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I have a rather simple question, sorry in advance if it's too easy. I'm working on a waveform generation project on MATLAB. When I look at the examples, I see that the time variables are defined like this;

t = 0:1/fs:T-1/fs;

What is the point of -1/fs?

Jdip
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spinmaster
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2 Answers2

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It's a matter of personal preference.

  • Some prefer that the length $N$ of the vector $t$ matches $$N = \lfloor f_sT\rfloor$$ where $\lfloor\rfloor$denotes the $\mathrm{floor}()$ function. This is done so that $$\frac{N}{f_s} \leq T$$

    Example 1: $f_sT$ is an integer: $f_s = 200\,\texttt{Hz}$, $T = 3\texttt{s}$ so $N = 600$:

    t = 0:1/fs:T % length(t) = 601, which would give 601/200 = 3.005s
t = 0:1/fs:T-1/fs % length(t) = 600, which gives 3s

    Example 2: $f_sT$ is not an integer: $f_s = 200\,\texttt{Hz}$, $T = 3.142\texttt{s}$ so $N = \lfloor628.4\rfloor = 628$:

    t = 0:1/fs:T % length(t) = 629, which would give 629/200 = 3.145s
t = 0:1/fs:T-1/fs % length(t) = 628, which gives 628/200 = 3.14s

  • However, others prefer that the last sample of $t$ be as close to $T$ as possible:
    With t = (0:1/fs:T), the very last sample of t will be closer to T:

    Example 1:

       t = 0:1/fs:T % t(end) = 3s
   t = 0:1/fs:T-1/fs % t(end) = 2.995s

    Example 2:

       t = 0:1/fs:T % t(end) = 3.14s
   t = 0:1/fs:T-1/fs % t(end) = 3.135s
Jdip
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2

It really depends what they are trying to do, but I assume they simulate several chunks of duration T with sampling interval 1/fs.

So they need to make sure that the last sample of each chunk is not computed again as the first sample of the next chunk.

Vito
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