Recent content by lrg

@Baluncore, I think that evaluating the ladder from left to right, instead of right to left, can be better (for ##\vert b \vert<1##). A forward continued fraction like that associated with the input impedance can be evaluated in the forward direction with the Euler continuants, but there is an...

Yes @Baluncore, that is what I found, and I put in the figure above. In fact, the continued fraction that computes the equivalent resistance of the original @bob012345's ladder should be: $$R = 1 + \cfrac{1}{\frac{1}{1} + \cfrac{1}{b + \cfrac{1}{b + \cfrac{1}{b^2 + \cfrac{1}{b^2 + \cfrac{1}{b^3...

Yes, you are right, very stupid mistake on my side. When I realized it, I decided to simulate the circuit (100 stages) with a Matlab/Octave script, changing the resistance at infinity (far right), and I got this graph: It shows that the input resistance can go from about 1.1 (short) to about...

Yes, it is very different, but I think the numerical evaluation above is misleading, because it starts with ##r = 10^{n+1}##, which means that the infinite ladder to the right of the last vertical resistor is completely cut off, which is the worst thing to do. If we start with ##r=0## (still...

Hello everyone, I think the exact value should be $$R_{\infty} = \sqrt{26} - 4 = 1.099019513592785,$$ that is the solution of the equation $$ x = 1+ \frac{1}{\frac{1}{1}+\frac{10}{x}},$$ or $$x^2+8x-10 = 0,$$ since the equivalent resistance at the right of the first vertical resistor is equal...