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!
Best Answer
The elliptic filter response is equi-ripple in the passband. The gain ripples between unity and 'design dB down', where the filter is designed for 0.1dB, or 1dB, or whatever ripple.
An odd order filter has unity gain at DC, and design dB down at the corner frequency.
An even order filter is still design dB down at the corner frequency, which means it must also be design dB down at DC, as the order of the filter defines the number of passband ripples.
For a lossless filter, if the source and load impedance were the same, there could not be a finite loss at DC. To create the DC loss, the source and load must be different impedances. The action of the filter 'tunes out' the difference in termination impedances at other frequencies to achieve the unity gain parts of the frequency response.