RF Bandwidth vs Data Rate (modulation)

Question

I was reading some articles about digital and analog modulating processes. One of them was this:

Modulation techniques

At a certain point is says:

The bandwidth produced is a function of the highest modulating frequency including harmonics and the modulation index, which is:

m = Δf(T)

Δf is the frequency deviation or shift between the mark and space frequencies, or:

Δf = fs – fm

T is the bit time interval of the data or the reciprocal of the data rate (1/bit/s).

I'm a bit confused with these terms. How exaclty the bandwidth relates do data rate? Let's take this image by example:

From what I've understood, if I want to modulate a digital signal by FSK process, I should just choose 2 different frequencies to represent 0 and 1. I should also choose two frequencies that are harmonics of a fundamental frequency to get "smooth" transitions (zero cross), is that correct? Besides that, how my data rate is related to this bandwidth? Isn't bandwidth the width of a channel?

Let's say I'm using a 2.4GHz transmission with a channel that goes from 2.4GHz until 2.450GHz, then I will have a channel of 50MHz, correct?

My bandwidth in this case isn't 50MHz? If so, my bandwidth should only be expressed by Δf, no? In other words, since I only need 2 differente frequencies, my channel width would be only the difference between those frequencies. With that said, I cannot see where data rates comes in. I think the only thing that would influence on that rate would be how long each bit is holded.

As I said I'm confused with these conecpts. So if someone please could clarify my ideas.

Answer 1

RF bandwidth and data rate are related by the modulation format. Different modulation formats will require different bandwidths for the same data rate. For FM modulation, the bandwidth is approximately 2*(df + fm) where df is the maximum frequency deviation and fm is the frequency of the message. FSK is basically FM where the message signal is a square wave. The highest frequency component of a binary bit sequence transmitted serially occurs when the sequence is 01010101. This component is one half of the bit rate. So for FSK, the bandwidth is approximately Δf + r where Δf is the separation between the two frequencies and r is the bit rate. The reason this is bigger than Δf is because whenever the frequency is changed, extra frequency components are generated. Switching between frequencies more often (higher data rate) results in more power in these extra frequency components. Now, these can be filtered out to some extent, but if you filter more of them than Δf + r, the result will be too distorted to reliably extract the original bitstream.

Think about it this way: a pure sinewave consumes zero bandwidth, but it also contains zero information. As soon as you start changing a characteristic of a pure sinewave (frequency, phase, amplitude, etc.) its bandwidth must increase accordingly. In the case of amplitude modulation, modulating the amplitidue of a sinewave of frequency fc at frequency fm will result in a signal with components at fc, fc+fm, and fc-fm. If the message contains components all the way down to DC, then the resulting modulated signal will have twice the bandwidth of the message signal. FSK is basically transmitting two AM signals at the same time on different frequencies, so the bandwidth will naturally be increased by the separation of these two carrier frequencies.

For FSK, the bit rate and the symbol rate are the same. But for higher order modulations like QPSK and QAM, each transmitted symbol can code for more than one bit so the bit rate can be significantly higher than the symbol rate. This means that the required transmit bandwidth is less than what would be required for AM or FSK. QPSK and QAM have higher spectral efficiency. However, QPSK and QAM are more susceptible to noise and distortion and therefore require a relatively higher SNR.

Also, for FSK, you want the two frequencies to be integer multiples of the data rate. This will result in an integer number of cycles in each bit period so that the carrier always ends up at the same level on data bit transitions. This probably won't be done at RF, though. Generally the FSK signal would be generated at an intermediate frequency which would then be mixed up to the actual RF carrier frequency.

Answer 2

It is not possible to switch frequencies and not occupy other frequencies. You might think that if you are switching between say, 100Hz and 120Hz, and if you are switching with continuous phase, then those are the only two frequencies you'd occupy. However, the math just doesn't work out that way.

Remember, the Fourier transform only works for periodic signals. There is a periodic function that consists of just two frequencies, but it doesn't look like FSK as you posted above. For example, \$\sin(t) + \sin(1.2t)\$ is periodic, and has just two frequency components. It looks like this:

Not very FSK-like, is it? Wolfram alpha will compute the Fourier transform for us:

This is, in so many symbols, just two impulses at 1.0 and 1.2. Two, pure frequency components, just as you'd expect.

FSK isn't two frequencies added together, but rather a piecewise function. I can't figure out how to make Wolfram Alpha compute the Fourier transform of this, but it's the two frequency components (1.0 and 1.2), plus some other components to make them not happen at the same time.

You can analyze this quite rigorously if you want, and people have done that. However, it boils down to this: frequency or phase or amplitude changes in the time domain make sidebands in the frequency domain. If you make those frequency changes abruptly, the sidebands are farther away and stronger, and you get reduced spectral efficiency. If you make them slowly, the sidebands are weaker and not so far away and weaker and you get better spectral efficiency.