Second Order Differential Equation

M

mario99

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\(\displaystyle M\frac{d^2x}{dt^2} + f_v\frac{dx}{dt} + kx = f(t)\)

I have that equation and when I apply laplace transform assuming all initial conditions are zero, I get this.

transfer function \(\displaystyle = \frac{X(s)}{F(s)} = \frac{1}{Ms^2 + f_vs + k}\)

Now I have to find the state-space representation of that.


According to this video, I get this.

\(\displaystyle \dot{X} = \begin{bmatrix}\dot{x_1} \\\dot{x_2} \end{bmatrix} = \begin{bmatrix}0 & 1 \\-k & -f_v \end{bmatrix} \begin{bmatrix}x_1 \\x_2 \end{bmatrix} + \begin{bmatrix}0 \\1 \end{bmatrix}f(t)\)

According to my notes, I get this.

\(\displaystyle \dot{X} = \begin{bmatrix}\dot{x_1} \\\dot{x_2} \end{bmatrix} = \frac{1}{M}\begin{bmatrix}0 & 1 \\k & f_v \end{bmatrix} \begin{bmatrix}x_1 \\x_2 \end{bmatrix} + \begin{bmatrix}0 \\1 \end{bmatrix}f(t)\)

Which method is the correct one?

Any help would be appreciated!
 
The video doesn't work. Without it I'm not sure what method is being used. Please post your work so we know what you are trying to do and so we can see where you might have made mistakes.

Though you've been told this countless times anyway...

-Dan
 
mario99 said:
\(\displaystyle M\frac{d^2x}{dt^2} + f_v\frac{dx}{dt} + kx = f(t)\)

I have that equation and when I apply laplace transform assuming all initial conditions are zero, I get this.

transfer function \(\displaystyle = \frac{X(s)}{F(s)} = \frac{1}{Ms^2 + f_vs + k}\)

Now I have to find the state-space representation of that.


According to this video, I get this.

\(\displaystyle \dot{X} = \begin{bmatrix}\dot{x_1} \\\dot{x_2} \end{bmatrix} = \begin{bmatrix}0 & 1 \\-k & -f_v \end{bmatrix} \begin{bmatrix}x_1 \\x_2 \end{bmatrix} + \begin{bmatrix}0 \\1 \end{bmatrix}f(t)\)

According to my notes, I get this.

\(\displaystyle \dot{X} = \begin{bmatrix}\dot{x_1} \\\dot{x_2} \end{bmatrix} = \frac{1}{M}\begin{bmatrix}0 & 1 \\k & f_v \end{bmatrix} \begin{bmatrix}x_1 \\x_2 \end{bmatrix} + \begin{bmatrix}0 \\1 \end{bmatrix}f(t)\)

Which method is the correct one?

Any help would be appreciated!
Click to expand...
In your solution, you have an "undefined" variable - (M).

Please define it.
 
topsquark said:
The video doesn't work. Without it I'm not sure what method is being used. Please post your work so we know what you are trying to do and so we can see where you might have made mistakes.

Though you've been told this countless times anyway...

-Dan
Click to expand...
I showed everything. What work you are talking about? We are supposed to get the state-space representation from the transfer function.

Here is the link again.

[math]https://www.youtube.com/watch?v=kipoOYKsEPI[/math]

Subhotosh Khan said:
In your solution, you have an "undefined" variable - (M).

Please define it.
Click to expand...
Mass.
 
mario99 said:
According to my notes, I get this.

\(\displaystyle \dot{X} = \begin{bmatrix}\dot{x_1} \\\dot{x_2} \end{bmatrix} = \frac{1}{M}\begin{bmatrix}0 & 1 \\k & f_v \end{bmatrix} \begin{bmatrix}x_1 \\x_2 \end{bmatrix} + \begin{bmatrix}0 \\1 \end{bmatrix}f(t)\)
Click to expand...
How did you get this?

Also: I'm not going to go through the whole video series, so please explain how the video's terminology,
[imath]z = \left [ \begin{matrix} x \\ \dot{x} \end{matrix} \right ] [/imath]

becomes
[imath]z = \left [ \begin{matrix} x_1 \\ x_2 \end{matrix} \right ] [/imath]

in your terminology. There seems to be a problem with the number of time derivatives.

-Dan
 
Last edited:
topsquark said:
How did you get this?

Also: I'm not going to go through the whole video series, so please explain how the video's terminology,
[imath]z = \left [ \begin{matrix} x \\ \dot{x} \end{matrix} \right ] [/imath]

becomes
[imath]z = \left [ \begin{matrix} x_1 \\ x_2 \end{matrix} \right ] [/imath]

in your terminology. There seems to be a problem with the number of time derivatives.

-Dan
Click to expand...
How did I get that? The matrix is fixed (almost). When you get the transfer function, you just have to plug in the coefficients inside the matrix while watching how many \(\displaystyle s\), you have. For example \(\displaystyle a_2s^2 + a_1s + a_0\), or \(\displaystyle a_3s^3 + a_2s^2 + a_1s + a_0\). You will alert the matrix according to that.

It is clear that \(\displaystyle x = x_1\) and \(\displaystyle \dot{x} = x_2\).

My notes prefer to do this: \(\displaystyle \dot{x_1} = \dot{x} = x_2\).

The notation is not the main issue. The main issue is that, do I insert the coefficients with negative signs while ignoring \(\displaystyle M\) or inserting \(\displaystyle M\) while ignoring the negative signs?
 
mario99 said:
How did I get that? The matrix is fixed (almost). When you get the transfer function, you just have to plug in the coefficients inside the matrix while watching how many \(\displaystyle s\), you have. For example \(\displaystyle a_2s^2 + a_1s + a_0\), or \(\displaystyle a_3s^3 + a_2s^2 + a_1s + a_0\). You will alert the matrix according to that.

It is clear that \(\displaystyle x = x_1\) and \(\displaystyle \dot{x} = x_2\).

My notes prefer to do this: \(\displaystyle \dot{x_1} = \dot{x} = x_2\).

The notation is not the main issue. The main issue is that, do I insert the coefficients with negative signs while ignoring \(\displaystyle M\) or inserting \(\displaystyle M\) while ignoring the negative signs?
Click to expand...
In order for us to help you the notation is critical! If we don't know what you are saying how can we help you? I don't know how you aren't seeing this.

Okay, so [imath]x_1 = x[/imath] and [imath]x_2 = \dot{x}[/imath]. The video's method used a leading coefficient of 1 and you have a leading coefficient of M so your whole equation, not just the part with A, has to be divided by M. The correction is to also divide f(t) by M in your equation.

-Dan
 
Thank you very much for helping me topsquark.

So, you mean the matrix must be like this?

\(\displaystyle \dot{X} = \begin{bmatrix}\dot{x_1} \\\dot{x_2} \end{bmatrix} = \frac{1}{M}\begin{bmatrix}0 & 1 \\k & f_v \end{bmatrix} \begin{bmatrix}x_1 \\x_2 \end{bmatrix} + \frac{1}{M}\begin{bmatrix}0 \\1 \end{bmatrix}f(t)\)
 
mario99 said:
Thank you very much for helping me topsquark.

So, you mean the matrix must be like this?

\(\displaystyle \dot{X} = \begin{bmatrix}\dot{x_1} \\\dot{x_2} \end{bmatrix} = \frac{1}{M}\begin{bmatrix}0 & 1 \\k & f_v \end{bmatrix} \begin{bmatrix}x_1 \\x_2 \end{bmatrix} + \frac{1}{M}\begin{bmatrix}0 \\1 \end{bmatrix}f(t)\)
Click to expand...
Yes. Start from [imath]\ddot{x} + \dfrac{f_v}{M} \dot{x} + \dfrac{k}{M} x = \dfrac{f(t)}{M}[/imath].

-Dan
 
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