# Why distance index between 2 frequencies in M-FSK has to be odd?

I'm learning about M-FSK modulation from A. Bruce Carlson book, Communication systems. I've seen that ak, an index to select the frequency of each symbol, has to take odd values, spaced by 2 fd (difference frequency). But I didn't find an explanation.

So, my question is: why can't do we use any value (even and odd, spaced by 1 fd), to reduce even more the bandwidth of our signal? Has it relation to interference between symbols in frequency spectrum?

Thanks for helping me ;)

• The convention keeps the spacing between all the signaling frequencies the same. The distance between $-1$ and $+1$ is $2$. The distance between $+1$ and $+3$ is $2$, etc. This way you aren't required to have better SNR for using the $\pm1$ frequencies vs. the other frequencies. May 29, 2022 at 11:26
• What is the value of $f_d$, and how, if at all, is the value of $f_d$ related to $D$, the duration of an M-FSK pulse? May 30, 2022 at 19:37

So, my question is: why can't do we use any value (even and odd, spaced by 1 fd), to reduce even more the bandwidth of our signal? Has it relation to interference between symbols in frequency spectrum?

Let's first consider binary or $$2$$-FSK. We have a carrier signal at fixed frequency $$f_c$$. Modulation by the binary data changes the frequency of the transmitted signal to $$f_c + f_d$$ or to $$f_c-f_d$$. Thus the change in the frequency of the transmitted signal which occurs when the transmitter transmits a $$1$$ followed by a $$0$$ is $$\pm 2f_d$$; the two possible FSK tones at frequencies $$f_c + f_d$$ and $$f_c-f_d$$ are spaced $$\mathbf 2f_d$$ apart in frequency; not $$f_d$$ apart as the OP apparently believes. (The "$$f_d$$ apart" model corresponds ta a different system where to send a $$0$$, a tone at $$f_c$$ is transmitted whereas to send a $$1$$, a tone at $$f_c+f_d$$ is transmitted). The transmitted signal consists of a succession of RF pulses of duration $$D$$. Let's assume that $$f_c \gg \dfrac 1D$$ so that each RF pulse has several periods of the RF sinusoid.

What we would like to arrange is for the two possible tones to be orthogonal over each signaling epoch, that is, the RF pulses $$\cos(2\pi(f_c + f_d)t+\theta_1)$$ and $$\cos(2\pi(f_c - f_d)t+\theta_0)$$ should satisfy $$\int_{kD}^{(k+1)D}\cos(2\pi(f_c + f_d)t+\theta_1)\cos(2\pi(f_c - f_d)t+\theta_0) \,\mathrm dt = 0$$ for all integers $$k$$. Why the insistence on orthogonality? Well, suppose that the receiver consists of two matched filters for the two RF pulses and a comparator to determine which filter is producing a larger response at the sampling instant. Then, the output of each matched filter is the cross-correlation function of the input RF pulse with the RF pulse to which the filter is matched, and at the sampling instant, this cross-correlation has value $$0$$ if the input is the other pulse, and large value (the peak value of the autocorrelation function, in fact) if the input is the one to which the filter is matched (cf. this answer for details.)

Now, assuming without loss of generality that $$f_1 > f_0$$, we have that \begin{align} & ~~~~~~~~\int_{kD}^{(k+1)D}2\cos(2\pi f_1t+\theta_1)\cos(2\pi f_0t+\theta_0) \,\mathrm dt\\ &= \int_{kD}^{(k+1)D}\cos(2\pi (f_1-f_0)t+\theta_1-\theta_0) +\cos(2\pi (f_1+f_0)t+\theta_1+\theta_0) \,\mathrm dt\\ &= \left.\frac{\sin(2\pi (f_1-f_0)t+\theta_1-\theta_0)}{2\pi (f_1-f_0)} + \frac{\sin(2\pi (f_1+f_0)t+\theta_1+\theta_0)}{2\pi (f_1+f_0)}\right\vert_{kD}^{(k+1)D}\\ &= ~~~~~\frac{\sin(2\pi (f_1-f_0)(k+1)D+\theta_1-\theta_0)-\sin(2\pi (f_1-f_0)kD+\theta_1-\theta_0)}{2\pi (f_1-f_0)}\\ &~~~~+\frac{\sin(2\pi (f_1+f_0)(k+1)D+\theta_1+\theta_0)-\sin(2\pi (f_1+f_0)kD+\theta_1+\theta_0)}{2\pi (f_1+f_0)} \end{align} The numerators of the two fractions in the last RHS above both have values in $$[-2,+2]$$ while in the denominators we have that $$2\pi (f_1+f_0)\gg 2\pi (f_1-f_0)$$. So the magnitude of the second fraction is small in comparison to the magnitude of the first. Now, bearing in mind that $$k$$ and $$k+1$$ are integers, let us note that if $$f_1$$ and $$f_0$$ are such that $$(f_1-f_0)D$$ is an integer, then the first fraction has value $$0$$. Similarly, if $$f_1$$ and $$f_0$$ are such that $$(f_1+f_0)D$$ is an integer, then the second fraction has value $$0$$. Now, if $$f_1D$$ and $$f_0D$$ both are distinct) integers ($$m$$ and $$n$$, say, with $$m> n$$) or both are half-integers ($$\frac{2m-1}{2}$$ and $$\frac{2n-1}{2}$$, say), then both $$(f_1-f_0)D$$ and $$(f_1+f_0)D$$ are integers, and the two possible RF pulses at frequencies $$f_1$$ and $$f_0$$ are orthogonal over the interval $$[kD, (k+1)D]$$. Note that in all symbol epochs $$[kD,(k+1)D]$$, it is either the case that both RF pulses have (different) integer number of periods of their respective sinusoids, or it is the case that both RF pulses have different half-integer number of periods of their respective sinusoids.

Applying the above notion to $$\int_{kD}^{(k+1)D}\cos(2\pi(f_c + f_d)t+\theta_1)\cos(2\pi(f_c - f_d)t+\theta_0) \,\mathrm dt = 0$$ we see that it must be that both $$(f_1-f_0)D = 2f_dD$$ and $$(f_1+f_0)D = 2f_cD$$ are distinct integers, or they both are distinct half-integers. The smallest possible value of $$2f_D$$ is thus $$\frac 12$$ which gives $$f_d = \dfrac{1}{4D}$$, which will readily be recognized as the frequency deviation in minimum-shift keying (MSK), which, though technically a special form of continuous-phase FSK or CPFSK, is not what is generally meant in general discussions of $$2$$-FSK or $$M$$-FSK which are often geared towards systems where noncoherent receivers are often used. In fact, coherent reception is generally not feasible in $$M$$-FSK systems because of the difficulty of tracking the various tones in the incoming signal in view of the fact that the tones appear and disappear from the input signal as time progresses. MSK of course is best viewed as offset QPSK with baseband pulses that are a half-period of a sinisuoid. Be that as it may, the canonical answer for $$2$$-FSK is $$f_d = \dfrac{1}{2D}$$ and so the two tones are at frequencies $$f_c \pm \dfrac{1}{2D}$$. The frequency spacing between the two tones in $$2$$-FSK is thus $$\mathbf 2f_d =\dfrac{1}{D}$$.

Turning to $$M$$-FSK with $$M>2$$, we have tones at $$f_c + \mathbf 1f_d$$ and $$f_c - \mathbf 1f_d$$ and so the next two tones must be at $$f_c + \mathbf 3f_d$$ and $$f_c - \mathbf 3f_d$$ in order to maintain the spacing of $$2f_d$$ between tones as required for orthogonality. Additional tones must be at $$f_c + \mathbf nf_d$$ and $$f_c - \mathbf nf_d$$ where $$n$$ is an odd number. And that's why we can't use even multiples of $$f_d$$ as the OP desires: the odd multiples are needed to provide the spacing of $$2f_d$$ between the tones.

To me it seems that you are asking why there are an even number of frequencies centered around $$f_c$$ (Hopefully this UNICODE art renders OK):

┻━━━━━━━┻━━━┳━━━┻━━━━━━━┻


and not an odd number of frequencies centered around $$f_c$$:

━━━━┻━━━┻━━━╋━━━┻━━━┻━━━━


I think the explanation is simple and doesn't require a frequency spectral analysis. Note that the frequency axis can be scaled freely, so we can easily make equal either the bandwidth or the spacing between any number of frequencies including any even or odd case.

The Wikipedia article Multiple frequency-shift keying puts it as:

$$M$$, the size of the alphabet, is usually a power of two so that each symbol represents $$\log_2M$$ bits.

If $$M$$ is a power of two then $$M$$ is even, not odd. So $$M$$ is usually even.

Binary logic is standard so working on binary-based signals and binary-based error correction codes comes more naturally than other bases like ternary.