Consider the network of Figure 5.2a with zero initial conditions. The inputs, the state variables, and the outputs of the network are shown in Figure 5.2b. Determine:
(a) The state equations of the network
(b) The transition matrix and the output vector of the network, when
$R_1 = 0 Ω,R_2=1.5 Ω, C= 1 F, L_1=L_2= 1 H$ , and $v_1(t) = v_2(t) = 1 V$ .

Question

Consider the network of Figure 5.2a with zero initial conditions. The inputs, the state variables, and the outputs of the network are shown in Figure 5.2b. Determine:
(a) The state equations of the network
(b) The transition matrix and the output vector of the network, when
[latex]R_1 = 0 Ω,R_2=1.5 Ω, C= 1 F, L_1=L_2= 1 H [/latex], and [latex]v_1(t) = v_2(t) = 1 V[/latex].

Accepted Answer

(a) The two loop equations are

[latex]R_1(i_1+i_2)+v_c+L_1\frac{di_1}{dt}=v_1(t)[/latex]

[latex]R_1(i_1+i_2)+v_c+L_2\frac{di_2}{dt}+R_2i_2=v_1(t)-v_2(t)[/latex]

Furthermore, we have that

[latex]C\frac{dv_c}{dt}=i_1+i_2[/latex]

Hence, the three first-order differential equations that describe the network are

[latex]\frac{dv_c}{dt}=\frac{1}{C}i_1+\frac{1}{C}i_2[/latex]

[latex]\frac{di_1}{dt}=-\frac{1}{L_1}v_c-\frac{R_1}{L_1}i_1-\frac{R_1}{L_1}i_2+\frac{1}{L_1}v_1(t)[/latex]

[latex]\frac{di_2}{dt}=-\frac{1}{L_2}v_c-\frac{R_1}{L_2}i_1-\frac{R_1+R_2}{L_2}i_2+\frac{1}{L_2}v_1(t)-\frac{1}{L_2}v_2(t)[/latex]

Note that [latex]i_1=i_{L_1}[/latex] and [latex]i_2=i_{L_2}[/latex] . Thus, the state equations of the network are

[latex]\dot{x}=Ax+Bu [/latex]

[latex]y=Cx[/latex]

where

[latex]x=\left[\begin{matrix}x_1(t)\x_2(t)\x_3(t)\end{matrix} ight]=\left[\begin{matrix}v_c\i_{L_1}\i_{L_2}\end{matrix} ight][/latex], [latex]u=\left[\begin{matrix}v_1(t)\v_2(t)\end{matrix} ight][/latex], [latex]y=\left[\begin{matrix}i_{L_1}\i_{L_2}\end{matrix} ight][/latex]

[latex]A=\left[\begin{matrix}0&\frac{1}{C}&\frac{1}{C}\ \-\frac{1}{L_1}&-\frac{R_1}{L_1}&-\frac{R_1}{L_1}\ \-\frac{1}{L_2}&-\frac{R_1}{L_2}&-\frac{R_1+R_2}{L_2}\end{matrix} ight][/latex], [latex]B=\left[\begin{matrix}0&0\ \-\frac{1}{L_1}&0\ \ \frac{1}{L_2}&-\frac{1}{L_2}\end{matrix} ight][/latex],

[latex]C=\left[\begin{matrix}0&1&0\0&0&1\end{matrix} ight][/latex]

(b) Substituting the given values for each element, we have

[latex]A=\left[\begin{matrix}0&1&1\-1&0&0\-1&0&-1.5\end{matrix} ight], B=\left[\begin{matrix}0&0\1&0\1&-1\end{matrix} ight], sI-A= \left[\begin{matrix}s&-1&-1\1&s&0\1&0&s+1.5\end{matrix} ight][/latex]

[latex]|sI-A|=s[s(s+1.5)]+s+1.5+s=s³+1.5s²+2s+1.5[/latex]

[latex]=(s+1)(s²+0.5s+1.5)[/latex]

Therefore

[latex]Φ(s)=(sI-A)^{-1}=\frac{1}{(s+1)(s²+0.5s+1.5)}[/latex]

[latex]\left[\begin{matrix}s(s+1.5)&s+1.5&s\-(s+1.5)&s²+1.5s+1&-1\-s&-1&s²+1\end{matrix} ight][/latex]

Hence

Φ(t)=[latex]L^{-1}[/latex]{[latex](sI-A)^{-1}[/latex]}[latex]=\left[\begin{matrix}φ_{11}(t)&φ_{12}(t)&φ_{13}(t)\φ_{21}(t)&φ_{22}(t)&φ_{23}(t)\φ_{31}(t)&φ_{32}(t)&φ_{33}(t)\end{matrix} ight][/latex]

where

[latex]φ_{11}(t)=-0.25e^{-t}+e^{-0.25t}(1.25\cos1.199t+0.052\sin1.199t)[/latex]

[latex]φ_{12}(t)=-0.5e^{-t}+e^{-0.25t}(0.5\cos1.199t+0.99\sin1.199t)[/latex]

[latex]φ_{13}(t)=-0.5e^{-t}+e^{-0.25t}(0.5\cos1.199t+0.521\sin1.199t)[/latex]

[latex]φ_{21}(t)=-φ_{12}(t)[/latex]

[latex]φ_{22}(t)=0.25e^{-t}+e^{-0.25t}(0.75\cos1.199t+0.365\sin1.199t)[/latex]

[latex]φ_{23}(t)=-0.5e^{-t}+e^{-0.25t}(0.5\cos1.199t-0.312\sin1.199t)[/latex]

[latex]φ_{31}(t)=-φ_{13}(t)[/latex]

[latex]φ_{32}(t)=φ_{23}(t)[/latex]

[latex]φ_{33}(t)=e^{-t}-0.417e^{-0.25t}\sin1.199t[/latex]

The output vector is

[latex]Y(s)=[C(sI-A)^{-1}B]U(s)=\frac{1}{(s+1)(s²+0.5s+1.5)}\left[\begin{matrix}s²+1.5s&1\s²&-s²-1\end{matrix} ight][/latex]

[latex]\left[\begin{matrix}1/s\1/s\end{matrix} ight][/latex]

[latex]=\frac{1}{s(s+1)(s²+0.5s+1.5)}\left[\begin{matrix}s²+1.5s+1\-1\end{matrix} ight][/latex]

Therefore

[latex]y(t)=L^{-1}[/latex]{Y(s)}=[latex]\left[\begin{matrix}y_1(t)\y_2(t)\end{matrix} ight][/latex]

[latex]=\left[\begin{matrix}0.67+0.08e^{-t}+0.75e^{-0.25t}\cos1.2t-0.82e^{-0.25t}\sin1.199t\-0.67+0.17e^{-t}+0.5e^{-0.25t}\cos1.2t+0.23e^{-0.25t}\sin1.199t\end{matrix} ight][/latex]

Question 5.3.2: Consider the network of Figure 5.2a with zero initial condit...

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