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The divider method

the two-element across-variable divider.
Figure 12.7: the two-element across-variable divider.

In Electronics, we developed the useful voltage divider formula for quickly analyzing how voltage divides among series electronic impedances. This can be considered a special case of a more general across-variable divider equation for any elements described by an impedance. After developing the across-variable divider, we also introduce the through-variable divider, which divides an input through-variable among parallel elements.

Across-variable dividers

First, we develop the solution for the two-element across-variable divider shown in figure 12.7. We choose the across-variable across \(Z_2\) as the output. The analysis follows the impedance method of , solving for \(\mathcal{V}_2\).

  1. Derive four independent equations.

    1. The normal tree is chosen to consist of \(\mathcal{V}_\text{in}\) and \(Z_2\).

    2. The elemental equations are

    3. The continuity equation is \(\mathcal{F}_2 = \mathcal{F}_1\).

    4. The compatibility equation is \(\mathcal{V}_1 = \mathcal{V}_\text{in} - \mathcal{V}_2\).

  2. Solve for the output \(\mathcal{V}_2\). From the elemental equation for \(Z_2\), \[\begin{aligned} \mathcal{V}_2 &= \mathcal{F_1} Z_2 \\ &= \frac{\mathcal{V}_1}{Z_1} Z_2 \\ &= \frac{Z_2} {Z_1} (\mathcal{V}_\text{in} - \mathcal{V}_2) \quad \Rightarrow \\ \mathcal{V}_2 &= \frac{Z_2} {Z_1+Z_2} \mathcal{V}_\text{in}.\end{aligned}\]

A similar analysis can be conducted for \(n\) impedance elements.

Through-variable dividers

By a similar process, we can analyze a network that divides a through-variable into \(n\) parallel impedance elements. For the output through-variable through \(Z_k\) in parallel with \(n\) impedance elements with input through-variable $\mathcal{

Transfer functions using dividers

An excellent shortcut to deriving a transfer function is to use the across- and through-variable divider rules instead of solving the system of algebraic equations, as in . An algorithm for this process is as follows.

  1. Identify the element associated with an output variable \(Y_i\). Call it the output element.

  2. Identify the source associated with an input variable \(U_j\). Set all other sources to zero.

  3. Transform the network to be an across- or through-variable divider that includes the “bare” (uncombined) output element’s output variable.1

    1. If necessary, form equivalent impedances of portions of the network, being sure to leave the output element’s output variable alone.

    2. If necessary, transform the source à la Norton or Thévenin.

  4. Apply the across- or through-variable divider equation.

  5. If necessary, use the elemental equation of the output element to trade output across- and through-variables.

  6. If necessary, use the source transformation equation of the input to trade input across- and through-variables.

  7. Divide both sides by the input variable.

It turns out that, despite its many “if necessary” clauses, very often this “shortcut” is easier than the method of for low-order systems if only a few transfer functions are of interest.

Example 12.6

Given the circuit shown with voltage source Vs and output vL,

  1. what is the transfer function $\dfrac{V_L} {V_s}$?

  2. Without transforming the source, find the transfer function $\dfrac{I_L} {V_s}$.

  3. Transforming the source, find $\dfrac{I_L} {V_s}$.

  1. We’ll use impedance methods, but with a voltage divider. The inductor is the output element, but we can combine the parallel capacitor and inductor without losing the output variable VL. This leaves us with a straightforward voltage divider! We can do this in one line: $$\begin{aligned} V_L &= \frac{\frac{Z_L Z_C} {Z_L + Z_C}} {Z_R + \frac{Z_L Z_C} {Z_L + Z_C}} V_s\quad \Rightarrow \\ \frac{V_L} {V_s} &= \frac{Z_L Z_C} {Z_R Z_L + Z_R Z_C + Z_L Z_C} \\ &= \frac{L/C} {R L s + R/(C s) + L/C} \\ &= \frac{L s} {R L C s^2 + L s + R}. \end{aligned}$$

  2. If we use the previous result and the inductor impedance (elemental equation), $$\begin{aligned} \frac{I_L Z_L} {V_s} &= \frac{L s} {R L C s^2 + L s + R} \quad \Rightarrow \\ \frac{I_L} {V_s} &= \frac{1} {R L C s^2 + L s + R} \end{aligned}$$

  3. Transforming the source puts R, C, and L in parallel with the new source Is, which is a straightforward through-variable divider: $$\begin{aligned} I_L &= \frac{1/Z_L} {1/Z_L + 1/Z_C + 1/Z_R} I_s \quad \Rightarrow \\ \frac{I_L} {I_s} &= \frac{1} {1+ L C s^2 + L s/R} \\ &= \frac{R} {R L C s^2 + L s + R}. \end{aligned}$$ Trade Is for Vs via the Norton/Thévenin transformation: $$\begin{aligned} \frac{I_L} {V_s/R} &= \frac{R} {R L C s^2 + L s + R} \quad \Rightarrow \\ \frac{I_L} {V_s} &= \frac{1} {R L C s^2 + L s + R}, \end{aligned}$$ which is the same result as in (b).

  1. In other words, if the across-variable of the output element is the output, do not combine it in series; if the through-variable is the output, do not combine it in parallel.↩︎

Online Resources for Section 12.6

No online resources.