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Unformatted text preview: ECE 155a Homework Assignment Dr. Anthony Weathers Assigned: llf7f2007 phone: 8583806803 Due: l 1/141'2007 email: [email protected] 1) Given the channel model f(D) = l + D — 2D 2 ,
(a) Draw the ﬁnite state machine representation which describes the inputr’output relationship of the channel,
(b) Draw one stage of the corresponding trellis. 2) Using the channel model f(D) = 1+ D — ZD2 , detect the sequence R = (0.3,l.8,0.4) by labeling the branches of the following diagram with the appropriate branch metrics and placing the results of the metric calculations at each stage in
the appropriate squares. (You may use Matlab or some other programming language to calculate these quantities in order to reduce the amount of work required.) Indicate the survivor paths at each stage and give the data sequence which corresponds to the survivor path that has the smallest metric after the last stage. 3) Find an optimal equalizer, using the technique discussed in class. Using delay values from d=0 to d=4, design an
equalizer for each delay value and plot the equalized signal overlaid with the ideal samples, then indicate the one that gives the best result. The parameters are given below along with the transition response. Assume the bit period is
T=l. randl'state‘,0l; % ini:ialize random number generator N = 20; % data block length a = roundlrand(1,N)); % data sequence h = [1 O —l]; % partial response polynomial coefficients
L = 5; % equalizer length k = OzN—l; % time indiees tor data t = {l—L:lengthla}—1l; % time indices for received samples function y=g(t)
pw50=2;
y = 1 ./ ll2*t/pw50l.*2 +1); end [a is — Small C01: pxiier 1 C T Sig T { RR? it 11/16/07 11:23 AM C:\Documents and Settings\Anthonx Weathers\My Documents\matlab\ece1553.m l of 2 function ece155a % Homework Solution, Prob 3.
% Fall Quarter, 2007 % Anthony Weathers % This part given in homework assignement
%%%%%%%%%%%%%%%%%%%%%% rand('state',0); % initialize random number generator N = 20; % data block length a = round(rand[l.NJ}: % data sequence h = [1 0 l]; % partial response polynomial coefficients
L = 5; % equalizer length R = 0:N—l; % time indices for data t = (1L:length(a)1); % time indices for received samples
%%%%%%%%%%%%%%%%%%%%%% % calculate received samples {no noise added}
r = zerostsize(t)};
for j=1:1ength(a) r r + a(j)*(g(t*j)*g(tijsl)}; end s = filter(h,1,a); % ideal output samples
toeplitz{r[L:—l:1),r{L:endJl; % form matrix
R'*inv(R*R'); % calculate psuedo—inverse NW
II II % find coefficients and error for each value of delay figure; err = zeros(l,L]; cﬂlist = zeros(L,L); for delay=0:4
sd = [zeros[1,delay),s(1:end—delay}]; % delayed s, with 3:0 for t<0
c = sd*Z; % calculate filter coefficients
y = 0*R; % calculate equalized received waveform
err(delay+1) = norm{ysd)“2; % save sqared error for each delay
c_list(delay+1,:) = c; % save coefficients % plot results
eubplott3,2,delay+1);
plot(k,y,'—x',k,a,'o',k,sd,'—x'):
title(sprintf(‘error=%g',err{delay+1))J;
ylabelisprintfi'delay=9d',delay)J; end % plot the coefficients for the best delay value
[mn,ix] = min(err}; subplot{3,2,6): stem(c_list(ix,:)); title('Best Coefficients']; end function y=g(t) pw50=2; y = 1 ./ [(2*t/pw50) .A2 +1);
end error=1 .36822 error=0.0280432 ﬂow =3 delay 2 Best Coefficients
1 r 0 ...
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 Winter '07
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