In my line of work, 110 GHz is considered 'the low end of our spectrum'.

Each layer is a frequency selective time delay and 90 degree phase shift. By delaying the lower frequencies first it has the effect of making the leading edge steeper. You will get a similar effect merely by making a symmetric (zero phase) waveform minimum phase. Bode showed that physically realizable filters are minimum phase. That only requires a 90 degree phase shift. The layered structure provides additional delay beyond just applying a convolution with a Hilbert operator aka quadrature operator.

A dispersive material used to compress a pulse and increase sharpness can only increase the sharpness of the edge insofar there is enough signal bandwidth already available to do so. To my knowledge this is what they do with femtosecond pulse lasers - but in order to do so, they need to start out with a wide-band pulse that contains lost of frequencies already.

Lets look at it this way: From the frequency-time-domain duality, we know that a pulse in the time-domain is \$\sin(x)\cdot x^{-1}\$ in the frequency domain --- in other words, we need infinite bandwidth to have an infinitly sharp pulse. If we band-limit this pulse, we chop off some of that bandwidth, and we get a lower rise time. How low a rise-time we get in relation to what we get doesn't matter for what follows:

A linear system \$H\$ cannot generate new frequencies. Let us prove this:

A system \$G\$ is linear if, for a given inputs \$x(f)\$ and \$y(f)\$ and scalars \$\alpha\$ and \$\beta\$ (if this holds in the frequency domain this also holds in the time domain, as the Fourier-transform itself is a linear operation), the following is true:

\$\alpha G{x(f)} + \beta G{y(f)} = G{\alpha x(f) + \beta y(f)}\$.

So let us say that for a given \$f_a\$, \$x(f_a)\$ and \$y(f_a)\$ equal zero (IE, signals \$x\$ and \$y\$ have no frequency content at \$f_a\$). Then if \$G\$ generates frequency components that weren't already there, \$ G{\alpha x(f_a) + \beta y(f_a)} = G{0} = c \neq 0\$, with \$c\$ a constant. We can now plug in this result into the LHS of our definition, and say that:

\$\alpha G{0} + \beta G{0} = \alpha c + \beta c\$

But the left-hand side of our equation must also equal \$G{\alpha x(f_a) + \beta y(f_a)}\$, which we already said was equal to \$G(0)=c\$. In other words, \$\alpha c + \beta c = c\$ for all possible \$\alpha\$ and \$\beta\$, which cannot be valid if \$c \neq 0\$. This proves by contradiction that if the input does not have any content at \$f_a\$, the output too does not have any content at \$f_a\$: a linear system cannot add frequencies - it can only 'reshuffle' them.

The problem in your suggested approach using is that your pulse source, that laser diode driver, will likely not have any spectral content high enough to make the pulse much faster. And even if it had: to get a 3 ps pulse, you would need spectral content up at 100 GHz - there is no way you are going to get that more than a few millimeters on an FR4 PCB. On a rogers substrate, maybe - but you would still need a connector that behaves itself past 100 GHz (only connectors that do that are 1 mm and 0.8 mm connectors), and that present an impedance you can work with. I also suspect that most micro-strip lines will be too dispersive to keep your pulse sharp very long. The difference in loss between frequencies will also cause issues, as this will reduce sharpness further and must be compensated for short rise-times.

In short: I think this won't work very well. A better solution to me would seem some kind of RF MEMS switch that is rated for many GHz, and try working with that. Creating a square wave can also be easier, since then you just need to play around with a discrete spectrum of harmonics. But I think in the end it is impossible to generate anything like this with discretes. This is hard enough to do on chip, with fancy technologies like SiGe, GaAs, GaN or InP (forget CMOS, won't slew fast enough me thinks).