# Is transformation from one state representation to other state representation changes the transfer function of the system

I was reading about designing of controller by state state model from book by Norman S. Nise

And author suggested that if state space representations given in problem are of form other than phase variable form then best method to design a controller would be

1.transform given system to phase variable form

2.then design the feedback gains

3.then again transform the designed system back to original state space form

So my question

1.when we transform from one of state space representations to another , isn't zeroes of system may changed?

Or in other words

isn't transfer function gets changed due to transformation

As I try to figure out myself here is my attempt

And after all this calculation to me it seems that both transfer function may not be equal always

but

example given in book shows that it has same transfer function in both transforms so I thought it(example) may be a special case

So can anyone explain

Mathematical expression I got can be solved further to obtain a relationship between both transfer function ?

Or

my conclusion is right that transfer function may or may not changed due to transformation?

• If the transfer function changes, then it is not the same system. The transformation is supposed to preserve the input output relationship; aka transfer function – AJN Aug 31 '20 at 13:43
• @AJN , can you please tell then where my mathematical expression is wrong or how to simplify it? – user215805 Aug 31 '20 at 13:53
• $sI-P^{-1}AP = sP^{-1}IP-P^{-1}AP = P^{-1}(sI-A)P$. From this try to see if the last expression in the second figure can be simplified into $C(sI-A)^{-1}B$. I think it will be. Check the derivation carefully. – AJN Aug 31 '20 at 14:11
• @AJN , thanks , very precise as always! – user215805 Aug 31 '20 at 14:22

Assuming that $$\P\$$ invertible,
\begin{align} \frac{Y(s)}{U(s)} ={} & CP\ (sI - P^{-1}AP)^{-1}\ P^{-1} B\\ {}={} & CP\ (sP^{-1}IP - P^{-1}AP)^{-1}\ P^{-1} B\\ {}={} & CP\ (P^{-1}\ (sI - A)\ P)^{-1} P^{-1} B\\ {}={} & CPP^{-1}\ (sI - A)^{-1}\ P P^{-1} B & \tiny{(ABC)^{-1} = C^{-1}B^{-1}A^{-1}}\\ {}={} & C (sI - A)^{-1} B\ \ \ \blacksquare\\ \end{align}