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For AWGN channel, 16-PSK requires less signal energy than the 16-FSK for the fixed noise power spectrum density and symbol error rate.

提问人：网友mick368 发布时间：2022-01-07

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第1题

the signal s（t) shown in figure is transmitted through an awgn channel with noise psd n0/2

The signal s(t) shown in figure is transmitted through an AWGN channel with noise PSD N0/2. Suppose a filter with impulse response h(t) is used at the receiver, the filter output should be:

A、

B、

C、

D、

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第2题

the signal s（t) shown in figure is transmitted through an awgn channel with noise psd n0/2

The signal s(t) shown in figure is transmitted through an AWGN channel with noise PSD N0/2. Suppose a matched filter with impulse response h(t) is used at the receiver, the matched filter output should be:

A、

B、

C、

D、

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第3题

For AWGN channel, QPSK with Gray mapping can achieve the same bit error rate as a coherent BPSK in white Gaussian noise, but the bandwidth efficiency of BPSK is twice that of QPSK.

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第4题

in a frequency selective fading channel, the received signal can be expressed by

In a frequency selective fading channel, the received signal can be expressed by, where x(t) and n(t) denote the transmit signal and AWGN, anddenotes the fading coefficient.

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第5题

the signal s（t) shown in figure is transmitted through an awgn channel with noise psd n0/2

The signal s(t) shown in figure is transmitted through an AWGN channel with noise PSD N0/2. Suppose a matched filter with impulse response h(t) is used at the receiver (coefficient K=1), the peak pulse signal-to-noise ratio is:

A、18/N0

B、12/N0

C、8/N0

D、4/N0

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第6题

In a flat fading channel, the received signal is w...

In a flat fading channel, the received signal is writen aswhere x(t) and n(t) denote the transmit signal and the AWGN, respectively, andis the fading coefficient. Assume that the receiver has got the ideal channel estimate denoted byalso. a) Design a linear equalizer. b) Write the output of the equalizer.

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第7题

The signal s（t) shown in figure is transmitted through an AWGN channel with noise PSD N0/2. Suppose a matched filter with impulse response h（t) is used at the receiver, the matched filter h（t) should

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第8题

Comparing QPSK and BPSK transmitted over AWGN channel, _______________.

A、QPSK can achieve the same bit error rate as BPSK.

B、QPSK can achieve the same symbol error rate as BPSK.

C、the bandwidth efficiency of QPSK is the same as that of BPSK.

D、the bandwidth efficiency of QPSK is twice that of BPSK.

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第9题

Regarding the ML decision rule for signal transmissions over AWGN channel, the statements _______________ are correct.

A、The ML decision rule is to choose the message point closest to the received signal point.

B、The ML decision rule always equals to the MAP rule.

C、The MAP and ML decision rules can be conditionally equivalent.

D、The ML decision rule can minimize the probability of error.

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第10题

write matlab-based codes to implement a 16qam system. the channel is ideal awgn channel. the modulation and demodulation should be implemented. the snr defined as

Write MATLAB-based codes to implement a 16QAM system. The channel is ideal AWGN channel. The modulation and demodulation should be implemented. The SNR defined asis assumed to change from 0 dB to 10 dB with step 2 dB. The BER-versus SNR curve based on statistics should be drawn, and should be compared with that by using the theoretical formula.

A、number_Frame=200; N_bits=400; Max_SNR=10; step=2; errs=zeros(1, fix(Max_SNR/step)+1); %___________________________________________________ for nframe=1:number_Frame nframe s_bits = round(rand(1, N_bits)); % info. bits Eav=1; d=sqrt(Eav./10); sig=qam16mapping(s_bits,d); A_WGN=randn(1,N_bits/4)+j.*randn(1,N_bits/4); %___________________________________________ nEN=0; for EbN0db= 0:step:Max_SNR snr=10.^(EbN0db/10); nEN=nEN+1; Eb_av=Eav/4; sigma =sqrt(Eb_av /(2*snr)); r = sig+sigma*A_WGN; de_bits=(hard_demap(r,d)+1)./2; err(1,nEN) = length(find(de_bits~=s_bits)); end % EbN0db %__________________________________________________ errs = errs + err; end %nframe ber= errs/number_Frame/N_bits; %________________________________________________________ snr_db=[0:step:Max_SNR]; snr=10.^(snr_db./10); M=16; xqam16=sqrt(3.*log2(M)./(M-1).*snr); SER_T=4.*(1-1./sqrt(M)).*Qfunct(xqam16); BER_T=SER_T./log2(M); %________________________________________________________ i=0:step:Max_SNR; figure semilogy(i,ber,’r-’,i,BER_T,’:*’); xlabel(’E_{b}/N_{0}’) ylabel(’BER’) legend(’Monte Claro’,’Theoretical result’); grid on function sig=qam16mapping(msg, d) N_bit=length(msg); mapping =[ d, d; d, 3*d; 3*d, d; 3*d, 3*d;... d, -d; d, -3*d; 3*d, -d; 3*d, -3*d;... -d, d; -d, 3*d; -3*d, d; -3*d, 3*d;... -d, -d; -d, -3*d; -3*d, -d; -3*d, -3*d]; %matrix 16 X 2 %--------------------------------------------------------------------- %qam modulation dsource=[]; sig=[]; for i=1:4:N_bit temp=[msg(i),msg(i+1),msg(i+2),msg(i+3)]; s_index = int_state(temp ); dsource=[dsource,s_index+1]; end %dsouce element in{1,...,16} N_QAM=length(dsource); %(N_bit/4) for i=1:N_QAM sig1=mapping(dsource(i),1); sig2=mapping(dsource(i),2); sig=[sig,sig1+j.*sig2]; % symbol row vector with N_bit/4 16QAM symbols end function hardbits=hard_demap(r,d) N_symbol=length(r); hardbits=[]; sbits=zeros(1,4); for k=1:N_symbol sk=r(k); sbits(1)=-real(sk); sbits(2)=-imag(sk); sbits(3)=abs(real(sk))-2.*d; sbits(4)=abs(imag(sk))-2.*d; hardbits=[hardbits,sign(sbits)]; end