I am solving a problem where I have to find the transfer function of a system. The system is
$$A = \begin{bmatrix}2&-1&2\\0&1&-2\\0&0&-1\end{bmatrix}, B = \begin{bmatrix}1\\1\\0\end{bmatrix}, C = \begin{bmatrix}1&-1&0\end{bmatrix}, D =0$$
When I solve for it's transfer function, it gives $$H(s)= 0.$$ What is the physical meaning of this?I am not exactly sure of the reason why is this happening, because from what I understand, the output is clearly dependent on input x as the matrix C is not null. But again the definition of transfer function is
$$H(s) = C(sI-A)^{-1}B+D,$$
so there might be other things in play here which I don't get.
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\$\begingroup\$ If your transfer function is constant it means there's no frequency response... But if it's zero it probably means there's not only no frequency response also no time response which means your system is... inactive? It would be like apply zero DC volts in your circuit. I'm not going to check your math but are you sure that your order of operations is correct? \$\endgroup\$– user103380Commented Oct 1, 2019 at 23:07
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\$\begingroup\$ Yes, I verified it via Matlab just to be sure \$\endgroup\$– Aakash GuptaCommented Oct 1, 2019 at 23:10
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\$\begingroup\$ I believe you should always get an \$s\$ term somewhere. I think you subtracted \$I-A\$ instead of multiplying \$sI\$ and then subtracting it from \$A\$. In other words, you should have a matrix containing \$s\$ values. \$\endgroup\$– user103380Commented Oct 1, 2019 at 23:13
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\$\begingroup\$ I did use sI instead of just I-A, I cross-verified with one of my friends just to be sure \$\endgroup\$– Aakash GuptaCommented Oct 1, 2019 at 23:16
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\$\begingroup\$ Homework? What's happening to the states? It's often informative to look at what the states are as a function of the input (i.e., set the output matrix to \$I\$). Then engage your brain, and think about what you're seeing. \$\endgroup\$– TimWescottCommented Oct 1, 2019 at 23:33
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1 Answer
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You have a system where not just some states, but the output itself is not controllable or observable. There are a couple of ways to look at what is happening.
Approach A
The third state equation is \$x_3'=-x_3\$. Since the conversion to transfer function assumes all initial conditions are zero, the solution for \$x_3\$ is \$x_3=0\$.
Now we are left with two equations: \$x_1'=2 x_1-x_2+u\$ and \$x_2'=x_2+u\$
So \$x_1'-x_2'=(2 x_1-x_2+u)-(x_2+u)= 2 (x_1- x_2)\$
The output is \$y = x_1- x_2 \$ and its equation becomes \$y'=2 y\$, whose solution is also 0.
Approach B
Another way of looking at it is to convert the model to Jordan form.
Now the first state equation is \$z_1'=-z_1\$ and the last one is \$z_3'=2 z_3\$. Again the solutions for these two equations are \$z_1=z_3=0\$.
The output in terms of the Jordan states is \$y=-4 z_1+ z_3=0\$.
The output is just a linear combination of the two uncontrollable states. The other state \$z_2\$ is controllable but it is not observable and does not show up in the output.
