PD Controller with Disturbances.

The questions below are due on Tuesday October 05, 2021; 09:00:00 AM.
 
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Closed-Loop Transfer Functions

Consider the following block diagrams, in the discrete-time and z, for an alternative model of the relation between rotational velocity and rotation angle.

Alternative model for relating rotational velocity to rotation angle.

Based on the above diagram, which of the following difference equations relates the sequence \omega_a to the sequence \theta_a?

Equation I: \theta_a[n] = \theta_a[n-1] + \Delta T \omega_a[n-1],

Equation II: \theta_a[n] = \theta_a[n-1] + \Delta T \omega_a[n],

Equation III: \theta_a[n] = \theta_a[n] + \Delta T \omega_a[n],

Which difference equation (I,II, or III) describes the above block diagram?

Based on the above diagram, which \mathcal{H}(z) (transfer function), I,II, or III, satisfies \Theta_a(z) = \mathcal{H}(z) \Omega_a(z) ?

Equation I: \mathcal{H}(z) = \frac{\Delta T}{1 - z^{-1}} ,

Equation II: \mathcal{H}(z) = \frac{\Delta T z^{-1}}{1 - z^{-1}} ,

Equation III: \mathcal{H}(z) = \Delta T (1 - z^{-1}),

Which transfer function, I, II, or III, describes the above block diagram?

Diagram of PD control of plant with the alternative model and disturbances.

From the above block diagram, we have the familiar transfer functions for the controller,

\mathcal{K}(z) = K_p + \frac{K_d}{\Delta T} (1-z^{-1}),
the top of the page. but the plant model \mathcal{H}(z) is different from any developed in lab, because of the alternative model shown above.

We can write the closed-loop transfer functions from desired input, \Theta_d(z), and from disturbance, \Alpha_{dist}(z), to the output, \Theta_a(z), in the form

\Theta_a(z) = \mathcal{G}(z) \Theta_d(z) + \mathcal{G}_{dist}(z) \Alpha_{dist}(z).

We can relate the two transfer closed-loop transfer functions, \mathcal{G}(z) and \mathcal{G}_{dist}(z) to \mathcal{K}(z) and \mathcal{H}(z), using variants of Black's formula. Which of the following statements is correct?

Statement I: \mathcal{G}_{dist}(z) = \frac{\mathcal{K}(z)}{1+\mathcal{K}(z) \mathcal{H}(z)} and \mathcal{G}(z) = \frac{\mathcal{K}(z)\mathcal{H}(z)}{1+\mathcal{K}(z)\mathcal{H}(z)}

Statement II: \mathcal{G}_{dist}(z) = \frac{\mathcal{H}(z)}{1+\mathcal{K}(z) \mathcal{H}(z)} and \mathcal{G}(z) = \frac{\mathcal{K}(z) \mathcal{H}(z)}{1+\mathcal{K}(z)\mathcal{H}(z)}

Statement III: \mathcal{G}_{dist}(z) = \frac{\mathcal{K}(z) \mathcal{H}(z)}{1+\mathcal{K}(z) \mathcal{H}(z)} and \mathcal{G}(z) = \frac{\mathcal{H}(z)}{1+\mathcal{K}(z)\mathcal{H}(z)}

Which of the above statements is correct?

From the above diagram, one can determine the transfer closed-loop transfer functions as rational functions in z. Be careful to note that the plant is a mixture of the alternative model above and the more typical model, and determine the closed-loop transfer functions in the forms,

\mathcal{G}(z) = \frac{B_1 z^{-1} + B_2 z^{-2}}{1 + A_1 z^{-1} + A_2 z^{-2}},
\mathcal{G}_{dist}(z) = \frac{Bd_1 z^{-1} + Bd_2 z^{-2}}{1 + A_1 z^{-1} + A_2 z^{-2}}.
Notice that the denominator is scaled with a particular normalization that simplifies the algebra.

The coefficients A_1, A_2, B_1, B_2, Bd_1 and Bd_2 are all functions of K_d, K_p, and \Delta T. Below please give the expressions for each using python syntax and use dt for \Delta T, Kp for K_p, and Kd for K_d. For example, to enter \frac{K_d}{\Delta T}, type Kd/dt, to enter -(1 + 2 K_p), type -(1+2*Kp).

A_1 =

A_2 =

B_1 =

B_2 =

Bd_1 =

Bd_2 =

Input and Disturbance Step Responses

For the questions below, we will be analyzing the behavior of the PD control system of section 1 for three different sets of feedback gains. For the cases we will analyze, we will use \Delta T = 0.01, K_p = 10 K_0, and K_d = 2 K_0, where K_0 is a gain scaling factor. We will consider three cases, K_0 = 1, K_0 = 10, and K_0 = 45, whose three step responses are plotted below (NOT RESPECTIVELY!). The plots in blue are the responses to a unit step in the input, \theta_d, the plots in red are the responses to a unit step in the disturbance, \alpha_{dist}.

Plots of responses to input(blue) and disturbance (red) unit steps.

Using partial fraction expansion, we've seen that the step response for a system with two natural frequencies, assuming they are different from each other and different from 1, will be of the form

\theta_a[n] = \alpha_1 \lambda_1^n + \alpha_2 \lambda_2^n + \alpha_{in} 1^n,

where \lambda_1, \lambda_2, are the natural frequencies, and \alpha_1, \alpha_2, and \alpha_{in} can be determined by partial fraction expansion. In addition, if the natural frequencies are all less than one in magnitude (the system is stable), then

\lim_{n\rightarrow \infty} \lambda_i^n = 0,

for all i, and

lim_{n \rightarrow \infty} \theta_a[n] = \alpha_{in}.

Use your formulas for the disturbance transfer function, to determine the answers to the questions below.

The step response plots are, in the order of lowest to highest K_0:

What is the steady-state value for the disturbance step response for the system with the best disturbance rejection?

Best disturbance steady-state =

What is the maximum magnitude natural frequency for the system whose maximum magnitude natural frequency is smallest?

Smallest Maximum Magnitude =

What is the maximum magnitude natural frequency for the system whose maximum magnitude natural frequency is largest?

For the system with the fastest settling step response, what is the value of \alpha_i for the natural frequency with the maximum magnitude?

**Clarification: You may approach this problem in several ways. (1) Computing the partial fraction decomposition of G(z)X(z) where X(z) is the z-transform of the unit step \frac{1}{1-z^{-1}} by hand (see the week 3 scribbles here and here) or (2) figure out how to use Matlab's residuez function. If you decide to use matlab's residuez function, you will need the numerator and denominator coefficients of G(z)X(z).

What you should have noticed from this problem is that the system with the highest K_0 (K_0 = 45) had the best disturbance rejection and the fastest settling time, but did NOT have the smallest maximum magnitude natural frequency. The reason the K_0=45 system settled faster even though one of its natural frequencies was larger than the K_0 = 10 case was because of a very small \alpha_i. So, natural frequencies tell you a LOT, but not everything!