System identification based on Step and Impulse response considering first and
second order transfer function models
These notes discuss system identification based on the step and impulse response. In the
following we consider linear, time-invariant systems of first and second order, as they
provide reasonable approximation for the description of the dominant behavior of most
linear time-invariant systems.
I. First order system
A first order system is described in frequency domain by the transfer function
k
G (s) =
t s +1
where the parameters k and t are the system gain ( k ) and the time constant of the
system ( t ).
The step response of this system, considering initial conditions equal with 0, is obtained
as follows:
a. the Laplace transform of the system input is
1
U (s) =
s
b. using the relation Y ( s ) = H ( s )U ( s ) one obtains the Laplace transform of the output
signal
k 1
Y (s) =
t s +1 s
c. we use partial fractions expansion and then inverse Laplace transform to obtain the
system response in time domain
A
B
Y (s) =
+
t s +1 s
k
k
A = Y ( s )(t s + 1) s =- 1 =
= -t k and B = Y ( s ) s s =0 =
=k
s s =- 1
t s + 1 s =0
t
t
tk
k
k
k
Y (s) = + =+
t s +1 s
s + 1/ t s
Inverse Laplace transform results in
y (t ) = - ke
-t
1
t
+ ku-1 (t )
ì0, t < 0
where u-1 (t ) denotes the step function u-1 (t ) = í
.
î1, t ³ 0
Figure 1 shows the step response of a system with the transfer function G ( s ) =
1
2
.
3s + 1
Step Response
2
1.8
System: sys1
Time (sec): 6
Amplitude: 1.73
1.6
1.4
System: sys1
Time (sec): 2
Amplitude: 0.971
Amplitude
1.2
1
0.8
0.6
0.4
0.2
0
0
2
4
6
8
10
12
14
16
18
Time (sec)
Figure 1. Step response of the system G ( s ) =
2
3s + 1
The identification procedure, i.e. finding the values of the two parameters k ,t describing
the system dynamics, amounts to solving a system of two equations defined by two
points on the system’s step response (t1 , y (t1 )) , (t2 , y (t2 )) .
1
ì
-t1
t
+ ku-1 (t1 )
ïï y (t1 ) = -ke
í
1
-t 2
ï
t
+ ku-1 (t2 )
ïî y (t2 ) = -ke
Dividing the two equations, and considering that u-1 (t ) = 1, t > 0 , one gets
1
- (t1 -t2 )
y (t1 ) - k
t
=e
y (t2 ) - k
which means
æ y (t ) - k ö 1
1
ln ç 1
÷=
t2 - t1 è y (t2 ) - k ø t
or
æ
æ y (t ) - k ö ö
t = (t2 - t1 ) çç1/ ln ç 1
÷ ÷÷ .
è y (t2 ) - k ø ø
è
2
This equation gives the value of the time constant of the system if y (t2 ), y (t1 ), t1, t2 , k are
known.
In order to determine the value of k one notes that the steady state value of the step
response is lim y (t ) = k which can be easily measured from the step response graph.
t ®¥
Question 1:
Use the graph in Fig. 1 and the relations given above to verify that the step response
2
presented in the figure is indeed the step response of a system described by G ( s ) =
.
3s + 1
Notes:
a. Remember also the final value theorem lim y (t ) = lim Y ( s ) , which means that the
t ®¥
s ®0
gain of the system at zero frequency, i.e. the DC gain, is k .
a
b. If the step input is not of amplitude 1, say U ( s ) =
where a denotes the
s
amplitude of the step input, then the step response of the system is
y (t ) = a (- ke
-t
1
t
+ ku-1 (t )) . Then the steady state value of the system’s step
response is lim y (t ) = ak . In this case, in order to determine the value of the
t ®¥
system’s DC gain one needs to divide the measured steady state value by the
amplitude of the step input.
c. Let t = t then y (t ) = k (1 - e-1 ) = k * 0.6321 thus the system’s time constant is the
time moment when the system step response reaches approximately 63% of the
steady state value.
d. Let t = 5t then y (t ) = k (1 - e-5 ) = k * 0.993 thus at t = 5t the system step
response reaches more than 99% of the steady state value.
k
k /t
q
e. G ( s ) =
can also be written as G ( s ) =
=
where - p is the pole
t s +1
s + 1/ t s + p
of the system.
The impulse response of this system, considering initial conditions equal with 0, is
1
k -t
y (t ) = e t u-1 (t )
t
Measuring the value of the impulse response at positive time moments, say t1 and t2 ,
one can determine the time constant of the system using
1
(t2 -t1 )
y (t1 )
t ,
=e
y (t2 )
3
æ y (t ) ö
thus t = (t2 - t1 ) / ln ç 1 ÷ .
è y (t2 ) ø
The time constant being known the system gain can be determined from
k = t y (t3 ) / e
-t3
1
t
for any t3 ³ 0 .
k
.
t
Figure 2 shows the impulse response of the system described by the transfer function
2
G (s) =
.
3s + 1
Looking at the system’s impulse response one sees that at t = 0 y (0) =
Impulse Response
0.7
System: sys1
Time (sec): 0
Amplitude: 0.667
0.6
Amplitude
0.5
System: sys1
Time (sec): 2
Amplitude: 0.343
0.4
0.3
System: sys1
Time (sec): 4
Amplitude: 0.176
0.2
0.1
0
0
2
4
6
8
10
12
14
16
Figure 2. Impulse response of the system G ( s ) =
2
.
3s + 1
18
Time (sec)
Question 2:
Use the graph in Fig. 2 and the relations given above to verify that the impulse response
2
presented in the figure is indeed the impulse response of the system G ( s ) =
.
3s + 1
4
II. Second order system (complex pole pair)
A second order system with no finite zeros is described in frequency domain by the
transfer function
kw 2
k0wn 2
k0wn 2
H ( s) = 2 0 n
=
º
D( s)
s + 2a s + wn 2 s 2 + 2zwn s + wn 2
The numerator is chosen to scale the transfer function so that the DC gain (which can be
calculated by lim H ( s ) ) is equal to k0 .
s ®0
The denominator is the Characteristic polynomial which can be written in several
canonical forms, including
2
2
D( s ) = s 2 + 2as + w n = s 2 + 2zw n s + w n .
One may write
2
D( s ) = s 2 + 2as + w n = ( s + a ) 2 + b 2
where
2
b 2 +a 2 = wn .
2
Thus if a 2 < w n , the polynomial D(s ) describes a complex pair of poles at
s = -a ± jb .
Figure 3 shows the location of the two poles in the s-plane. The real part of the poles is
-a and the imaginary part is jb. The norm of the vector from the origin to the pole is wn,
which is known as the natural frequency.
Figure 3. Location of the complex poles of a second order system in the s-plane
Notice that the system dynamics is completely described by the triplets (k0 , a , b ) or
( k 0 , z , wn ) .
To link the two forms of the characteristic polynomial we define the damping ratio as
a
z =
,
wn
which defines it as
5
a
.
wn
For complex poles in the left-half plane one has 0 < z < 1 . If 0 > z > -1 then one has a
complex pair in the right-half plane (e.g. unstable complex pair). Note that one may
write
z = cos q = - cos j =
2
b = wn -a 2 = wn 1-z 2 .
The impulse response of the system with transfer function
kw 2
H ( s ) = 2 0 n2
(s + a ) + b 2
is given by
wn 2 -a t
y (t ) = k0
e sin b t u-1 (t ) ,
b
which is plotted for 0 < z < 1 in the figure. This is known as the underdamped case. The
figure clearly shows the meaning in the time domain of the real part -a of the poles,
which provides the exponential decay term. Having in mind the standard form for
sinusoids sin 2p t , the period of the oscillation is given by
T
2p
T=
.
b
The variable b is known as the oscillation frequency.
Figure 4. Sketch of the impulse response of a second order system with complex poles
If z = 0 then a = 0, b = w n and the poles are at s = ± jb on the imaginary axis. This is
known as the undamped case. If z = 1 then b = 0,a = w n and the poles are on the real
axis, both at s = -a . In this overdamped case, the impulse response has the form te -at .
If z > 1 then there are two real poles and we can split the quadratic factor
2
D( s ) = s 2 + 2as + w n into two real linear factors.
6
In order to determine the parameters of the system, one can simply determine the period
2p
of oscillation T and subsequently use T =
to determine b .
b
Then by measuring the values of the impulse response at two moments in time when
sin b t = 1 one has
wn 2 -a t1
w 2
e , y (t2 ) = k0 n e -a t2 .
b
b
y (t1 )
Dividing the two one gets
= ea (t2 -t1 ) and from here
y (t2 )
y (t1 ) = k0
æ y (t ) ö
a = ln ç 1 ÷ / (t2 - t1 ) .
è y (t2 ) ø
Knowing the values a , b and y (t3 ) = k0
wn 2 -a t3
e
sin b t3 one can also determine the
b
system DC gain k0 .
The step response of this second order system is given by inverse transforming
k0wn 2
1
Y ( s ) = H ( s )U ( s ) = H ( s ) =
s s [( s 2 + a )2 + b 2 ]
to obtain
w
a
y (t ) = k0 (1 - e-a t [cos b t + sin b t ])u-1 (t ) = k0 (1 - n e-a t sin( b t + q ))u-1 (t )
b
b
where the angle q = arctan b is shown in Figure 3.
a
The step response of a underdamped second order system is presented in Figure 5.
Figure 5. Sketch of the step response of a second order system with complex poles
The steady state value of the system’s step response is yss = lim y (t ) = k0 .
t ®¥
7
An important quantity for characterizing the performance of systems is the percent
overshoot (POV) in the step response. This is defined as
y
- yss
POV = max
´ 100%
yss
where ymax is the maximum value of the step response and yss is its steady-state value.
The rise time, tr, is the time required for the step response to rise from 0.1 to 0.9 of its
steady-state value.
The settling time ts is the time required for the signal to effectively reach its steady-state
value.
Note that for a first order system one has t r = 2.2t and t s = 5t (some take t s = 4t ).
For the underdamped pole pair, the time constant is t = 1 and one may use t s = 5t to
a
calculate the settling time. In this case however the signal rises faster and one may
2.16z + 0.6
.
approximate, for 0.3 £ z £ 0.8 , using t r =
wn
The percent overshoot is a function of damping ratio
POV = 100e -p z /
and conversely
(
1-z 2
)
ù
é ln 2 POV
100 ú
z =ê 2
ê ln POV
+p 2 ú
100
û
ë
(
)
1/ 2
.
To determine the parameters of the system, from the system’s step response, one can
simply calculate the POV and then determine the damping factor z . Then by measuring
the settling time one can use t s = 5t to determine the time constant of the system and
then a . Measuring the steady state value of the step response one gets the system DC
gain k0 = yss .
a
Note that in the case in which the system input is not a unit step, e.g. U ( s ) = , then
s
y
yss = ak0 and thus k0 = ss .
a
In Figure 6 are compared the step responses of the two systems H1 ( s ) =
H 2 (s) =
6
2
s + 5s + 6
1
and
0.4 s + 1
which have the same time constant t = 0.4 . (Notice the difference.)
8
Step Response
1
0.9
0.8
0.7
Amplitude
0.6
0.5
0.4
0.3
0.2
second order system
first order system
0.1
0
0
0.5
1
1.5
2
2.5
3
3.5
Time (sec)
Figure 6. Step response comparison of a first order system and a second order,
overdamped, system. Both systems have the same time constant thus the same settling
time.
9
You can consider solving some of the following questions.
1. The following figures present the step responses (to a unit step) of two first order
systems. Determine the transfer functions of the two systems. To help with the
calculations, some points were given explicitly on the graph.
Step Response
2
System: tff
Time (sec): 25.1
Amplitude: 1.99
1.8
System: tff
Time (sec): 6.07
Amplitude: 1.41
1.6
1.4
Amplitude
1.2
1
0.8
0.6
0.4
0.2
0
0
5
10
15
20
25
30
1
1.2
Time (sec)
Figure 1. Step response of a system
Step Response
0.4
0.35
System: x
Time (sec): 0.2
Amplitude: 0.253
0.3
System: x
Time (sec): 0.5
Amplitude: 0.367
Amplitude
0.25
0.2
0.15
0.1
0.05
0
0
0.2
0.4
0.6
0.8
Time (sec)
Figure 2. Step response of a system
10
2. The following figure presents the step responses (to a unit step) of a system.
Determine the model transfer function of the system.
To help with the calculations, some points were given explicitly on the graph.
Step Response
1.6
System: x
Time (sec): 0.32
Amplitude: 1.53
1.4
System: x
Time (sec): 1.59
Amplitude: 1.04
1.2
Amplitude
1
System: x
Time (sec): 2.5
Amplitude: 0.994
System: x
Time (sec): 1.35
Amplitude: 0.95
0.8
System: x
Time (sec): 0.624
Amplitude: 0.717
0.6
0.4
0.2
0
0
0.5
1
1.5
2
2.5
3
Time (sec)
Figure 3. Step response of a system
Consider that the steady state is obtained when the signal enters in the ±1% of the steady
state value (this is related with the value of the settling time).
11
3. The following figure presents the impulse response of a second order system.
Determine the model transfer function of the system.
To help with the calculations, some points were given explicitly on the graph.
Impulse Response
8
6
System: syst
Time (sec): 0.155
Amplitude: 7.68
Amplitude
4
2
System: syst
Time (sec): 0.865
Amplitude: 3.83
System: syst
Time (sec): 2.26
Amplitude: 0.947
0
-2
-4
-6
0
System: syst
Time (sec): 1.93
Amplitude: -1.32
System: syst
Time (sec): 0.51
Amplitude: -5.43
1
System: syst
Time (sec): 4.19
Amplitude: 0.00273
2
3
4
Time (sec)
Figure 4. Impulse response of a system
12
5
6