The classic answer to this question must be "Zverev". But that might be overkill, unless you have access to a really good library.
A simpler and non-mathematical answer to some of your questions is possible, which may help:
R1 and R2 provide impedance matching; the original filter is designed to accept a signal driven from a specific source impedance, and deliver its output to a specific load impedance (R1,R2 are also mentioned later). These impedances are:
- normally the same
- known as the "characteristic impedance" of the circuit
- usually the same as the characteristic impedance of the application's standard cables (e.g. coaxial cable in RF applications)
- commonly (but not always) 50 ohms. (you will see 75 ohms in video applications, and (rarely nowadays) 600 ohms in audio and telephony.
Check the original filter info for its characteristic impedance, but 50 ohms is most likely. So - the impedance of the L-C network was not exactly 50 ohms, and R1,R2 reduced the input and output impedances to match.
C5,C6,C7 ... Consider that C5 and L1 on their own form a parallel L/C resonant circuit. This acts as an inductor (L1) at low frequencies, and as a pure capacitor at high frequencies (VERY high since it is 1 pf!)
But at the resonant frequency, the impedance is infinite. Therefore at this frequency, the filter will have infinite attenuation. (Over-simplification! all the components interact with each other, so the actual frequency is slightly different from this calculation)
There are three such notches in the frequency response; and you can learn a bit about this filter by calculating C5/L1, C6/L2, C7/L3. Usually 2 are quite close together and the third will be significantly higher; without doing the math I can already see that here.
That makes this a 7th order Cauer filter (or Cauer/Chebyshev) and the art of getting good stopband rejection (or the reason for 592 pages of Zverev) is the art of tuning C5-C7 to place those notches (last picture on Wiki page) the right distance apart so the peaks between them are all the same height.
Theory apart, circuit tolerances virtually guarantee tweaking trimmer caps or inductor cores while watching a spectrum analyzer for best results!
C1 to C4 also resonate with L1 to L3; in this case, the main effect is on the passband flatness as well as the actual cutoff frequency (which must be below the first notch!) It can be understood as a cascade of 2nd-order sections with different characteristics and one first-order section. Look at Figure 3 in that article (embedded below, hope that's OK)
It shows underdamped sections (with peaks) and overdamped ones (which just roll off). A skilful combination of these will give an (approximately) flat response up to the cutoff. Again, I cannot cover the details here, but I hope it is clear how different values of inductor forming different 2nd-order filters are part of the puzzle. Getting R1 and R2 wrong will principally affect the passband flatness, by affecting the Q (damping) of the input and output sections (L1 etc and L3 etc).
Here is a more typically mathematical explanation
Now to the most important part of the question:
How do I select part values for 100 MHz?
Given all the above, usually not from scratch...
You can take an existing filter, and simply scale it.
Given Xl=jwL and Xc=1/jwC,
assuming the current filter is set for 50MHz,
assuming you want the new filter set for 100 MHz
and assuming the characteristic impedance is to remain the same,
you can simply halve all the inductances and capacitances, so that Xl is the same at twice the frequency, and ditto for Xc. Resistances remain the same, since the characteristic impedance is the same, and a resistor's impedance is not a function of frequency. (Check both versions in simulation!)
Best Answer
It's possible you're overthinking this. \$s\$ is the Laplace notation for the complex frequency, \$j2\pi f\$, or \$j\omega\$, and it spans the entire frequency range. When you're saying that \$|s|\rightarrow0\$ or \$|s|\rightarrow\infty\$, what you're probably referring to is the frequency, itself, as it approaches DC or as it goes to infinity, not \$s\$.
The transfer function you're showing is that of a 1st order system, and it can only have simple, real roots. Therefore using the absolute notation, \$|p|\$, or \$|z|\$, is unnecessary.
The lead or lag compensators are also called shelf filters, because of the way their magnitude looks like when plotted. Since you, yourself, said just before that the gains converge towards \$z/p\$ or \$p/z\$, then you know that there is no infinite attenuation towards that frequency. You can't talk about "recovering" some "underlying" filter from it. The topologies for lowpass and highpass are ones, and for lead and lag filters are anothers. You can't mix the two. The only similarity between a lead compensator and a highpass is the part of the spectrum situated at the geometrical mean between the zero and the pole, and until the very end; similar for lowpass. Note: similar, not identical. See below a comparison between \$\dfrac{s+0.1}{s+1}\$ and \$\dfrac{s}{s+1}\$, and notice how the magnitudes have similar slopes until around \$\sqrt{0.1\cdot 1}\$:
If you want to talk about how the limit of the pole or the zero approaches zero or infinity, then chances are you are no longer talking about lead/lag compensators, but of low-/highpass filters, instead. Otherwise, the same conclusion could be drawn about Cauer/elliptic filters that converge towards a Butterworth when the ripples are set to zero, and the zeroes to infinity: you may end up with a Butterworth, but it's not one. Fun fact: the same Cauer/elliptic also converges towards both a Chebyshev, or an inverse Chebyshev, with the proper settings; it's still not either of them.
Therefore, lead compensators do not have the behaviour of highpass filters, in the same way the same lead compensators do not have the behaviour of lowpass filters -- they have them both and clearly defined. It's what makes them shelf filters. Similar for lag compensator. Similar for xomplex poles/zeroes.
I'll add here the responses to your comments, since they might be a bit longer. For your first comment, the table in the linked OP is awfully confusing. As noted in the answer, those are plain high-/lowpass filters, so yes, they are (very much) misnomers.
For the second comment, w.r.t. OP's awfully confusing table, you are referring to the case where \$0<z<p\$, which is labeled "lag high-pass", but the answerer is referring to the "lead low-pass", which is the 3rd entry, or \$0<p<z\$ (blue is the compensator, orange the lowpass):
So, no, the answerer is right, and I'll venture to say it's most probable that he is not wrong (as his profile on dsp.ee might make it more clear). And since I fully agree that the table is horrible, I'll put this minor mishap on the account of that utterly repulsive monstrosity of a table (did I mention it's awfully confusing?). Unfortunately, you will see in the literature various people trying to impose their names through the usage of "new and improved" ways of re-defining already defined definitions, seemingly not counting the effect that it may have on the readers.
As for the topologies, there is no "standard" way to build these things, though some schematics may make more sense than others (i.e. using RC instead or RL), but if the name is not clear, think of the topology as the transfer function, and then all things should become clear.
To avoid the information going in the comments, I'll add this to the answer, but diverting too much from the topic means a new question is needed.
A pole in the right-half plane would make the system unstable (the real axis is positive). The zero could be there without affecting stability, but you're showing \$s+z\$, not \$s-z\$. In either case, when you say \$z<p\$, I take it to understand the absolute value (what I said earlier still applies, unless comparing strictly values). But even with a RHP zero, the transfer function becomes:
$$\dfrac{s-1}{s+2}\;\rightarrow\;\dfrac{j\omega-1}{j\omega+2}\cdot\dfrac{-j\omega+2}{-j\omega+2}\;\rightarrow\;\dfrac{(\omega^2-2)+j3\omega}{\omega^2+4}$$
For completeness, for a LHP zero:
$$\dfrac{s+1}{s+2}\;\rightarrow\;\dfrac{j\omega+1}{j\omega+2}\cdot\dfrac{-j\omega+2}{-j\omega+2}\;\rightarrow\;\dfrac{(\omega^2+2)+j\omega}{\omega^2+4}$$
For even more completeness, the phases of the LHP zero (blue), RHP zero (orange), plus a lowpass with a pole at \$-2\$ (red), showing a more positive phase than the lowpass: