Error probability of binary ASK extended to M-ary

Binary ASK

ASK (on–off keying) signals over $0\le t\le T_b$:

$$s_1(t)=A\cos\omega_ct\ \ (\text{bit }1),\qquad s_2(t)=0\ \ (\text{bit }0)$$

Coherent receiver (correlator = matched filter):

 r(t) -->(x)--> integrate --> sample --> y > lam ? 1 : 0
          ^       0..Tb       t=Tb
          |
       cos(wc t)  (local carrier, in phase)

Correlator output at $t=T_b$, with $r(t)=s_i(t)+w(t)$:

$$y=\int_0^{T_b}r(t)\cos\omega_ct\,dt=a_i+n$$

• Bit 1: $a_1=\int_0^{T_b}A\cos^2\omega_ct\,dt=\frac{AT_b}{2}$ (for $f_c$ an integer multiple of $1/T_b$).
• Bit 0: $a_2=0$.
• Noise: $n=\int_0^{T_b}w(t)\cos\omega_ct\,dt$ is Gaussian, mean 0, variance

$$\sigma^2=\int_0^{T_b}\!\!\int_0^{T_b}\frac{N_0}{2}\delta(t-u)\cos\omega_ct\cos\omega_cu\,dt\,du=\frac{N_0}{2}\cdot\frac{T_b}{2}=\frac{N_0T_b}{4}$$

Threshold midway: $\lambda=\frac{AT_b}{4}$. Using the binary result $P_e=\frac12\operatorname{erfc}\left(\frac{a_1-a_2}{2\sqrt2\sigma}\right)$:

$$\begin{aligned} \frac{a_1-a_2}{2\sqrt2\,\sigma} &= \frac{AT_b/2}{2\sqrt2\sqrt{N_0T_b/4}}=\frac{A\sqrt{T_b}}{2\sqrt{2N_0}}=\sqrt{\frac{A^2T_b}{8N_0}} \end{aligned}$$

Energy of a "1" pulse is $E_1=\frac{A^2T_b}{2}$; the average energy per bit (half the bits are 0) is $E_b=\frac{E_1}{2}=\frac{A^2T_b}{4}$. Hence $\frac{A^2T_b}{8N_0}=\frac{E_b}{2N_0}$ and

$$P_e(\text{ASK})=\frac12\operatorname{erfc}\left(\sqrt{\frac{E_b}{2N_0}}\right)=Q\left(\sqrt{\frac{E_b}{N_0}}\right)$$

(In terms of peak energy $E_1$: $P_e=\frac12\operatorname{erfc}\sqrt{E_1/4N_0}$.)

Extension to M-ary ASK

In M-ary ASK the carrier takes $M$ amplitudes $A_i=iA$, $i=0,1,\dots,M-1$, each symbol of duration $T_s$ carrying $\log_2M$ bits. The same coherent correlator gives

$$y=a_i+n,\qquad a_i=\frac{iAT_s}{2},\qquad \sigma^2=\frac{N_0T_s}{4}$$

Levels are spaced $\Delta=\frac{AT_s}{2}$ apart, and thresholds are placed midway. The two end levels can be mistaken on one side only; the $M-2$ inner levels on both sides:

$$\begin{aligned} P_e &= \frac1M\left[(M-2)\cdot2Q\left(\frac{\Delta}{2\sigma}\right)+2\cdot Q\left(\frac{\Delta}{2\sigma}\right)\right]=2\left(1-\frac1M\right)Q\left(\frac{\Delta}{2\sigma}\right) \\ \frac{\Delta}{2\sigma} &= \frac{AT_s/2}{2\sqrt{N_0T_s/4}}=\sqrt{\frac{A^2T_s}{4N_0}} \end{aligned}$$

Average symbol energy: $E_i=\frac{i^2A^2T_s}{2}$, so

$$E_s=\frac{A^2T_s}{2M}\sum_{i=0}^{M-1}i^2=\frac{A^2T_s(M-1)(2M-1)}{12}$$

Substituting $A^2T_s=\frac{12E_s}{(M-1)(2M-1)}$:

$$P_e(\text{M-ASK})=2\left(1-\frac1M\right)Q\left(\sqrt{\frac{3E_s}{(M-1)(2M-1)N_0}}\right)$$

Check $M=2$: $P_e=Q\left(\sqrt{E_s/N_0}\right)$ with $E_s=E_b$, the binary ASK result. For larger $M$ the levels crowd together for the same average energy, so $P_e$ grows, while the bandwidth per bit falls by $\log_2M$.