Communication Systems: Solved Assignments and Essays
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Added on 2023/06/04
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This document contains solved assignments and essays on various topics related to Communication Systems. It covers topics such as QPSK, binary communication systems, M-ary PSK modulation, and more. The assignments also include MATLAB implementations and probability of error performance analysis.
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ASSIGNMENT 2 COMMUNICATION SYSTEMS INSTITUTIONAL AFFILIATION LOCATION INSTRUCTOR (PROFESSOR) STUDENT NAME STUDENT REGISTRATION NUMBER DATE OF SUBMISSION 1
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QUESTION 1 Part A (i)H(t) as a function of A and T s(t)={A,for0≤t≤T 0,otherwise Introducing the noise signal, H(t)=s(t)+n(t)0≤t≤T η=|s0(t)|2 E[n2(t)] h(t)=F−1[H(f)] 2K N0[∫ −∞ ∞ S¿(f)ejwt0 e−jwtdf]¿ ¿2K N0 [s(t0−t)]¿ From the impulse response output, the constant value is given as, Cs=2K N0 Therefore, h(t)=Cs(t0−t) (ii)H(f), the Fourier transform of h(t) 2
H(f)=2K N0 S¿(f)e−jwt0 (iii)Maximum signal to noise ratio at the output of h(t). (iv)Show The impulse response is demonstrated by plotting the images below, Part B Section 1 QPSK transmitted with a required probability of bit error, PB 3
051015202530354045 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 Sm(t)=√(2Pr)cos[2πf0t+ϕ(m)]0≤t<T ϕ(m)ϵ{π 4,3π 4,5π 4,7π 4} Pr−receivedaveragesignalpower r(t)=sm(t)+n(t) Pr N0 ,Pr=1 T∫ 0 T sm 2(t)dt SQPSK={√Escos[(i−1)π 2]ϕ1(t)−√Essin[(i−1)]ϕ2(t)} The bit rate in either I or Q channel are ½ input data rate. Section 2 Matlab Implementation 4
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051015 SNR in db 10-10 10-8 10-6 10-4 10-2 100 probability of error Probability of Error Performance QUESTION 2 Binary communication system Part a P0∧P1=1−P0bearbitrary V→rv(conditionalprobabilitydensity) fv∨u{v|am}−finite∧non−zero∀v∈R m∈{0,1} To determine the a-posteriori probability for the input, pu∨v(am|v)=(Pmfv∨u(v|am)) fv(u) Part b It is possible to obtain the likelihood ratio of the MAP decision, 5
A(v)=fv∨u(v|a0) fv∨u(v|a1),p1 p0 =η Part c The overall probability of the error is given as, Pr{e}=p0Pr{e|U=a0}+p1{e|U=a1} P(error)=0.5Q (√Es N0 2)+0.5Q (√Es N0 2) P(error)=Q(√2Es N0)=0.5erfc(√Es N0) Part d Replacing with real values, fR∨0=√1 2πe−(5y2) fR∨1=√1 2πe−¿¿ fY/Hi(y/Hi)=∏ k=1 m 1 √2πσexp[−(yk−si,k) 2 2σ2] To obtain the error probability, Perror=(0.5)(0.86675)+(0.5)(0.56703)=0.716578 QUESTION 3 Communication system: binary information transmission 6
Part A ωc=2π Ts =N(2π Tb) ρ=A2Tb 2η Part B π0=Pr{transmitss0(t)}=1 2 π1=Pr{transmitss1(t)}=1 2 H0:r(t)=S0(t)+n(t),0≤t<T H1:r(t)=S1(t)+n(t),0≤t<T Sn(f)=N0 2,−∞<f<∞ s0(t)={A,0≤t≤3 4T 0,otherwise s1(t)={A,1 4T≤t≤T 0,otherwise ωc=2π Ts =N(2π Tb) 7
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ρ=A2Tb 2η SNRe=10log10[A2Tb 2η] The maximum data rate is given as, Eb=π0∫ 0 T s0 2(t)dt+π1∫ 0 T s1 2(t)dt To obtain the probability of error, Pb=erfc(√Eb N0)[1−1 2erfc(√Eb N0)] Part C It is observed that an increased in the rb value causes a pb value decrease. The illustration below expounds on the relationship, QUESTION 4 Part A For an M-ary PSK modulation ϕi(t)=Acos(ωct+θi)0<t≤Ts θi=0,2π M,…,2(M−1)π M 8
sϕ(ω)=A2TsSa2 [(ω−ωc)Ts 2] Tb=Ts log2M Pϵ≈2erfc√2Es ηsin2π Mm>2 Pϵ≈2erfc√2Es ηsin2π √2M Γ= sin2π M sin2π √2M -505101520 Es/No, dB 0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.2 Symbol Error Rate Symbol error probability curve for QPSK aplha=0 aplha=0.15 aplha=0.25 aplha=0.35 aplha=0.45 Part B 9
-505101520 Es/No, dB 0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.2 Symbol Error Rate Symbol error probability curve for QPSK aplha=0 aplha=0.15 aplha=0.25 aplha=0.35 aplha=0.45 QUESTION 5 Efficiency of 0.55 EIRPtransmit=10log(Ptransmitter)−Transmitterfeederloss +TransmitterAntennagain −Antennapointingloss (dBW) EIRPtransmit=10log10−0.5dB+45.21−0.70 EIRPtransmit=54.01dB Gr Ts[dB K]=−4dB/K 10
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Pr=Pt(1−|Γt|2 .gt.(λ 4πd)2 .1 αm.Aeff(1−|Γr|2 )) Lbf=20log10(4πd λ) Lbfuplink =20log10 (4π∗36000 c f)=20log10(144e6π 0.01) Lbfdownlink =20log10 (4π∗36000 c f)=20log10(144e6π 0.015) Lbfdownlink =209.5884dB Part B P1=Pr{1biterror|s1(t)} ¿Pr{s2(t)|s1(t)}+Pr{s4(t)|s1(t)}+Pr{s8(t)|s1(t)} The autocorrelation is given as, Rc(τ)=1 Tc ∫ −Tc 2 Tc 2 c(t)c(t−τ)dt 11
051015 SNR in db 10-10 10-8 10-6 10-4 10-2 100 probability of error Probability of Error Performance The receiver output is given as, d(t)c2(t)=d(t) Sk(t)=√2Eb Tmk(t)pk(t)cos(2πfct+φk) 12