Single Degree of Freedom, Free Undamped Vibration
 Single Degree of Freedom, Free Undamped Vibration Single DOF, Undamped Torsional Vibration Modeling In order to prevent spam, users must register before they can edit or create articles.

## 1 Introduction to Single Degree of Freedom, Free Undamped Vibration

A body of mass m is free to move along a fixed horizontal surface. A spring of stiffness k is fixed at one end and attached to the mass at the other end. The horizontal force F can be used to disturb the mass or control it.

At equilibrium the spring force (kx) is equal to 0. Define x = 0 to be at the spring's equilibirum. Moving the mass to the right (+x) will cause the spring to pull the mass to the left. Moving the mass to the left of x = 0 causes the spring to push the mass to the right. In this ideal system there is no damping so moving the mass to position x0 will cause the mass to oscillate between -x0 and +x0.

## 2 Equations for Single Degree of Freedom, Free Undamped Vibration

Table 1: System Givens
Givens

Constant Mass

Spring stiffness is Constant

Spring mass is negligible

Modeling Stages
Stage I: Devise a mathematical or physical model of the system to be analyzed.
Stage II: From the model, write the equations of motion.
Stage III: Evaluate the system response to relevant specific excitation.

Upon examination of the free body diagram (FBD) the equation of motion for the system is $LaTeX: m\ddot{x}=ma=-kx$ or $LaTeX: \ddot{x}+\left(\frac{k}{m}\right)x=0$ (2.1)

The solution to this simple harmonic motion equation is $LaTeX: x=A\mbox{cos}\omega t + B\mbox{sin}\omega t$ (2.2)

where $LaTeX: A$ and $LaTeX: B$ are constants.

A and B can be found by considering the initial conditions and the natural frequency $LaTeX: \omega$. Substituting (2.2) into (2.1) $LaTeX: -\omega^2\left(A\mbox{cos}\omega t + B\mbox{sin}\omega t\right) + \left(\frac{k}{m}\right)\left(A\mbox{cos}\omega t + B\mbox{sin}\omega t\right)=0$

Since $LaTeX: \left(A\mbox{cos}\omega t + B\mbox{sin}\omega t\right) \ne 0$ (otherwise no motion), $LaTeX: \omega=\sqrt{\frac{k}{m}}$ rad/s

and $LaTeX: x=A\mbox{cos}\sqrt{\frac{k}{m}}t + B\mbox{sin}\sqrt{\frac{k}{m}}t$

Now $LaTeX: x=x_{0}$ at $LaTeX: t=0$

thus $LaTeX: x_{0}=A\mbox{cos}0 + B\mbox{sin}0$, and therefore $LaTeX: x_{0}=A$

and $LaTeX: \dot{x}=0$ at $LaTeX: t=0$

thus $LaTeX: 0=-A\sqrt{\frac{k}{m}}\mbox{sin}0 + B\sqrt{\frac{k}{m}}\mbox{cos}0$, and therefore $LaTeX: B=0$

that is, $LaTeX: x=x_{0}\mbox{cos}\sqrt{\frac{k}{m}}t$ (2.3)

System parameters control ω and the type of motion but not the amplitude. The mass is important but the weight is not; so for a given system the natural frequency, ω, is independent of the local gravitational field.

The frequency of vibration , f, is given by $LaTeX: f=\frac{\omega}{2\pi}$, or $LaTeX: f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}$ Hz (2.4)

The period of oscillation, τ, is the time taken for 1 complete cycle. $LaTeX: \tau=\frac{1}{f}=2\pi\sqrt{\frac{m}{k}}$ seconds (2.5)

## 3 Control of a Single Degree of Freedom, Free Undamped Vibration System

### 3.1 State-Space Formulation

Using the equations of motion we can create a state-space model of the system. $LaTeX: \ddot{x}+\left(\frac{k}{m}\right)x=0$ $LaTeX: \ddot{x}=-\left(\frac{k}{m}\right)x$

Use the following substitution $LaTeX: q = \begin{Bmatrix}\dot{x} \\ x \end{Bmatrix}$

then $LaTeX: \dot{q}=\begin{bmatrix}0 & -\frac{k}{m} \\ 1 & 0\end{bmatrix}q+Bu$

where $LaTeX: B$ is $LaTeX: u$ is the control force.

and $LaTeX: y=Cq+Du$

where $LaTeX: C$ is $LaTeX: D$ is $LaTeX: y$ is the output.

For this example we want to control position so $LaTeX: B=\begin{bmatrix}1 \\ 0\end{bmatrix}$ $LaTeX: C=\begin{bmatrix}0 & 1\end{bmatrix}$ and $LaTeX: D=0$

### 3.2 Transfer Function Formulation

The transfer function equivalent to this state-space example is $LaTeX: \frac{1}{s^2+0.1}$

This is equivalent to a standard 2nd order system where $LaTeX: \zeta = 0$, $LaTeX: \omega_{n} = \sqrt{0.1}$ and $LaTeX: k = \frac{1}{\omega_{n}^2}$. Considering a second order system with a damping coefficient of 0 we would expect a large spike at the natural frequency of $LaTeX: \omega_{n}=\sqrt{0.1}$.

### 3.3 Controller Design

This simple single DOF undamped system will be used to explore a couple of different controller design techniques.

#### 3.3.1 Simple Integrator

The following MATLAB code will create a state space object

 >> A = [0, -k/m ; 1, 0];
>> B = [1 ; 0];
>> C = [0, 1];
>> D = 0;
>> sys = ss(A, B, C, D);


Adding an integrator (transfer function object) for control produces the following open loop (as a transfer function object)

 >> int = tf(, [1 0]);
>> ol = int * tf(sys);


To form the closed loop

 >> cl = ol / (1 + ol);
>> [z, p, k] = zpkdata(cl, 'v')
z = 0
0 +/- 0.3162i
p = 0
0.4833 +/- 0.8949i
-0.9667
0.0000 +/- 0.3162i
k = 1


Note that the closed loop system is unstable - 2 poles in right half plane (RHP).

This author (Gabe) has seen a lot of 2nd order systems at work. Typically these systems are well controlled by a PI-Lead controller. This type of controller combines a PI controller with a Lead Controller to improve the phase margin of the system under control.

The PI Controller has the following form $LaTeX: C_{PI}=K_{P} + K_{I}\frac{1}{s}=\frac{K_{P}s+K_{I}}{s}$

The Lead Controller has the following form $LaTeX: C_{Lead}=\frac{s+a}{s+b}$

where $LaTeX: a so that the phase "hump" created by the controller is positive.

I haven't found many resources on how to decide where to place the PI controller zero. My thinking is that you place the zero where you need to in order to achieve the disturbance rejection that the system requires. In other words if you are controlling a 2nd order system that requires 40 dB of disturbance rejection at 1 Hz then your PI zero needs to be no lower than 100 Hz; this is because the integrator will give you approximately 20 dB per decade of rejection. Let's assume that a PI zero at 100 Hz will provide ample disturbance rejection.

The Lead controller is incorporated to improve phase margin. Phase margin is calculated at the open loop crossover frequency so the peak phase improvement from the Lead controller should occur at approximately the open loop crossover. The peak phase improvement occurs at the geometric mean. For the sake of this example let's place the open loop crossover at 200 Hz.

The MATLAB commands for the PI-Lead are

 >> f = 200 * (2*pi);       % Open Loop Crossover
>> z = 100 * (2*pi);       % Zero for PI
>> X = 10;                    % Separation factor for Lead
>> a = f / sqrt(X);         % Zero for Lead
>> b = f * sqrt(X);         % Pole for Lead
>> cntl = tf([1 z], [1 0]) * tf([1 a], [1 b]);


The open loop system is formed by

 >> ol = cntl * tf(sys);


The gain adjustment to achieve an open loop crossover at f

 >> [mag] = bode(ol, f);
>> ol = 1/mag * ol;


The closed loop system is formed like this

 >> cl = ol / (1 + ol);


MATLAB's isstable command returns false for the closed loop system. The margin command will return a plot with 28° of phase margin. However, the pole command will show the only unstable poles to be approximately 2E-16. This may well be a truncation error. The result of the step command does not look unstable.

## 4 Final Notes on Control of Single DOF Undamped System

We know that stable 2nd order systems can be created. This system, without damping, is not really stable. The integrator and PI-Lead controls were designed to control the position of the mass m.

The main part of the 2nd order systems that is missing is the damping. Damping operates on the velocity of the mass m. This example could be reformulated as a velocity control problem. When you wish to control position but do it through control of the velocity some alterations are required.