# Why is it impossible to realize elliptic even-order filters with $R_S = R_L$?

Referring to RLC continuous-time prototypes of filters, I have read that an elliptic filter of even order can not have equal source and load termination resistances.

The trasfer function for such filters is

$$|H(j \omega)|^2 = \frac{1}{1 + \epsilon^2 R^2_n(\omega)}$$

which for $n$ even and $\omega \to 0$ is

$$|H(j 0)|^2 = \frac{1}{1 + \epsilon^2} \neq 1$$

for $\epsilon > 0$.

1) Why this circuit with $R_S = R_L$ can't realize the above transfer function for $\omega \to 0$?

simulate this circuit – Schematic created using CircuitLab

2) Are there some modified types of elliptic filters of even order feasible with $R_S = R_L$?

If the answers would be too long, and you would like to post some link about these questions, I'll anyway appreciate it!

• I'm not 100% up on elliptic filters but doesn't your circuit need to be a bit different to those shown i.e. series tuned shunt elements OR parallel tuned series elements rather than straight C or L? Maybe you can offer some insight here? Commented Dec 17, 2014 at 14:50
• So you're meaning some capacitance in parallel with the inductors and some inductor series with the capacitors? Commented Dec 17, 2014 at 16:27
• One or the other is relevant for an elliptical filter I believe. Commented Dec 18, 2014 at 8:26
• @Andyaka (I was asking because I was not sure to having understood your comment). Anyway, I don't have a specific circuit: almost certainly such series (or parallel) elements should be added. But my question was more general: why an elliptic transfer function for a filter doesn't allow to have a load resistance equal to the source resistance? This statement is in fact always true and it doesn't depend on the particular topology of the circuit, but just on the fact that it is an elliptic filter. Commented Dec 18, 2014 at 10:51
• I know I'm too late, but I'd just like to point out that it's not "impossible" to have equal terminating impedances, it's just that the response will not be the same, it will not blow up or anything. It will force the response to follow the DC ratio given by resistors, in this case 0.5, so the magnitude will look like an odd order filter, with a slight droop, that accentuates towards the corner frequency, to compensate for the loss. Commented Sep 1, 2016 at 19:48