Second order transfer functions
1
2nd Order transfer function - Summary of results
jω
The canonical 2nd order transfer function is expressed as
ωn2
H(s) = 2
s + 2ζ ω n s + ω n2
ζ=sinθ
…1
s
jωn
ζ<1
-ωn
ωn is the natural frequency; ζ is the damping coefficient.
…2
The Poles, s1, s2 are
ζ2 – 1 )
s1,s2 = ωn ( – ζ ± √

ζ=0
θ
ζ>1
jωn
σ
ζ>1
ζ=1
…3
ζ<1
On the right is the root locus for fixed ωn and varying ζ.
-jωn
ζ=0
The transfer function in the frequency domain is
H(ω) =
ωn2
=
ωn2 – w2 + 2ζωnw
ωn2
(ωn2 – ω2)2 + 4 ζ2 ωn2 ω2

√
with tanΦ = –
ejΦ
2ζωn ω
ω n2 – ω 2
The Bode plot is :–
H, dB
1
For ζ = √2 we have the sharpest
corner without a maximum.
10
ζ= 0.1
0
1
For ζ < √2 there is a peak in the plot of -10
magnitude
-20
1
|H|max =
…5
1–ζ2
2ζ √

-30
This is above the level at ω = 0.
ζ=2
ζ=
-40
-0.1
1
Frequency, rad s-1
For the H(s) given at t= 0 then |H| = 1
The roll-off rate at high frequencies
is 40 dB per decade ( in frequency).
Phase, Φ
degrees
0
The overall phase change is -180˚.
-45
For low ζ the swing in phase is all
close to the frequency at the maximum
|H|.
-90
The peak itself is at
ωp = ωn√
1 – 2 ζ2

Beware; later we introduce
1 – ζ2
ωd = ω √

1
√2
10
ζ = 0.1
ζ=
1
√2
ζ=2
-135
-180
0.01
© w t norris 2nd order transfer functions Monday 3 March 2008
0.1
10
1
50
Frequency, rad s-1 ( log scale )
Second order transfer functions
2
Impulse Responses
y(t)
1/ωn
ζ = 0 Undamped
2π/ωn
y(t) = ωn sin( ωnt)
t
-1/ωn
ζ > 0 Damped
y(t)
exp( s1t)
0.3
y(t) =
1
( exp(s1t) – exp(s2t))
ζ2 – 1
2ωn√

ζ = 1 Critically damped
y(t) = ωn2t exp( -ωn t)
0.2
0.1
t
2
3
4
5
-0.1
-0.2
exp( s2t)
-0.3
Curve is very similar to damped response
ζ < 0 Underdamped
y(t)
ωn2
y(t) = ω exp(-ζωnt) sin(ωdt)
d
The logarithm of the ratio of successive
peaks is the logarithmic decrement or logdec for short
δ=
2πζ
1–ζ2

√
≈ 2πζ if ζ << 1
© w t norris 2nd order transfer functions Monday 3 March 2008
0.8
0.6
0.4
0.2
-0.2
-0.4
-0.6
25 t
10
5
15
20
Second order transfer functions
3
Step responses
y(t)
ζ = 0 Undamped
y(t) =
2
( 1 – cos(ωn t))
1
t
4
2
6
8
10
y(t)
ζ > 0 Damped
1

1

y(t) =  1 +
[ s2 exp(s1t) – s1 exp(s2t) ]


ζ2 – 1

√
0.8
0.6
0.4
0.2
t
10
ζ = 1 Critically damped
15
y(t)
1
y(t) = ( 1 – exp( – ωnt ) [1 + ωnt] )
t
4
2
ζ < 0 Underdamped
y(t) =
ζωn



1 – exp( – ζωn t)  cos( ωd t) + ω sin(ωd t)  
d



6
8
10
y(t)
MP
Overshoot,
OP
1
First peak* ,
 πζ 
Μ P = 1 + exp –


1–ζ2

√
δ
Period,
T = 2π
ω
t
d
Proportional overshoot:
 πζ 
ΟP = y –1 = exp –

1–ζ2
 √

ln(δ)
Time to be within δ , Ts = – ζω
n
4
For δ = 0,02, i.e., 2% : Ts ≈ ζω
n
Root Total square error =
√
1+ 4ζ2
4 ζ ωn
* Some authors call MP the overshoot .
© w t norris 2nd order transfer functions Monday 3 March 2008
Tp
10
Ts
20
Frequency of oscillation:
1 – ζ2
ωd = ωn √

Period of oscillation:
2π
2π
T=ω =
d
1-ζ2
ωn√

For least root square total deviation ζ = 1 / 2.
Then OP ≈ 16%
Second order transfer functions
Ramp response
illustrated.
Now set
f(t) = t as
4
f= t
1
F= 2
s
t
Perhaps unkindly I leave derivations to the reader. The general response is: –
ωn2
1
Y = H(s) × 2 = 2 2
s
s (s + 2ζ ω n s + ω n2 )
ζ = 0 Undamped
y(t)
1
1

Y =  2 – 2

2
 s s +ω n 
10
Response
8
6

sin(ωnt ) 
y=t– ω


n

4
y=t
2
t
Plot is for ωn =1; y = t – sin(t)
Note: s1s2= ωn2 ;
2
4
6
8
10
s1 + s2 = – 2ζ ωn
ζ > 0 Damped : s1, s2 = ωn ( -ζ ± √
ζ2 – 1 )

 1 s +s
1  s2
s1  
Y =  2 + 1 22 +
–

s1–s2 s-s1 s–s2 
s
sωn


2ζ
1
y =  t – ω +s –s (s2es1t – s1es2t)


n
1 2
Plot is with ωn= 1, ζ = 2
ζ = 1 Critically Damped
ωn 
1 ω 2
2
Y = ω  2n – s + s+ω +

n s
b (s+ωn)2
y(t)
6
Response
4
2
t
2
-2
4
6
8
10
t - 2ζ
ωn
-4
Y(t)
8
6
4
Response
2
1
y = ω (ωn t – 2 + (2 + ωnt ) e–ωnt)
n
The plot is for ωn = 1.
© w t norris 2nd order transfer functions Monday 3 March 2008
t
4
2
tωn
2
-2
6
8
10
Second order transfer functions
ζ < 0 Underdamped: ωd = ωn √
1 – ζ2

ω

) – ωn(1–2ζ2)ωd
2ζ(s+ζω
n

2ζ
ω
1
d

Y = ω  2n – s +
n s
(s+ζωn)2s+ωn2(1–ζ2) 
5
y(t)
8
6
Response
4
1
y= ω ×
n
ωn


 ωnt –2ζ+e–ωnζt[2ζcos(ωdt)–ωd (1–2ζ2) sin( ωdt)]


t - 2ζ
ωn
2
t
2
4
6
8
10
The plot is with ωn = 1 and ζ = 0·2
Second order transfer function analysis
As an example consider the mass, damper and spring acted on by a force, f(t). We had for
the 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
s2Y +
or
which with ωn =
√m and
k
ζ=
c
k
m s Y + m Y = F/m
c
becomes If we redefine F ≡ F/ωn2m = F/k:–
2√
km
s2 Y + 2ζ ωn s Y + ωn2 = Fωn2 :
or
(s2 + 2ζ ωn s + ωn2 ) Y = Fωn2
ωn2
F
H(s) = Y = 2
s + 2ζ ω n s + ω n2
and so
…1
This equation 1 is a standard form for any second order transfer function. Whatever the
physical 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
Poles
Poles, s1, s2 are the values of s for which the denominator is zero.
–2ζωn ± √
4 ζ2ω n2 – 4 ω n2

2
= ωn ( – ζ ± √
ζ2 – 1 )

s1,s2 =
© w t norris 2nd order transfer functions Monday 3 March 2008
…3
Second order transfer functions
6
Cases;
jω
(1) ζ = 0: poles at s = ± jωn,
ζ=0
(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 for
constant ωn.
jωn
ζ<1
-ωn
jωn
ζ>1
ζ>1
σ
ζ=1
ζ<1
-jωn
ζ=0
© w t norris 2nd order transfer functions Monday 3 March 2008
s
Second order transfer functions
Bode Plot
ωn2
H(s) = 2
s + 2ζ ω n s + ω n2
ωn2
H(ω) = 2
ωn – ω2 + j 2ζωnω
From eq. 1
and putting s → jω
…4a
= |H(ω)| × exp( jΦ)
with
|H(ω)| =
and
…4b
ωn2
…4c
(ω n2 – ω 2)2 + 4 ζ2 ω n2 ω 2

√
tanΦ = –
2ζωn ω
ω n2 – ω 2
…4d
ω→0
ω = ωn
ω→∞
|H|
1
1
2ζ
ω 2
→ n2
ω
Φ
0
π
–2
7
–π
1
For ζ < √2 there is a peak in the | H | curve. It is easier to find the peak of |H|2 which is at
1
the same value of ω as the peak of |H|. And easier still to find the minimum of 2 whic
|H|
occurs at teh same value of ω as the peak in |H|
d  1  d
ωn4 dω  2  = dω ((ω n2 – ω 2)2 + 4 ζ2 ω n2 ω 2) = 2(ωn2 – ω2)(-2ω) + 8ζ2ωn2ω
 |H| 
= 4ω ( ω2 – ωn2 ( 1–2ζ2))
d  1 
so dω  2  = 0
 |H| 
if ω = 0;
In the latter case |H| =
=
or if ω = ωn√
1 – 2 ζ2

1
( 1– 1+ 2ζ2)2 + 4 ζ2( 1–2 ζ2)
√
1
1
=
4 ζ2( 1–ζ2) 2ζ √
1–ζ2
√

…5
Note that if ζ = 0 then the peak of | H | → ∞. This is the undamped case. Note also that the
1
peak is at a frequency close to but slightly less than ωn. There is no peak if ζ > 2 since then
the frequency is imaginary.
This is illustrated in the next figures with ωn = 1. Note that the frequency scale has been
extended 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.
© w t norris 2nd order transfer functions Monday 3 March 2008
Second order transfer functions
8
(ii) the peak is at about 14 dB for ζ = 0·1. Eq. 5 gives a peak of
a factor of 5: so 20 log10(5) = 13.97. The theory pans out correctly.
H, dB
10
ζ= 0.1
0
ζ=2
-10
ζ=
-20
1
√2
-30
-40
-0.1
1
Frequency, rad s-1
Phase, Φ
degrees
0
10
ζ = 0.1
-45
ζ=
-90
1
√2
ζ=2
-135
-180
0.01
0.1
10
1
50
Frequency, rad s-1 ( log scale )
Impulse Responses
Consider the Dirac impulse, δ(t), as an input.
ζ = 0.
ω 2
ω ×ω n
Y = 2 n 2 × 1 = 2n
s + ωn
s + ω n2
f(t)
δ(t)
so y(t) = ωn sin( ωnt)
t
There is no damping and the system rings
forever.
y(t)
ωn
2π/ωn
t
-ωn
© w t norris 2nd order transfer functions Monday 3 March 2008
Second order transfer functions
9
ζ > 1 The roots are real. We had eq. 3: –
s1,s2 = ωn ( – ζ ± √
ζ2 – 1 ) ; s1 > s2

…3
ωn2
Y = (s–s )(s-s ) × 1
1
2
Thus the response is
=
and
ωn2  1
1 
–

s1– s2  s-s1 s–s2 
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 exponentials
is omitted in the figure. The formula is :– y = 0·289( e–0·27 t – e–3·7 t)
y(t)
exp( s1t)
0.3
0.2
0.1
t
2
3
4
5
-0.1
-0.2
exp( s2t)
-0.3
ζ=1
Y=
so
ωn2
ωn2
×
1
=
(s + ωn)2
(s + ωn)2
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.
ωn2
Y = H(s) × 1 = 2
s + 2ζ ω n s + ω n2
with
=
ωn2
(s + ζ ωn)2 – ζ2ωn2 + ωn2
=
ωn2
(s + ζ ωn)2 + ωd2
1 – ζ2
ωd = ωn √

This frequency appears often and should be noted.
© w t norris 2nd order transfer functions Monday 3 March 2008
…6
…7
Second order transfer functions
10
ω2
ωd
Y = ωn ×
d
(s + ζ ωn)2 + ωd2
Thus
so, from the tables
ω2
y(t) = ωn exp(-ζωnt) sin(ωdt)
d
=
ωn
1–ζ2

√
…8a
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)
y(t)
0.8
0.6
0.4
0.2
25 t
10
5
-0.2
-0.4
-0.6
15
20
2π
The period of the oscillation, T = ω . The ratio of succeeding positive peaks is then: –
d
 2πζω 
exp(-ζωnt)
= exp ω n
 d 

 2π 
exp–ζωn t + ω  
d 


δ=
The logarithm of this ratio is
so
2πζωn
2πζωn
ωd = ω 1–ζ2
δ=
n√

2πζ
1–ζ2

√
≈ 2πζ if ζ << 1
…9
δ is called the logarithmic decrement or log-dec for short. It is a useful way of
measuring the damping in a highly resonant system.
Note we use the symbol δ with another meaning later.
© w t norris 2nd order transfer functions Monday 3 March 2008
Second order transfer functions
11
Step function responses
The input is f = u(t)
f= u(t)
so
1
1
F=s
t
The response is, using H(s) from eq. 1
Y=
ωn2
1
×s
2
2
s + 2ζ ω n s + ω n
ζ=0
Y=
ωn2
A
Bs + C
≡ + 2
s( s2 + ω n2 ) s
s + ω n2
Thus
1 ≡ A(s2 + ω n2) + s (Bs+C)
ω n2 = A ω n2 : ∴ A = 1
0=A+B: ∴ B=–A=–1
0=C:
∴C=0
…10
Cases
Put s = 0:
Coeff. s2 :
Coeff, s :
and
1
s

Y =  s – 2
2
s + ω n 

y(t) = ( 1 – cos(ωn t))
The plot for ωn = 1 is on the right.
The system oscillates about the mean
indefinitely.
y(t)
2
1
t
2
4
6
8
10
ζ > 1 The transfer function is factorisable and the roots are real. As before eq. 3 gives the
roots: –
s1,s2 = ωn ( – ζ ± √
ζ2 – 1 ) ; s1 > s2

and the response is
ωn2
1
Y = (s–s )(s-s ) × s
1
2
We have
ω 2`
A
B
B
Y = s(s–s n)(s-s ) ≡ s + s-s + s-s
1
2
1
2
© w t norris 2nd order transfer functions Monday 3 March 2008
…3
Second order transfer functions
1 ≡ A(s – s1)(s – s2) + Bs(s–s2) + Cs(s – s1)
ω n2
s2
ω n2 = B s1(s1 – s2) ∴ B =
=
s 1 (s 1 – s 2 ) s 1 – s 2
ω n2
s1
ω n 2 = C s2 ( s 2 – s1 ) ∴ C = –
=s 2 (s 1 – s 2 )
s1 – s2
ω n2
ω n 2 = A s1 s 2
∴ A=
=1
s1s2
whence
Put s = s1:
Put s = s2 :
Put s = 0 :
1
Y= s +
ω
So
12
1
 s2
s1 
ζ2– 1
 s–s – s–s  ) ; since s1 s2 = ωn2 ; s1 – s2 = ωn√

2
1
2


ζ
–
1
n√

1
[ s2 exp(s1t) – s1 exp(s2t) ])
2
ωn√
ζ
–
1

and
y(t) = 1 +
noting
ζ2 – 1 ) ; s1 > s2
s1,s2 = ωn ( – ζ ± √

y(t)
With ωn =1 and ζ = 2 this becomes
1
y(t) = 1 + 0·077
e–3·73 t
– 1·077
e–0·27 t
0.8
0.6
0.4
0.2
which is illustrated on the right.
t
10
15
ωn2
ωn2
1
×
=
(s +ωn)2 s
s(s + ωn)2
A
B
C
≡ s + s+ω +
, say
n (s+ωn)2
ζ=1
Y=
Thus
ω n 2 ≡ Α (s + ω n)2 + Bs(s + ω n) + Cs
2
2
Put s = 0 : ω n = A ω n ∴ A =1. Put s = – ω n : ω n2 = – C ω n ∴ C = – ω n
Coeff. s2 :
0=A+B
∴ B = – A = –1
ωn
1
1
Y = s – s+ω –
n
(s + ωn)2
Thus from the Tables:
y(t)
1
y(t) = 1 – exp( – ωnt ) [1 + ωnt ]
For ωn = 1 y(t) = 1 – e–t( 1+t)
t
This is illustrated on the right.
2
4
6
8
ζ < 1 The more complicated but more interesting case. From eq. 1 we have
ωn2
1
×s
Y= 2
2
s + 2ζ ω n s + ω n
© w t norris 2nd order transfer functions Monday 3 March 2008
10
Second order transfer functions
=
ωn2
s(s2 + 2ζ ω n s + ω n2 )
A
=s +
Thus
and
13
Bs + C
, say.
s2 + 2ζ ω n s + ω n2
1 ≡ A (s2 + 2ζ ω n s + ω n 2 ) + s (B s
Put s = 0 : ω n 2 = A ω n 2
∴
Coeff. s2 : 0 = A + B
∴
Coeff. s :
0 = A 2 ζ ωn + C
∴
s
+
2ζω
1
n
Y= s – 2
s + 2ζ ω n s + ω n2
+C)
A=1
B = – A = –1
C = – 2ζωn A = – 2ζωn
s + 2ζωn
1
= s–
: completing the square as in eq. 6
(s + ζ ωn)2 + ωd2
ζωn
(s + ζωn) + ω ωd
1
d
: making the numerator to suit the Tables
= s–
2
(s + ζ ωn) + ωd2
with
Thus
1 – ζ2
ωd = ωn √

ζω


y(t) = 1 – exp( – ζωn t)  cos( ωd t) + ω n sin(ωd t) 
d


For ωn = 1 and ζ = 0·2
we have
…11
y(t)
2
1+ e-t
y(t) ≈ 1 – e– 0·2 t [ cos( 0·98 t ) + 0·2
sin( 0·98 t) ]
This is plotted on the right.
1
Note that the gradient is zero at t = 0
1- e-t
t
5
© w t norris 2nd order transfer functions Monday 3 March 2008
10
15
20
Second order transfer functions
14
There are properties of interest of the curve.
y(t)
MP
Overshoot,
OP
1
δ
Period,
T = 2π
ω
t
d
Tp
10
Ts
20
Overshoot
dy ω n2
= ω exp(–ζωnt) sin(ωdt)
dt
d
The derivative
is remarkably simple. It is equal to the impulse response. Indeed it is generally the case that
the derivative of the step response of any linear system is the impulse response.
dy
dt = 0 when
nπ
ωd t = nπ or when t = ω =
d
2π
T=ω =
where the period of the oscillations is
The value of y at these extrema is
d
nπ
nT
= 2
1-ζ2
ωn√

2π
1-ζ2
ωn√

 nπ ζ 
Y = 1 + exp –


1–ζ2

√
The first of these when n = 1 gives the peak of the overshoot ( often simply called the
overshoot.
 πζ 
ΜP = 1 + exp –

1–ζ2
 √

Overshoot, Op
1
Express proportional overshoot as the
excess as a proportion of the final steady
state value.
 πζ 
ΟP = y –1 = exp –


1–ζ2

√
0.6
0.4
…12a
0.2
occurring at
TP =
0.8
0
π
π
T
= ω =2
d
1-ζ2
ωn√

…12b
© w t norris 2nd order transfer functions Monday 3 March 2008
0
0.2
0.4
0.6
0.8
1
Damping, ζ
Second order transfer functions
15
We may be interested in the time to reach and stay within a certain 'distance' from the final
value. The deviation of the nth extremum is
 nπζ 
δ = exp –
 which occurs at
2
1–ζ
 √
 
tn =
nπ
or
2
1
–
ζ
ωn√

so
 nπζ 
δ = exp –
 = exp( – ωn ζ tn)

2
1–ζ
 
√
and
ln(δ)
Ts = tn = – ζω
n
nπ
= tnωn
2
1
–
ζ

√
Strictly tn should be an integral number of half periods but the discrepancy is usually
ignored.
4
Ts ≈ ζω
For δ = 0·02, i.e., 2%
n
3
Ts ≈ ζω
and for δ = 0·05, i.e., 5%
n
We may also consider an overall expression of deviation from the final value. The root
mean square seems a suitable choice.
(y-1)2
y(t) - 1
1
1
0.8
0.6
t
5
10
15
0.4
20
0.2
t
-1
5
15
10
20
The deviation and its square are illustrated above for the previously shown response. We
take then as a measure of the overall deviation as
Erts = the total square deviation ( integrated from t = 0 to t = ∞ )
Erts
We require
 ∞
 1/2

Erts =  ∫ ( y –1)2 dt 
 0

=
√

1+4 ζ2
4ωnζ
This rather simple expression is
valid for all ζ, i.e., 0 < ζ < ∞. It has
1
a minimum value of 1 when ζ = 2 .
2.2
2
1.8
1.6
1.4
1.2
ζ
1.0
© w t norris 2nd order transfer functions Monday 3 March 2008
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2
3
4
5
Second order transfer functions
This minimum square deviation
condition is illustrated on the right
with a 16% overshoot..
16
y(t)
1.16
1
t
5
© w t norris 2nd order transfer functions Monday 3 March 2008
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15
Second order transfer functions
17
There are other measures of the error. To put emphasis on the approach to the final value on
might weight erros by muptiplying by time. The integral of this de-emphasises the initial
approach to the final value.. Thus we have
t (y - 1)2
The root of the integral of time weighted
total square erroe is
0.8
 ∞
 1/2

Erts =  ∫ t ( y –1)2 dt 

 0
0.6
0.4
0.2
1
= 2ω ζ
n
√

1+ 8 ζ4
2
1
This has a minimum of 1/4
when
2 ωn
 1  1/4
ζ = 8 
≈ 0·595
 
© w t norris 2nd order transfer functions Monday 3 March 2008
t
5
10
15
20
The overshoot is
 ζπ 
π 
exp –
 = exp –
 ≈ 9·8%
 1 – ζ2
 √
2√2–1


√
Second order transfer functions
18
Keeping factors constant
1–ζ2 ) :
s =ωn( – ζ ± √

Poles are at
Constant imaginary part.
y(t)
jω
ωd = ωn 
1–ζ2 = constant.
√
So Constant frequency in
response
s
σ
1
t
Constant real part:–
ωn ζ is constant.
y(t)
2
jω
s
Fixed envelope
y =1 – exp( – ζωn t)
σ
1
and so time to get within a
given limit is deermined by
ωnζ.
4
E.g., Tδ=2% = ζω
n
t
y(t)
Constant damping ratio, ζ
Thus
also
overshoot:
jω
θ
constan
s
1
σ
 πζ 
OP = exp –

1–ζ2
 √

t
Note: ζ = sinθ where θ is
marked in the figure.
There is scarcely any
overshoot when θ > 60˚.
2
Fractional overshoot,
OP
1.8
θ
jω
s
σ
1.6
1.4
1.2
θ
20
© w t norris 2nd order transfer functions Monday 3 March 2008
40
60
90
Second order transfer functions
© w t norris 2nd order transfer functions Monday 3 March 2008
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