CONTROL
SYSTEMS II
Root Locus
Eng. AHMED AL-NAHHAL
Sketching the Root Locus
7. Determine a pair of dominant complex-conjugate closed-loop poles
of a certain damping ratio ƞ
– Closed-loop poles with ƞ=0.5 lie on lines passing through the origin
and making the angles ± cos −1 ƞ = ± cos −1 0.5 = ±60 with the
negative real axis.
– From root locus plot, such closed loop poles having ƞ =0.5 are 𝑠 =
− 0.3337 ± 𝑗0.5780.
– The value of K that yields such poles is found from the magnitude
condition as follows:
𝑘 = 𝑠(𝑠 + 1)(𝑠 + 2) 𝑠=−0.3337+𝑗0.5780 = 1.0383
Sketching the Root Locus
– Using this value of K, the third pole
is found at s = –2.3326.
– Finally, note that, if necessary, the
root loci can be easily graduated in
terms of K by use of the magnitude
condition.
Example #1
– Sketch the root-locus plot of the negative feedback system
shown.
– Find the value of K at which the complex-conjugate closed-
loop poles have the damping ratio ƞ=0.7
Sketching the Root Locus
1. Determine the root loci on the real axis
– The open-loop poles are 𝑠 = −1 ± 𝑗 2 and open-loop zero is 𝑠 =
− 1.
– The root locus segment lies between two zeros (s= –2 and s= – ꝏ). In
other words, two root loci break in the part of the negative real axis
between –2 and –ꝏ.
2. Determine the asymptotes of the root loci
– Since there are two complex open-loop poles and one zero on real
axis, there is one asymptote, which coincides with the negative real
axis.
Sketching the Root Locus
3. Determine the breakaway and break-in points
– There is no breakaway point
while the break-in point can
be found as follows:
– Point s= – 3.7320 is on the
root locus. Hence this point
is an actual break-in point.
The corresponding gain
value is K= 5.4641.
Sketching the Root Locus
4. Determine the points where the root loci cross the
imaginary axis.
– There is no points where the root locus cross the imaginary axis.
5. Angles of Departure and Arrival
– The angle of departure from these poles
Sketching the Root Locus
6. Sketch the root loci
– Based on the information
obtained in the foregoing
steps, as shown.
Sketching the Root Locus
7. Find the value of K at a
certain damping ratio
– The value of K (ƞ=0.7) can be found
by locating the roots, as shown.
Example #2
Given a unity feedback system that has the following forward
transfer function. Sketch the root locus and identify all
points.
𝐾 (𝑠 + 2)
𝐺𝐻 =
(𝑆 + 1)(𝑆 + 3 + 𝑗)(𝑆 + 3 − 𝑗)
Example #2
1. The open loop poles are at -1, -3+j and -3-j. the open loop zero is
at -2.
−5
2. Asymptote intercept is 𝜎𝑎 = = −2.5. While the asymptote
2
angles are 𝜃𝑎 = 90° 𝑎𝑛𝑑 270° . (No jw-axis crossing)
3. The breakaway and break-in point.
(𝑠 + 1)(𝑠 2 + 6𝑠 + 10)
𝑘=−
(𝑠 + 2)
𝑑𝑘 𝑑 (𝑠 + 1)(𝑠 2 + 6𝑠 + 10)
=−
𝑑𝑠 𝑑𝑠 (𝑠 + 2)
Example #2
𝑑𝑘 𝑑 𝑠 + 1 𝑠 2 + 6𝑠 + 10
=− =0
𝑑𝑠 𝑑𝑠 𝑠+2
2𝑠 3 + 13𝑠 2 + 28𝑠 + 22 = 0
𝑠 = 8.33 𝑎𝑛𝑑 − 0.92 ± 𝑗0.67
There are neither breakaway nor break-in points.
4. Departure angles.
∅𝑑 = 180 − 90 + 153.43 − 135 = 71.57
There are no arrival angles.
Example #2
6. Sketch the root locus.
Example #3
Given a unity feedback system that has the
following forward transfer function.
a. Sketch the root locus.
b. Find the point where the locus crosses the 0.5 damping ratio line.
c. Find the gain at the point where the locus crosses the 0.5 damping
ratio line.
d. Find the range of gain, K, for which the system is stable.
Example #3
Example #3
Transient Response Design via Gain
Adjustment
– The formulas describing percent overshoot, settling time, and
peak time were derived only for a system with two closed-loop
complex poles and no closed-loop zeros.
– The conditions justifying a second-order approximation are:
1. The response that results from a higher order pole does not appreciably
change the transient response expected from the dominant second-order
poles.
2. Closed-loop zeros near the closed-loop second-order pole pair are nearly
canceled by the close proximity of higher-order closed-loop poles.
Transient Response Design via Gain
Adjustment
Figure (b) would yield a much better second-order approximation than
Figure (a), since closed-loop pole p3 is farther from the dominant,
closed-loop second-order pair, p1 and p2
Transient Response Design via Gain
Adjustment
Figure (d) would yield a much better second-order approximation than
Figure (c), since closed-loop pole p3 is closer to canceling the closed-
loop zero.
Transient Response Design via Gain
Adjustment
Summarizing the design procedure for higher-order systems:
1. Sketch the root locus for the given system.
2. Assume the system is a second-order system without any zeros and
then find the gain to meet the transient response specification.
3. Justify your second-order assumption by finding the location of all
higher-order poles and evaluating the fact that they are much farther
from the jw-axis than the dominant second-order pair. Also, verify that
closed-loop zeros are approximately canceled by higher-order poles.
Example #4
Consider the system shown. Design the value of gain, K, to
yield 1.52% overshoot. Also estimate the settling time, peak
time, and steady-state error.
Example #4
– Breakaway points on the real axis can occur between 0 and -1 and
between -1.5 and -10, where the gain reaches a peak, breakaway
points are found at -0.62 with a gain of 2.511 and at -4.4 with a
gain of 28.89.
– A break-in point on the real axis can occur between -1.5 and -10,
where the gain reaches a local minimum, a break-in point is found
at -2.8 with a gain of 27.91.
Example #4
Example #4
– Assume that the system can be approximated by a second-order,
underdamped system without any zeros. A 1.52% overshoot
corresponds to a damping ratio of 0.8. Sketch this damping ratio
line on the root locus.
– The points where the root locus crosses the 0.8 damping ratio or
1.52 percent overshoot line are −0.87 ± 𝑗0.66, −1.19 ± 𝑗0.9 and
− 4.6 ± 𝑗3.45 with respective gains of 7.36, 12.79, and 39.64.
Example #4
– To test our assumption of a second-order system, we must calculate
the location of the third pole. For each of the three crossings of the
0.8 damping ratio line, the third closed-loop pole is at -9.25, -8.6, and
-1.8, respectively.
– For each point the settling time and peak time are evaluated using
Example #4
The results for each case are shown in following table.
Cases 1 and 2 yield third closed-loop poles that are relatively far from the
closed-loop zero. For these two cases there is no pole-zero cancellation,
and a second-order system approximation is not valid.
Example #4
In Case 3, the third closed-loop pole and the closed-loop zero are
relatively close to each other, and a second-order system approximation
can be considered valid.
Root Locus for Positive-Feedback
Systems
A positive-feedback system can be thought of as a negative-
feedback system with a negative value of H(s).
Root Locus for Positive-Feedback
Systems
– Using this concept, we find that the transfer function for the positive-
feedback system shown is:
– We now retrace the development of the root locus for the denominator.
Obviously, a pole, s, exists when:
Root Locus for Positive-Feedback
Systems
– To illustrate the root-locus plot for the positive-feedback system, we
shall use the following transfer functions G(s) and H(s) as an
example.
– The general rules for constructing root loci for negative-feedback
systems given must be modified.
Root Locus for Positive-Feedback
Systems
1. Plot the open-loop poles (s=-1+j, s=-1-j, s=-3) and zero (s=-2) in the
complex plane. The closed-loop poles start at the open-loop poles
and terminate at the open-loop zeros (finite or infinite).
2. Determine the root loci on the real axis. Root loci exist on the real
axis between -2 and +ꝏ and between -3 and -ꝏ.
3. Determine the asymptotes of the root loci.
This simply means that asymptotes are on the real axis.
Root Locus for Positive-Feedback
Systems
4. Determine the breakaway and break-in points. Since the characteristic
equation is:
Point s= - 0.8 is on the root locus. It is an actual break-in point. Points 𝑠 = −2.35 ±
𝑗0.77 do not satisfy the angle condition and, therefore, they are neither breakaway
nor break-in points.
Root Locus for Positive-Feedback
Systems
5. There are not points where the loci cross the jw axis except w =0.
6. Find the angle of departure of the root locus from a complex pole. For
the complex pole at s=–1+j, the angle of departure ϴ is:
• The angle of departure from the complex pole at s= -1-j is 72°.
• There are no angles of arrival as there are no repeated zeros.
Root Locus for Positive-Feedback
Systems
7. Sketch the root loci plot based on the information obtained in the
foregoing steps, as shown.
IF
Then one real root enters the
right-half s plane. Hence, for
values of K >3, the system
becomes unstable.
Root Locus for Positive-Feedback
Systems
– The closed-loop transfer function for the positive-feedback system is
given by:
– The closed-loop transfer function for the negative-feedback system is
given by:
Root Locus for Positive-Feedback
Systems
– To compare this root-locus plot with that of the corresponding
negative-feedback system
Negative-feedback system Positive-feedback system
Root Locus for Positive-Feedback
Systems
– Root-Locus Plots of Negative-Feedback and Positive-Feedback
Systems.
Root locus plot of positive feedback systems enters the right hand side
of s plane. Therefore the system is unstable
Pole Sensitivity
– Any change in the parameter changes the closed-loop poles and,
subsequently, the performance of the system.
– Along some sections of the root locus, very small changes in gain yield
very large changes in pole location and hence performance. The
system has a high sensitivity to changes in gain.
– Along other sections of the root locus, very large changes in gain
yield very small changes in pole location. The system has a low
sensitivity to changes in gain.
– We prefer systems with low sensitivity to changes in gain.
Pole Sensitivity
– Root sensitivity is the ratio of the fractional change in a closed-loop
pole to the fractional change in a system parameter. The sensitivity of
a closed-loop pole, s, to gain, K:
s is the current pole location, and K is the
current gain.
– The actual change in the closed-loop poles can be approximated as:
Δs is the change in pole location, and Δ
K/K is the fractional change in the gain, K.
Example #5
Find the root sensitivity of the shown system at s =-9.47 and
-5 + j5. Also calculate the change in the pole location for a
10% change in K.
Example #5
The system’s characteristic equation is 𝑠 2 + 10𝑠 + 𝑘 = 0. Differentiating
with respect to K, we have:
The sensitivity is found to be
– For s= -9.47 (K = 5), yields Ss:K= - 0.059 and then Δs= 0.056 or the pole
will move to the right by 0.056 units for a 10% change in K.
– For s= -5+j5 (K = 50), yields Ss:K= 1/(1+ j)and then Δs= j0.5 or the pole
will move vertically by 0.5 units for a 10% change in K.
Homework #1
Control Systems Engineering,6th Edition, Norman Nise.
– Chapter 8/ Page 432
– 1, 3, 13, 18, 24, 27, 42, 44
– Facebook/ Ahmed Elnahhal
– Email/ a.nahal@up.edu.ps
