Determine The Values Of K And Β Such That The System With Open Loop Transfer Function G(s) = K(s+1)/(s³ + Βs² + 5s + 1) Oscillates At 2 Rad/sec With Unity Feedback.
Introduction
In control systems engineering, a critical task is ensuring system stability and achieving desired performance characteristics. One common requirement is to design a system that oscillates at a specific frequency. This often involves tuning parameters within the system's transfer function. This article will delve into the process of determining the values of K and β for a given open-loop transfer function, G(s) = K(s+1)/(s³ + βs² + 5s + 1), such that the closed-loop system oscillates at a frequency of 2 rad/sec. We will assume a unity feedback configuration, a standard setup in control systems. The approach involves leveraging the characteristic equation of the closed-loop system and the Routh-Hurwitz stability criterion to establish conditions for sustained oscillations. Understanding these principles is fundamental in control system design, allowing engineers to tailor system responses for various applications. Designing stable control systems is very crucial. The concepts which will be discussed can be applied to different systems and oscillations. By the end of this comprehensive guide, readers will gain a solid understanding of how to calculate the gain K and parameter β to achieve the desired oscillatory behavior.
Problem Statement
Given the open-loop transfer function:
G(s) = K(s+1)/(s³ + βs² + 5s + 1)
Determine the values of K and β such that the closed-loop system oscillates at a frequency of 2 rad/sec, assuming unity feedback.
This problem is a classic example of how to tune control system parameters to achieve specific performance goals. The desired oscillation frequency of 2 rad/sec provides a key constraint that we will use to solve for the unknowns K and β. The unity feedback configuration simplifies the analysis, but the principles can be extended to systems with more complex feedback structures. To solve this, we will use the characteristic equation of the closed-loop system. The Routh-Hurwitz stability criterion is a powerful tool that allows us to determine the stability of a system based on the coefficients of its characteristic equation. By applying this criterion, we can establish the conditions under which the system will exhibit sustained oscillations. This article will provide a step-by-step approach to solving this problem, making it a valuable resource for students and practicing engineers.
Methodology: Routh-Hurwitz Stability Criterion
To solve this problem, we will employ the Routh-Hurwitz stability criterion. This method allows us to determine the stability of a system by examining the coefficients of its characteristic equation. The characteristic equation is obtained from the closed-loop transfer function. For a unity feedback system, the closed-loop transfer function is given by:
T(s) = G(s) / (1 + G(s))
Substituting the given G(s), we have:
T(s) = [K(s+1)/(s³ + βs² + 5s + 1)] / [1 + K(s+1)/(s³ + βs² + 5s + 1)]
Simplifying the expression, we get:
T(s) = K(s+1) / (s³ + βs² + 5s + 1 + K(s+1))
T(s) = K(s+1) / (s³ + βs² + (5+K)s + (1+K))
The denominator of the closed-loop transfer function is the characteristic equation:
q(s) = s³ + βs² + (5+K)s + (1+K) = 0
The Routh-Hurwitz criterion involves constructing a table based on the coefficients of the characteristic equation. This table provides information about the location of the system's poles in the complex plane, which in turn determines the system's stability. A system is stable if all the poles lie in the left-half plane. The Routh-Hurwitz table helps us identify if any poles are in the right-half plane, indicating instability. For sustained oscillations, we need to find the conditions where the system is marginally stable, meaning the poles lie on the imaginary axis. This is a crucial step in determining the values of K and β that will result in oscillations at the desired frequency.
Constructing the Routh Table
The Routh table is constructed using the coefficients of the characteristic equation:
q(s) = s³ + βs² + (5+K)s + (1+K) = 0
The coefficients are arranged in the first two rows of the table as follows:
| s³ | 1 | 5+K |
|---|---|---|
| s² | β | 1+K |
| s¹ | ||
| s⁰ |
The subsequent rows are calculated based on the elements in the previous rows. For the s¹ row, the elements are calculated as:
- Element 1:
[(β * (5+K)) - (1 * (1+K))] / β=(5β + βK - 1 - K) / β - Element 2: 0
For the s⁰ row, the element is calculated as:
- Element 1:
[( (5β + βK - 1 - K) / β ) * (1+K) - (β * 0)] / [(5β + βK - 1 - K) / β]=1+K
The completed Routh table is:
| s³ | 1 | 5+K |
|---|---|---|
| s² | β | 1+K |
| s¹ | (5β + βK - 1 - K) / β | 0 |
| s⁰ | 1+K |
The Routh-Hurwitz criterion states that for the system to be stable, all the elements in the first column of the Routh table must be positive. For sustained oscillations, we need a row of zeros in the Routh table, which indicates the presence of roots on the imaginary axis. This condition will help us determine the values of K and β that lead to oscillations at 2 rad/sec.
Condition for Sustained Oscillations
For sustained oscillations, the s¹ row must be zero. This means:
(5β + βK - 1 - K) / β = 0
Which simplifies to:
5β + βK - 1 - K = 0
Additionally, for the system to be marginally stable and oscillate, all elements in the first column must be positive except for the row that is zero. Therefore, we also need:
- β > 0
- 1 + K > 0 => K > -1
From the condition for sustained oscillations, we have:
β(5 + K) = 1 + K
β = (1 + K) / (5 + K)
Since we want the system to oscillate at 2 rad/sec, we need to find the auxiliary equation. The auxiliary equation is formed from the row above the row of zeros (the s² row in this case):
βs² + (1 + K) = 0
Substituting the expression for β, we get:
[(1 + K) / (5 + K)]s² + (1 + K) = 0
Determining K and β
Divide the auxiliary equation by (1 + K), assuming K ≠ -1:
s² / (5 + K) + 1 = 0
s² = -(5 + K)
Since the oscillation frequency is 2 rad/sec, the roots are s = ±j2. Substituting s = j2 into the equation:
(j2)² = -(5 + K)
-4 = -(5 + K)
4 = 5 + K
K = -1
However, we assumed K ≠ -1 when dividing by (1 + K). This indicates a potential issue with this solution. Let's revisit the condition for sustained oscillations:
5β + βK - 1 - K = 0
And the auxiliary equation:
βs² + (1 + K) = 0
Since s = ±j2, we have:
β(j2)² + (1 + K) = 0
-4β + (1 + K) = 0
1 + K = 4β
We now have two equations:
- 5β + βK - 1 - K = 0
- 1 + K = 4β
Substitute the second equation into the first:
5β + βK - 4β = 0
β + βK = 0
β(1 + K) = 0
Since we are looking for oscillations, β cannot be zero. Therefore:
1 + K = 0
K = -1
Again, we encounter K = -1. This suggests that our initial assumption of K ≠ -1 might be problematic. Let's go back to the two equations:
- 5β + βK - 1 - K = 0
- 1 + K = 4β
Instead of substituting, let's rearrange the first equation:
5β - 1 = K(1 - β)
K = (5β - 1) / (1 - β)
Now, substitute this into the second equation:
1 + (5β - 1) / (1 - β) = 4β
(1 - β + 5β - 1) / (1 - β) = 4β
4β / (1 - β) = 4β
If β ≠ 0, we can divide by 4β:
1 / (1 - β) = 1
1 = 1 - β
β = 0
However, if β = 0, the characteristic equation becomes:
s³ + (5 + K)s + (1 + K) = 0
And the Routh table is:
| s³ | 1 | 5+K |
|---|---|---|
| s² | 0 | 1+K |
| s¹ | ||
| s⁰ |
This case is more complex because of the zero in the first column. We need to use a different approach or consider a slightly perturbed value for β.
Let's reconsider the equations:
- 5β + βK - 1 - K = 0
- 1 + K = 4β
From equation (2), K = 4β - 1. Substitute this into equation (1):
5β + β(4β - 1) - 1 - (4β - 1) = 0
5β + 4β² - β - 1 - 4β + 1 = 0
4β² = 0
β = 0
If β = 0, then K = 4(0) - 1 = -1. We already analyzed this case and found it problematic. It appears there might be an inconsistency or no solution for sustained oscillations at exactly 2 rad/sec with these constraints.
However, let’s look at the limiting case as β approaches a small positive value. If β is a very small positive number, then K will be slightly less than -1, but close to it. This means the system will be very close to instability and might exhibit near-oscillatory behavior at a frequency close to 2 rad/sec.
Approximate Solution:
Given the mathematical challenges in finding an exact solution, we can conclude that achieving sustained oscillations at precisely 2 rad/sec might not be feasible with the given system and unity feedback. However, we can get close to the desired behavior by choosing a very small positive value for β (approaching 0) and setting K slightly less than -1.
Conclusion
Determining the values of K and β to achieve sustained oscillations in a control system requires careful analysis using tools like the Routh-Hurwitz stability criterion. In this case, while an exact solution for sustained oscillations at 2 rad/sec proved elusive due to mathematical constraints, the analysis provided valuable insights into the system's behavior. We found that setting β to a very small positive value and K slightly less than -1 would result in near-oscillatory behavior. The problem highlights the importance of understanding the limitations and trade-offs in control system design. While theoretical calculations provide a strong foundation, practical implementation often involves fine-tuning parameters through experimentation and simulation. This article showcased a detailed approach to analyzing system stability and oscillation conditions, emphasizing the critical role of parameter selection in achieving desired system performance. The findings underscore the complex interplay between system parameters and their impact on stability and oscillatory behavior, offering a valuable learning experience for control system engineers and students alike.