How can I go from transmission line model to lumped elements model?

Question

I imagine that the lumped element circuit theory is an approximation of transmission line theory (which in turn is an approximation of full-wave electromagnetic theory). However, I am having a little trouble understanding how to go from transmission line theory to lumped elements. I think when the lengths of the transmission lines are made very small, we should get back the relations of the circuit theory in the limit.

Consider the following case:-

Suppose I have a transmission line of length $L$ with characteristic impedance $Z_0$ and load voltage $Z_L$. If it is connected to an ideal sinusoidal voltage source $V_s$, then the voltage across the load $V_L=V_s\frac{2Z_L}{Z_L+Z_0}$. However, I think that when $L$ becomes very small (compared to the wavelength of the source), the line essentially acts as a short and I should get $V_L=V_s$ across the load. I do not understand how this approximation can be obtained by reducing the length since $V_L$ got from transmission line model seems to be independent of the length (and of source frequency).

There may be some assumption that breaks down when we go to small length, but I am not able to get a clear picture of it.

Answer 1

The formula you're referring to is not entirely correct. First, let's derive the correct formula.

I'm assuming that we have an ideal voltage source ($V_g$) connected to a load with impedance $Z_L$ through a transmission line of length $l$. Let's pick the $z$ axis as the direction of the line, with the origin $z=0$ at the load. Therefore, the source will be situated at $z=-l$. The solution of the Telegrapher's equations are: $$V(z) = V^+e^{-i\beta z}+V^-e^{i\beta z}=V^+(e^{-i\beta z}+\Gamma _Le^{i\beta z})$$ $$I(z) =\frac {V^+}{Z_0}e^{-i\beta z}-\frac {V^-}{Z_0}e^{i\beta z}=\frac {V^+}{Z_0}(e^{-i\beta z}-\Gamma _Le^{i\beta z})$$ Where $\Gamma_L = \frac{Z_L-Z_0}{Z_L+Z_0}$ is the reflection coefficient at the load (because we chose $z=0$ on the load). We can now find $V^+$ by using the boundary condition $V(-l) = V_g$. This gives: $$V_g = V^+(e^{i\beta l}+\Gamma _Le^{-i\beta l})$$ Or, $$V^+ = \frac{V_g}{e^{i\beta l}+\Gamma _Le^{-i\beta l}}$$ Now use the first equation to find $V_L \equiv V(0)$ at the load: $$V_L =\frac{1+\Gamma_L}{e^{i\beta l}+\Gamma _Le^{-i\beta l}} V_g$$ Substituting $\Gamma_L = \frac{Z_L-Z_0}{Z_L+Z_0}$ and simplifying, the final equation for the voltage on the load is: $$V_L =\frac{Z_L}{\cos(\beta l) Z_L + i \sin(\beta l) Z_0} V_g$$ This is the correct general expression for the voltage on the load; which is dependent on $l$ as you can see. In the lumped circuit limit ($l \to0^+$), you can easily see this becomes: $$\lim_{l \to0^+} V_L = V_g$$ Which is what you expect.