tarea osman 18_11
TRANSCRIPT
syms t n
>> Xt=cos(4000*pi*t)+cos(10000*pi*t)+cos(18000*pi*t);
>> Xn=cos(4000*pi*n/18000) + cos(10000*pi*n/18000) + cos(18000*pi*n/18000);
>> T=1/1000;N=18;
>> t=linspace(0,T,1000);
>> Xt=Xt;
>> Xt=cos(4000*pi*t)+cos(10000*pi*t)+cos(18000*pi*t);
>> plot(t,Xt);
>> hold on;
>> plot(t,cos(4000*pi*t),'r');plot(t,cos(10000*pi*t),'m');plot(t,cos(18000*pi*t),'g ');
>> title('Xt=cos(4000\pit)+cos(10000\pit)+cos(18000\pit)');xlabel('t');
>>legend({'cos(4000\pit)+cos(10000\pit)+cos(18000\pit)','cos(4000\pit)','cos(10000\pit)','cos(1800
0\pit)'});
>> n=0:N;
>> Xn=cos(4000*pi.*n/18000) + cos(10000*pi.*n/18000) + cos(18000*pi.*n/18000);
>> hold off
>> stem(n,Xn);
>> title('cos(2\pin/9) + cos(5\pin/9) + cos(\pin)');xlabel('n');legend({'cos(2\pin/9) + cos(5\pin/9) +
cos(\pin'});
>> Xn=cos(2*pi*n/9) + cos(5*pi*n/9) + cos(pi*n);
>> Xn=1/2*(exp(i*2*pi*n/9)+exp(-i*2*pi*n/9))+1/2*(exp(i*5*pi*n/9)+exp(-
>> (pi);N/2*(2*pi/N);18/2*(2*pi/N);9*(2*pi/N);
Se acomodan los argumentos de los exponentes para realizar inspección con:
𝑥[𝑛] = ∑ 𝑎𝑘𝑒𝑗𝑘(
2𝜋𝑁)𝑛
𝑘=<𝑁>
;
>> [(2*pi/9)/(2*pi/N) (5*pi/9)/(2*pi/N) pi/(2*pi/N)]
ans =
2 5 9
%se compara y se llega a la conclusión de que los términos son:
𝑎𝑘 → 𝑎−2 = 𝑎2 = 𝑎−5 = 𝑎5 = 𝑎−9 = 𝑎9 =1
2
>> n_FXn=-9:9;
>> FXn=1/2*[1 0 0 0 1 0 0 1 0 0 0 1 0 0 1 0 0 0 1];
>> stem(n_FXn,FXn)
>> title('a_k de x[n]');xlabel('k');
>> Xt=cos(4000*pi*t)+cos(10000*pi*t)+cos(18000*pi*t);
>> %se descompone con formula de Euler;
>> Xt=1/2*(exp(i*4000*pi*n)+exp(-i*4000*pi*n))+1/2*(exp(i*10000*pi*n)+exp(-
i*10000*pi*n))+1/2*(exp(i*18000*pi*n)+exp(-i*18000*pi*n));
>> pretty(Xt)
exp(-4000 pi n i) exp(4000 pi n i) exp(-10000 pi n i) exp(10000 pi n i)
----------------- + ---------------- + ------------------ + ----------------- +
2 2 2 2
exp(-18000 pi n i) exp(18000 pi n i)
------------------ + -----------------
2 2
>> %se acomodan los exponentes para que queden de la forma
𝑥(𝑡) = ∑ 𝑎𝑘𝑒𝑗𝑘(
2𝜋𝑇)𝑡
+∞
𝑘=−∞
;
>> [(4000*pi)/(2*pi/T) (10000*pi)/(2*pi/T) (18000*pi)/(2*pi/T)]
ans =
2.0000 5.0000 9.0000
%se compara y se llega a la conclusión de que los términos son:
𝑎𝑘 → 𝑎−2 = 𝑎2 = 𝑎−5 = 𝑎5 = 𝑎−9 = 𝑎9 =1
2
>> FXt=1/2*[1 0 0 0 1 0 0 1 0 0 0 1 0 0 1 0 0 0 1];
>> n_FXt=-9:9;
>> stem(n_FXt,FXt)
>> title('a_k de x(t)');xlabel('k');
%Comparacion:
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Ya que