Cracking Control Systems: What You Need to Know for the FE Electrical Exam
Apr 05, 2024Ever wonder how autopilot works or how assembly robots move so quickly yet accurately? Control Systems. Not only does this apply to automation or specialized industries but also to our refrigerators, our cars, and everything where a device or system needs to have a desired output, control systems are implemented.
The FE requires examinees to understand and be able to solve questions regarding control systems. In this blog, we will go over a brief overview of what control systems are and what equations are expected for test takers to know for the exam.
What are Control Systems?
A control system is a system of devices that regulates the different behaviors of other devices to achieve a desired result. There are two types of control loops: openloop and closedloop.
Openloop control systems have the control action totally independent of the outcome of their systems. A simple example of this is an electric hand dryer; the hot air (output) remains on for as long as an object is detected under its sensor (input), regardless of how dry the object may be.
Closedloop control systems, on the other hand, have the input affected by the output in order to achieve the desired output. This behavior is referred to as feedback. To bring back a previous comparison, refrigerators are an example of a closedloop system, where the cooling devices have their output throttled to maintain a certain temperature inside the refrigerator. This not only ensures that the temperature in the device is constant but also a balance between cooling time and power efficiency.
Control system stability refers to the behavior of a system over time and its ability to maintain a desired state or equilibrium. A stable control system is one that, after experiencing a disturbance or change, eventually returns to its original state or reaches a new stable state.
In the Stability section of the Control System topic for the Fundamental of Engineering (FE) exam, there are two methods to determine the stability of the system:
1.Stability Theorem
2.Coefficient Test
3.RouthHurwitz Criterion
Stability Theorem
We can use the Stability Theorem to determine the system’s stability.
1.Stable System: A system is considered stable when the poles of its transfer function reside on the left side of the splane. However, if the poles move closer to the origin, the system’s stability decreases (Stefani, 2002, 71).
Ex: H (s) = 2 / ((s+1) (s+2))
a. Set the denominator equal to zero, we have s = 1 and s = 2 We know the poles lie on the lefthand side of the splane. Therefore, the system is stable.
Figure 1
2. Marginally Stable System: A system is considered marginally stable when the poles are present in the imaginary axis (Stefani, 2002, 71).
Ex: H (s) = 2/ (s^{2}+2) = 2/((s+√2 i) (s√2 i))
a. Set the denominator equal to zero, we have s = ±√2 i. We know the poles lie on the imaginary axis of the splane. Therefore, the system is marginally stable.
Figure 2
3. Unstable System: A system is considered unstable when there are repeated poles on the imaginary axis of the splane and/or poles in the right half of the splane (Stefani, 2002, 71).
Ex: H(s) = 2 / (〖(s〗^{2}+2) (s1)) = 2 / ((s+√2 i)(s√2 i)(s1))
a. Set the denominator equal to zero, we have s = ±√2 i and s = 1. We know the poles lie on the imaginary axis of the splane. But there is a pole on the righthand plane. Therefore, the system is unstable.
Figure 3
Coefficient Test
For a highorder polynomials transfer function, if there is a missing s term, it is safe to say that the system is unstable (Stefani, 2002, 144145).
Example, a transfer function T(s) = 6s^{5} + 5s^{4} + 3s^{2} + 7s + 10. The transfer function is missing the s^{3} term.
RouthHurwitz Criterion
The RouthHurwitz stability criterion is a mathematical method used to determine the stability of a linear control system. It provides a systematic approach to analyze the stability of a system by examining the coefficients of its characteristic equation.
The characteristic equation of a linear control system is obtained by setting the denominator of its transfer function equal to zero (Eqn. 1). It is typically a polynomial equation in terms of the system's parameters. The RouthHurwitz criterion uses the coefficients of the characteristic equation to construct a table called the Routh array or Routh table. The Routh array organizes the coefficients into rows and columns, allowing for a systematic evaluation of stability (Stefani, 2002, 145).
a_{n} s^{n }+ a_{(n1)} s^{(n1)} + a_{(n2)} s^{(n2)} + ... + a_{0} = 0 (1)
Arrange all the coefficients into the table below:
Table 1
s_{n}

a_{n}

a_{n2}

a_{n4}

s_{n1}

a_{n1}

a_{n3}

a_{n5}

s_{n2}

b^{1}

b^{2}

b^{3}

s_{n3}

c^{1}

c^{2}

c^{3}

:




s^{0}




b_{1} = (a_{(n1)} a_{(n2)}  a_{n} a_{(n3)}) / a_{(n1)} (2)
b_{2} = (a_{(n1)} a_{(n4)}  a_{n} a_{(n5)}) / a_{(n1)} (3)
c_{1} = (b_{1} a_{(n3)}  b_{2} a_{(n1)}) / b_{1} (4)
c_{2} = (b_{1} a_{(n5)}  b_{3} a_{(n1)}) / b_{1} (5)
When all the coefficients a_{n }, a_{n1 }, b_{1 }, c_{1} are all of the same positive sign, and none is zero ,then the system is stable. Else, the system is unstable.
Let’s examine this T(s) transfer function stability:
T(s) = s^{4} + 2s^{3} + 6s^{2} + 4s + 1 (6)
Table 2
s^{4}

1

6

1

s^{3}

2

4


s^{2}

b^{1}

b^{2}


s^{1}

c^{1}



s^{0}

d^{1}



b_{1} = 2(6)  4(1) / 2 = 4
b_{2} = 2(1)  1(0) / 2 = 1
c_{1} = b_{1}(4) 2(b_{2}) / b_{1} = 4(4)  2(1) /4 = 3.5
d_{1} = c_{1}(b_{2})  b_{1}(0) / c_{1} = 3.5(1) 4(0) / 3.5 = 1
Therefore, the complete table is:
Table 3
s^{4}

1

6

1

s^{3}

2

4


s^{2}

4

1


s^{1}

3.5



s^{0}

1



By looking at the highlighted region, there is no change in sign, and all coefficients are positive. Therefore, the system is stable.
The quantity of right halfplane (RHP) roots in T(s) corresponds to the count of sign changes in the elements of the leftmost column of the array, progressing from the top to the bottom (Stefani, 2002, 147). We will change the value sign of s_{1} value row in Table 3 to demonstrate this aspect.
Table 4
s^{4}

1

6

1

s^{3}

2

4


s^{2}

4

1


s^{1}

3.5



s^{0}

1



From Table 4, there are two signs of changes in the highlighted column (4 to 3.5 and 3.5 to 1); therefore, T(s) has two RHP roots.
Conclusion
We’ve gone over what control systems are and how to use the Stability Theorem, Coefficient Test, and RouthHurwitz Criterion, but for a more thorough and comprehensive study plan, I recommend checking out School of PE’s
FE Electrical exam review course, which offers an extensive and detailed curriculum to ensure a comprehensive understanding of the subject matter. Good luck!
References
Stefani, R. T. (2002). Design of Feedback Control Systems. Oxford University Press.
About the Author: Khoa Tran
Khoa Tran is an electrical engineer working at the Los Angeles Department of Water and Power and is currently pursuing his master's in electrical Power from the University of Southern California. He is fluent in both Vietnamese and English and is interested in outdoor activities and exploring new things.