Problems

1. This is a straightforward across-variable divider: \[\begin{aligned} \frac{P_C(s)} {P_S(s)} &= \frac{Z_C} {Z_I+Z_R+Z_C} \\ &= \frac{1} {I C s^2 + R C s + 1}. \end{aligned}\]

2. The input impedance for the system consisting of three series elements is \[\begin{aligned} Z_R + Z_I + Z_C = R + I s + \frac{1} {C s}. \end{aligned}\]

We will use an across-variable (voltage) divider. The elements \(C_1\), \(R_2\), and \(C_2\) can all be combined (in series and parallel) to form an equivalent impedance \[\begin{aligned} Z_e &= \frac{1} {\frac{1} {Z_{C_1}} + \frac{1} {Z_{R_2} + Z_{C_2}}} \\ &= \frac{1} {C_1 s + \frac{1} {R_2 + \frac{1} {C_2 s}}} \\ &= \frac{1} {C_1 s + \frac{C_2 s} {R_2 C_2 s + 1}} \\ &= \frac{R_2 C_2 s + 1} {(R_2 C_2 s + 1) C_1 s + C_2 s} \\ &= \frac{R_2 C_2 s + 1} {R_2 C_1 C_2 s^2 + (C_1 + C_2) s}. \end{aligned}\]

The voltage divider rule gives \[\begin{aligned} \frac{V_{R_1}}{V_S} &= \frac{Z_{R_1}}{Z_{R_1} + Z_e} \\ &= \frac{(R_2 C_1 C_2 s^2 + (C_1 + C_2) s) R_1} {(R_2 C_1 C_2 s^2 + (C_1 + C_2) s) R_1 + R_2 C_2 s + 1} \\ &= \frac{R_1 R_2 C_1 C_2 s^2 + R_1 (C_1 + C_2) s} {R_1 R_2 C_1 C_2 s^2 + R_1 (C_1 + C_2) s + R_2 C_2 s + 1} \\ &= \frac{(R_1 R_2 C_1 C_2) s^2 + R_1 (C_1 + C_2) s} {(R_1 R_2 C_1 C_2) s^2 + \left(R_1 (C_1 + C_2) + R_2 C_2\right) s + 1}. \end{aligned}\]

But we want \(I_{R_1}/V_S\). The impedance elemental equation for \(R_1\) lets us trade \(V_{R_1} \mapsto I_{R_1}\): \[\begin{aligned} V_{R_1} &= I_{R_1} Z_{R_1} \\ &= R_1 I_{R_1}. \end{aligned}\] Substituting this into the transfer function above, \[\begin{aligned} \frac{R_1 I_{R_1}}{V_S} &= \frac{(R_1 R_2 C_1 C_2) s^2 + R_1 (C_1 + C_2) s} {(R_1 R_2 C_1 C_2) s^2 + \left(R_1 (C_1 + C_2) + R_2 C_2\right) s + 1} \Rightarrow \\ \frac{I_{R_1}}{V_S} &= \frac{(R_2 C_1 C_2) s^2 + (C_1 + C_2) s} {(R_1 R_2 C_1 C_2) s^2 + \left(R_1 (C_1 + C_2) + R_2 C_2\right) s + 1}. \end{aligned}\]

The transfer function can be derived using impedance methods, as required, in two different manners:

1. with the systematic approach of or

2. with the divider method of .

We choose the latter because it is a relatively simple system and the across-variable divider looks promising: (a) the source \(P_S\) is across-variable and (b) the linear graph can be reduced to the source \(P_S\) in series with elements that include \(C\), the element associated with the output variable \(Q_C\).

Before we begin with the divider method, let’s list out the impedances associated with each element: \[\begin{gathered} Z_{R_1} = R_1,\ Z_{R_2} = R_2, \\ Z_{I_1} = I_1 s,\ Z_{I_2} = I_2 s,\ Z_C = \frac{1} {C s}. \end{gathered}\]

We must first find an equivalent impedance of the two upper \(R\)-\(I\) “branches.” The fluid resistors combine in series with the inertances as \[\begin{aligned} Z_{R_1} + Z_{I_1} \quad\text{and} Z_{R_2} + Z_{I_2} \end{aligned}\] to form two parallel elements that can be combined to form an element with equivalent impedance \[\begin{aligned} Z_e &= \frac{1} {\frac{1} {Z_{R_1} + Z_{I_1}} + \frac{1} {Z_{R_2} + Z_{I_2}}}. \end{aligned}\] The corresponding simplified graph is shown in .

We can now apply the across-variable divider rule: \[\begin{aligned} \label{eq:concrete-divider} \frac{P_C(s)} {P_S(s)} &= \frac{Z_C} {Z_C + Z_e}. \end{aligned}\]

This is nearly what we want, \(Q_C/P_S\). For all passive elements, the impedance or “generalized Ohm’s law” elemental equation allows us to easily trade through- and across-variable. The \(P_C\) impedance elemental equation is \[\begin{aligned} P_C = Q_C Z_C. \end{aligned}\] Substituting this into , \[\begin{aligned} \label{eq:concrete-divider2} \frac{Q_C(s) Z_C} {P_S(s)} &= \frac{Z_C} {Z_C + Z_e}. \end{aligned}\] This is simply solved for \[\begin{aligned} \label{eq:concrete-divider3} \frac{Q_C(s)} {P_S(s)} &= \frac{1} {Z_C + Z_e}. \end{aligned}\]

The transfer function can be derived using impedance methods, as required, in two different manners:

1. with the systematic approach of or

2. with the divider method of .

We choose the latter because the across-variable divider looks promising: (a) the source \(P_S\) is across-variable and (b) the linear graph can be reduced to the source \(P_S\) in series with elements that include \(R_3\) and \(C\), the elements associated with the output variables \(Q_{R_3}\) and \(Q_C\).

Before we begin with the divider method, let’s list out the impedances associated with each element: \[\begin{gathered} Z_{R_1} = R_1,\ Z_{R_2} = R_2,\ Z_{R_3} = R_3,\\ Z_{I_1} = I_1 s,\ Z_{I_2} = I_2 s, Z_C = \frac{1} {C s}. \end{gathered}\]

We must first find an equivalent impedance of the two upper \(R\)-\(I\) “branches.” The fluid resistors combine in series with the inertances as \[\begin{aligned} Z_{R_1} + Z_{I_1} \quad\text{and } Z_{R_2} + Z_{I_2} \end{aligned}\] to form two parallel elements that can be combined to form an element with equivalent impedance \[\begin{aligned} Z_e &= \frac{1} {\frac{1} {Z_{R_1} + Z_{I_1}} + \frac{1} {Z_{R_2} + Z_{I_2}}}. \end{aligned}\] The corresponding simplified graph is shown in .

The across-variable divider also requires we combine the impedances of \(R_3\) and \(C\) in parallel. This is associated with the two output elements, so we call it \(Z_o\), and define it as follows: \[\begin{aligned} Z_o &= \frac{1} {\frac{1} {Z_{R_3}} + \frac{1} {Z_{C}}}. \end{aligned}\]

We can now apply the across-variable divider rule: \[\begin{aligned} \label{eq:rhine-divider} \frac{P_o(s)} {P_S(s)} &= \frac{Z_o} {Z_o + Z_e}. \end{aligned}\]

This is nearly what we want, \(Q_{R_3}/P_S\) and \(Q_C/P_S\). We recognize that the pressure \(P_o = P_{R_3} = P_C\), so we have \(P_{R_3}/P_S\) and \(P_C/P_S\). For all passive elements, the impedance or “generalized Ohm’s law” elemental equation allows us to easily trade through- and across-variable.

The \(P_{R_3}\) impedance elemental equation is \[\begin{aligned} P_{R_3} = Q_{R_3} Z_{R_3}. \end{aligned}\] Substituting this into , \[\begin{aligned} \label{eq:rhine-divider2} \frac{Q_{R_3}(s) Z_{R_3}}{P_S(s)} &= \frac{Z_o} {Z_o + Z_e}. \end{aligned}\] This is simply solved for \[\begin{aligned} \label{eq:rhine-divider3} \frac{Q_{R_3}(s)}{P_S(s)} &= \frac{Z_o/Z_{R_3}}{Z_o + Z_e}. \end{aligned}\]

The \(P_C\) impedance elemental equation is \[\begin{aligned} P_C = Q_C Z_C. \end{aligned}\] Substituting this into , \[\begin{aligned} \label{eq:rhine-divider2} \frac{Q_C(s) Z_C} {P_S(s)} &= \frac{Z_o} {Z_o + Z_e}. \end{aligned}\] This is simply solved for \[\begin{aligned} \label{eq:rhine-divider3} \frac{Q_C(s)} {P_S(s)} &= \frac{Z_o/Z_C} {Z_o + Z_e}. \end{aligned}\]