I have a sinusoidal signal that is sampled at $f_s=10$ Hz. I am asked to find the absolute value of the DFT for different data lengths (1 sec., 2 sec. etc.)

I do not understand where will I use the information about the data lengths.

Does the question want me to change the time vector for each attempt? as such:

** original time vector **

f=10;         % sampling freq


x = cos(2*pi*3*t)

** time vector for each attempt **

t1=0:1/f:f*1; % data length = 1 sec
t2=0:1/f:f*2; % data length = 2 sec

thus rendering

x1 = cos(2*pi*3*t1);
x2 = cos(2*pi*3*t2);

and the rest would be how we usually get the FFT.

Did I understand the request correctly or not?

----Edit ------

The question is specifically for matlab simulation: enter image description here

  • $\begingroup$ In t1=0:1/f:f*1, you are mixing times (1/f) and frequencies (f*1). Use times for everything: t1 = 0:1/f:1 for samples going from 0 to 1 second. $\endgroup$
    – Juancho
    Feb 6, 2017 at 16:31

1 Answer 1


The information about the frequency of the sine seems to be missing. This could be an hint for you to compute the DFT exactly (not with Matlab simulation), for a given frequency $f$ and a number $s$ of seconds. Since the sampling frequency $f_0$ gives a integer multiple of samples inside one second, you should be able to find manageable formulas.

As commented by @Juancho, there is a lack of homogeneity in your time base: a frequency times a duration in seconds is not homogeneous to a time (it is unitless). A flexible solution is:

fs=10; % sampling freq
lFreq = [3 3.5 4]; % list of frequencies
lDuration = [1 2 4 10]'; % list of durations

for iDuration = 1:length(lDuration)
    timeBase = [0:1/fs:lDuration(iDuration)]';
    dataCosine = cos(timeBase*lFreq*2*pi); % Three cosines
    data = sum(dataCosine,2); % sum the three cosine components
    ylabel('Amplitude (a.u.)')
    xlabel('Time (seconds)')
    title(['Duration: ',num2str(lDuration(iDuration)),' s'])
    grid on
  • $\begingroup$ Hi, thank you. But the question is meant to be for matlab. I'm not sure how to incorporate the different "data lengths" into the solution. Was my attempt correct? $\endgroup$
    – HappyBee
    Feb 7, 2017 at 6:47
  • $\begingroup$ Not really, $f*1$ is a number of samples, not a length in seconds $\endgroup$ Feb 7, 2017 at 7:08

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