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--- ps:laboratoare:04 [2019/10/20 19:05]
ionut.gorgos
+++ ps:laboratoare:04 [2020/10/07 18:26] (current)
ionut.gorgos
@@ Line 68: / Line 68: @@
   - scădeți pătratele termenilor de la $c_k, k \in \{-K...K\}$ obținând suma termenilor necesari pentru calcularea rms.
  </note>
-<hidden>
-<note tip>
-O posibilă soluție:
-<code matlab ex1.m>
-A = 1;
-T = 100;
-x = 1:T;
-s = -A*ones(1, T);
-s(1:(T/2)) = A;
-h = figure;
-plot(x,s);
-ylabel('Amplitude');
-xlabel('Time');
-ylim([-A-0.5, A+0.5]);
-title('Original signal s(t)');
-print(h, '-dpng', 'original_signal.png');
-% Compute some of the Fourier coefficients
-cmax = 21;
-kv = -cmax:cmax;
-N = length(kv);
-coef = zeros(N,1);
-for i=1:N
-    k = kv(i);
-    if mod(k,2) ~= 0
-        coef(i) = (2*A) / (j*pi*k);
-    end
-end
-% Plot power of coefficients
-h = figure;
-csquare = coef.*coef;
-stem(kv,abs(csquare));
-xlabel('Frequency component (k)');
-ylabel('Magnitude of component');
-title('Fourier coefficients squared');
-print(h, '-dpng', 'squared_coefficients.png');
-% Plot rms of error signal
-kmax = 500;
-kv = 0:kmax;
-N = length(kv);
-coef = zeros(N,1);
-for i=1:N
-    k = kv(i);
-    if mod(k,2) ~= 0
-        coef(i) = (2*A) / (j*pi*k);
-    end
-end
-rmse = zeros(kmax, 1);
-for i=1:kmax
-    ss = A^2 - 2 * coef(2:i)'*coef(2:i) - coef(1)^2;
-    rmse(i) = sqrt(ss);
-end
-h=figure;
-plot(0:(kmax-1), rmse); % Look with plot, semilogy and loglog
-title('RMS of e_k');
-ylabel('RMS(e_k)');
-xlabel('K');
-grid on;
-print(h, '-dpng', 'rms_ek.png');
-% Reconstruct signal with only some of the components
-% Note: need to use both positive and negative k's
-% The fourier coeficients  c(-k) = conj(c(k)), for real signals
-% As the original signal is odd, the fourier coeficient are completly imaginary
-% so c(-k) = -c(k)
-fk = find(rmse < 0.05);
-N = fk(1);
-sr = zeros(1, T);
-for t=1:T
-    for i=1:N
-        sr(t) = sr(t) + coef(i)*exp(j*2*pi*kv(i)*t/T) ...
-                      - coef(i)*exp(-j*2*pi*kv(i)*t/T);
-    end
-end
-h=figure;
-plot(1:T, sr);
-ylim([-A-1, A+1]);
-title('Reconstructed signal s(t)');
-ylabel('Amplitude');
-xlabel('Time');
-print(h, '-dpng', 'reconstructed_signal.png');
-</code>
-</note>
-</hidden>
 === Exercițiul 2 - filtrare [4p] ===
@@ Line 205: / Line 112: @@
     - $f_c = 10/T$ (frecvența de cut-off mult mai mare decât frecvența fundamentală => filtrare slabă)
   - Reconstruiți semnalul de output cu ajutorul seriei Fourier (folosind formula de la exercițiul 1) [1p]
-<hidden>
-<note tip>
-Ideea este să folosiți seriile Fourier pentru a determina ieșirea filtrului linear.
-<code matlab ex3.m>
-A = 1;
-D = 20;
-T = 100;
-x = 1:T;
-s = zeros(1,T);
-s(1:D) = A;
-h = figure;
-plot(x,s);
-ylabel('Amplitude');
-xlabel('Time');
-ylim([-A-0.5, A+0.5]);
-title('Original signal s(t)');
-print(h, '-dpng', 'pulse_signal.png');
-% Compute some of the Fourier coefficients
-cmax = 20;
-kv = 0:cmax;
-N = length(kv);
-coef = zeros(N,1);
-for i=1:N
-    k = kv(i);
-    coef(i) = (D/T)*A*exp(-j*pi*k*D/T)*sinc(k*D/T);
-end
-% Plot magnitude of pulse's coefficients
-h = figure;
-stem(kv, abs(coef));
-xlabel('Frequency component (k)');
-ylabel('Magnitude of component');
-title('Input Fourier coefficients');
-hold on;
-plot(kv, abs(coef), 'r--');
-print(h, '-dpng', 'coef_pulse.png');
-% Compute coefficients of output after lp filter
-fcoef = zeros(N,1);
-%fc = 0.1/T;
-%fc = 1/T;
-fc = 10/T;
-RC = 1/(2*pi*fc);
-for i=1:N
-    k = kv(i);
-    fcoef(i) = coef(i) / (1+j*2*pi*RC*k/T);
-end
-% Plot magnitude of output's coefficients
-h = figure;
-stem(kv, abs(fcoef));
-xlabel('Frequency component (k)');
-ylabel('Magnitude of component');
-title('Output Fourier coefficients');
-hold on;
-plot(kv, abs(fcoef), 'r--');
-print(h, '-dpng', 'coef_out.png');
-%Construct output signal with the computed coefficients
-sr = zeros(1, T);
-for t=1:T
-    for i=1:N
-        if kv(i) ~= 0
-            sr(t) = sr(t) + fcoef(i)*exp(j*2*pi*kv(i)*t/T) ...
-                    + conj(fcoef(i))*exp(-j*2*pi*kv(i)*t/T);
-        else
-            sr(t) = sr(t) + fcoef(i)*exp(j*2*pi*kv(i)*t/T);
-        end
-    end
-end
-h=figure;
-plot(1:T, sr);
-ylim([-A, A]);
-title('Output filtered signal s(t)');
-ylabel('Amplitude');
-xlabel('Time');
-print(h, '-dpng', 'output_signal.png');
-</code>
-<note>
-If time allows and some of the students are interested, it would be nice to make a SPICE simulation
-(e.g. on ngspice.com) with a pulse signal and an RC low-pass filter to check that the output
-of the simulation is similar to what we did in MATLAB.
-Here is an example of what to write in ngspice to get the "shark teeth" (for fc=ft):
-<code>
-*RC Filter Transient example
-V1 in 0 PULSE(0 1 1ns 1ns 1ns 40us 200us)
-R1 in out 3.2k
-C1 out 0 10n
-.TRAN 10n 1m
-.end
-</code>
-Change the capacitor value to modify the effect of the filter (smaller values mean a weaker filtering, while larger values mean a stronger filtering).
-</note>
-</note>
-</hidden>
 === Exercițiul 3 - comunicație digitală [Bonus] ===
@@ Line 325: / Line 128: @@
 Notă: pentru a genera o secvență random de valori întregi inspectați funcția 'randi' din MATLAB.
-<hidden>
-<note tip>
-Posibila solutie:
-<code matlab>
-%% Frquency modulation using two frequencies
-close all;
-k1 = 1;
-k2 = 2;
-T = 1;
-f1 = k1/T;
-f2 = k2/T;
-t = 0:T/100:T;
-s1 = sin(2*pi*f1*t);
-s2 = sin(2*pi*f2*t);
-values = randi([0 3], 1, 10);
-nv = length(values);
-s = [];
-for i=1:nv
-    si = zeros(1,length(s1));
-    if mod(values(i), 2) == 1
-        si = si + s1;
-    end
-    if values(i) > 1
-        si = si + s2;
-    end
-    s = [s, si];
-end
-tt = length(s);
-x = ((1:tt) - 1) * nv / tt;
-h1 = figure;
-plot(x, s);
-grid on;
-xlabel('Time (s)');
-ylabel('Amplitude');
-title('Frequency modulated signal');
-print(h1, '-dpng', 'freqmod.png');
-fprintf('The transmitted values are:\n');
-values
-</code>
-</note>
-</hidden>
 /*</hidden>*/

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