Control Systems/Examples/Second Order Systems

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Second Order Systems: Examples[edit]

Example 1[edit]

A damped control system for aiming a hydrophonic array on a minesweeper vessel has the following open-loop transfer function from the driveshaft to the array.

G(s)=\frac{K}{Js^2 + K_ds}

The gain parameter K can be varied. The moment of inertia, J, of the array and the force due to viscous drag of the water, Kd are known constants and given as:

  • J = 9 \, N \, m \, s^2 \, rad^{-1}
  • K_d = 2 \, N \, m \, s \, rad^{-1}

Tasks[edit]

  1. The system is arranged as a closed loop system with unity feedback. Find the value of K such that, when the input is a unit step, the closed loop response has at most a 50% overshoot (approximately). You may use standard response curves. Should K be greater or less than this value for less overshoot?
  2. Find the corresponding time-domain response of the system.
  3. The system is now given an input of constant angular velocity, V. For the limiting value of K found above, calculate the maximum value of V such that the array follows the input with at most 5° error.

Task 1[edit]

First, let us draw the block diagram of the system. We know the open-loop transfer function, and that there is unit feedback. Therefore, we have:

Control Theory Example 1.svg

The closed-loop gain is given by:

H(s) =\frac{G(s)}{1+G(s)}
  =\frac{\dfrac{K}{Js^2 + K_ds}}{\dfrac{Js^2 + K_ds+K}{Js^2 + K_ds}}
  =\frac{K}{Js^2 + K_ds+K}

We now need to express the closed-loop transfer function in the standard second order form.

\frac{\omega_n^2}{s^2+2\zeta\omega_n s + \omega_n^2} =\frac{K}{Js^2 + K_ds+K}
  =\frac{\frac{K}{J}}{s^2 + \frac{K_d}{J}s+\frac{K}{J}}

We can now express the natural frequency ωn and damping ratio, ζ:

\omega_n^2 = \frac{K}{J} = \frac{K}{9}
2 \zeta \omega_n = \frac{K_d}{J}
\zeta = \frac{K_d}{2J} \sqrt{\frac{J}{K}} = \frac{K_d}{2} \sqrt{\frac{1}{KJ}}=\frac{1}{3\sqrt{K}}

We now look at the standard response curves for second order systems.

2nd Order Damping Ratios.svg

We see that for 50% overshoot, we need ζ=0.2 or more.

\zeta = \frac{1}{3 \sqrt{K}} \le \frac{1}{5}
K \ge \frac{25}{9}

This is the maximum permissible value, thus K should be less than this value for less overshoot. We can now evaluate the natural frequency fully:

\omega_n = \frac{5}{9}

Task 2[edit]

The output of the second order system is given by the following equation:

y(t) = 1 - \frac{1}{\sqrt{1-\zeta^2}} e^{-\zeta \omega_n t} \sin \left( \sqrt{1-\zeta^2} \omega_n t + \sin^{-1} \sqrt{1-\zeta^2} \right)
  = 1 - \frac{1}{\sqrt{0.96}} e^{-t/9} \sin \left( \frac{5\sqrt{0.96}}{9} t + \sin^{-1} \sqrt{0.96} \right)
  = 1 - 1.02 e^{-t/9} \sin \left( 0.544 t +0.436 \pi \right)

We can plot the output of this system:

Control Theory Example 1a.svg

Task 3[edit]

The tracking error signal, E(s), is equal to the output's deviation from the input.

E(s) = R(s) - Y(s)\,

Now, we can find the gain from the reference input, R(s) to the error tracking signal:

\frac{E(s)}{R(s)} = \frac{R(s)-Y(s)}{R(s)}

The gain from the input to the error tracking signal of a unity feedback system like this is simply \frac{1}{1+G(s)}.

\frac{E(s)}{R(s)} = \frac{1} { 1 + \frac{K}{9s^2 + 2s}}
  = \frac{9s^2 + 2s}{9s^2 + 2s + 2.78}

Now, R(s) is given by the Laplace transform of a ramp of slope V:

R(s) = \frac{v}{s^2}

We now use the final value theorem to find the value of E(s) in the steady state:

\lim_{t \to \infty}{e(t)} = \lim_{s \to 0}{sE(s)}
  =\lim_{s \to 0}{s \frac{9s^2 + 2s}{9s^2 + 2s + 2.78} \frac{V}{s^2} }
  =\frac{2V}{2.78}

We require this to be less than 5^\circ = \frac{5\pi}{180} \mbox{rad}

V \le \frac{2.78}{2}\times \frac{5\pi}{180} = 0.12 \,\mbox{rad/s}