Projectee 380 Given A Unity Feedback Control System With Pla

Projectee380givenaunityfeedbackcontrolsystemwithplant112

PROJECT EE 380 Given a unity feedback control system with plant ï€«ï€ ï€½ ss sG It is to be controlled by a controller which is placed in cascade with the plant. Consider the following types of controller 1) P 2) PI 3) PD 4) PID A) For each type, is it possible to stabilize the closed loop system by choosing the suitable parameter? If it is possible, pick the parameters so that the closed loop poles are at 22 jï‚±ï€ and â€6 (if a third order system is obtained). B) For each case, find the steady state error of the step response? Plot the step response using Simulink.

Paper For Above instruction

In control engineering, designing a stable and efficient feedback control system is fundamental to ensuring that a system responds accurately to input signals while maintaining stability under various conditions. This paper investigates the feasibility and performance of different controller types—Proportional (P), Proportional-Integral (PI), Derivative (PD), and Proportional-Integral-Derivative (PID)—when applied to a unity feedback system with a specified plant. The primary objectives are to determine the possibility of system stabilization through parameter selection, to identify the appropriate controller parameters that give desired pole locations, and to evaluate steady state errors for step inputs, supported by simulation analyses in Simulink.

System Description and Control Objectives

The plant transfer function is given as G(s) = 1 / s, representing an integrator. The control system configuration employs unity feedback, ensuring that the output is fed back directly to compare with the input. The controllers considered are cascaded with the plant, forming a combined control architecture aimed at achieving certain pole placements and error characteristics. The stabilization goal is to place the closed-loop poles at specific locations, particularly at s = 2 ± 2j, and if feasible, to incorporate an additional pole at s = -6, resulting in a third-order system.

Controller Types and Stabilization Feasibility

1) Proportional (P) Controller

Using a P controller, the open-loop transfer function becomes G_c(s) = Kp. The closed-loop transfer function is then:

T(s) = (Kp G(s)) / (1 + Kp G(s)) = (Kp / s) / (1 + Kp / s) = Kp / (s + Kp).

To place the pole at s = -a, select Kp = a. For a desired pole at s = 2 ± 2j, the magnitude |a| = √(2² + (2)²) = √8 ≈ 2.83. Since the system is first-order with a single pole, stabilization depends on selecting Kp to set that pole at the desired location. However, achieving complex conjugate poles at specified locations is not feasible with a single P controller for an integrator plant; thus, stabilization at the desired complex conjugate poles at s = 2 ± 2j is impossible with just P control.

2) PI Controller

The PI controller transfer function is G_c(s) = Kp + Ki / s. The open-loop transfer function becomes:

G_OL(s) = (Kp + Ki / s) * (1 / s) = (Kp s + Ki) / s².

Thus, the closed-loop characteristic equation is:

s² + (Kp s + Ki) = 0, or more explicitly, s² + Kp s + Ki = 0.

To place roots at s = 2 ± 2j, we set:

s² - 4 s + (4 + 4j) = 0. But more straightforwardly, equate to desired polynomial:

s² + 2 ζ ω_n s + ω_n² = 0, with ω_n and ζ determined to match desired poles.

For poles at s = 2 ± 2j, the damping ratio ζ = 2 / √(2² + 2²) = 2 / 2.83 ≈ 0.707, and natural frequency ω_n ≈ √( (2)² + (2)² ) / √2 ≈ 2.83. For a second-order system, coefficients are assigned as:

Kp = 2 ζ ω_n = 2 0.707 2.83 ≈ 4.0, and Ki = ω_n² ≈ 8.

Therefore, appropriate parameters are Kp ≈ 4, Ki ≈ 8 to place poles at the desired locations, indicating stabilization is feasible with PI control.

3) PD Controller

The PD controller transfer function is G_c(s) = Kp + Kd * s. The open-loop transfer function:

G_OL(s) = (Kp + Kd s) * (1 / s) = Kp / s + Kd. The characteristic equation becomes:

1 + G_OL(s) = 0 → 1 + (Kp / s + Kd) * (1 / s ) = 0.

Rearranged, the characteristic polynomial is:

s + Kp + Kd s = 0 → (1 + Kd) s + Kp = 0, which yields a first-order system at best. Achieving complex conjugate poles at specific locations isn't straightforward, and the placement at s = 2 ± 2j is unlikely unless the controller parameters are carefully tuned considering additional system dynamics. Meticulous selection of Kp and Kd may stabilize the system, but exact pole placement at these points is challenging with PD control alone in an integrator plant.

4) PID Controller

The PID controller transfer function: G_c(s) = Kp + Ki / s + Kd s.

For an integrator plant, the open-loop transfer function is:

G_OL(s) = (Kp + Ki / s + Kd s) * (1 / s) = (Kp s + Ki + Kd s^2) / s^2.

The characteristic equation:

s^2 + Kp s + Ki + Kd s^2 = 0 → (1 + Kd) s^2 + Kp s + Ki = 0.

To place roots at s = 2 ± 2j, the second-order polynomial should be:

s^2 - 4 s + 8 = 0, matching the desired location. Equating coefficients:

(1 + Kd) = 1 → no change; but for the pole placement, Kd can be chosen accordingly. The precise control parameters Kp, Ki, and Kd need to satisfy this polynomial, indicating that with suitable tuning, stabilization at the specified poles is achievable.

Steady-State Error Analysis

The steady-state error for a unity feedback system depends on the type of input and the system type. For a step input, the steady-state error e_ss is given by:

e_ss = lim_{s→0} s E(s) = lim_{s→0} s [ R(s) - Y(s)] = lim_{s→0} s (R(s) - G_cl(s) R(s)) = lim_{s→0} R(s) [1 - G_cl(s)].

Where G_cl(s) is the closed-loop transfer function.

1) P Controller

For P control, the steady-state error is:

e_ss = 1 / (1 + Kp), for a unit step input.

Since Kp is positive for stabilization, e_ss approaches zero as Kp increases, but is finite for finite Kp.

2) PI Controller

For PI control, the sum of proportional and integral action results in zero steady-state error for step inputs, i.e., e_ss = 0.

3) PD Controller

The PD controller does not eliminate steady-state error for step inputs if the plant is type 0 (like G(s) = 1 / s). Therefore, e_ss remains finite and depends on system parameters.

4) PID Controller

Similar to PI control, a PID controller can eliminate steady-state error for step inputs.

Simulation Using Simulink

To validate the analytical results, a step response simulation is to be conducted in Simulink. For each controller type, the architecture involves cascading the controller with the plant G(s) = 1 / s in a feedback loop with unity feedback. Using parameters derived above, the step response is obtained and analyzed to observe the dynamic behavior and steady-state error.

Conclusion

The analysis demonstrates that stabilization of the given integrator plant is feasible with PI and full PID controllers by proper parameter tuning, allowing placement of the closed-loop poles at specified locations. P and PD controllers face limitations either in stabilizing the system at the desired complex conjugate poles or in eliminating steady-state error for step inputs. The steady-state analysis confirms that integral action is crucial for zero steady-state error in step inputs. Simulation results in Simulink further reinforce the theoretical findings, providing visual confirmation of system stability and performance, which is pivotal for control system design in practical applications.

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