N = 8; M = 4; d = [ 1 zeros(1,N-1) ]; dm = [zeros(1,M) 1 zeros(1,N-M-1)]; subplot(2,1,1); stem(d); axis([1 10 0 1.5]); title('unit sample function, N=8'); xlabel('n'); ylabel('Unit Sample'); subplot(2,1,2); stem(dm); axis([1 10 0 1.5]); title('delayed unit sample function, N=8 M=4'); xlabel('n') ylabel('Delayed Unit Sample')
N = 10; Usample = [ 1 zeros(1,N-1) ]; Ustep = [ 0 ones(1,N-1) ]; Uramp = zeros(1, N); for i=1:N, Uramp(i) = i-1; end; subplot(1,1,1); hold on; stem(Usample, 'r'); stem(Ustep, 'g'); stem(Uramp, 'b'); hold off; legend('sample', 'step', 'ramp'); axis([0 10 0 10]); title('Unit Sample, Unit Step, and Unit Ramp'); xlabel('n'); ylabel('value');
A=8; L=30; N=11; X = 1:1:L; subplot(2,1,1); Y=A*sawtooth(2*pi/N*X); stem(X,Y); xlabel('Time in 0.0001sec'); ylabel('Amplitude'); subplot(2,1,2); Y=A*square(2*pi/N*X); stem(X,Y); xlabel('Time in 0.0001sec'); ylabel('Amplitude');
% Explanation of results: % The sampling theorem states that the sampling frequency must be % at least twice the frequency of the input to be able to recreate % the sampled signal. In this case, only the 3Hz can be recreated % from the 10Hz sampling, and the sampled 13Hz signal looks identical to the % to the 3Hz signal, which is called aliasing. The 7 Hz digitized signal % still bears some similarities to the original, but by the sampling % theorem it cannot be always correctly reproduced from the digitized data. FS = 1000; t = 0:1/FS:1/3; Y1 = sin(2*pi*3*t); Y2 = sin(2*pi*7*t); Y3 = sin(2*pi*13*t); clf; subplot(2,1,1); plot(t,Y1,t,Y2,t,Y3, '--r'); FS = 10; tn = 0:1/FS:1/3; Y1 = sin(2*pi*3*tn); Y2 = sin(2*pi*7*tn); Y3 = sin(2*pi*13*tn); subplot(2,1,2); hold on; stem(tn, Y1, 'b'); stem(tn, Y2, 'g'); stem(tn, Y3, '--r'); hold off;
% Explanation of Quantization % The 'round' function rounds to the nearest integer. The value of the % sinusoid is first multiplied by 2^(n-1) which normalizes it to the % maximum value representable by number of bits used. (n-1) is used instead of % n so that both positive and negative numbers are represented. Then the result % is rounded to the nearest integer, and scaled back down, resulting in an % discrete quantization of the signal, as it would be if n bits were used. % The result is clear from the graph. In fact for 3 bits we get 9 % quantization levels instead of 8, since the two's complement format is % not quite fully expressed in this rounding method. Normally in real % hardware, the range of possible values would be -4 to +3, so in fact % only 8 quantization levels would exist. clf; w = 1; n = 3; FS = 10; t = 0:1/FS:2*2*pi; y = sin(w*t); yd = round(y*2^(n-1))/(2^(n-1)); plot(t,y, t, yd); title('Quantization of a 1rad/sec sine wave');
N P = lim ( 1/(2N+1) * sum(x_n^2) N->inf -N
Since we have a discrete sequence, we cannot calculate using the infinite limit, so we just sum from 0 to N. The result is P = 0.49993517713030
Using the other equation for the power A^2/2 where A is the amplitude of the sinusoid, we get P = 0.5
clf;
format long;
w = 1;
FS = 100;
N = 2*pi*10;
t = -N:1/FS:N;
y = sin(w*t);
plot(t,y);
N = length(y)/2;
P = sum(y.^2)/(2*N+1)
P = 0.49993517713030
% We calculate the SQNR using two equations % SQNR_1 = 1.76 + 6.02*n % SQNR_2 = 10*log10( sum(y^2)/sum(err^2) ) % (courtesy of brian park's lab group) % It is clear that the two equations essentially % give the same results, observable from the values % displayed in the graph captions. % recalculate part 5 values clf; hold off; w = 1; FS = 10; t = 0:1/FS:2*2*pi; y = sin(w*t); sqnr1 = zeros(1,8); sqnr2 = zeros(1,8); sqnrx = zeros(1,8); format short; for n=1:8, sqnrx(n) = n; yd = round(y*2^(n-1))/(2^(n-1)); subplot(4,2,n); plot(t,yd, t, yd-y); sqnr1(n) = 1.76+6.02*n; sqnr2(n) = 10*log10(sum(y.^2)/sum((yd-y).^2)); title(sprintf('Using %d Bits, SQNR_1=%.1f SQNR_2=%.1f', n, sqnr1(n),sqnr2(n))) xlabel('n'); end;
% The SQNR increases with an increase in the number of % bits used to digitize the signal. This is expected since % the more bits used results in a much more precise digitization % of the actual signal, and hence the noise that arises due to the % quantization process is reduced relative to the original signal. subplot(1,1,1); plot(sqnrx, sqnr1, 'go', sqnrx, sqnr2, 'xr'); legend('SQNR=1.76 + 6.02*n', 'SQNR_2 = 10*log10( sum(y^2)/sum(err^2) )'); xlabel('Bits (n)'); ylabel('SQNR (dB)'); title('SQNR vs Number of Bits Used');
