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Chapter 11: Problem 9

Suppose the closed-loop poles of a feedback system satisfy \\[\frac{(s+1)(s+3)}{(s+2)(s+4)}=-\frac{1}{K}.\\] Use the root-locus method to determine whether there are any values of the adjustable gain \(K\) for which the system's impulse response has an oscillatory component of the form \(e^{-a t} \cos \left(\omega_{0} t+\phi\right),\) where \(\omega_{0} \neq 0\).

Short Answer

Expert verified
Yes, if \(K\) results in complex conjugate poles with \(\omega_0 \neq 0\).

Step by step solution

01

Analyze the Characteristic Equation

The given equation is \(\frac{(s+1)(s+3)}{(s+2)(s+4)}=-\frac{1}{K}\). By rearranging, we have the characteristic equation \((s+1)(s+3) + K(s+2)(s+4) = 0\). This equation represents the closed-loop poles that determine the stability of the system.
02

Determine Breakaway and Break-in Points

Identify the locations on the real axis where the poles of the system will break away from or break into the real axis. Set the derivative of the characteristic equation with respect to \(s\), denoted as \(\frac{d}{ds}(K(s+2)(s+4)-(s+1)(s+3)) = 0\), equal to zero and solve for \(s\). This tells us the points where the root locus will move towards or away from on the real axis.
03

Evaluate the Imaginary Axis Crossings

Set the imaginary part equal to zero in the characteristic equation to determine where the root locus crosses the imaginary axis, if at all. Substitute \(s = j\omega\) into the characteristic equation and solve \((j\omega+1)(j\omega+3) + K(j\omega+2)(j\omega+4) = 0\) separating the real and imaginary parts. Both parts should separately equal zero to find \(\omega_0\).
04

Check Conditions for Oscillatory Response

For an oscillatory component, the poles must have a non-zero imaginary part. Hence, verify from Step 3 if the characteristic equation has roots with a non-zero imaginary part. This is crucial to check whether the impulse response has cosine terms indicative of oscillations.
05

Conclusion on the Existence of Oscillations

If there is a value of \(K\) that results in complex conjugate poles with non-zero imaginary parts, the system can have an oscillatory response. Then, confirm if \(\omega_0\) is not zero from previous calculations to conclude the existence of oscillations.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Root Locus Method
The Root Locus Method is a graphical approach used in control systems to analyze how the roots of the characteristic equation change with variations in system parameters, specifically the gain, denoted by \( K \). This method provides insights into the stability and transient response of a control system.
Key points of the Root Locus Method include:
  • The plot begins at the poles of the open-loop transfer function and ends at the zeros as \( K \) approaches infinity.
  • It helps in identifying breakaway and break-in points on the real axis. These are points where the roots of the characteristic equation move towards or away from the real axis as \( K \) changes.
  • By analyzing the point where the imaginary axis is intersected, one can assess potential oscillations within the system.
By employing this method, it becomes feasible to predict the system's reaction to variations in \( K \), making it an invaluable tool for designing stable feedback control systems.
Impulse Response
The impulse response of a system describes how the system reacts over time to a brief input signal, usually represented by a Dirac delta function (an impulse at time zero). It's a fundamental concept for understanding the behavior of linear time-invariant (LTI) systems.
  • The impulse response often reveals the stability and dynamics of the system.
  • If the system has complex conjugate poles with a non-zero imaginary part, the impulse response will include oscillations in the form \( e^{-a t} \cos(\omega_0 t + \phi) \) where \( \omega_0 \) indicates the frequency of oscillation.
  • Assessing the poles of the system's characteristic equation can determine if they lie in a region that produces such an oscillatory response.
Understanding the impulse response helps engineers predict how the system will behave when subjected to sudden changes or disturbances.
Feedback System
Feedback systems are integral components in maintaining desired states within various control systems. They loop a portion of the system's output back as input to control its dynamic behavior.
Key characteristics of feedback systems include:
  • Stability: To maintain steady performance, the feedback system should stabilize the output to settle at the desired value despite external disturbances.
  • Accuracy and Robustness: Feedback aids in adjusting the system to compensate for unpredicted changes, allowing for more precise operation.
  • Control of Transient Behavior: Feedback is crucial in governing the system's reaction over time, ensuring minimal deviation from desired trajectories.
Ultimately, feedback systems enhance performance by regulating the system response, minimizing errors, and ensuring consistent outputs.
Characteristic Equation
The characteristic equation is a fundamental equation derived from a system's transfer function. It is pivotal in determining the system's behavior and stability by describing its poles, which are the roots of the equation.
Important aspects of the characteristic equation include:
  • It identifies the system's poles, which dictate response characteristics such as damping, oscillation, and stability.
  • The equation is found by setting the denominator of the closed-loop transfer function to zero.
  • Real poles often lead to exponential decay or growth, while complex conjugate poles result in oscillations given the non-zero imaginary parts.
  • System tuning often involves adjusting the parameters within the characteristic equation to achieve optimal performance.
Grasping how the characteristic equation operates is key to mastering control systems design.

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Most popular questions from this chapter

Consider a feedback system, either in continuous-time or discrete-time, and suppose that the Nyquist plat for the system passes through the point \(-1 / K\). Is the feedback system stabie or unstable for this value of the gain? Explain your answer.

cConsider a continuous-time feedback system whose closed-loop poles satisfy \\[G(s) H(s)=\frac{1}{(s+1)}=-\frac{1}{K}.\\] Use the Nyyuist plot and the Nyquist stability criterion to determine the range of values of \(K\) for which the closed-loop system is stable. Hin: In sketching the Nyquist plot, you may find it useful to sketch the corresponding Bode plot first. It also is helpful to determine the values of \(\omega\) for which \(G(j \omega\\} H(j \omega)\) is real.

Consider a continuous-time feedback system whose closed-loop poles satisfy \\[G(s) H(s)=\frac{1}{(s+1)(s / 10+1)}=-\frac{1}{K}.\\] Use the Nyquist plot and the Nyguist stability criterion to determine the range of yaiues of \(K\) for which the closed-loop system is stable.

Suppose the closed-loop poles of a discrete-time feedback system satisfy \\[\frac{z^{-2}}{\left(1-\frac{1}{2} z^{-1}\right)\left(1+\frac{1}{2} z^{-1}\right)}=-\frac{1}{K}.\\] Using the root-locus method, determine the positive values of \(K\) for which this system is stable.

Consider a feedback system with \\[H(s)=\frac{s+2}{s^{2}+2 s+4}, \quad G(s)=K.\\] (a) Sketch the root locus for \(K > 0\) (b) Sketch the root locus for \(K < 6 \) (c) Find the smallest positive value of \(K\) for which the elosed-loop impulse response does not exhibit any oscillatory behavior.
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