--- title: "Coherent ASK and PSK error; ASK needs double power" subject: "Communication Systems" chapter: 9 key: "semester_6/communication_systems:ask-psk-error-double-power" source: https://www.bctengineer.com/q/semester_6/communication_systems/ask-psk-error-double-power site: BCT Engineering --- # Coherent ASK and PSK error; ASK needs double power Communication Systems · Chapter 9: Noise in Communication Systems - Asked in: 2072 Chaitra (CS II), 4+3 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/communication_systems/ask-psk-error-double-power - Whole chapter: https://www.bctengineer.com/md/semester_6/communication_systems/chapter-9 ## Question Find the error probability in coherent ASK and PSK detections and show that ASK requires double the average signal power than PSK for same error probability. ## Short answer Coherent ASK has Pe = 0.5 erfc(√(Eb/2N0)), while coherent PSK has Pe = 0.5 erfc(√(Eb/N0)). Equating arguments gives Eb,ASK = 2 Eb,PSK for equal error, so ASK requires twice (3 dB more) average signal power than PSK. ## Answer #### 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):** ```text 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 $\lambda=0$. ```text 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. ```text ASK: *0 ------------ *sqrt(2Eb) d = sqrt(2Eb) PSK: *-sqrt(Eb) -- 0 -- *+sqrt(Eb) d = 2 sqrt(Eb) ``` ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. The coherent ASK error probability in terms of E_b (average energy per bit) is: A. (1/2) erfc( sqrt(E_b/N0) ) B. (1/2) erfc( sqrt(E_b/2N0) ) C. (1/2) erfc( sqrt(2E_b/N0) ) D. (1/2) erfc( sqrt(E_b/4N0) ) Answer: B. The argument is sqrt(A^2T_b/8N0), and with E_b = A^2T_b/4 this is sqrt(E_b/2N0). (Key point: ASK error probability) 2. For coherent BPSK (E_b = A^2 T_b/2), the error probability is: A. (1/2) erfc( sqrt(E_b/N0) ) = Q( sqrt(2E_b/N0) ) B. (1/2) erfc( sqrt(E_b/2N0) ) C. Q( sqrt(E_b/N0) ) D. (1/2) exp(-E_b/N0) Answer: A. Levels ±AT_b/2 with σ^2 = N0T_b/4 give erfc argument sqrt(A^2T_b/2N0) = sqrt(E_b/N0). (Key point: PSK error probability) 3. Equating the erfc arguments for equal P_e gives which relation? A. E_b,ASK = E_b,PSK / 2 B. E_b,ASK = E_b,PSK C. E_b,ASK = 2 E_b,PSK D. E_b,ASK = 4 E_b,PSK Answer: C. E_b,ASK/2N0 = E_b,PSK/N0 gives E_b,ASK = 2E_b,PSK, i.e. 3 dB more average power for ASK. (Key point: Why ASK needs double power) 4. For the same average energy E_b, the distance between signal points is: A. Larger for ASK: 2 sqrt(E_b), versus sqrt(2E_b) for PSK B. The same for both C. Smaller for ASK: sqrt(2E_b), versus 2 sqrt(E_b) for PSK D. Zero for PSK Answer: C. PSK points at ±sqrt(E_b) are 2sqrt(E_b) apart; ASK points 0 and sqrt(2E_b) are closer, so ASK errs more. 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