Coherent ASK and PSK error; ASK needs double power

Coherent 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 $E_1$: $P_e=\frac12\operatorname{erfc}\sqrt{E_1/4N_0}$.)

Coherent PSK

BPSK signals over $0\le t\le T_b$ (antipodal):

$$s_1(t)=A\cos\omega_ct\ \ (1),\qquad s_2(t)=-A\cos\omega_ct\ \ (0),\qquad E_b=\frac{A^2T_b}{2}$$

Coherent receiver: multiply by the locally generated $\cos\omega_ct$, integrate over $T_b$, sample, compare with threshold $\lambda=0$.

 r(t) -->(x)--> integrate --> sample --> y > 0 ? 1 : 0
          ^       0..Tb       t=Tb
          |
       cos(wc t)  (from carrier recovery)

Correlator output $y=a_i+n$ with

$$a_1=+\frac{AT_b}{2},\qquad a_2=-\frac{AT_b}{2},\qquad \sigma^2=\frac{N_0T_b}{4}$$

The pdfs of $y$ are Gaussian centred at $\pm AT_b/2$. Error when 0 is sent ($y>0$):

$$\begin{aligned} P_e &= \int_0^{\infty}\frac{1}{\sqrt{2\pi}\sigma}e^{-(y+AT_b/2)^2/2\sigma^2}dy=\frac12\operatorname{erfc}\left(\frac{AT_b/2}{\sqrt2\,\sigma}\right) \\ &= \frac12\operatorname{erfc}\left(\frac{AT_b/2}{\sqrt{N_0T_b/2}}\right)=\frac12\operatorname{erfc}\left(\sqrt{\frac{A^2T_b}{2N_0}}\right) \end{aligned}$$

By symmetry the error for a sent 1 is the same, so

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

ASK needs double the power of PSK

Comparison (same $P_e$). ASK: $P_e=\frac12\operatorname{erfc}\sqrt{E_b/2N_0}$; PSK: $P_e=\frac12\operatorname{erfc}\sqrt{E_b/N_0}$. For equal $P_e$ the arguments must be equal:

$$\frac{E_{b,ASK}}{2N_0}=\frac{E_{b,PSK}}{N_0}\ \Rightarrow\ E_{b,ASK}=2E_{b,PSK}$$

Since average power $=E_b/T_b$, ASK needs twice (3 dB more) average signal power than PSK for the same error probability. Physically, the PSK signal points are $2\sqrt{E_b}$ apart, while ASK points are only $\sqrt{2E_b}$ apart for the same average energy.

 ASK:  *0 ------------ *sqrt(2Eb)   d = sqrt(2Eb)
 PSK:  *-sqrt(Eb) -- 0 -- *+sqrt(Eb) d = 2 sqrt(Eb)