Why Shannon Theorem has nothing to do with frequency?

According to Shannon theorem $$C= B \log _2(1 + S/N)$$ My question is, why there is no frequency in this formula?

Let's say we have 2 channels, both of them have same 20 MHz bandwidth,
one is from 100 MHz to 120 MHz,
another is from 1 GHz to 1.02 GHz.

Does this formula mean that these two channels have the same transmission speed?

Cause I think the one from 1 GHz to 1.02 GHz is faster, since it has a higher frequency.

• The higher the frequency the faster the signal moves through space or a wire or optical fibre or whatever? You have to jump to hyperspace and increase speed to Warp 8 to do that.... – Dilip Sarwate Aug 21 '18 at 15:46

Does this formula mean that these two channels have the same transmission speed?

Yes. That's exactly what you should take away from this: Channel capacity has nothing to do with center frequency; only bandwidth and SNR are relevant.

Cause I think the one from 1 GHz to 1.02 GHz is faster, since it has a higher frequency.

But that's plain wrong. It's not based on an understanding of what it means to use the channel:

To transport a specific amount of data, you will need to transmit something; the faster you change what you're transmitting, the more data you can send per second.

But the more often you change what you're transmitting, the higher the bandwidth. Whether that happens in a channel around 100 MHz or around 72 GHz makes no difference whatsoever.

• But 10 Hz transmit 10 bits per second, 1000 Hz transmit 1000 bits per second, higher the frequency, more bits can be transmitted in a unit time, isn't it? What's wrong with this? – Wulfric Lee Aug 21 '18 at 23:19
• Please reread my answer. Everything is wrong with what you claim: the frequency doesn't matter. What matters is the bandwidth. I don't know why you think the frequency matters; it simply doesn't. Just to give a silly example: you could as well say "I have more ducks in my garden than my neighbor, so my transmission must be faster! What's wrong with that?", and it would make as much sense as your frequency statement ;) – Marcus Müller Aug 21 '18 at 23:23
• Sorry I still couldn't understand this, a simple example is, if I want to send 0xAA(b10101010), and I send them out in 1 Hz, which is 1 bit per second, then I have to spend 8 seconds to send them out, and if send them out in 8 Hz, which is 8 bits per second, then I only need 1 second to send them out, am I wrong? – Wulfric Lee Aug 21 '18 at 23:48
• You use 1 Hz of bandwidth or 8 Hz of bandwidth. Not carrier frequency. You seem to be confusing the two. – Marcus Müller Aug 21 '18 at 23:49
• @GlennPallad: you don’t transmit data at 8MHz just because your carrier frequency is 8MHz. You modulate a signal onto your carrier frequency. The bandwidth of the signal determines the transmission speed. And that signal typically has a much lower bandwidth than the carrier frequency. – Cris Luengo Aug 22 '18 at 1:20

This picture may help clear up the OP's confusion. On the right is the baseband signal of interest, with finite bandwidth centered at DC as shown in the frequency spectrum plot. This is a complex signal, so it has I (in-phase, on the real axis) and Q (quadrature, on the imaginary axis) components. We implement these signals in hardware using two real signals, one representing I and and the other Q. The I-Q plot at the top right is showing what these I and Q components do versus time, representing a vector I + jQ whose magnitude and phase changes with time.

Now we introduce the concept of bandwidth: The bandwidth of our channel limits how quickly the vector can move around. If higher frequency components are filtered out, as represented in the spectrum shown below the I-Q plot, then the vector can not move quickly. To get a sense of this consider the rule of thumb for the 10%-90% rise time (or fall time) of a first-order low-pass system:

$t_r = \frac{0.35}{BW}$

Where $t_r$ is the rise time in seconds, and BW is the bandwidth of the low-pass system in Hz. The rise time representing the transition time from one position to another is similar to the time required (within a given BW) to move between any two points on the IQ plane (given an input stimulus of a step between those two points). Bottom line-- narrow bandwidth means slow transitions.

The BW we can use, and the signal power we can transmit (and receive relative to our noise floor) will define how much data we can transmit, which is the essence of Shannon's Capacity theorem.

If we up-convert or frequency shift our signal to another carrier, the bandwidth will be exactly the same! The I-Q vector now represents the magnitude and phase change of that carrier with time, but will follow the exact same trajectory we originally had at DC (or any other frequency we choose to move it to!), at the exact same rate (BW). The "magnitude and phase change of that carrier" is relative to the original carrier itself, just like when we were at DC it was relative to 0 but in all other regards is identical.

If you have seen any of my other postings I much prefer to think it terms of complex signals rather than cosines and sines for representing frequencies. One reason is in this case, DC is treated no differently than any other carrier. Many other computations and thought processes are greatly simplified once you convert to thinking directly with complex frequency terms. For more info on that specifically please see my post at this link:

Frequency shifting of a quadrature mixed signal