Fourier Transform

hey guys i am kinda new to matlab, and would really appreciate it if you guys help me thanx so much

Generate the following function

      y1(t)=2*sin(2*pi*1*t)+4*sin(2*pi*2*t)+3*sin(2*pi*4*t)
      y2(t)=2*cos(2*pi*1*t)+4*cos(2*pi*2*t)+3*cos(2*pi*4*t)
      First, required by sampling theorem, the step size of t  (delta t) has to be smaller than a value. Figure out this value and set the delta t at least 3 times smaller than this value and calculated y1(t) and y2(t). Increase the step size and recalculate y1(t) and y2(t). See the difference. 
      Calculate Fourier Transform of y1(t) and y2(t) using ‘fft’ and plot your results. Using frequency as the x axis data and amplitude as the y axis data. Also, plot the phase, too.
f=[0:1/length(y1):1/2]*(1/delta_t)
figure; plot(f, abs(y1(1:length(f))), ‘-*’)

      Use function of ‘abs’ to calculate the amplitude  and ‘angle’ to calculate phase.

[Paulo Silva]

doc fft

[zaini]

please use code writing mode, to make it more readable

[khalid]

y1(t)=2*sin(2*pi*1*t)+4*sin(2*pi*2*t)+3*sin(2*pi*4*t)

y2(t)=2*cos(2*pi*1*t)+4*cos(2*pi*2*t)+3*cos(2*pi*4*t)

does this make it better?

Answer 1

[Sulaymon Eshkabilov]

delta_t=0.001;   % Sampling time
fs=1/delta_t;    % Sampling frequency
t = 0:delta_t:pi;% Time space
N = 2048;        % Block size
y1=2*sin(2*pi*1*t)+4*sin(2*pi*2*t)+3*sin(2*pi*4*t);
y2=2*cos(2*pi*1*t)+4*cos(2*pi*2*t)+3*cos(2*pi*4*t);
Y1 = abs(fft(y1, N));
Y2 = abs(fft(y2, N));
f=(0:length(Y1)-1)*fs/length(Y1);
plot(f, Y1, 'r-', f, Y2, 'b'), grid on, legend('Y1', 'Y2', 'location', 'SouthEast')
title 'FFT of y_1(t) and y_2(t)'
xlim([0, 10])
shg

Answer 2

[Yazan Almasri]

x(t)=2*sin(2*pi*t)