Millores en les t` ecniques i els resultats

1. In order to generate the signal plots, we use the following MATLAB program

%---Chap. 19 Prob 1---s=[1 4 3 2 1 3 1 2]

h=[2 0 2 -1];

N=length(s); %# of OFDM tones L=length(h)-1; % length of CP sc=[s(N-L+1:N),s]; % adding CP yc=conv(sc,h); % channel y=yc(L+1:L+N); % CP deletion Y=fft(y);

%---Get H(k)’s---H=zeros(1,8);

for k=1:N H(k)=0;

for i=1:length(h)

H(k)=H(k)+h(i)*exp(-j*2*pi*(i-1)*(k-1)/N);

end end

S_hat=Y./H;

s_hat=real(ifft(S_hat));

figure(1)

subplot(2,2,1);stem(s);ylabel(’s[n]’);

subplot(2,2,2);stem(sc);ylabel(’sc[n]’);

subplot(2,2,3);stem(yc);ylabel(’yc[n]’);

subplot(2,2,4);stem(y);ylabel(’y[n]’);

figure(2)

subplot(2,2,1);stem(abs(Y));ylabel(’Y[k]’);

subplot(2,2,2);stem(abs(S_hat));ylabel(’S-hat[k]’);

subplot(2,2,3);stem(s_hat);ylabel(’s-hat[n]’);

1 2 3 4 5 6 7 8

Figure 19.1: Signal vectors for OFDM system from Exercise 19.1: original signal; transmit signal with cyclic prefix, received signal; received signal after stripping off cyclic prefix.

1 2 3 4 5 6 7 8

Figure 19.2: Signals for OFDM system from Exercise 19.1: received signal in frequency domain; signal in frequency domain after equalization; final signal in time domain.

(a) The block diagram of a baseband OFDM system can be found in Fig. B-19.4.

(b) The minimum CP length should be 4 − 1 = 3. Observing that ˆs[n] = s[n], the transmitted data can be fully recovered, because of the implementation of cyclic prefix. In another word, there is no intercarrier interference due to frequency selectivity. At the baseband, the receiver discards the first 3 received samples.

(c) The signal vectors are plotted in Figs. 19.1-19.2.

2. We need to find the backoff level that limits the probability of exceeding the desired amplitude level to x%. As mentioned in Sec. 19.6, the distribution of in-phase and quadrature-phase amplitudes are Gaussian. The distribution of the total amplitude is thus Rayleigh

pdf (r) = r

Since the average mean power is unity, the σ in each of the branches must be σ = 1/√

2, and the cdf can be written as

cdf (r) = 1 − exp

Since we require that the amplitude does not exceed a level A0, we find that

which is 1.517, 2.146, 2.628, respectively.

3. Let the block X_k⁽ⁱ⁾ be the i-th block of the transmission, with k indexing the N complex modulation symbols. The transmit signal in the time domain x⁽ⁱ⁾[n] can then be written as

x⁽ⁱ⁾[n] = 1 N

NX−1 k=0

X_k⁽ⁱ⁾e^j2πNnk , −G ≤ n < N (19.5) where G is the length of the CP; if the CP is absent, we simply set G = 0. The channel performs the convolution with the impulse response of the channel h[n, l], which consists of L multipath components, 0 ≤ l < L, and is time-variant, as reflected by the double indexing. The received signal y⁽ⁱ⁾[n] (without noise added by the channel) is then

y⁽ⁱ⁾[n] = (h ∗ x⁽ⁱ⁾)[n] , (19.6)

As the CP is discarded by the receiver, we henceforth consider only 0 ≤ n < N. The first L − G − 1 samples of y⁽ⁱ⁾[n] are distorted by ISI from the previous symbol x⁽ⁱ⁻¹⁾, thus y⁽ⁱ⁾ consists of two parts,

y⁽ⁱ⁾[n] =

where σ is the Heaviside function, and y^(i,i)[n] and y^(i,i_i ⁻¹⁾[n] are the contributions to y⁽ⁱ⁾[n] stemming from the i−th and i − 1-th block, respectively. To reconstruct the sent symbols, the receiver performs a DFT on y⁽ⁱ⁾[n], Y_k⁽ⁱ⁾= DFT {y⁽ⁱ⁾[n]}[k] where again k is the subcarrier index. Throughout the paper we assume perfect synchronization of carriers and blocks, and further, if not otherwise stated, absence of noise.

We now write the Discrete Fourier Transform of Eq. (19.7) in the receiver as Y_k⁽ⁱ⁾=

Y_k^(i,i) and Y_i,k^(i,i−1) are again the parts of Y_k⁽ⁱ⁾ that originate from the own symbol i and the previous symbol i − 1, respectively. If we use (19.5), these parts can then be written as

Y_k^(i,i) =

Figure 19.3: Signal waveforms for an 8-point MC-CDMA system (Exercise 19.4).

where

H_k,m^(i,i)= 1 N

XN n=0

LX−1 l=0

h[n + iN, l]e^{j2πN(nm−lm−nk)}σ[n − l + G] (19.11) and

H_k,m^(i,i−1)= 1 N

XN n=0

L−1

X

l=G+1

h[n + (i − 1)N, l]e^{j2πN(nm−lm−nk)} (19.12) This can be written in a compact form as a vector-matrix-product Y⁽ⁱ⁾ = Y^(i,i)+ Y^(i,i−1) = H^(i,i)· X⁽ⁱ⁾+ H^(i,i−1)· X⁽ⁱ⁻¹⁾ .where Y^(i,i−1)is the ISI term and Y^(i,i)contains the desired data disturbed by ICI. Note that if X⁽ⁱ⁻¹⁾ was detected successfully (and the channel is completely known, as we always assume), Y^(i,i−1) can be computed and subtracted from Y⁽ⁱ⁾.

4. (a) The plots of the various signal components can be seen in Fig. 19.3.

(b) The average noise enhancement with a zero-forcing receiver can be computed as µ1

8 X 1

|Hⁱ|²

¶ µ1 8

X|Hⁱ|²

¶

= 11.27 = 10.5 dB (19.13)

(c) Simulation of the system with zero-forcing equalization is straightforward, though care must be taken about the correct normalizations in the spreading and despreading operation. A simulation plot is shown in Fig.

5. For the evaluation of this problem, we apply Eq. (19.14) SIN R =

ES

N0P_sigN^N

c p+N ES

N0Psig N Nc p+N

PI S I+PI C I

Ps i g + 1 . (19.14)

where Psig, PICI, and PISIare given by Eqs. (B-19.21) - (B-19.23). Since the Doppler effect is negligible, the function R(k, l) becomes the power delay profile, sampled at the OFDM sampling instants k · 1µs

R(k, l) = exp (−k/16) (19.15)

Inserting this, and Eq. (B-19.20) into Eqs. (B-19.21) - (B-19.23), we obtain the values for Psig, PICI, and PISI. The result is shown in Fig. 19.5.

Figure 19.4: BER of the OFDM system with (solid) and without (dotted) WH transformation. See Exercise 19.4.

Figure 19.5: SINR as a function of the duration of the cyclic prefix for SN R = 8 dB (Exercise 19.5).

6. The capacity for waterfilling is given by Eq. (B-19.31), with the power constraint Eq. (B-19.32). We find by inspection that with α²_n= 1, 0.1, 0.01, then ε = 55.5, so that P1= 54.5, P2= 45.5, and P3= 0.

Thus, the third channel is not selected by the system. The SNRs in the two used channels are then 54.5 and 45.5. From this, we find that that the capacity in the first channel is

log₂(1 + 54.5) = 5.79 bit/s/Hz (19.16)

while for the second channel it is

log₂(1 + 54.5) = 5.54 bit/s/Hz (19.17)

and the total capacity is 11.33 bit/s/Hz.

When using BPSK, each of the channels can achieve at most 1 bit/s/Hz. Thus, the total capacity is limited to 2 bit/s/Hz if waterfilling is used. When equal power allocation is used, 3 bit/s/Hz can be used.

7. The key to this exercise is to recognize that the different carriers are fading independently, and thus the behavior of the coded system is the same as a properly interleaved block-coded single-carrier system in a flat-fading channel. Thus, Eq. (B-14.52) and the subsequent discussion applies without modifications.

A diversity order of 4 can be achieved, so that the BER is proportional to 1/γ⁴.

8. The spectral efficiency of BPSK with rasied-cosine spectrum follows from Eq. (B-11.5). We find that the spectrum occupies a band of width

∆B = (1 + α)/TB (19.18)

where α is the rolloff factor; no energy exists outside that band (of course, that is an idealization; in practical circumstances, the rolloff is never perfect). Spacing the carriers at ∆B then ensures complete orthogonality. For OFDM, the spacing between the carriers is 1/TB, which also guarantees perfect orthogonality, even though the bands on the different carriers do overlap. It thus seems that OFDM shows the better spectral efficiency. For this comparison, we disregard the possible introduction of a cyclic prefix: measures fighting delay dispersion have to be implemented in any system.

Chapter 20