1. Effects of poles and zeros on performance of control
systems
1 Dominantly first order systems. Effects of additional poles
and zeros on performance of a first order control system
1.1 Step response of a first order system
Consider a first order system of the form
1
x = −ax + r,
˙ or, equivalently, X(s) = R(s).
s+a
Let r(t) be a unit step input, i.e, R(s) = 1 . Then, assuming x(0) = 0,
s
1 1/a 1/a
X(s) = = − ,
s(s + a) s s+a
and
1 1
x(t) = (1 − e−at ), and x(t) → as t → ∞.
a a
The output of the system monotonically approaches to the final value of 1/a; the rate of con-
vergence is exponential and is determined by the pole of the system at s = −a. Note that the
time constant of the transient is τ = 1/a is also related to the location of the pole. To achieve a
desired rate of convergence, the designer must place the pole at a corresponding location; e.g.,
using the state or output feedback; see Figure 1.
Example Suppose the plant is quite slow, its time constant is τ orig = 100 sec,
1
x = −0.01x + u,
˙ or, equivalently, X(s) = U (s).
s + 0.01
To speed up it to τdesired = 1 sec, use the feedback controller
u = −kx + r.
Then, the closed loop system is
1
x = −(0.01 + k)x + r,
˙ or, equivalently, X(s) = R(s).
s + 0.01 + k
To achieve the desired time constant of 1 sec, the pole of the closed loop system must be placed
at s = −1/τdesired = −1, i.e., we need a = (0.01 + k) = 1, which is achieved using the gain of
k = 0.99.
1

2. R U Y
+
Plant
−
K
PSfrag replacements
Feedback Controller
Figure 1: Static feedback control system.
1.2 The Effect of an Additional Pole
A first order system often serves as an approximation of a system of higher order. Therefore, it
is important to study the error of approximation. To this end, we examine the effect of additional
poles and zeros on a first-order system. This will explain the difference between an original
system and its first-order approximation.
We now examine the step response of a system whose transfer function is given by
Y (s) p 1
= = 1 . (1)
R(s) (s + 1)(s + p) (s + 1)( p s + 1)
That is, we are concerned with a 2nd order system which is approximated by the first order
approximation s+a , where a = 1.
1
Note on the form of the system. Eq. (1) is selected because its value at s = 0 equals 1 for
any p. We will see that the step response of the system will approach the value of 1 for large
values of t, independent of the value of p. This is convenient, since we wish to examine the step
response for different values of p. However, the conclusions drawn from the example will apply
to any second order system with two real poles.
Let R(s) = 1/s and perform a partial-fraction expansion on the resulting step response
transform Y (s):
p 1
p 1
Y (s) = = − p−1 + p−1 .
s(s + 1)(s + p) s s+1 s+p
The step response y(t) is given by
p −t 1 −pt
y(t) = 1 − e + e . (2)
p−1 p−1
p
We will consider this response as the sum of two terms. The first term is given by 1 − p−1
e−t ,
and the second term is p−1 e−pt .
1
Let us examine the step response of Eq. (2) for different values of p:
2

3. 1
0.9
0.8
0.7
0.6
0.5
0.4
0.3
p=infinity
p=10
0.2 p=1
p=0.5
0.1
0
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5
Figure 2: Step response of the system with the additional pole at −p.
pv=[.5 1 10 inf];
t=[0:.1:5];
yv=[];
for i=4:-1:1
p=pv(i);
G=tf(1,conv([1 1],[1/p 1]));
[y,x]=step(G,t);
yv=[yv y];
end
plot(t,yv(:,1),’-’,t,yv(:,2),’--’,t,yv(:,3),’-.’,t,yv(:,4),’.’)
legend(’p=infinity’,’p=10’,’p=1’,’p=0.5’);
grid
This code produces the step response of Eq. (2) for p = 0.5, 1, 10 and ∞; see Figure 2. For
p = ∞, the step response is the same as the step response of the system with the first-order
transfer function s+1 . As p reduces, the plot moves away from the step response of this first-
1
order system.
Indeed p/(p − 1)e−t → e−t as p → ∞, and 1/(p − 1)e−pt → 0 as p → ∞. In the latter limit,
two factors matter: 1/(p − 1) p/(p − 1), and also e−pt has a time constant 1/p decreases as
p → ∞, hence this is a fast term. The conclusion can be drawn from Figure 2 that if p 1, the
term p−1 e corresponds to the fast part of the step response. The effect of this term (and hance
1 −pt
the effect of the additional pole at s = −p is negligibly small as t → ∞. Furthermore, the term
p
1 − p−1 e−t describes the dominant part of the steop response. Since the response is similar to
the response of the first-order system, the system with p 1 is said to be dominantly first-order.
Also, we say that the pole at s = −1 dominates on the pole at s = −p; see Figure 3
3

4. Im
Re
-p -1
Dominant pole
Figure 3: Pole locus of a dominantly first-order system.
Additional zero Im
Re
-p -z -1
Dominant pole
Figure 4: Zero-pole locus of the system with the additional zero at s = −z.
1.3 The Effect of a Zero on a Dominantly First-order System
We have seen that an additional pole retards a dominantly first-order system as the pole moves
along the negative real axis. We now examine the effect of moving a zero in along the negative
real axis. The system to be studied is given by
1
Y (s) s+1
= z
1 . (3)
R(s) (s + 1)( 10 s + 1)
If we set z = ∞, then we obtain the system of the previous section in which p = 10. We have
seen that for the example in the previous section, this value of the parameter p was sufficently
large in order to say that the system was a dominantly first-order system. The zero-pole locus
of the system (3) is shown in Figure 4. We are again interested in the step response, that is
R(s) = 1/s,
1
z
s+1
Y (s) = 1 .
s(s + 1)( 10 s + 1)
Take the partial fraction expansion of Y (s):
10 z−1 1 z−10
1
Y (s) = − 9 z + 9 z .
s s+1 s + 10
The corresponding step response is then found to be:
10 z − 1 −t 1 z − 10 −10t
y(t) = 1 − e + e .
9 z 9 z
4

5. 4.5
z=∞
4 z=10
z=2.
z=1.
3.5
z=.5
z=.2
3
2.5
2
1.5
1
0.5
0
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5
Figure 5: Step response of the dominantly first-order system with the additional zero at s = −z.
Effects of the additional zero:
z 10. We can see that if z 10, then 10 z−1
9 z
≈ 10/9, 1 z−10
9 z
≈ 1/9, and
10 −t
y(t) ≈ yd (t) = 1 − e for sufficiently large t.
9
Thus, the pole at s = −1 remains dominant.
z = 10. This zero cancels the pole at s = −10. The system becomes a first-order system.
1 < z < 10. In this case, the additional zero speeds up the system; see Figure 5.
z = 1. This cancels the pole at s = −1.
z < 1. The additional zero becomes dominant. It speeds up the system and at the same time
leads to occuring an overshoot; see Figure 5.
From this analysis, one can see the general effect that the speed of the response increases as zero
moves from +∞ to 0 along the negative real axis. When zero becomes dominant, an overshoot
occurs.
To obtain Figure 5, the following Matlab code was used:
zv=[.2 .5 1. 2. 10 inf];
t=[0:.1:5];
yv=[];
for i=6:-1:1
z=zv(i);
G=tf([1/z 1],conv([1 1],[1/10 1]));
5

6. [y,x]=step(G,t);
yv=[yv y];
end
plot(t,yv(:,1),’--’,t,yv(:,2),’-’,t,yv(:,3),’-.’,t,yv(:,4),’-o’,t,yv(:,5),
’-x’,t,yv(:,6),’-*’)
legend(’z=infty’,’z=10’,’z=2.’,’z=1.’,’z=.5’,’z=.2’);
grid;
1.3.1 The Effect of a Right Half-plane Zero
We now examine the step response for the system given by Eq. (3) for the case when z is negative.
A negative value for z corresponds to a zero in the right half-plane.
The found expression fot the step response does not depend on whether z is negative or
positive. Let us find the derivative y(t) at t = 0:
˙
dy 10 z − 1 −t 10 z − 10 −10t 10
= e − e = .
dt t=0 9 z 9 z t=0 z
When z is positive, corresponding to a left half-plane zero, the derivative is also positive;
when z is negative, corresponding to a right half-plane zero, the derivative is negative, indicating
that the response starts off in the negative direction. As the magnitude of a positive z decreases,
corresponding to a zero moving along the negative real axis toward the origin, the increasing
magnitude of the derivative indicates, first a more rapid response caused by the zero and, then,
an increase in overshoot. For right half-plane zeros moving along the positive real axis toward
the origin, the increase in the magnitude of the derivative indicates an increase in the magnitude
of the overshoot going in the opposite direction (“undershoot”) and an increase in the delay
before the response approaches its final value. This conclusion is illustrated in Figure 6. The
corresponding Matlab code is the following:
zv=[-.2 -.5 -2. -inf];
t=[0:.1:5];
yv=[];
for i=4:-1:1
z=zv(i);
G=tf([1/z 1],conv([1 1],[1/10 1]));
[y,x]=step(G,t);
yv=[yv y];
end
plot(t,yv(:,1),’-’,t,yv(:,2),’--’,t,yv(:,3),’-.’,t,yv(:,4),’-x’)
legend(’z=-infty’,’z=-2.’,’z=-0.5’,’z=-0.2’);
grid;
1.4 Summary
1. We have seen that if there is a single pole that is significantly closer to the origin than
other poles and zeros of a transfer function with all its poles in the left halfplane, the time
constant of that pole closest to the origin dominates the response of the system.
6

7. 1
0.5
0
−0.5
−1
−1.5
−2
z=−∞
−2.5 z=−2.
z=−0.5
z=−0.2
−3
−3.5
−4
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5
Figure 6: Effect of an additional zero in the right half-plane.
2. The response from the dominant pole is modified from a pure first-order system response
by the presence of other poles and zeros. Additional poles delay the response of the system
while left half-plane zeros speed up the response. Right halfplane zeros cause the response
to start off in the wrong direction before recovering. The effect increases as either a pole
or zero moves toward the origin.
3. If there is a zero and a pole modifying the effect of another dominant pole and the modi-
fying zero is closer to the origin than the modifying pole, the response from the dominant
pole is modified more by the zero than the pole and the response is slightly advanced or
sped up in time. If the modifying pole is closer to the origin than the modifying zero, the
response is modified more by the pole than the zero and the response is slightly delayed.
7

8. 2 Effects of additional poles and zeros on performance of a
second order underdamped system
It often happens that the poles closest to the origin are a complex pole pair. Hence, the dominant
response will be from those poles which correspond to a second-order system. We will see in
this section that additional poles and zeros modify the response of a second-order system in a
similar way to how the response of a first-order system is modified.
The discussion is based upon the second-order differential equation
d2 y dy
2
2 2
+ 2ζωn + ωn y(t) = ωn r(t), (4)
dt dt
where r(t) is a step-function input and the initial conditions are assumed to be zero. The constant
ζ is called the damping ratio, and ωn is referred to as the undamped natural frequency.
There are numerous examples in which the above differential equation occurs such as a series
electrical circuit containing resistors, inductors and capacitors, a system consisting of a spring,
mass, and viscous friction etc.
2.1 The derivation of the unit step response of the prototype second order
system with two complex conjugate poles
The transfer function of the system of Eq. (4) is
2
Y (s) ωn
= 2 2
(5)
R(s) s + 2ζωn s + ωn
The transfer function (5) has poles located at
s = −ζωn ± ωn (ζ 2 − 1) (6)
These poles are real or complex conjugate depending on ζ.
If ζ < 0, the poles are in the right halfplane, hence the system is unstable.
If ζ > 1 then the system has two real poles. This case has already been studied. We know
that the step response is determined to a large extent by a dominant pole in this case. The system
is “overdamped”, its step response approaches the final value.
If ζ = 1 then the system has a double pole at −ζωn . This is a critical case (“critically damped
system”). The step response has the form
R1 + R2 e−ζωn + R3 te−ζωn ;
R1 , R2 , R3 are constants.
If 0 < ζ < 1, poles are complex conjugate and located in the left half-plane:
s = −ζωn ± jωn (1 − ζ 2 ) (7)
We focus on the last case. The location of the poles with respect to the damping ratio and the
undamped natural frequency is indicated in Fig. 7. To investigate the response of the system, we
8

9. Pole jω
s−plane
ωn √
ωd = ω n 1 − ζ 2
θ
0
Re
ωn ζ
PSfrag replacements
Pole
Figure 7: Location of the poles with respect to ζ and the ωn
chose the input as a step input. For R(s) = 1/s, Y(s) is given by
2
ωn
Y (s) = .
s [(s + ζωn )2 + ωn (1 − ζ 2 )]
2
The partial fraction expansion is found for the case of the pair of complex conjugate poles and a
single real pole:
R1 R R¯
Y (s) = + √ + √
s s + ζωn + jωn 1 − ζ 2 s + ζωn − jωn 1 − ζ 2
We use the residues method:
R1 = 1
2
ωn
R = √ √
s(s + ζωn − jωn 1 − ζ 2 ) s=−ζωn −jωn 1−ζ 2
−j
= √ √
2(ζ + j 1 − ζ 2 ) 1 − ζ 2
1 ζ
= − 1 + j√
2 1 − ζ2
1 √
2
ejtan (ζ/ 1−ζ )
−1
= − √
2 1 − ζ2
1
= − √ ej(π/2−θ)
2 1 − ζ2
9

10. where
θ = cos−1 ζ.
We now use the table of Laplace transforms and write y(t) as
1 π
y(t) = 1 − √ 2
e−ζωn t cos(ωn t 1 − ζ 2 + θ − )
1−ζ 2
1
= 1− √ 2
e−ζωn t sin(ωn t 1 − ζ 2 + θ)
1−ζ
√
The value ωd = ωn 1 − ζ 2 is the actual frequency of oscillation in radians per second, it is
referred to as the damped frequency. The period of oscillation, T d , associated with the damped
frequency ωd is
2π 2π
Td = = √ .
ωd ωn 1 − ζ 2
Typical responce of the system is shown in Fig. 8. In addition to the period and frequency of
Step Response
1.5
Maximum overshoot
Td
105%
1
95%
90%
Amplitude
0.5
Settling time
10%
0
0 5 10 15 20 25
Time (sec)
Rise time
Figure 8: Typical unit step response of an underdamped (0 < ζ < 1) second-order system.
damped oscillations, other important transient performance characteristics of a stable 2nd order
underdamped system are
10

11. − √ πζ
• Percent maximum overshoot, P O = e 1−ζ 2 × 100%;
• Rise time shows how long it takes for the response to rise from from 10% of the final value
to 90% of the final value. There is no exact equation to express the rise time.
• Settling time shows how long it takes for transients to settle. The settling time is measured
as a time required for the response to settle to within ±5%, or in some cases ±2% of the
final value. In the first case, the settling time is approximately equal to t setlle = ζωn .
3
Effect of varying the damping ratio ζ Let ωn = 1 be fixed. Step responces for various
damping ratios are given in the Figure 9. The overshoot decreases as ζ increases. However
the decrease in the overshoot is at the expense of rise time. While the response becomes more
sluggish, its settles quicker as ζ increases.
Effect of varying the undamped natural frequency ωn Varying the undamped natural fre-
quency ωn simply scales the time axis, see the Figure 10. It affects the period of damped oscilla-
tions, rise and settling time, but has no effect on the percent maximum overshoot.
2.2 The effect of additional poles and zeros on a dominantly second order
system
The effects that added poles and zeros have on dominantly second-order systems are similar to
the effects that added poles and zeros have on dominantly first-order systems.
2.2.1 An additional zero in the closed loop transfer function
Figure 11 displays the step responses of the system given by
ωn 2
Y (s) (s + z)
= 2 z 2
R(s) s + 2ζωn s + ωn
This system has the same poles as the system in equation (5). Also, it has a single zero at s = −z.
The effect of additional left-halfplane zero can be seen from the following analysis. The unit
step response of the above transfer function can be written as follows
ωn2
z
(s
+ z)
Y (s) =
s(s2 2
+ 2ζωn s + ωn )
ωn 2
2 s
ωn z
= 2 + 2ζω s + ω 2 )
+ 2 + 2ζω s + ω 2 )
s(s n n s(s n n
2
ωn dy2nd order (t)
y(t) = y2nd order (t) + .
z dt
Hence, the additional zero in the left half-plane speeds up transients, making rises and falls
sharper. Smaller values of z make this effect more prominent. As a result, an additional zero
11

12. Pole−Zero Map
1.2
0.74 0.6 0.42 0.2
0.83
ζ=0.1
1 ζ=0.3
ζ=0.5
0.8
0.91 ζ=0.707
0.6
Imag Axis
0.96
0.4
0.99
0.2
1.6 1.4 1.2 1 0.8 0.6 0.4 0.2
0
−0.2
−1.6 −1.4 −1.2 −1 −0.8 −0.6 −0.4 −0.2 0
Real Axis
1.8
ζ=0.1
1.6
ζ=.3
ζ=.5
ζ=.707
1.4
1.2
1
0.8
0.6
0.4
0.2
0
0 5 10 15 20 25
Step responses
Figure 9: Second order system with various damping ratios. Locations of poles on the complex
plane are obtained using Matlab function pzmap. The (ζ, ω n )-grid can be added using grid or
sgrid.
12

13. Pole−Zero Map
1.5
0.74 0.6 0.42 0.22
ω =2
0.84 n
1 System: sys
Pole: −0.707 + 0.707i
0.91 Damping: 0.707
Overshoot (%): 4.32
Imag Axis
Frequency (rad/sec): 1
ω =1
n
0.96
0.5
ωn=0.5
0.99
2 1.75 1.5 1.25 1 0.75 0.5 0.25
0
−2 −1.8 −1.6 −1.4 −1.2 −1 −0.8 −0.6 −0.4 −0.2 0
Real Axis
1
0.8
0.6
ωn=0.5
ωn=1.
ωn=2.
0.4
0.2
0
0 5 10 15 20 25
Figure 10: Second order system with various undamped natural frequencies.
13

14. zv=[0.5 1 2 10 inf];
t=[0:.1:5];
omega_n=1;
zeta=sqrt(2)/2;
yv=[];
for i=5:-1:1
z=zv(i);
G=tf([omega_nˆ2/z omega_nˆ2],[1 2*zeta*omega_n omega_nˆ2]);
[y,x]=step(G,t);
yv=[yv,y];
end
plot(t,yv(:,1),’-’,t,yv(:,2),’--’,t,yv(:,3),’-.’,t,yv(:,4),’-o’,t,yv(:,5),’-
grid;
legend(’z=infty’,’z=10’,’z=2’,’z=1’,’z=0.5’);
Effect of an addtitional zero, ζ=0.707, ωn=1
1.5
1
0.5
z=∞
z=10
z=2
z=1
z=0.5
0
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5
Figure 11: Effect of an additional left halfplane zero on a dominantly second-order system
14

15. in the left half-plane makes the system faster and more oscillatory. This can be seen from the
simulations.
As the zero moves along the negative real axis toward the origin, the time to the first peak of
the step response decreases monotonically while the percent overshoot increases monotonically.
Also, it takes longer for the system to settle to the final value of the response.
The zero in the right half-plane retards the system and produces an undershoot. The persent
undershoot decreases as the zero moves along the positive real axis toward the infinity, see Fig-
ure 12. Again the system oscillates for a longer time.
zv=[-0.5 -1 -2 -10 -inf];
t=[0:.1:5];
omega_n=1;
zeta=sqrt(2)/2;
yv=[];
for i=5:-1:1
z=zv(i);
G=tf([omega_nˆ2/z omega_nˆ2],[1 2*zeta*omega_n omega_nˆ2]);
[y,x]=step(G,t);
yv=[yv,y];
end
plot(t,yv(:,1),’-’,t,yv(:,2),’--’,t,yv(:,3),’-.’,t,yv(:,4),’-o’,t,yv(:,5),’-
grid;
legend(’z=-infty’,’z=-10’,’z=-2’,’z=-1’,’z=-0.5’);
Effect of an addtitional zero in the right half−plane, ζ=0.707, ωn=1
1.2
1
0.8
0.6
0.4
0.2
0
z=−∞
z=−10
−0.2 z=−2
z=−1
z=−0.5
−0.4
−0.6
−0.8
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5
Figure 12: Effect of an additional zero in the right half-plane on a dominantly second-order
system
15

16. 2.2.2 An additional zero in the forward path transfer function
Consider an example. Let the forward path transfer function be
Y (s) 6(zs + 1)
=
R(s) s(s + 1)(s + 10)
The effect of the pole at s = −10 is insignificant as transients due to this pole should die out
fairly quickly. The closed loop transfer function is
Y (s) 6(zs + 1)
= 3
R(s) s + 11s2 + (10 + 6z)s + 6
AS we can see, this time z affects not only the numerator, but the denominator also contains z. In
this case, a larger z (and a zero closer to the imaginary axis) has an effect of improving damping
and reducing maximum overshoot.
2.2.3 The effect of an additional closed loop pole
Conversely to the effect of an additional closed loop zero, as an added pole moves along the
negative real axis toward the origin, the time to the first peak of the step response increases
and the percent overshoot decreases. Also, in the presence of an additional poles the amount of
oscillations reduces, i.e., the system becomes ‘better damped’. This can be seen on Figure 13
where the step responses of the system
2
Y (s) ωn
=
R(s) (s/p + 1)(s2 + 2ζωn s + ωn )
2
are shown for different values of p. This system has the additional real pole located at s =
−p. The most dramatic effect occurs as the additional pole becomes dominant, i.e. when −p
becomes greater than −ζωn . In this case, the overshoot “disappears” and the system slows down
considerably, becoming effectively a dominantly first-order system.
2.2.4 The effect of an additional forward-path pole
When the open loop transfer function is of the form
2
Y (s) ωn
=
R(s) s(s + 2ζωn )(1 + s/p)
(i.e., the pole at s = −p is added to the prototype forward-path transfer function), the closed
loop system becomes
2
Y (s) ωn
=
R(s) (1/p)s3 + (1 + 2ζωn /p)s2 + 2ζωn s + ωn
2
When p is large, the added pole has little effect on the system step response as the system remains
dominantly second-order. As p reduces, and the pole moves closer to the origin, the overshoot
increases, i.e, the effect is opposite to that seen in the case of an added closed-loop pole. If the
pole gets to close to the origin, the closed loop system can even go unstable; see Figure 14
16

17. pv=[0.1 .5 2 10 inf];
t=[0:.3:15];
omega_n=1;
zeta=sqrt(2)/2;
yv=[];
for i=5:-1:1
p=pv(i);
G=tf(omega_nˆ2,conv([1/p 1],[1 2*zeta*omega_n omega_nˆ2]));
[y,x]=step(G,t);
yv=[yv,y];
end
plot(t,yv(:,1),’-’,t,yv(:,2),’--’,t,yv(:,3),’-.’,t,yv(:,4),’-o’,t,yv(:,5),’-
grid;
legend(’p=infty’,’p=10’,’p=2’,’p=0.5’,’p=0.1’);
1.4
p=∞
p=10
1.2 p=2
p=0.5
p=0.1
1
0.8
0.6
0.4
0.2
0
0 5 10 15
Figure 13: Effect of an additional pole on a dominantly second-order system
17

18. Step Response
2.5
p=10
p=1
2
1.5
Amplitude
1
0.5
0
−0.5
p=0 (no added pole)
−1
0 5 10 15
Time (sec)
Figure 14: The effect of an additional forward-path pole.
2.3 Summary
As was in the case of a dominantly first-order system, the response of a dominantly second-
order system is sped up by an additional zero and is slowed down by an additional pole. In
the dominantly second-order system the added closed loop zero also has the important effect of
increasing the amount of oscillation in the system while an added closed loop pole has the effect
of decreasing the amount of oscillation. Added forward path zeros and added forward path poles
have an opposite effect on the overshoot. A forward path pole which is too close to the origin
may turn the closed loop system unstable.
A right half-plane zero also causes a ‘wrong way’ response. All effects become more pro-
nounced as the additional zero or pole approach the origin and become dominant.
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