–ζ2) =

1

2ζ √1–ζ2…5

Note that if ζ = 0 then the peak of | H | → ∞. This is the undamped case. Note also that the

peak is at a frequency close to but slightly less than ωn. There is no peak if ζ > 12 since then

the frequency is imaginary.

This is illustrated in the next figures with ωn = 1. Note that the frequency scale has beenextended in the phase plot to show the approach to the limits.

In the magnitude plot note (i) the roll of rate is 40 dB per decade.



(ii) the peak is at about 14 dB for ζ = 0·1. Eq. 5 gives a peak ofa factor of 5: so 20 log10(5) = 13.97. The theory pans out correctly.

-0.1 1 10-40

-30

-20

-10

0

10

H, dB

Frequency, rad s-1

ζ=2

ζ= 0.1

ζ = 1√2

0.01 0.1 1 10-180

-135

-90

-45

0

50

Frequency, rad s-1 ( log scale )

Phase, Φdegrees

ζ=2

ζ = 1√2

ζ = 0.1

Impulse Responses

Consider the Dirac impulse, δ(t), as an input.

ζ = 0. Y = ωn2

s2 + ωn2 × 1 = ωn ×ωn

s2 + ωn2

so y(t) = ωn sin( ωnt)

There is no damping and the system ringsforever.

t

f(t)

δ(t)

ωn

-ωn

y(t)

t

2π/ωn



ζ > 1 The roots are real. We had eq. 3: –

s1,s2 = ωn ( ) – ζ ± √ζ2 – 1 ; s1 > s2 …3

Thus the response is Y = ωn2

(s–s1)(s-s2) × 1

= ωn2

s1– s2

1

s-s1 –

1s–s2

and y(t) = ωn

2√ζ2 – 1 ( ) exp(s1t) – exp(s2t)

The response below is for ωn = 1 and ζ = 2. The coefficients of the component exponentialsis omitted in the figure. The formula is :– y = 0·289( e–0·27 t – e–3·7 t)

2 3 4 5

-0.3

-0.2

-0.1

0.1

0.2

0.3

t

y(t)

exp( s2t)

exp( s1t)

ζ = 1 Y = ωn2

(s + ωn)2 × 1 =

ωn2

(s + ωn)2

so y(t) = ωn2 t exp( -ωn t)

The plot is very like the previous one.

ζ < 1 This is a bit more complicated. We need to complete squares in the denominator.

Y = H(s) × 1 = ωn2

s2 + 2ζ ωn s + ωn2

= ωn2

(s + ζ ωn)2 – ζ2ωn2 + ωn2

= ωn2

(s + ζ ωn)2 + ωd2 …6

with ωd = ωn √1 – ζ2 …7

This frequency appears often and should be noted.



Thus Y = ωn2ωd

× ωd

(s + ζ ωn)2 + ωd2

so, from the tables y(t) = ωn2ωd

exp(-ζωnt) sin(ωdt) …8a

= ωn

√1–ζ2 exp(-ζωnt) sin(ωdt) …8b

The response for ωn = 1 and ζ = 0·1 is below. It is an exponentially decaying sine wave.

The formula is : – y = 1.005 e–0.1 t sin( 0·995 t)

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15 2025

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0.20.40.60.8

t

y(t)

The period of the oscillation, T = 2πωd

. The ratio of succeeding positive peaks is then: –

exp(-ζωnt)

exp

–ζωn

t + 2πωd

= exp

2πζωn

ωd

The logarithm of this ratio is δ = 2πζωn

ωd =

2πζωn

ωn√1–ζ2

so δ = 2πζ

√1–ζ2 ≈ 2πζ if ζ << 1 …9

δ is called the logarithmic decrement or log-dec for short. It is a useful way ofmeasuring the damping in a highly resonant system.

Note we use the symbol δ with another meaning later.



Step function responses

The input is f = u(t)

so

F = 1s

f= u(t)

t

1

The response is, using H(s) from eq. 1

Y = ωn2

s2 + 2ζ ωn s + ωn2 ×

1s …10

Cases

ζ = 0 Y = ωn2

s( s2 + ωn2 ) ≡

As +

Bs + C s2 + ωn2

Thus 1 ≡ A(s2 + ωn2) + s (Bs+C)Put s = 0: ωn2 = A ωn2 : ∴ A = 1Coeff. s2 : 0 = A + B : ∴ B = – A = – 1Coeff, s : 0 = C : ∴ C = 0

Y =

1s –

s s2 + ωn2

and y(t) = ( ) 1 – cos(ωn t)

The plot for ωn = 1 is on the right.

The system oscillates about the meanindefinitely.

2 4 6 8 10

1

2

t

y(t)

ζ > 1 The transfer function is factorisable and the roots are real. As before eq. 3 gives theroots: –

s1,s2 = ωn ( ) – ζ ± √ζ2 – 1 ; s1 > s2 …3

and the response is Y = ωn2

(s–s1)(s-s2) × 1s

We have Y = ωn2`

s(s–s1)(s-s2) ≡ As + B

s-s1 +

Bs-s2



whence 1 ≡ A(s – s1)(s – s2) + Bs(s–s2) + Cs(s – s1)

Put s = s1: ωn2 = B s1(s1 – s2) ∴ B = ωn2

s1(s1– s2) =

s2s1 – s2

Put s = s2: ωn2 = C s2( s2 – s1) ∴ C = – ωn2

s2(s1– s2) = -

s1s1 – s2

Put s = 0 : ωn2 = A s1s2 ∴ A = ωn2

s1s2 =1

So Y = 1s +

1

ωn√ζ2 –1

s2

s–s1 –

s1

s–s2 ) ; since s1 s2 = ωn2 ; s1 – s2 = ωn√ζ2– 1

and y(t) = 1 + 1

ωn√ζ2 –1 [ s2 exp(s1t) – s1 exp(s2t) ])

noting s1,s2 = ωn ( ) – ζ ± √ζ2 – 1 ; s1 > s2

With ωn =1 and ζ = 2 this becomes

y(t) = 1 + 0·077 e–3·73 t – 1·077 e–0·27 t

which is illustrated on the right.

10 15

0.2

0.4

0.6

0.8

1

t

y(t)

ζ = 1 Y = ωn2

(s +ωn)2 ×

1s =

ωn2

s(s + ωn)2

≡ As + B

s+ωn +

C(s+ωn)2

, say

Thus ωn2 ≡ Α (s + ωn)2 + Bs(s + ωn) + Cs

Put s = 0 : ωn2 = A ωn2 ∴ A =1. Put s = – ωn : ωn2 = – C ωn ∴ C = – ωn

Coeff. s2 : 0 = A + B ∴ B = – A = –1

Y = 1s –

1s+ωn

– ωn

(s + ωn)2

Thus from the Tables:

y(t) = 1 – exp( – ωnt ) [1 + ωnt ]

For ωn = 1 y(t) = 1 – e–t( 1+t)

This is illustrated on the right. 2 4 6 8 10

1

t

y(t)

ζ < 1 The more complicated but more interesting case. From eq. 1 we have

Y = ωn2

s2 + 2ζ ωn s + ωn2 ×

1s



= ωn2

s(s2 + 2ζ ωn s + ωn2 )

= As +

Bs + C s2 + 2ζ ωn s + ωn2

, say.

Thus 1 ≡ A (s2 + 2ζ ωn s + ωn2 ) + s (B s +C)Put s = 0 : ωn2 = A ωn2 ∴ A = 1Coeff. s2 : 0 = A + B ∴ B = – A = –1Coeff. s : 0 = A 2 ζ ωn + C ∴ C = – 2ζωn A = – 2ζωn

and Y = 1s –

s + 2ζωn

s2 + 2ζ ωn s + ωn2

= 1s –

s + 2ζωn

(s + ζ ωn)2 + ωd2 : completing the square as in eq. 6

= 1s –

(s + ζωn) + ζωnωd

ωd

(s + ζ ωn)2 + ωd2 : making the numerator to suit the Tables

with ωd = ωn √1 – ζ2

Thus y(t) = 1 – exp( – ζωn t)

cos( ωd t) + ζωnωd

sin(ωd t) …11

For ωn = 1 and ζ = 0·2

we have

y(t) ≈ 1 – e– 0·2 t [ cos( 0·98 t ) + 0·2sin( 0·98 t) ]

This is plotted on the right.

Note that the gradient is zero at t = 0

5 10 15 20

1

2

t

y(t)

1+ e-t

1- e-t



There are properties of interest of the curve.

10 20

1

t

y(t)

MP

Tp Ts

Overshoot,OP

T = 2πωd

δ

Period,

Overshoot

The derivative dydt =

ωn2

ωd exp(–ζωnt) sin(ωdt)

is remarkably simple. It is equal to the impulse response. Indeed it is generally the case thatthe derivative of the step response of any linear system is the impulse response.

dydt = 0 when ωd t = nπ or when t =

nπωd

= nπ

ωn√1-ζ2 =

nT2

where the period of the oscillations is T = 2πωd

= 2π

ωn√1-ζ2

The value of y at these extrema is Y = 1 + exp

– nπ ζ

√1–ζ2

The first of these when n = 1 gives the peak of the overshoot ( often simply called theovershoot.

ΜP = 1 + exp

– π ζ

√1–ζ2

Express proportional overshoot as theexcess as a proportion of the final steadystate value.

ΟP = y –1 = exp

– π ζ

√1–ζ2…12a

occurring at

TP = π

ωn√1-ζ2 =

πωd

= T2 …12b

0 0.2 0.4 0.6 0.8 10

0.2

0.4

0.6

0.8

1

Overshoot, Op

Damping, ζ



We may be interested in the time to reach and stay within a certain 'distance' from the finalvalue. The deviation of the nth extremum is

δ = exp

– n π ζ

√1–ζ2 which occurs at tn =

nπωn√1 – ζ2

or nπ

√1 – ζ2 = tnωn

so δ = exp

– n π ζ

√1–ζ2 = exp( – ωn ζ tn)

and Ts = tn = – ln(δ)ζωn

Strictly tn should be an integral number of half periods but the discrepancy is usuallyignored.

For δ = 0·02, i.e., 2% Ts ≈ 4

ζωn

and for δ = 0·05, i.e., 5% Ts ≈ 3

ζωn

We may also consider an overall expression of deviation from the final value. The rootmean square seems a suitable choice.

5 10 15 20

-1

1

t

t

y(t) - 1

5 10 15 20

0.2

0.4

0.6

0.8

1(y-1)2

The deviation and its square are illustrated above for the previously shown response. Wetake then as a measure of the overall deviation as

Erts = the total square deviation ( integrated from t = 0 to t = ∞ )

We require

Erts =

∫0

∞ ( y –1)2 dt

1/2

= √1+4 ζ2

4ωnζ

This rather simple expression isvalid for all ζ, i.e., 0 < ζ < ∞. It has

a minimum value of 1 when ζ = 12 . 1 2 3 4 5

1.2

1.0

1.4

1.6

1.8

2

2.2

Erts

ζ



This minimum square deviationcondition is illustrated on the rightwith a 16% overshoot..

105 15

1

t

y(t)

1.16



There are other measures of the error. To put emphasis on the approach to the final value onmight weight erros by muptiplying by time. The integral of this de-emphasises the initialapproach to the final value.. Thus we have

The root of the integral of time weightedtotal square erroe is

Erts =

∫0

∞ t ( y –1)2 dt

1/2

= 1

2ωnζ √1+ 8 ζ4

2

This has a minimum of 1

21/4 ωn when

ζ =

1

8 1/4

≈ 0·595

5 10 15 20

0.2

0.4

0.6

0.8

t

t (y - 1)2

The overshoot is

exp

–ζ π

√1 – ζ2 = exp

–

π√2√2–1

≈ 9·8%



Keeping factors constant

Poles are at s =ωn( ) – ζ ± √1–ζ2 :

Constant imaginary part.

ωd = ωn √1–ζ2 = constant.

So Constant frequency inresponse

1

t

y(t) jω

σ

s

Constant real part:–

ωn ζ is constant.

Fixed envelope

y =1 – exp( – ζωn t)

and so time to get within agiven limit is deermined byωnζ.

E.g., Tδ=2% = 4

ζωn

1

2

t

y(t)

jω

σ

s

Constant damping ratio, ζ

Thus a lso constanovershoot:

OP = exp

– π ζ

√1–ζ2

Note: ζ = sinθ where θ ismarked in the figure.

1

t

y(t)jω

σ

sθ

There is scarcely anyovershoot when θ > 60˚.

20 40 60 90

1.2

1.4

1.6

1.8

2Fractional overshoot,

OP

θ

θ

σ

jωs

2nd order transfer function - Katedra výkonových …€¦ · · 2013-12-042nd Order transfer...

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Transcript of 2nd order transfer function - Katedra výkonových …€¦ · · 2013-12-042nd Order transfer...