How to find time period of a discrete time signal?
π₯[π] = cos (π/8β π)
Getting the value of cos(0)=1. For that we get n=8π and for again getting cos(x)=1, we can put n=24π. So the value is repeating itself in the interval of 16π. How do i state it mathematically? Or there is some another answer?
Can anyone please demonstrate by solving the above problem?
$\endgroup$ 31 Answer
$\begingroup$Note that the normal cosine function has a period of $2\pi$. To find the period of this function, we want to find the smallest positive $n$ such that $\frac{n}{8} - \pi$ is a multiple of $2\pi$ off of $\frac{0}{8} - \pi$. Note that setting $\frac{n}{8} - \pi - (0 - \pi) = 2 \pi k$ for arbitrary integer $k$ gives us $\frac{n}{8} = 2\pi k$ so $n = 16\pi k$. Then the minimum positive value at $k=1$ gives us a period of $16 \pi$.
In general, for sinusoids of the form: $$f(x) = a\cos (bx - h) + k$$
- $a$ represents amplitude
- $b$ represents frequency (so the period is given by $\frac{2\pi}{b}$)
- $h$ represents horizontal shift (to the right)
- $k$ represents vertical shift (upwards)
