# Time domain representation of a modulated signal when symbol rate >> carrier frequency?

I need to understand the relationship between carrier vs. bits with PSK modulation, and what the signal in the time domain would look like if my symbol rate is far greater than my carrier frequency. How can I put more symbols on a carrier which is at a set frequency? For example how would I put 1000000000000000 bits on a 1 Hz carrier wave, is this possible and what would that look like?

tl;dr: you can't. That's basics!

longer story:

When the symbol rate is much larger than the carrier frequency, you've basically lost, and cannot transport that many symbols per second; you don't have the bandwidth; you get at most twice the carrier frequency in bandwidth, namely from 0 Hz to twice the carrier frequency.

However, you're asking how to modulate 10¹⁵ bits on a 1 Hz carrier, and that's a different question: theoretically, you can take forever to transport these 10¹⁵ bits, so there's no problem here. Realistically, we'll all be dead by the time you've finished your transfer, but whether or not that is a problem is of mere philosophical interest.

If you, however, mean you want to transport 10¹⁵ bits per second on a 1 Hz carrier wave, you have less than two channel access per second, so you'll need enough SNR that you get more than $$2^{5·10^{14}}$$ different constellation points.

If you reasonably try this, I hope someone stops you in time: as physics dictates the difference between two detectable constellation points needs to be at least the Planck constant, in fairly good approximation, the action needed to get from the all-0 bitword to the all-1 bitword would be at the very least,

$$\frac{2^{\left(5\cdot 10^{14}\right)}\cdot h}{\pi} =\frac{2^{\left(5\cdot 10^{14}\right)}\cdot 2^{-110.2}\frac{\text{kg·m²}}{\text{s}}}{\pi} \approx 2^{\left(4\cdot 10^{12}\right)} \frac{\text{kg·m²}}{\text{s}}\approx 10^{4\cdot10^{11}} \text{J·s}.$$

That is incredibly much action. This is nuts. Your photon of 1 Hz would have to have an energy of $$10^{4\cdot10^{11}}\,\text{J}$$ to be able to walk a path of that action; that's as much energy in a single photon as the sun emits in 487 years. This photon would end the universe as we know it. There'd be only heat and chaos. And then, who the hell will care about your 10¹⁵ bits?

• There used to be a joke that the fastest way to send 1 TB from New York to Los Angeles was to ship a hard drive via FedEx! Mar 2, 2021 at 20:48
• Thanks for the answer. Nothing like ending the known universe. Mar 3, 2021 at 15:53