FSK modulator, demodulator and signal space

Binary FSK (BFSK) sends a bit by switching the carrier between two frequencies: one frequency for symbol 1 and another for symbol 0, while the amplitude stays constant.

$$s_i(t)=\begin{cases}\sqrt{\dfrac{2E_b}{T_b}}\cos(2\pi f_i t), & 0\le t\le T_b\\[4pt] 0, & \text{elsewhere}\end{cases}\qquad i=1,2$$

with $f_i=\dfrac{n_c+i}{T_b}$ ($n_c$ an integer), so that the two tones are orthogonal over one bit. Symbol 1 is sent by $s_1(t)$ (frequency $f_1$) and symbol 0 by $s_2(t)$ (frequency $f_2$).

Modulator

The binary data in on-off form is applied to one product modulator, and its inverse to the other. Exactly one oscillator output reaches the adder in each bit interval.

 b(t) on-off
   |---------------->(X)<---- cos 2*pi*f1*t
   |                  |
   |                  v
   |                 (+)------> BFSK s(t)
   |                  ^
   v                  |
 [Inverter]-------->(X)<---- cos 2*pi*f2*t

• Bit 1: the upper channel is ON and the lower OFF, so the output is tone $f_1$.
• Bit 0: the inverter turns the lower channel ON, so the output is tone $f_2$.
• If the two oscillators are phase-synchronised (or one VCO is used), the phase is continuous (CPFSK).

Coherent demodulator

The receiver uses two correlators, one matched to each basis function, and compares their outputs.

         +->(X)->[int 0..Tb]--x1--+
         |   ^ phi1(t)            |+
 x(t) ---+                       (S)--> l = x1 - x2
         |   v phi2(t)            |-      |
         +->(X)->[int 0..Tb]--x2--+       v
                              [l > 0 -> 1, l < 0 -> 0]

$\phi_1(t)=\sqrt{2/T_b}\cos 2\pi f_1t$ and $\phi_2(t)=\sqrt{2/T_b}\cos 2\pi f_2t$. If $l=x_1-x_2>0$ the receiver decides 1, otherwise 0. A non-coherent version uses band-pass filters at $f_1$ and $f_2$ followed by envelope detectors and a comparator.

Signal space diagram

BFSK is a two-dimensional scheme ($N=2$, $M=2$). The signal vectors are

$$\mathbf{s}_1=\begin{bmatrix}\sqrt{E_b}\\0\end{bmatrix},\qquad \mathbf{s}_2=\begin{bmatrix}0\\\sqrt{E_b}\end{bmatrix}$$

   phi2
    |              / decision boundary
    |   Z2       /   (x1 = x2 line)
 s2 *          /
    |(0,sqrt Eb)
    |       /
    |     /      Z1
    |   /
    | /
    +-----------*---------> phi1
  0          s1 (sqrt Eb, 0)

• Distance between points: $d_{12}=\sqrt{2E_b}$ (smaller than $2\sqrt{E_b}$ for BPSK).
• The decision boundary is the line $x_1=x_2$ through the origin at 45 degrees (the perpendicular bisector of the line joining $\mathbf{s}_1$ and $\mathbf{s}_2$).
• Bit error probability (coherent): $P_e=\tfrac12\,\text{erfc}\sqrt{E_b/2N_0}$, so BFSK needs 3 dB more $E_b/N_0$ than BPSK for the same $P_e$.
• Bandwidth is about $\lvert f_1-f_2\rvert+2/T_b$, larger than BPSK.