# Two state space models represent the same transfer function

Exercise

Show that two state space models ($A_1,b_1,c_1$) and ($A_2,b_2,c_2$) represent the same transfer function if a matrix $T$ exists ($\det T \ne 0$) such that $$T^{-1}A_1T=A_2,~~~~T^{-1}b_1=b_2,~~~~c_1T=c_2$$

Solution

The 'if' part of the proof follows by algebraic substitution: $$G_1(s)=c_1[sI-A_1]^{-1}b_1$$ $$G_2(s)=c_2[sI-A_2]^{-1}b_2$$ $$G_2(s)=c_1T[sI-T^{-1}A_1T]^{-1}T^{-1}b_1$$ Then, because $I = T^{-1}T$ $$G_2(s)=c_1T[T^{-1}sT-T^{-1}A_1T]^{-1}T^{-1}b_1$$

The factors $T^{-1}$ (left) and $T$ (right) can be divided $$G_2(s) = c_1 T[T^{-1}(sI-A_1)T]^{-1}T^{-1}b_1$$ because $[X_1YX_2]^{-1}=X_2^{-1}Y^{-1}X_1^{-1}$ $$G_2(s)=c_1TT^{-1}[sI-A_1]^{-1}TT^{-1}b_1 = c_1 [sI-A_1]^{-1}b_1 =G_1(s)$$

Questions

1. How could I have known that the addresser of the task wants to see the format $G_i(s)=c_i[sI-A_i]^{-1}b_i$? And what is this format?

2. I know that $T^{-1}T$ is the identity matrix $I$ but why does the solution surround the $T$ and $T^{-1}$ around the bracket and in reverse order around the $A$ in $c_1T[sI-T^{-1}A_1T]^{-1}T^{-1}b_1$? Why is this mathematically allowed?

3. What steps are carried out to get from $[T^{-1}sT-T^{-1}A_1T]$ to $[T^{-1}(sI-A_1)T]$?

Exercise

Show that two state space models ($A_1,b_1,c_1$) and ($A_2,b_2,c_2$) represent the same transfer function if a matrix $T$ exists ($\det T \ne 0$) such that $$T^{-1}A_1T=A_2,~~~~T^{-1}b_1=b_2,~~~~c_1T=c_2$$

1. How could I have known that the addresser of the task wants to see the format $G_i(s)=c_i[sI-A_i]^{-1}b_i$? And what is this format?

First, we need to understand the format of a transfer function. Given the state equations

$$x'(t) = A x(t) + B u(t) \\ y(t) = C x(t) + D u(t)$$

We may take it's Laplace transform and rearrange it as follows

$$sX(s) = A X(s) + BU(s) \implies (sI - A) X(s) = BU(s) \implies X(s) = (sI-A)^{-1}BU(s)$$

If we combine this with the transform of the output equation, $Y(s) = C X(s) + DU(s)$, we get

$$Y(s) = C X(s) + DU(s) = C(sI-A)^{-1}B U(s) + DU(s)$$

or equivalently,

$$\dfrac{Y(s)}{U(s)} = G(s) = C (sI - A)^{-1} B + D$$

We were given no $D$, which is zero in this case. So, now we know the format of the transfer function is

$$G_i(s) = C_i (sI - A)^{-1} B_i$$

Next, from the given conditions, we need to show that the two transfer functions are equal, that is

$$G_1(s) = C_1 (sI - A_1)^{-1} B_1 = C_2 (sI - A_2)^{-1} B_2 = G_2(s)$$

This must be done by a series of "legal" algebraic manipulations.

1. I know that $T^{-1}T$ is the identity matrix $I$ but why does the solution surround the $T$ and $T^{-1}$ around the bracket and in reverse order around the $A$ in $c_1T[sI-T^{-1}A_1T]^{-1}T^{-1}b_1$? Why is this mathematically allowed?

From the problem statement, we are given that $A_2 = T^{-1}A_1T,b_2 = T^{-1}b_1,c_2 = c_1T$, and we substitute these in to the transfer function and arrive at

$$G_2(s) = c_2 (sI - A_2)^{-1} b_2 = c_1 T(sI - T^{-1}A_1T)T^{-1}b_1$$

Next, we have set things up to do our algebraic manipulations to arrive at showing $G_2(s) = G_1(s)$. Note that we chose to use the transfer function that let us use the givens as-is, but we could have also chosen $G_1(s)$ with slight modifications to the givens.

1. What steps are carried out to get from $[T^{-1}sT-T^{-1}A_1T]$ to $[T^{-1}(sI-A_1)T]$?

From above, we have

$$\tag 1 G_2(s)=c_1T[sI-T^{-1}A_1T]^{-1}T^{-1}b_1$$

From the givens, we know that $T$ and $T^{-1}$ exist, so we can write $sI = sT^{-1}T = T^{-1}sT$ and write $(1)$ as

$$\tag 2 G_2(s)=c_1T[T^{-1}sT-T^{-1}A_1T]^{-1}T^{-1}b_1$$

Now, let's do the algebraic manipulations with justification to see how we arrive at the result. We notice that inside the $[~~]$, we can factor a $T^{-1}$ on the left and then factor a $T$ on the right and arrive at

$$\tag 3 G_2(s)=c_1T[T^{-1}(sT-A_1T)]^{-1}T^{-1}b_1 = c_1T[T^{-1}(sI-A_1)T]^{-1}T^{-1}b_1$$

Next, using the matrix property $[X_1YX_2]^{-1}=X_2^{-1}Y^{-1}X_1^{-1}$, we can rewrite $(3)$ as

$$\tag 4 G_2(s)= c_1T[T^{-1}(sI-A_1)T]^{-1}T^{-1}b_1 = c_1T[T^{-1}(sI-A_1)^{-1}T]T^{-1}b_1$$

Next, we can distribute and have two occurrences of $TT^{-1} = I$, so can reduce $(4)$ to

$$\tag 5 G_2(s)= c_1T T^{-1}(sI-A_1)^{-1}TT^{-1}b_1 = c_1 [sI-A_1]^{-1}b_1$$

Now we conclude as we have successfully shown:

$$G_2(s) = c_2 (sI - A_2)^{-1} b_2 = c_1 (sI - A_1)^{-1} b_1 = G_1(s)$$