# Why bit rate is same in a lower frequency spectrum and higher spectrum?

If there is a spectrum between 0 Mhz - 1 Mhz, bandwidth would be 1 Mhz here and for example with baseband we can send 2000 bps. What I can't understand is if spectrum is 5-6 Mhz instead of 0 - 1 Mhz, although the spectrum is the same 1 Mhz, shouldn't we be able to send more data because we're sending it with much higher frequencies? Why is bit rate fixated on spectrum instead frequencies itself? What am I missing here?

## 3 Answers

If you are a bass and sing A 220 Hz and B 247 Hz at 120 beats per minute, you can sing data at a certain rate.

If you are a soprano and sing two quarter notes A 880 Hz and B 988 Hz at the same BPM, your data rate isn't any higher.

If you sing more notes (higher bandwidth) you could communicate a more complex score (e.g. a higher data rate). So bandwidth (range of notes) matters, not being a tenor or alto-soprano (spectral frequency).

Same at RF frequencies.

• But isn't this a waste if we can sample 880 Hz can't we send more meaningful notes than just A ? I really don't get it. – Abdullah Akçam Jul 15 '19 at 17:29
• Only if you can tell the notes apart. But that might work just as well at low frequencies as high. A single held notes is quite boring, no matter low or high. – hotpaw2 Jul 15 '19 at 17:34
• In terms of digital signals this doesn't make sense to me. One single frequency can carry different types of data with different modulation techniques. – Abdullah Akçam Jul 15 '19 at 18:07
• If you modulate a pure sinewave in any way, it's no longer a single frequency: the Fourier transform will show a non-zero bandwidth, more than one single frequency. – hotpaw2 Jul 15 '19 at 19:48

Because bit rate depends on the bandwidth and not on the carrier frequency. Of course at higher frequencies you have more bandwidth, and thus you can transmit more data. But 1 MHz in lower frequencies and 1 MHz at higher frequencies have no difference on the data rate. Other effects may need to be taken into consideration though at higher frequencies. For example at mmWave frequencies the path loss is very high over distance, and thus supporting a specific data rate for a given SNR or BER requirement maybe different than that of microwave frequencies for the same SNR or BER.

To complement the other quite appropriate answers:

Say you build a real signal $$s(t)$$ with bandwidth $$B$$ that transmits $$R$$ bits per second. Its spectrum $$S(f)$$ extends from $$-B$$ to $$B$$.

The modulation property of the Fourier Transform tells you that the spectrum of $$s(t)\cos(2\pi f_c t)$$ is $$\frac{1}{2} \large[ S(f+f_c)-S(f-f_c) \large].$$ Note that nothing fundamental has changed; the spectrum shape is the same as before, except shifted.

(Note that now the spectrum has support $$2B$$, which seems to imply that, if anything, going up in frequency reduces the data rate, since now you need $$2B$$ to transmit the same $$R$$ bits per second. However, this is easily fixed using quadrature modulation.)

In other words: the carrier frequency $$f_c$$ is not fundamental; its value is largely irrelevant (unless you make practical considerations like those BlackMath points out).

What is fundamental is the bandwidth; in fact, "classical" communications theory tells us that we can transmit up to $$2B$$ symbols per second over a bandwidht $$B$$, irrespective of $$f_c$$.