periodic - 2

[logistic_guy]

Determine whether or not each of the following signals is periodic. In case a signal is periodic, specify its fundamental period.

\(\displaystyle \bold{(a)} \ x_a(t) = 3\cos(5t + \pi/6)\)
\(\displaystyle \bold{(b)} \ x(n) = 3\cos(5n + \pi/6)\)
\(\displaystyle \bold{(c)} \ x(n) = 2e^{j(n/6 - \pi)}\)
\(\displaystyle \bold{(d)} \ x(n) = \cos(n/8)\cos(\pi n/8)\)
\(\displaystyle \bold{(e)} \ x(n) = \cos(\pi n/2) - \sin(\pi n/8) + 3\cos(\pi n/4 + \pi/3)\)

 

[logistic_guy]

\(\displaystyle \bold{(a)} \ x_a(t) = 3\cos(5t + \pi/6)\)

This is not a discrete-time signal. It is continuous.

A continuous signal is periodic when its \(\displaystyle \omega\) is rational and multiple of \(\displaystyle \pi\). Its fundamental period is \(\displaystyle T_p = \frac{2\pi}{\omega}\).

The signal \(\displaystyle x_a(t)\) has the form \(\displaystyle A\cos(\omega t + \phi)\).

Then, its angular frequence \(\displaystyle \omega = 5\).

\(\displaystyle T_p = \frac{2\pi}{\omega} = \frac{2\pi}{5}\)

This says it is \(\displaystyle \textcolor{blue}{\text{periodic}}\) with \(\displaystyle T_p = \textcolor{blue}{\frac{2\pi}{5}}\)

 

[logistic_guy]

A continuous signal is periodic when its \(\displaystyle \omega\) is rational and multiple of \(\displaystyle \pi\). Its fundamental period is \(\displaystyle T_p = \frac{2\pi}{\omega}\).

This definition needs to be corrected.

A continuous signal of the form \(\displaystyle \cos(\omega t + \phi)\) or \(\displaystyle \sin(\omega t + \phi)\) is periodic as long as \(\displaystyle \omega\) is a real nonzero constant. Its fundamental period is \(\displaystyle T_p = \frac{2\pi}{\omega}\).

 

[logistic_guy]

\(\displaystyle \bold{(b)} \ x(n) = 3\cos(5n + \pi/6)\)

\(\displaystyle \omega = 2\pi f\)

\(\displaystyle f = \frac{\omega}{2\pi} = \frac{5}{2\pi} \neq \frac{r}{N_p}\)

where \(\displaystyle r, N_p \in \mathbb{Z}\)

The discrete-time signal \(\displaystyle \textcolor{blue}{\text{is not periodic}}\) as \(\displaystyle \frac{5}{2\pi} \ \textcolor{red}{\text{is not rational}}\)

 

[logistic_guy]

\(\displaystyle \bold{(c)} \ x(n) = 2e^{j(n/6 - \pi)}\)

\(\displaystyle x(n) = 2e^{j(n/6 - \pi)} = 2[\cos(n/6 - \pi) + j\sin(n/6 - \pi)]\)

\(\displaystyle \omega = 2\pi f\)

\(\displaystyle f = \frac{\omega}{2\pi} = \frac{\frac{1}{6}}{2\pi} = \frac{1}{12\pi} \neq \frac{r}{N_p}\)

where \(\displaystyle r, N_p \in \mathbb{Z}\).

Then, the discrete-time signal \(\displaystyle \textcolor{blue}{\text{is not periodic}}\) as \(\displaystyle \frac{1}{12\pi} \ \textcolor{red}{\text{is irrational}}\)

 

[logistic_guy]

\(\displaystyle \bold{(d)} \ x(n) = \cos(n/8)\cos(\pi n/8)\)

My signal here is the product of two signals. If one of them is non-periodic, then their product is non-periodic too.

Let us check \(\displaystyle \cos n/8\)

\(\displaystyle \omega = 2\pi f\)

\(\displaystyle f = \frac{\omega}{2\pi} = \frac{\frac{1}{8}}{2\pi} = \frac{1}{16\pi} \neq \frac{r}{N_p}\)

where \(\displaystyle r,N_p \in \mathbb{Z}\).

Then the discrete-time signal \(\displaystyle x(n)\) is \(\displaystyle \textcolor{blue}{\text{non-periodic}}\).

 

[logistic_guy]

\(\displaystyle \bold{(e)} \ x(n) = \cos(\pi n/2) - \sin(\pi n/8) + 3\cos(\pi n/4 + \pi/3)\)

I think that this is the most beautiful problem of all above. From the previous exercises, we have realized that if \(\displaystyle \omega\) contains \(\displaystyle \pi\), then the discrete-time signal is periodic. And this signal is the sum of three signals each one of them its \(\displaystyle \omega\) contains \(\displaystyle \pi\). This says that the sum is periodic and in turn the signal \(\displaystyle x(n)\) is periodic, but the tough part is how to find its fundamental period!

\(\displaystyle \cos(\pi n/2) \rightarrow N_{p_1} = 4\)

\(\displaystyle \sin(\pi n/8)) \rightarrow N_{p_2} = 16\)

\(\displaystyle \cos(\pi n/4 + \pi/3) \rightarrow N_{p_3} = 8\)

We found their fundamental periods by the previous method we learnt. The trick is that the least common multiple of the three periods is the fundamental period of \(\displaystyle x(n)\). That is:

\(\displaystyle N_p = \text{lcm}(4,8,16)\)

We are now experts in finding the least common multiple and we have different ways to use. But since the numbers are small we don't need to use a fancy way. We will use the trivial method (prime factorization method). That is:

\(\displaystyle 4 \rightarrow 2^2\)
\(\displaystyle 8 \rightarrow 2^3\)
\(\displaystyle 16 \rightarrow 2^4\)

Since they all share the same base, the one with the highest exponent is the least common multiple. That is:
\(\displaystyle N_p = \text{lcm}(4,8,16) = 2^4 = 16\)

And we conclude that the discrete-time signal is \(\displaystyle \textcolor{blue}{\text{periodic}}\) with \(\displaystyle N_p = \textcolor{blue}{\text{16}}\)