Time vector to generate a sinusoid signal with 0.5 seconds duration , Fs =200

Hi , please I'm confused with the following time vectors t1 and t2 to generate 0.5 second duration sinusoid signal :

Vector t1 has equally spaced samples at Ts intervals , but the signal x1 has duration equal to ( duration+Ts ) seconds , I checked x1 length .

Vector t2 has equally spaced samples not Ts intervals , signal x2 dutaion = ( duration ) secons excatly , which time vector is correct ?

%% Generates a sine wave of 0.5 second duration 
F0=10; % Sinusoid frequency in Hz
Fs=200; % sampling frequency samples / sec
duration=0.5; % signal duration is seconds
Ts=1/Fs; % sampling Interval
t1=0:Ts:duration; % time vector with equally spaced samples at Ts intervals 
t2=linspace(0,duration,duration*Fs); % time vector with samples not at Ts intervals 
x1=sin(2*pi*F0*t1); % signal duration = ( duration+Ts ) seconds , length = 101 points
x2=sin(2*pi*F0*t2); %  signal duration = ( duration ) secons , length = 100 points

[Mathieu NOE]

Hi

simply checking the actual dt of t1 and t2 gives you the reason why it bugs

>> mean(diff(t1))
ans =
    0.0050
>> mean(diff(t2))
ans =
    0.0051

so to get the correct result, correct t2 as follows (lenght incresed by +1)

t2=linspace(0,duration,duration*Fs+1); % time vector with samples not at Ts intervals 

    

[Mohamad]

Hi

Yes this is clear , but I need to know how to make the signal duration 0.5 Seconds with sampling Instatnts exctaly = 1/Fs .

t1=0:Ts:duration; % time vector with equally spaced samples at Ts intervals
t2=linspace(0,duration,duration*Fs); % time vector with samples not at Ts intervals

Because if I use the 1st line I get signal duration = 0.501 seconds not 0.5 Seconds .

If I used the 2ed line I get signal duration = 0.5 Seconds , but sampling Instatnts = 0.0051 ( i.e. not 1/Fs) .

Thanks

[Mathieu NOE]

Hi

I don't agree with your statement

Because if I use the 1st line I get signal duration = 0.501 seconds not 0.5 Seconds .

max value of t1 = duration so it cannot reach 0.501 s if you have declared duration = 0.5 s

(and that's what I get too)

now , maybe you're confused by the fact that t1 has 101 values - simply when you define a time vector going from 0 to n*Ts you will have n+1 samples.

[Mohamad]

t1 has 101 points , this means signal duration = ( 101/Fs ) = 0.505 seconds not 0.5 seconds .

I need the signal duration to be 0.5 seconds with sampling instants at 1/Fs (0.005) .

So I think using the following will be correct , right ?

t2=linspace(0,0.5,duration*Fs+1);

[Mathieu NOE]

yes , if you prefer take the second option

t2=linspace(0,0.5,duration*Fs+1);

but the first method works also. example you just take 2 samples with dt = 0.005

first sample correspond to t = 0

second sample correspond to t = 1 * dt = 0.005

you have 2 samples for a signal duration = 0.005 (= 1 x dt)

so this is what I tried to explain

Answer 1

[Cris LaPierre]

When using linspace, don't forget to account for 0. For example, how many numbers are there between 0 and 10? 11.

You should add 1 to your calculated number of points. This way, the colon operator and linspace give the same result, and both t1 and t2 are vectors with 101 points.

t1 = 0:1/200:0.5;
mean(diff(t1))

ans = 0.0050

t2 = linspace(0,0.5,0.5*200+1);
mean(diff(t2))

ans = 0.0050

[Mohamad]

Thanks a lot , now I understand it , actually I was confused because the following example from matlab to generate a signal with a sampling frequency of 1 kHz and a signal duration of 1.5 seconds :

https://www.mathworks.com/help/matlab/ref/fft.html

Fs = 1000;            % Sampling frequency                    
T = 1/Fs;             % Sampling period       
L = 1500;             % Length of signal
t = (0:L-1)*T;        % Time vector

The example says signal duration 1.5 seconds , but I see signal duration it is 1.499 seconds .