53. Consider a system S represented in state space as ...

Q.53 Consider a system S represented in state space as

$$\frac{dx}{dt}=\left[\begin{array}{cc}0& -2\\ 1& -3\end{array}\right]x+\left[\begin{array}{c}1\\ 0\end{array}\right]r,y=\left[\begin{array}{cc}2& -5\end{array}\right]x$$

Which of the state space representations given below has/have the same transfer function as that of S?

1. $$\frac{dx}{dt}=\left[\begin{array}{cc}0& 1\\ -2& -3\end{array}\right]x+\left[\begin{array}{c}0\\ 1\end{array}\right]r,y=\left[\begin{array}{cc}1& 2\end{array}\right]x$$
2. $$\frac{dx}{dt}=\left[\begin{array}{cc}0& 1\\ -2& -3\end{array}\right]x+\left[\begin{array}{c}1\\ 0\end{array}\right]r,y=\left[\begin{array}{cc}0& 2\end{array}\right]x$$
3. $$\frac{dx}{dt}=\left[\begin{array}{cc}-1& 0\\ 0& -2\end{array}\right]x+\left[\begin{array}{c}-1\\ 3\end{array}\right]r,y=\left[\begin{array}{cc}1& 1\end{array}\right]x$$
4. $$\frac{dx}{dt}=\left[\begin{array}{cc}-1& 0\\ 0& -2\end{array}\right]x+\left[\begin{array}{c}1\\ 1\end{array}\right]r,y=\left[\begin{array}{cc}2& 2\end{array}\right]x$$

Answer: Option A, C

Solution

To solve this question, compute the transfer function

$$G\left(s\right)=C{(sI-A)}^{-1}B+D$$

for the original system and compare it with the transfer functions obtained from each option.

Step 1: Original System

Given

$$A=\left[\begin{array}{cc}0& -2\\ 1& -3\end{array}\right],B=\left[\begin{array}{c}1\\ 0\end{array}\right],C=\left[\begin{array}{cc}2& -5\end{array}\right],D=0$$

The transfer function is

$$G\left(s\right)=C{(sI-A)}^{-1}B$$

First compute

$$sI-A=\left[\begin{array}{cc}s& 2\\ -1& s+3\end{array}\right]$$

Determinant:

$$|sI-A|=s(s+3)+2=s^2+3s+2=(s+1)(s+2)$$

Hence

$${(sI-A)}^{-1}=\frac{1}{s^2+3s+2}\left[\begin{array}{cc}s+3& -2\\ 1& s\end{array}\right]$$

Multiplying by B:

$${(sI-A)}^{-1}B=\frac{1}{s^2+3s+2}\left[\begin{array}{c}s+3\\ 1\end{array}\right]$$

Now multiply by C:

$$G\left(s\right)=\frac{2(s+3)-5}{s^2+3s+2}$$ $$G\left(s\right)=\frac{2s+1}{(s+1)(s+2)}$$

Therefore,

$$G\left(s\right)=\frac{2s+1}{(s+1)(s+2)}$$

Step 2: Check Option (A)

This is in controllable canonical form. The denominator is

$$s^2+3s+2$$

and the numerator is

$$2s+1$$

Therefore,

$$G$$_A(s) = 2s+1 s2 + 3s + 2

Option A matches.

Step 3: Check Option (B)

The numerator becomes

$$2s$$

Therefore,

$$G$$_B(s) = 2s s2 + 3s + 2

Option B does not match.

Step 4: Check Option (C)

Since A is diagonal,

$${(sI-A)}^{-1}=\left[\begin{array}{cc}\frac{1}{s+1}& 0\\ 0& \frac{1}{s+2}\end{array}\right]$$

Thus,

$$G$$_C(s) = -1s+1 + 3s+2 $$=\frac{2s+1}{(s+1)(s+2)}$$

Option C matches.

Step 5: Check Option (D)

$$G$$_D(s) = 2s+1 + 2s+2 $$=\frac{4s+6}{(s+1)(s+2)}$$

Option D does not match.

Final Answer

The systems in Option A and Option C have the same transfer function as the original system.

$$\text{Answer:} \text{(A) and (C)}$$

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