I understand the whole theory behind bypass capacitors, how they are necessary to deliver high-frequency currents to the IC, as the PCB in general has high-inductance, thus high-impedance in high-frequencies. This graph shows the PCB impedance on the frequency-domain, showing how only the bypass capacitors can deliver the MHz currents the IC needs.
I understand this all on the frequency-domain, but not so much on the time-domain.
If an IC has a high-frequency I/O operation, switching from 5V to 0V, this current will come from the bypass capacitor, but as soon as the bypass capacitor starts to deplete, the circuit will try to recharge it.
The problem is, since all other parts of the circuit (bulk capacitors, VRM, etc) can't respond to such a higher-frequency, then how will the capacitors recharge at all?
As soon as they start to recharge, the IC will try to deplete them again and again.
An analogy I can think of is as if the bypass capacitors are a bucket full of water, the IC is a drain pipe connected to the bucket and the rest of the circuit (bulk capacitors, VRM and etc.) are a dripping faucet, trying to fill the bucket.
The dripping faucet will never be able to keep the bucket filled, as it can't fill the bucket as fast as the drain pipe is pulling water from it.