Question 8.14: Response of a bridged-T networknIn Figure 8.18 the excitatio......

We can write nodal equations.

C_{1}\frac{d}{d t}[v_{x}(t)-v_{i}(t)]+C_{2}\frac{d}{d t}[v_{x}(t)-v_{R L}(t)]+G_{1}v_{x}(t)=0

C_{2}\frac{d}{d t}[\mathbf{v}_{R L}(t)-\mathbf{v}_{x}(t)]+G_{L}\mathbf{v}_{R L}(t)+G_{2}[\mathbf{v}_{R L}(t)-\mathbf{v}_{i}(t)]=0

where G_{1}=1/R_{1}=10^{-4}\ S,G_{2}=1/R_{2}=10^{-4}\ S and G_{L}=10^{-3}\,{S}. Laplace transforming the equations

C_{1}\{s\,V_{x}(s)-v_{x}(0^{-})-[s\,V_{i}(s)-v_{i}(0^{-})]\}+C_{2}\{s\,V_{x}(s)-v_{x}(0^{-})-[s\,V_{R L}(s)-v_{R L}(0^{-})]\} {}+G_{1}\,V_{x}(s)=0

C_{2}\{s\,V_{R L}(s)-v_{R L}(0^{-})-[s\,V_{x}(s)-v_{x}(0^{-})]\}+G_{L}\,V_{R L}(s)+G_{2}[V_{R L}(s)-v_{i}(s)]=0

Since we seek the zero-state response, all the initial conditions are zero and the equations
simplify to

s C_{1}[V_{x}(s)-v_{i}(s)]+s C_{2}[V_{x}(s)-v_{R L}(s)]+G_{1}V_{x}(s)=0

s C_{2}[{{\bf V}_{R L}(s)}-{{\bf V}_{x}(s)}]+G_{L}{{\bf V}_{R L}(s)}+G_{2}[{{\bf V}_{R L}(s)}-{{\bf V}_{i}(s)}]=0

The Laplace transform of the excitation is \mathrm{V}_{i}(s)=10/s. Then

\left [ \begin{matrix} s(C_{1}+C_{2})+G_{1} & -s C_{2} \\-s C_{2} & s C_{2}+(G_{L}+G_{2}) \end{matrix} \right ] \left [ \begin{matrix} V_{x}(s) \\ V_{R L}(s) \end{matrix} \right ] =\left [ \begin{matrix} 10C_{1} \\ 10G_{2}/s \end{matrix} \right ]

The determinant of the 2 by 2 matrix is

\Delta=[s(C_{1}+C_{2})+G_{1}][s C_{2}+(G_{L}+G_{2})]-s^{2}C_{2}^{2}=s^{2}C_{1}C_{2}+s[G_{1}C_{2}+(G_{L}+G_{2})(C_{1}+C_{2})]+G_{1}(G_{L}+G_{2})

and, by Cramer’s rule, the solution for the Laplace transform of the response is

V_{RL}(s)=\frac{\begin{vmatrix} s(C_{1}+C_{2})+G_{1} & 10C_{1} \\ -s C_{2} & 10G_{2}/s \end{vmatrix} }{s^{2}C_{1}C_{2}+s[G_{1}C_{2}+(G_{L}+G_{2})(C_{1}+C_{2})]+G_{1}(G_{L}+G_{2})}

\mathrm{V}_{R L}(s)=10\frac{s^{2}C_{1}C_{2}+s G_{2}\left(C_{1}+C_{2}\right)+G_{1}G_{2}}{s\left\{s^{2}C_{1}C_{2}+s\left[G_{1}C_{2}+\left(G_{L}+G_{2}\right)(C_{1}+C_{2} )\right]+G_{1}\left(G_{L}+G_{2}\right)\right\}}

or

\mathrm{V}_{R L}(s)=10\frac{s^{2}+s G_{2}(C_{1}+C_{2})/C_{1}C_{2}+G_{1}G_{2}/C_{1}C_{2}}{s\{s^{2}+s[G_{1}/C_{1}+(G_{L}+G_{2})(C_1 +C_2)/C_{1}C_{2}]+G_{1}(G_{L}+G_{2})/C_{1}C_{2}\} }

Using the component numerical values,

\mathrm{V}_{R L}(s)=10\frac{s^{2}+200s+10,000}{s(s^{2}+2300s+110,000)}.

Expanding in partial fractions,

V_{R L}(s)={\frac{0.9091}{s}}-{\frac{0.243}{s+48.86}}+{\frac{9.334}{s+2251}}.

Inverse Laplace transforming,

\mathrm v_{ R L}(t)=[0.9091-0.243e^{-48.36t}+9.334e^{-2251t}]\mathbf{u}(t).

As a partial check on the correctness of this solution the response approaches 0.9091 as t → ∞.

This is exactly the voltage found using voltage division between the two resistors, considering the capacitors to be open circuits. So the fi nal value looks right. The initial response at time t=0^{+} is 10 V. The capacitors are initially uncharged so, at time t=0^{+} , their voltages are both zero and the excitation and response voltages must be the same. So the initial value also looks right. These two checks on the solution do not guarantee that it is correct for all time, but they are very good checks on the reasonableness of the solution and will often detect an error.