The electrical system shown in the figure converts input source current is(t) to output voltage v0(t)...

 

1. At node X

The source current $i_s(t)$ splits into capacitor current and resistor current:

$$i_s(t)=i_C(t)+i_R(t)$$

For the capacitor,

$$i_C=C\frac{dv_C}{dt} =\frac{dv_C}{dt}$$

because $C=1F$.

For the $1\Omega$ resistor,

$$i_R=\frac{v_C}{1}=v_C$$

Therefore,

$$i_s=\frac{dv_C}{dt}+v_C$$

so

$$\boxed{\dot v_C=i_s-v_C}$$

This gives the second state equation.

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2. For the inductor branch

The inductor $1H$ and resistor $1\Omega$ are parallel.

Therefore, they have the same voltage.

For the inductor,

$$v_L=L\frac{di_L}{dt}$$

Since $L=1H$,

$$v_L=\frac{di_L}{dt}$$

For the parallel $1\Omega$ resistor,

$$v_R=i_R$$

Since their voltages are equal,

$$\frac{di_L}{dt}=i_R$$

Now the source current splits between the inductor and resistor:

$$i_s=i_L+i_R$$

Therefore,

$$i_R=i_s-i_L$$

Hence,

$$\boxed{\dot i_L=i_s-i_L}$$

This is exactly the step in the solution.

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3. State equations

Taking

$$x_1=i_L,\qquad x_2=v_C$$

we get

$$\dot i_L=-i_L+i_s$$

$$\dot v_C=-v_C+i_s$$

Therefore,

$$\begin{bmatrix} \dot i_L\\ \dot v_C \end{bmatrix} = \begin{bmatrix} -1&0\\ 0&-1 \end{bmatrix} \begin{bmatrix} i_L\\ v_C \end{bmatrix} + \begin{bmatrix} 1\\ 1 \end{bmatrix}i_s$$

So,

$$\boxed{ A= \begin{bmatrix} -1&0\\ 0&-1 \end{bmatrix}, \qquad B= \begin{bmatrix} 1\\ 1 \end{bmatrix}}$$

$$i_L(s)=\frac{i(s)}{s+1}$$

comes from current division and assumes zero initial inductor current. For state-space derivation, the time-domain equation

$$\boxed{\dot i_L=i_s-i_L}$$

is the more fundamental step.

We have

$$A= \begin{bmatrix} -1&0\\ 0&-1 \end{bmatrix}, \qquad B= \begin{bmatrix} 1\\ 1 \end{bmatrix}$$

and since the output is

$$V_o=V_C$$

we have

$$C= \begin{bmatrix} 0&1 \end{bmatrix}, \qquad D=0.$$

1. Controllability

The controllability matrix is

$$\mathcal C=[B\ AB]$$

Now,

$$AB= \begin{bmatrix} -1&0\\ 0&-1 \end{bmatrix} \begin{bmatrix} 1\\ 1 \end{bmatrix} = \begin{bmatrix} -1\\ -1 \end{bmatrix}$$

Therefore,

$$\mathcal C= \begin{bmatrix} 1&-1\\ 1&-1 \end{bmatrix}$$

Its rank is

$$\boxed{\operatorname{rank}(\mathcal C)=1<2}$$

Hence,

$$\boxed{\text{Not completely controllable}}$$

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2. Observability

The observability matrix is

$$\mathcal O= \begin{bmatrix} C\\ CA \end{bmatrix}$$

$$CA= \begin{bmatrix} 0&1 \end{bmatrix} \begin{bmatrix} -1&0\\ 0&-1 \end{bmatrix} = \begin{bmatrix} 0&-1 \end{bmatrix}$$

Therefore,

$$\mathcal O= \begin{bmatrix} 0&1\\ 0&-1 \end{bmatrix}$$

Its rank is

$$\boxed{\operatorname{rank}(\mathcal O)=1<2}$$

Hence,

$$\boxed{\text{Not completely observable}}$$

Final answer

Property	Rank	Result
Controllability	1 < 2	Not completely controllable
Observability	1 < 2	Not completely observable

Summary: Since $AB=-B$, the columns of the controllability matrix are dependent → not controllable. Similarly, $CA=-C$, so the observability matrix has dependent rows → not observable.