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Bumped by Community user
Bumped by Community user
Bumped by Community user
Bumped by Community user
included a factor s in the denominator of the transfer function + the left hand side of the backdoor is G(s), not y
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The following system is given and I'm asked to find the transfer function $$\frac{Y(s)}{U(s)}=G(s)$$ $$\bar {\dot x}=\begin{bmatrix} 0 & 1 & 0 & 0 &0 \\0 & 0 & 1 & 0 & 0 \\ 0 & -1 & -2 & 0 & 0 \\ 0 & 0 & 0 & -2 & 0\\ 0 & 0 & 0 & 0 & -3\end{bmatrix} \bar x + \begin{bmatrix} 0\\0\\1\\1\\0\end{bmatrix}u=A\bar x + Bu \\ $$

$$y=\begin{bmatrix}0 & 1 & 0 & 0 & 1\end{bmatrix} \bar x $$

I haven't practised that much on state space models and I don't remember that much from matrix algebra but here's what I've thought.

I found the eigenvalues seeing the matrix as a 2x2 diagonal matrix with matrix elements therefore getting the denominator of my transfer function $$(s+1)^2(s+2)(s+3)$$$$s(s+1)^2(s+2)(s+3)$$.

I also have my backdoor $$y=C(sI-A)^{-1}B $$$$G(s)=C(sI-A)^{-1}B $$ but this includes many calculations and I guess that there is a faster solution.

The following system is given and I'm asked to find the transfer function $$\frac{Y(s)}{U(s)}=G(s)$$ $$\bar {\dot x}=\begin{bmatrix} 0 & 1 & 0 & 0 &0 \\0 & 0 & 1 & 0 & 0 \\ 0 & -1 & -2 & 0 & 0 \\ 0 & 0 & 0 & -2 & 0\\ 0 & 0 & 0 & 0 & -3\end{bmatrix} \bar x + \begin{bmatrix} 0\\0\\1\\1\\0\end{bmatrix}u=A\bar x + Bu \\ $$

$$y=\begin{bmatrix}0 & 1 & 0 & 0 & 1\end{bmatrix} \bar x $$

I haven't practised that much on state space models and I don't remember that much from matrix algebra but here's what I've thought.

I found the eigenvalues seeing the matrix as a 2x2 diagonal matrix with matrix elements therefore getting the denominator of my transfer function $$(s+1)^2(s+2)(s+3)$$.

I also have my backdoor $$y=C(sI-A)^{-1}B $$ but this includes many calculations and I guess that there is a faster solution.

The following system is given and I'm asked to find the transfer function $$\frac{Y(s)}{U(s)}=G(s)$$ $$\bar {\dot x}=\begin{bmatrix} 0 & 1 & 0 & 0 &0 \\0 & 0 & 1 & 0 & 0 \\ 0 & -1 & -2 & 0 & 0 \\ 0 & 0 & 0 & -2 & 0\\ 0 & 0 & 0 & 0 & -3\end{bmatrix} \bar x + \begin{bmatrix} 0\\0\\1\\1\\0\end{bmatrix}u=A\bar x + Bu \\ $$

$$y=\begin{bmatrix}0 & 1 & 0 & 0 & 1\end{bmatrix} \bar x $$

I haven't practised that much on state space models and I don't remember that much from matrix algebra but here's what I've thought.

I found the eigenvalues seeing the matrix as a 2x2 diagonal matrix with matrix elements therefore getting the denominator of my transfer function $$s(s+1)^2(s+2)(s+3)$$.

I also have my backdoor $$G(s)=C(sI-A)^{-1}B $$ but this includes many calculations and I guess that there is a faster solution.

deleted 1 character in body
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John Katsantas
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The following system is given and I'm asked to find the transfer function $$\frac{Y(s)}{U(s)}=G(s)$$ $$\bar {\dot x}=\begin{bmatrix} 0 & 1 & 0 & 0 &0 \\0 & 0 & 1 & 0 & 0 \\ 0 & -1 & -2 & 0 & 0 \\ 0 & 0 & 0 & -2 & 0\\ 0 & 0 & 0 & 0 & -3\end{bmatrix} \bar x + \begin{bmatrix} 0\\0\\1\\1\\0\end{bmatrix}u=A\bar x + Bu \\ $$

$$y=\begin{bmatrix}0 & 1 & 0 & 0 & 1\end{bmatrix} \bar x $$

I haven't practised that much on state space models and I don't remember that much from matrix algebra but here's what I've thought.

I found the eigenvalues seeing the matrix as a 2x2 diagonal matrix with matrix elements therefore getting the denominator of my transfer function $$(s+1)^2(s+2)(s+3)$$.

I also have my backdoor $$y=C(sI-A)^{-1}B $$ but this includes many calculations and I guess that there is a simplerfaster solution.

The following system is given and I'm asked to find the transfer function $$\frac{Y(s)}{U(s)}=G(s)$$ $$\bar {\dot x}=\begin{bmatrix} 0 & 1 & 0 & 0 &0 \\0 & 0 & 1 & 0 & 0 \\ 0 & -1 & -2 & 0 & 0 \\ 0 & 0 & 0 & -2 & 0\\ 0 & 0 & 0 & 0 & -3\end{bmatrix} \bar x + \begin{bmatrix} 0\\0\\1\\1\\0\end{bmatrix}u=A\bar x + Bu \\ $$

$$y=\begin{bmatrix}0 & 1 & 0 & 0 & 1\end{bmatrix} \bar x $$

I haven't practised that much on state space models and I don't remember that much from matrix algebra but here's what I've thought.

I found the eigenvalues seeing the matrix as a 2x2 diagonal matrix with matrix elements therefore getting the denominator of my transfer function $$(s+1)^2(s+2)(s+3)$$.

I also have my backdoor $$y=C(sI-A)^{-1}B $$ but this includes many calculations and I guess that there is a simpler solution.

The following system is given and I'm asked to find the transfer function $$\frac{Y(s)}{U(s)}=G(s)$$ $$\bar {\dot x}=\begin{bmatrix} 0 & 1 & 0 & 0 &0 \\0 & 0 & 1 & 0 & 0 \\ 0 & -1 & -2 & 0 & 0 \\ 0 & 0 & 0 & -2 & 0\\ 0 & 0 & 0 & 0 & -3\end{bmatrix} \bar x + \begin{bmatrix} 0\\0\\1\\1\\0\end{bmatrix}u=A\bar x + Bu \\ $$

$$y=\begin{bmatrix}0 & 1 & 0 & 0 & 1\end{bmatrix} \bar x $$

I haven't practised that much on state space models and I don't remember that much from matrix algebra but here's what I've thought.

I found the eigenvalues seeing the matrix as a 2x2 diagonal matrix with matrix elements therefore getting the denominator of my transfer function $$(s+1)^2(s+2)(s+3)$$.

I also have my backdoor $$y=C(sI-A)^{-1}B $$ but this includes many calculations and I guess that there is a faster solution.

edited body
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John Katsantas
  • 901
  • 4
  • 16
  • 35

The following system is given and I'm asked to find the transfer function $$\frac{Y(s)}{X(s)}=G(s)$$$$\frac{Y(s)}{U(s)}=G(s)$$ $$\bar {\dot x}=\begin{bmatrix} 0 & 1 & 0 & 0 &0 \\0 & 0 & 1 & 0 & 0 \\ 0 & -1 & -2 & 0 & 0 \\ 0 & 0 & 0 & -2 & 0\\ 0 & 0 & 0 & 0 & -3\end{bmatrix} \bar x + \begin{bmatrix} 0\\0\\1\\1\\0\end{bmatrix}u=A\bar x + Bu \\ $$

$$y=\begin{bmatrix}0 & 1 & 0 & 0 & 1\end{bmatrix} \bar x $$

I haven't practised that much on state space models and I don't remember that much from matrix algebra but here's what I've thought.

I found the eigenvalues seeing the matrix as a 2x2 diagonal matrix with matrix elements therefore getting the denominator of my transfer function $$(s+1)^2(s+2)(s+3)$$.

I also have my backdoor $$y=C(sI-A)^{-1}B $$ but this includes many calculations and I guess that there is a simpler solution.

The following system is given and I'm asked to find the transfer function $$\frac{Y(s)}{X(s)}=G(s)$$ $$\bar {\dot x}=\begin{bmatrix} 0 & 1 & 0 & 0 &0 \\0 & 0 & 1 & 0 & 0 \\ 0 & -1 & -2 & 0 & 0 \\ 0 & 0 & 0 & -2 & 0\\ 0 & 0 & 0 & 0 & -3\end{bmatrix} \bar x + \begin{bmatrix} 0\\0\\1\\1\\0\end{bmatrix}u=A\bar x + Bu \\ $$

$$y=\begin{bmatrix}0 & 1 & 0 & 0 & 1\end{bmatrix} \bar x $$

I haven't practised that much on state space models and I don't remember that much from matrix algebra but here's what I've thought.

I found the eigenvalues seeing the matrix as a 2x2 diagonal matrix with matrix elements therefore getting the denominator of my transfer function $$(s+1)^2(s+2)(s+3)$$.

I also have my backdoor $$y=C(sI-A)^{-1}B $$ but this includes many calculations and I guess that there is a simpler solution.

The following system is given and I'm asked to find the transfer function $$\frac{Y(s)}{U(s)}=G(s)$$ $$\bar {\dot x}=\begin{bmatrix} 0 & 1 & 0 & 0 &0 \\0 & 0 & 1 & 0 & 0 \\ 0 & -1 & -2 & 0 & 0 \\ 0 & 0 & 0 & -2 & 0\\ 0 & 0 & 0 & 0 & -3\end{bmatrix} \bar x + \begin{bmatrix} 0\\0\\1\\1\\0\end{bmatrix}u=A\bar x + Bu \\ $$

$$y=\begin{bmatrix}0 & 1 & 0 & 0 & 1\end{bmatrix} \bar x $$

I haven't practised that much on state space models and I don't remember that much from matrix algebra but here's what I've thought.

I found the eigenvalues seeing the matrix as a 2x2 diagonal matrix with matrix elements therefore getting the denominator of my transfer function $$(s+1)^2(s+2)(s+3)$$.

I also have my backdoor $$y=C(sI-A)^{-1}B $$ but this includes many calculations and I guess that there is a simpler solution.

added 5 characters in body
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John Katsantas
  • 901
  • 4
  • 16
  • 35
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Source Link
John Katsantas
  • 901
  • 4
  • 16
  • 35
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