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State space representation

If u(t)=my¨(t)u(t)=m\ddot{y}(t),

x(t)=(y(t)y˙(t))x(t)=\begin{pmatrix} y(t) \\ \dot{y}(t) \end{pmatrix}

And x˙(t)=(y˙(t)y¨(t))=(0100)(y(t)y˙(t))+(01m)u(t)\dot{x}(t)=\begin{pmatrix} \dot{y}(t) \\ \ddot{y}(t) \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} \begin{pmatrix} y(t) \\ \dot{y}(t) \end{pmatrix} + \begin{pmatrix} 0 \\ \frac{1}{m} \end{pmatrix} u(t)

Then what is x˙(t)\dot{x}(t) if we set x(t)=(y˙(t)y(t))x(t)=\begin{pmatrix} \dot{y}(t) \\ y(t) \end{pmatrix} instead?
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TeeEm
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Original post by RamocitoMorales
If u(t)=my¨(t)u(t)=m\ddot{y}(t),

x(t)=(y(t)y˙(t))x(t)=\begin{pmatrix} y(t) \\ \dot{y}(t) \end{pmatrix}

And x˙(t)=(y˙(t)y¨(t))=(0100)(y(t)y˙(t))+(01m)u(t)\dot{x}(t)=\begin{pmatrix} \dot{y}(t) \\ \ddot{y}(t) \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} \begin{pmatrix} y(t) \\ \dot{y}(t) \end{pmatrix} + \begin{pmatrix} 0 \\ \frac{1}{m} \end{pmatrix} u(t)

Then what is x˙(t)\dot{x}(t) if we set x(t)=(y˙(t)y(t))x(t)=\begin{pmatrix} \dot{y}(t) \\ y(t) \end{pmatrix} instead?


I do not know what is more scary

your avatar, your reputation or your question ...

I wish I could help you but this question brought back some bad memories ...
Original post by TeeEm
I wish I could help you but this question brought back some bad memories ...


Vector calculus isn't really my cup of tea either. :indiff:
Original post by RamocitoMorales
If u(t)=my¨(t)u(t)=m\ddot{y}(t),

x(t)=(y(t)y˙(t))x(t)=\begin{pmatrix} y(t) \\ \dot{y}(t) \end{pmatrix}

And x˙(t)=(y˙(t)y¨(t))=(0100)(y(t)y˙(t))+(01m)u(t)\dot{x}(t)=\begin{pmatrix} \dot{y}(t) \\ \ddot{y}(t) \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} \begin{pmatrix} y(t) \\ \dot{y}(t) \end{pmatrix} + \begin{pmatrix} 0 \\ \frac{1}{m} \end{pmatrix} u(t)

Then what is x˙(t)\dot{x}(t) if we set x(t)=(y˙(t)y(t))x(t)=\begin{pmatrix} \dot{y}(t) \\ y(t) \end{pmatrix} instead?


Methinks...

For x(t)=(y˙(t)y(t))\mathbf{x}(t)=\begin{pmatrix} \dot{y}(t) \\ y(t) \end{pmatrix},

x˙(t)=(y¨(t)y˙(t))=(0010)(y˙(t)y(t))+(1m0)u(t)=(0010)x+(1m0)u(t)\dot{\boldsymbol{x}}(t) = \begin{pmatrix} \ddot{y}(t) \\ \dot{y}(t) \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} \dot{y}(t) \\ y(t) \end{pmatrix} + \begin{pmatrix} \frac{1}{m} \\ 0 \end{pmatrix} u(t) = \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix} \boldsymbol{x} + \begin{pmatrix} \frac{1}{m} \\ 0 \end{pmatrix} u(t)

Neither equation says a lot, they are just trivially true. y(t),u(t)y(t), u(t) are scalar functions.
(edited 9 years ago)
Original post by dingdongbat
Methinks...

For x(t)=(y˙(t)y(t))\mathbf{x}(t)=\begin{pmatrix} \dot{y}(t) \\ y(t) \end{pmatrix},

x˙(t)=(y¨(t)y˙(t))=(0010)(y˙(t)y(t))+(1m0)u(t)=(0010)x+(1m0)u(t)\dot{\boldsymbol{x}}(t) = \begin{pmatrix} \ddot{y}(t) \\ \dot{y}(t) \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} \dot{y}(t) \\ y(t) \end{pmatrix} + \begin{pmatrix} \frac{1}{m} \\ 0 \end{pmatrix} u(t) = \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix} \boldsymbol{x} + \begin{pmatrix} \frac{1}{m} \\ 0 \end{pmatrix} u(t)

Neither equation says a lot, they are just trivially true. y(t),u(t)y(t), u(t) are scalar functions.


So if we say that,

x˙(t)=Ax(t)+Bu(t)\dot{x}(t)=Ax(t)+Bu(t), y(t)=Cx(t)+Du(t)y(t)=Cx(t)+Du(t)

Then we'd have A=(0010)A=\begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}, B=(1m0)B=\begin{pmatrix} \frac{1}{m} \\ 0 \end{pmatrix}, then what would CC and DD be?
Original post by RamocitoMorales
So if we say that,

x˙(t)=Ax(t)+Bu(t)\dot{x}(t)=A\boldsymbol{x}(t)+Bu(t), y(t)=Cx(t)+Du(t)y(t)=C\boldsymbol{x}(t)+Du(t)

Then we'd have A=(0010)A=\begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}, B=(1m0)B=\begin{pmatrix} \frac{1}{m} \\ 0 \end{pmatrix}, then what would CC and DD be?


I don't know whether that is true but assuming it is, then since y(t),u(t)y(t), u(t) are scalars, CC must be a row matrix of length 2, so:-


y(t)=(c1c2)x(t)+Du(t)=(c1c2)(y˙(t)y(t))+Dmy¨(t)=c1y˙(t)+c2y(t)+Dmy¨(t)y(t) = \begin{pmatrix} c_1 & c_2 \end{pmatrix} \mathbf{x}(t) + Du(t) = \begin{pmatrix} c_1 & c_2 \end{pmatrix} \begin{pmatrix} \dot{y}(t) \\ y(t) \end{pmatrix} + Dm\ddot{y}(t) \\ = c_1 \dot{y}(t) + c_2 y(t) + Dm\ddot{y}(t)

which should result in a second-order ODE:-
0=c1y˙(t)+(c2−1)y(t)+Dmy¨(t)0 = c_1 \dot{y}(t) + (c_2 - 1) y(t) + Dm\ddot{y}(t)

that is readily solvable.

But it's been years since I messed with this stuff so you'll need to check whether the above is sound.
Original post by dingdongbat
which should result in a second-order ODE:-
0=c1y˙(t)+(c2−1)y(t)+Dmy¨(t)0 = c_1 \dot{y}(t) + (c_2 - 1) y(t) + Dm\ddot{y}(t)

that is readily solvable.


The characteristic equation is

Dmr2+c1r+(c2−1)=0Dmr^{2}+c_{1}r+(c_{2}-1)=0.

What does that tell me, as I know neither if c12−4Dm(c2−1)c^{2}_{1}-4Dm(c_{2}-1) is less than, greater than or equal to 0?

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