Bit error probability for binary ASK

Binary ASK (on-off keying) sends a carrier burst for "1" and nothing for "0". With coherent (matched-filter/correlator) detection in AWGN, the bit error probability is $P_e = Q\left(\sqrt{E_b/N_0}\right)$, where $E_b$ is the average energy per bit.

Assumptions

1. Signals over $0 \le t \le T_b$:

$$s_1(t) = A\cos\omega_c t \ \ (\text{"1"}), \qquad s_0(t) = 0 \ \ (\text{"0"})$$

with $f_c$ an integer multiple of $1/T_b$. 2. Channel adds white Gaussian noise $n(t)$ of two-sided PSD $N_0/2$. 3. Symbols are equally likely; carrier phase and bit timing are known (coherent detection).

Receiver

 r(t) --->(X)--->[ integrate 0..Tb ]--> y --> [ y > Vth ? ] --> 1/0
           ^                                   sample at Tb
     cos(wc t)  (local coherent carrier)

Correlator output

$$y = \int_0^{T_b} r(t)\cos\omega_c t\,dt$$

• For "1": signal part $= \int_0^{T_b}A\cos^2\omega_c t\,dt = \dfrac{AT_b}{2}$
• For "0": signal part $= 0$

Noise at the output: $n_o = \int_0^{T_b} n(t)\cos\omega_c t\,dt$ is Gaussian, zero mean, with variance

$$\sigma^2 = \frac{N_0}{2}\int_0^{T_b}\cos^2\omega_c t\,dt = \frac{N_0T_b}{4}$$

Threshold. For equal priors and equal variances, the optimum threshold is midway: $V_{th} = AT_b/4$.

Error probabilities

$$\begin{aligned} P(e|0) &= P\left(n_o > \tfrac{AT_b}{4}\right) = Q\!\left(\frac{AT_b/4}{\sigma}\right) \\ P(e|1) &= P\left(n_o < -\tfrac{AT_b}{4}\right) = Q\!\left(\frac{AT_b/4}{\sigma}\right) \end{aligned}$$ $$P_e = \tfrac12 P(e|0) + \tfrac12 P(e|1) = Q\!\left(\frac{AT_b/4}{\sqrt{N_0T_b/4}}\right) = Q\!\left(\sqrt{\frac{A^2T_b}{4N_0}}\right)$$

In terms of energy. Energy of the "1" pulse $E_1 = A^2T_b/2$; the "0" has zero energy, so average energy per bit $E_b = A^2T_b/4$. Hence

$$P_e = Q\!\left(\sqrt{\frac{E_b}{N_0}}\right) = Q\!\left(\sqrt{\frac{E_1}{2N_0}}\right) = \frac12\,\text{erfc}\!\left(\sqrt{\frac{E_b}{2N_0}}\right)$$

Remarks

• Coherent BPSK gives $Q(\sqrt{2E_b/N_0})$, so ASK needs 3 dB more average energy for the same $P_e$.
• Non-coherent (envelope) ASK gives approximately $P_e \approx \tfrac12 e^{-E_b/2N_0}$, slightly worse but simpler.