Context: Digital Signal Processing
Given a signal and have to determine whether is it periodic, if so then find the fundamental period. I must justify either conclusion.
So I am struggling with two signals I was given:
1) x[n] = e^(j*0.75*pi*(n^2))
I attempted to try and use the formula for checking periodicity: x[n] = x[n + N] but the n raise tot he power of two in the exponential is throwing me off.
2) x[n] = cos(sqrt(2)pi(n) + pi)
My initial approach for this one was to do wo/2pi which ends up being 1/sqrt(2) and thus I want to conclude that the signal is not periodic because the number is irrational, but I'm not sure if the rule applies for cosine functions of which I thought were always considered periodic?
Any help with this would be greatly appreciated, thank you!
Best Answer
I assume that exercise is meant to demonstrate the difference in periodicity between functions from a continuous domain (like the real or complex numbers) and from a discrete domain (like the natural numbers/integers).
1) This function is periodic when discretized as it happened here. Try to prove that $$x[n+4]=x[n]$$ by taking into account that the sine and cosine functions are periodic with any integer multiple of $2\pi$.
2) This function is not periodic. $x[0]=\cos(\pi)=0$. If there was some $N>0$ with $x[n+N]=x[n]$ for all $n$, then it must be true that $x[N]=0$. The cosine function is $0$ only for integer multiples of $\pi$ (more accurately, it is $0$ for exactly the odd integer multiples of $\pi$, but the less accurate version is enough). $x[N]=0$ implies there is some integer $k$ that
$$\sqrt{2}\pi N +\pi = k\pi$$
Dividing by $\pi$ results in
$$\sqrt{2}N +1 = k,$$
which can be transformed into
$$\sqrt{2}=\frac{k-1}{N}$$
since $N > 0$. But this is impossible since $\sqrt{2}$ is an irrational number. $\blacksquare$
As you correctly wrote the fundamental period of the $\mathbb R \rightarrow \mathbb R$ function $f(t) = \cos(\sqrt{2}\pi t + \pi)$ is $\sqrt{2}$. That means for question 2) the underlying $\mathbb R \rightarrow \mathbb R$ function is periodic, while the discrete version is not.
It is the other way around for question 1). Here the discretized version is periodic, while the $\mathbb R \rightarrow \mathbb C$ function $f(t)=e^{j\frac{3}{4}\pi t^2}$ is not. There are many ways to visualize that. Let's just look at the real part of $f: Re(f(t)) = \cos(\frac{3}{4}\pi t^2)$. The derivative of it is
$$ (Re(f(t)))' = -\frac{3}{2}\pi t\sin(\frac{3}{4}\pi t^2)$$
which is unbounded when $t$ becomes large. But the derivative of a periodic function is itself periodic (say with period $p$), and the value range over all reals is the same as the value range over the interval $[0,p]$. Since $(Re(f(t)))'$ is continuous, its value range over a finite closed interval must be bounded.