1
Relationship Between Transfer Function Pole
Locations and Time-Domain Step Responses
A. Introduction
This example looks at the relationship between the locations of the poles of the transfer function H(s) for a
linear, time-invariant (LTI) system and the response of the system in the time domain to a unit step input u(t).
The important characteristics of the step response that will be considered are the general shape of the response,
the value of the response as t → ∞, and the time it takes the response to settle near its final value if it does settle.
These characteristics will be related to the real and imaginary parts of the closed-loop poles.
For a system modeled by a transfer function with distinct poles p1 , p2 , . . . , pn subjected to a step input, the
transform of the zero-state response is
Yzs (s) = H(s) ·
N(s)
A0
A1
A2
An
1
=
=
+
+
+ ··· +
s
s (s − p1 ) (s − p2 ) · · · (s − pn )
s
(s − p1 ) (s − p2 )
(s − pn )
(1)
and the output signal in the time domain is
£
¤
yzs (t) = A0 + A1 ep1 t + A2 ep2 t + · · · + An epn t u(t)
(2)
If a pole pi is real, then the corresponding term in (2) is a decaying exponential if pi < 0 and a growing exponential
if pi > 0. If a pole is complex, pi = α + jβ , then the corresponding term in (2) is a sinusoid of frequency β rad/sec
multiplied by an exponential which is decaying if α < 0 or growing if α > 0. If α = 0, then the term represents a
sinusoid of constant amplitude. For either real or complex poles, if Re [pi ] < 0, then the pole lies in the left half
of the complex s-plane (LHP). If Re [pi ] > 0, then the pole lies in the right half of the complex s-plane (RHP),
and if Re [pi ] = 0, then the pole lies on the jω axis which separates the left and right halves of the s-plane.
If the transfer function has a repeated pole, such as (s − p1 )2 , then in addition to the exponential terms in
(2), there will also be a term involving tep1 t . For repeated poles with multiplicity m, (s − p1 )m , the time-domain
expression will contain terms in the time variable up to tm−1 , each multiplied by the exponential ep1 t .
The transfer function in this example has 2 poles (p1 , p2 ) and no zeros. The locations of the poles for the example
are such that p1 × p2 = 1. The transfer function has the form
H(s) =
1
1
1
= 2
= 2
(s − p1 ) (s − p2 )
s − (p1 + p2 ) s + p1 p2
s − (p1 + p2 ) s + 1
(3)
The locations of the poles are shown below.
Trial
p1
p2
1
−10
−0.1
2
−4
−0.25
3
−1
−1
4
−0.8 + j0.6
−0.8 − j0.6
5
−0.5 + j0.866
−0.5 − j0.866
6
−0.375 + j0.927
−0.375 − j0.927
7
j1
−j1
8
0.5 + j0.866
0.5 − j0.866
9
1
1
10
4
0.25
For each of these trials, the pole locations will be shown in the s-plane and the corresponding step response will
be plotted. A discussion of the relationships between the pole locations and the step response will be given.
2
B. Trial 1
1
1
=
2
s (s + 0.1) (s + 10)
s (s + 10.1s + 1)
£
¤
−0.1t
yzs (t) = 1 − 1.0101e
+ 0.0101e−10t u(t)
Yzs (s) =
(4)
(5)
The system is bounded-input, bounded-output (BIBO) stable since the poles of H(s) are in the left-half plane
(LHP). The system’s step response settles at a final value of 1, which is the final value of the unit step input. The
output asymptotically approaches the final value, with the length of time to get “close” to the final value being
controlled by the pole of H(s) that is the closest to the jω axis, that is, the pole at s = p1 = −0.1. The exponential
decay of e−0.1t is much slower than the decay of e−10t . At t = 40 seconds, the output is within 2% of the final
value. The response is well behaved, but slow. There is no overshoot of the final value, and there are no oscillations
in the response. Fig. 1 shows the pole locations of H(s) and the step response yzs (t) for Trial 1.
C. Trial 2
1
1
=
s (s + 0.25) (s + 4)
s (s2 + 4.25s + 1)
£
¤
yzs (t) = 1 − 1.0667e−0.25t + 0.0667e−4t u(t)
Yzs (s) =
(6)
(7)
This system is also BIBO stable since the poles of H(s) are in the left-half plane. The system’s step response
settles at a final value of 1, which is the final value of the unit step input. The output asymptotically approaches
the final value, with the length of time to get close to the final value being controlled by the pole of H(s) that is
the closest to the jω axis, that is, the pole at s = p1 = −0.25. At t = 16 seconds, the output is within 2% of the
final value. As the pole closest to the jω axis moves to the left, the response gets faster. The response is still well
behaved. There is no overshoot of the final value, and there are no oscillations in the response. Fig. 2 shows the
pole locations of H(s) and the step response yzs (t) for Trial 2.
D. Trial 3
1
1
2 = s (s2 + 2s + 1)
s (s + 1)
£
¤
yzs (t) = 1 − e−t (t + 1) u(t)
Yzs (s) =
(8)
(9)
This system is also BIBO stable since the poles of H(s) are in the left-half plane. The system’s step response
settles at a final value of 1, which is the final value of the unit step input. The output now has a term involving
te−t since there is a repeated pole. The response is faster than the previous case, with the output getting close to
its final value in approximately 6 seconds. There is no overshoot of the final value, and there are no oscillations in
the response. Fig. 3 shows the pole locations of H(s) and the step response yzs (t) for Trial 3.
E. Trial 4
1
1
=
2
s (s + 0.8 − j0.6) (s + 0.8 + j0.6)
s (s + 1.6s + 1)
£
¤
−0.8t
−0.8t
yzs (t) = 1 − e
cos(0.6t) − 1.3333e
sin(0.6t) u(t)
Yzs (s) =
(10)
(11)
This system is also BIBO stable since the poles of H(s) are in the left-half plane. The system’s step response
settles at a final value of 1, which is the final value of the unit step input. The output no longer asymptotically
approaches the final value, but now has a slight bit of overshoot of the final value. The maximum value of the
output is 1.0152. The length of time to get close to the final value is controlled by the real part of the complex
poles of H(s), that is, by s = Re [p1 ] = −0.8. The output is very close to its final value at approximately t = 3.75
seconds, and stays close to that value, even with the overshoot. Fig. 4 shows the pole locations of H(s) and the
step response yzs (t) for Trial 4.
3
Step Response: H(s) = 1/[s2 + 10.1s + 1]
1
0.8
0.6
Imag Axis
0.4
0.2
0
−0.2
−0.4
−0.6
−0.8
−1
−10
−8
−6
−4
Real Axis
−2
0
Step Response: H(s) = 1/[s2 + 10.1s + 1]
1
0.9
0.8
Amplitude
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
Fig. 1.
0
5
10
15
H(s) pole locations and step response for Trial 1.
20
25
Time (s)
30
35
40
45
50
4
Step Response: H(s) = 1/[s2 + 4.25s + 1]
1
0.8
0.6
Imag Axis
0.4
0.2
0
−0.2
−0.4
−0.6
−0.8
−1
−5
−4
−3
−2
Real Axis
−1
0
1
14
16
Step Response: H(s) = 1/[s2 + 4.25s + 1]
1
0.9
0.8
Amplitude
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
Fig. 2.
0
2
4
6
H(s) pole locations and step response for Trial 2.
8
10
Time (s)
12
18
20
5
Step Response: H(s) = 1/[s2 + 2s + 1]
1
0.8
0.6
Imag Axis
0.4
0.2
0
−0.2
−0.4
−0.6
−0.8
−1
−1.5
−1
−0.5
Real Axis
0
0.5
Step Response: H(s) = 1/[s2 + 2s + 1]
1
0.9
0.8
Amplitude
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
Fig. 3.
0
1
2
3
H(s) pole locations and step response for Trial 3.
4
5
Time (s)
6
7
8
9
10
6
Step Response: H(s) = 1/[s2 + 1.6s + 1]
1
0.8
0.6
Imag Axis
0.4
0.2
0
−0.2
−0.4
−0.6
−0.8
−1
−1
−0.5
0
Real Axis
0.5
1
Step Response: H(s) = 1/[s2 + 1.6s + 1]
1.4
1.2
Amplitude
1
0.8
0.6
0.4
0.2
0
Fig. 4.
0
1
2
3
H(s) pole locations and step response for Trial 4.
4
5
Time (s)
6
7
8
9
10
7
F. Trial 5
1
1
=
2
s (s + 0.5 − j0.866) (s + 0.5 + j0.866)
s (s + s + 1)
£
¤
−0.5t
−0.5t
yzs (t) = 1 − e
cos(0.866t) − 0.57737e
sin(0.866t) u(t)
Yzs (s) =
(12)
(13)
This system is also BIBO stable since the poles of H(s) are in the left-half plane. The system’s step response
settles at a final value of 1, which is the final value of the unit step input. The output now has a noticeable amount
of overshoot of the final value. The maximum value of the output is 1.1631, that is, the output overshoots its final
value by 16.31% of the size of the step input. The length of time to get close to the final value is controlled by
the real part of the complex poles of H(s), that is, by s = Re [p1 ] = −0.5. The output is very close to its final
value at approximately t = 8 seconds. This time is longer than in the previous case since the real part of the poles
is closer to the jω axis. The overshoot and oscillations are more prominent since the imaginary part of the poles
is larger in Trial 5 than in Trial 4, particularly with respect to the real part of the poles. Fig. 5 shows the pole
locations of H(s) and the step response yzs (t) for Trial 5.
G. Trial 6
1
1
=
2
s (s + 0.375 − j0.927) (s + 0.375 + j0.927)
s (s + 0.75s + 1)
£
¤
−0.375t
−0.375t
yzs (t) = 1 − e
cos(0.927t) − 0.40453e
sin(0.927t) u(t)
Yzs (s) =
(14)
(15)
This system is also BIBO stable since the poles of H(s) are in the left-half plane. The system’s step response
settles at a final value of 1, which is the final value of the unit step input. The output now has even more overshoot
of the final value. The maximum value of the output is 1.2806, that is, the output overshoots its final value by
28.06% of the size of the step input. The length of time to get close to the final value is controlled by the real
part of the complex poles of H(s), that is, by s = Re [p1 ] = −0.375. The output is very close to its final value at
approximately t = 10 seconds. As before, this time is longer than in the previous case since the real part of the
poles is closer to the jω axis. The overshoot and oscillations are more prominent since the imaginary part of the
poles is larger in Trial 6 than in Trial 5, particularly with respect to the real part of the poles. The absolute value
of the ratio of the imaginary parts of the pole to the real parts of the pole increases from 0.75 to 1.732 to 2.472
in going from Trial 4 to Trial 5 to Trial 6. Fig. 6 shows the pole locations of H(s) and the step response yzs (t)
for Trial 6.
H. Trial 7
1
1
=
2
s (s − j1) (s + j1)
s (s + 1)
yzs (t) = [1 − cos(t)] u(t)
Yzs (s) =
(16)
(17)
This system is not BIBO stable since the poles of H(s) are on the jω axis. If the input signal contained a sine
or cosine term with frequency 1 rad/sec, then Yzs (s) would contain repeated roots, which always give rise to a
term involving t. Since the poles are on the jω axis, there would be no exponential attenuation of that term, and
it would become unbounded, that is, yzs (t) → ∞ as t → ∞. The system’s step response does not settle at a final
value of 1, rather it oscillates from 0 to 2 with an average value of 1 for all t ≥ 0.. Fig. 7 shows the pole locations
of H(s) and the step response yzs (t) for Trial 7.
8
Step Response: H(s) = 1/[s2 + s + 1]
1
0.8
0.6
Imag Axis
0.4
0.2
0
−0.2
−0.4
−0.6
−0.8
−1
−1
−0.5
0
Real Axis
0.5
1
Step Response: H(s) = 1/[s2 + s + 1]
1.4
1.2
Amplitude
1
0.8
0.6
0.4
0.2
0
Fig. 5.
0
1
2
3
H(s) pole locations and step response for Trial 5.
4
5
Time (s)
6
7
8
9
10
9
Step Response: H(s) = 1/[s2 + 0.75s + 1]
1
0.8
0.6
Imag Axis
0.4
0.2
0
−0.2
−0.4
−0.6
−0.8
−1
−1
−0.5
0
Real Axis
0.5
1
Step Response: H(s) = 1/[s2 + 0.75s + 1]
1.4
1.2
Amplitude
1
0.8
0.6
0.4
0.2
0
0
5
10
Time (s)
Fig. 6.
H(s) pole locations and step response for Trial 6.
15
10
Step Response: H(s) = 1/[s2 + 1]
1
0.8
0.6
Imag Axis
0.4
0.2
0
−0.2
−0.4
−0.6
−0.8
−1
−1
−0.5
0
Real Axis
0.5
1
Step Response: H(s) = 1/[s2 + 1]
2
1.8
1.6
Amplitude
1.4
1.2
1
0.8
0.6
0.4
0.2
0
Fig. 7.
0
2
4
6
H(s) pole locations and step response for Trial 7.
8
10
Time (s)
12
14
16
18
20
11
I. Trial 8
1
1
=
2
s (s − 0.5 − j0.866) (s − 0.5 + j0.866)
s (s − s + 1)
£
¤
0.5t
0.5t
yzs (t) = 1 − e cos(0.866t) + 0.57737e sin(0.866t) u(t)
Yzs (s) =
(18)
(19)
The poles of H(s) in this trial are the mirror images about the jω axis of the poles in Trial 5; they are in
the right-half of the s-plane rather than the left, with the same magnitudes for the real and imaginary parts. The
time-domain expression is also a mirror image of the response in Trial 5. The exponential is now growing rather
than decaying, and the sine function in yzs (t) has the opposite sign. The system is not BIBO stable since the poles
of H(s) are in the RHP, resulting in a response that is unbounded. The output will continue to oscillate with a
frequency of 0.866 rad/sec as the amplitude grows exponentially, controlled by the e0.5t term. Fig. 8 shows the
pole locations of H(s) and the step response yzs (t) for Trial 8.
J. Trial 9
1
1
2 = s (s2 − 2s + 1)
s (s − 1)
£
¤
yzs (t) = 1 + et (t − 1) u(t)
Yzs (s) =
(20)
(21)
This system is also not BIBO stable due to the RHP poles in H(s). The pole locations and step response are
mirror images of Trial 3. The exponential is growing rather than decaying because of the positive exponent. The
response will not oscillate since the poles are real; however, the amplitude of the response grows much more quickly
than in Trial 8 since the poles are farther into the right-half plane. Fig. 9 shows the pole locations of H(s) and the
step response yzs (t) for Trial 9.
K. Trial 10
1
1
=
2
s (s − 0.25) (s − 4)
s (s − 4.25s + 1)
£
¤
0.25t
yzs (t) = 1 − 1.0667e
+ 0.0667e4t u(t)
Yzs (s) =
(22)
(23)
This system is also not BIBO stable due to the RHP poles in H(s). The pole locations and step response are
mirror images of Trial 2. The exponential is growing rather than decaying because of the positive exponents. The
response will not oscillate since the poles are real; however, the amplitude of the response grows much more quickly
than in Trial 8 or Trial 9 since the poles are farther into the right-half plane. At t = 5 seconds, the response in
Trial 9 had reached a magnitude of 600 while in Trial 10, the magnitude at t = 5 seconds is over 3 × 107 . For both
the stable and unstable systems, the speed of response is seen to be controlled by the pole in H(s) that is farthest
to the right in the s-plane. Fig. 10 shows the pole locations of H(s) and the step response yzs (t) for Trial 10.
12
2
Step Response: H(s) = 1/[s − s + 1]
1
0.8
0.6
Imag Axis
0.4
0.2
0
−0.2
−0.4
−0.6
−0.8
−1
−1
−0.5
0
Real Axis
0.5
1
Step Response: H(s) = 1/[s2 − s + 1]
50
40
30
Amplitude
20
10
0
−10
−20
−30
−40
0
1
2
3
4
5
Time (s)
Fig. 8.
H(s) pole locations and step response for Trial 8.
6
7
8
9
13
Step Response: H(s) = 1/[s2 − 2s + 1]
1
0.8
0.6
Imag Axis
0.4
0.2
0
−0.2
−0.4
−0.6
−0.8
−1
−0.5
0
0.5
Real Axis
1
1.5
Step Response: H(s) = 1/[s2 − 2s + 1]
600
500
Amplitude
400
300
200
100
0
Fig. 9.
0
0.5
1
1.5
H(s) pole locations and step response for Trial 9.
2
2.5
Time (s)
3
3.5
4
4.5
5
14
Step Response: H(s) = 1/[s2 − 4.25s + 1]
1
0.8
0.6
Imag Axis
0.4
0.2
0
−0.2
−0.4
−0.6
−0.8
−1
−1
0
2
Real Axis
3
4
5
Step Response: H(s) = 1/[s2 − 4.25s + 1]
7
3.5
1
x 10
3
Amplitude
2.5
2
1.5
1
0.5
0
Fig. 10.
0
0.5
1
1.5
2
H(s) pole locations and step response for Trial 10.
2.5
Time (s)
3
3.5
4
4.5
5