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Altera_Forum's avatar
Altera_Forum
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14 years ago

dft problem

greetings,

hi i have typed followin code to verify samplng theorem but the answer is coming something different.may i know the reason.

clc;

fs=800;

t=0:1/fs:1

x=cos(2*pi*400*t);

xm=(fft(x));

k=0:length(xm)-1;

subplot(2,2,1);

stem(k,xm)

xlabel('hertz')

ylabel('magnitude')

title('sampling')

the result should come at only one frequency(i.e one line at frequency of cosine) as the signal is cosine but spectrum is continious. why ?

the spectrum (o/p)is attached here

thanks in advance

4 Replies

  • Altera_Forum's avatar
    Altera_Forum
    Icon for Honored Contributor rankHonored Contributor
    14 years ago

    You're doing:

    t=0:1/fs:1

    x=cos(2*pi*400*t);

    The step is 1/fs, which is 1/800 = 0.00125

    This means, that from range 0 to 1, every 0.00125 step You'll get a new frequency (because argument 't' in cosine function is changing), that is: 800 results. So what do You expect in frequency domain to see?
  • Altera_Forum's avatar
    Altera_Forum
    Icon for Honored Contributor rankHonored Contributor
    14 years ago

    your frequency is half sampling rate and is one line mirrored

    plot(k,abs(xm));
  • Altera_Forum's avatar
    Altera_Forum
    Icon for Honored Contributor rankHonored Contributor
    14 years ago

    --- Quote Start ---

    You're doing:

    t=0:1/fs:1

    x=cos(2*pi*400*t);

    The step is 1/fs, which is 1/800 = 0.00125

    This means, that from range 0 to 1, every 0.00125 step You'll get a new frequency (because argument 't' in cosine function is changing), that is: 800 results. So what do You expect in frequency domain to see?

    --- Quote End ---

    but my signal is cosine and the sampling rate is also twice of highest frequency ,so corresponding i must gate only two frequency component.