How do you derive an expression for the output voltage of a ladder network of N sections?

Question

I am really stuck with a result from a circuit lecture slide. I understand how the thevenin equivalent circuit for this section shows the general result. But I am stuck on how to prove it generally for N sections.

Any help much appreciated.

Answer 1

By inspection, adding a further \$\small 1 \:\Omega /2\: \Omega\$ stage always results in the same Thevenin resistance, but half the Thevenin voltage, of the preceding stage. So the output resistance never changes, and the Thevenin voltage source halves each time a stage is added; hence, for the \$\small N'th\$ stage the voltage is \$\frac{10}{2^N}\$

Answer 2

^{simulate this circuit – Schematic created using CircuitLab}

It's the case that \$R_i=\frac{2R\left(R+R_{i-1}\right)}{3R+R_{i-1}}\$, where \$R_{i-1}\$ is the prior source impedance leading up to section \$i\$. It seems easy to see that this is bounded such that: \$\frac23R\le R_i\le2R\$.

It's also the case that \$V_i=V_{i-1}\frac{2R}{3R+R_{i-1}}\$, where \$V_{i-1}\$ is the prior ideal voltage source leading up to section \$i\$. It seems easy to see that this is bounded such that: \$0\:\text{V}\le V_i\le\frac23V_{i-1}\$.

$$\begin{align*} R_0 &= R & V_0&=V\\\\ R_1 &= \frac{2R\left(R+R_{i-1}\right)}{3R+R_{i-1}}=R & V_1&=V_{i-1}\frac{2R}{3R+R_{i-1}}=\frac12V\\\\ R_2 &= \frac{2R\left(R+R_{i-1}\right)}{3R+R_{i-1}}=R & V_2&=V_{i-1}\frac{2R}{3R+R_{i-1}}=\frac14V\\\\ R_3 &= \dots & V_3&=\dots \end{align*}$$

The above represents two recurrences: \$R_i\$ and \$V_i\$. The recurrence \$R_i\$ only depends upon itself. But the recurrence \$V_i\$ depends upon both itself as well as upon the recurrence \$R_i\$.

It's not terribly hard to prove the solution for the recurrence \$R_i\$ by induction, as Eugene suggested:

1. For \$i=0\$, it is the case that \$R_0=R\$. This is axiomatically true.

2. For any \$i\gt 0\$, the inductive hypothesis is that if you assume \$R_{i-1}=R\$ then \$R_{i}=\frac{2R\left(R+R_{i-1}\right)}{3R+R_{i-1}}=R\$.

   And therefore \$R_i=R_{i-1}\$.

3. For \$i=1\$ it is the case that the assumption of \$R_{i-1}=R\$, made in #2 above, is true (given #1 above.) Given that fact, it therefore also follows from #2 above that \$R_{i}=R\$.

   As \$R_0=R\$ and all \$R_i=R_{i-1}\$ for \$i\gt 0\$, the solution to this recurrence is that \$R_i=R\$ for all \$i\ge 0\$.

In the case of the voltage, it's more like this:

1. For \$i=0\$, it is the case that \$V_0=V\$. This is axiomatically true.

2. For any \$i\gt 0\$, the inductive hypothesis is that \$R_{i-1}=R\$ has already just proven so it follows that \$V_{i}=V_{i-1}\frac{2R}{3R+R_{i-1}}\$.

   And therefore \$V_i=\frac12 V_{i-1}\$.

3. For \$i=1\$ it is the case that the assumption of \$R_{i-1}=R\$, made in #2 above, is true (given #1 above.) Given that fact, it therefore also follows from #2 above that \$R_{i}=R\$.

I'm going to leave this here, for now, simply by way of extending on Eugene's comment and Chu's answer. (I may choose to come back and add more, illustrating it better via generating functions.)