Fundamental Frequency of Discrete Signals

For a discrete complex exponential $x[n]=e^{j\omega_1 n}$ to be periodic with period $N$, it has to satisfy

$$e^{j\omega_1(n+N)}=e^{j\omega_1n},\;\;\;\;\;\mbox{i.e.,} \;\;\;\;\; e^{j\omega_1N}=1=e^{j2\pi k}$$

that is, $\omega_1 N$ has to be a multiple of $2\pi$:

$$\omega_1N=2\pi k,\;\;\;\;\;\mbox{i.e.,}\;\;\;\;\;\; \frac{\omega_1}{2\pi}=\frac{k}{N}$$

As $k$ is an integer, $\omega_1/2\pi$ has to be a rational number (a ratio of two integers). In order for the period

$$N=k\frac{2\pi}{\omega_1}$$

to be the fundamental period, $k$ has to be the smallest integer that makes $N$ an integer, and the fundamental angular frequency is

$$\omega_0=\frac{2\pi}{N}=\frac{\omega_1}{k}$$

The original signal can now be written as:

$$x[n]=e^{j\omega_1 n}=e^{jk\omega_0 n}=e^{jk\frac{2\pi}{N}n}$$

Example 2: Show that a discrete signal

$$x[n]=e^{jm(2\pi/N)n}$$

has fundamental period

$$N_0=N/gcd(N,m)$$

According to the discussion above, the fundamental period $N_0$ should satisfy

$$m\frac{2\pi}{N}N_0=k2\pi,\;\;\;\;\mbox{or}\;\;\;\;\; N_0=\frac{kN}{m}=\frac{N}{m/k}$$

We see that for $N_0$ to be an integer, $l\stackrel{\triangle}{=}m/k$ has to divide $N$. But since $k=m/l$ is an integer, $l$ also has to divide $m$. Moreover, since $k$ needs to be the smallest integer satisfying the above equation, $l=m/k$ has to be the greatest common divisor of both $N$ and $m$, i.e., $l=gcd(N,m)$, and the fundamental period can be written as

$$N_0=\frac{N}{m/k}=\frac{N}{gcd(N,m)}$$

Example 2: Find the fundamental period of the following discrete signal:

$$x[n]=e^{j(2\pi/3)n}+e^{j(3\pi/4)n}$$

We first find the fundamental period for each of the two components.

• Assume the period of the first term is $N_1$, then it should satisfy
  
  $$e^{j(2\pi/3)(n+N_1)}=e^{j(2\pi/3)n}\cdot 1=e^{j(2\pi/3)n}\cdot e^{jk2\pi}=e^{j(2\pi n/3+k2\pi)}$$
  
  
  where $k$ is an integer. Equating the exponents, we have
  
  $$\frac{2\pi}{3}n+\frac{2\pi}{3}N_1=\frac{2\pi}{3}n+k2\pi$$
  
  
  which can be solved to get $N_1=3k$. We find the smallest integer $k=1$ for $N_1=3$ to be an integer, the fundamental period.

• Assume the period of the second term is $N_2$, then it should satisfy
  
  $$e^{j(3\pi/4)(n+N_2)}=e^{j(3\pi/4)n}\cdot 1=e^{j(3\pi/4)n}\cdot e^{jk2\pi}=e^{j(3\pi n/4+k2\pi)}$$
  
  
  where $k$ is an integer. Equating the exponents, we have
  
  $$\frac{3\pi}{4}n+\frac{3\pi}{4}N_2=\frac{3\pi}{4}n+k2\pi$$
  
  
  which can be solved to get $N_2=8k/3$. We find the smallest integer $k=3$ for $N_2=8$ to be an integer, the fundamental period. Now the second term can be written as
  
  $$e^{j\frac{3\pi}{4}n}=e^{j2\frac{3\pi}{8}n}$$
  
  

Given the fundamental periods $N_1=3$ and $N_2=8$ of the two terms, the fundamental period $N_0$ of their sum is easily found to be their least common multiple

$$N_0=lcm(3, 8)=24$$

and the fundamental frequency is

$$\omega_0=\frac{2\pi}{24}=\frac{\pi}{12}$$

Now the original signal can be written as

$$x[n]=e^{8\frac{2\pi}{24}n}+e^{9\frac{2\pi}{24}n}$$

i.e., the two terms are the 8th and 9th harmonic of the fundamental frequency $\omega_0=\pi/12$.

Similar to the continuous case, to find the fundamental frequecy of a signal containing multiple terms all expressed as a fraction multiplied by $\pi$, we can rewrite these fractions in terms of the least common multiple of all the denominators.

Example 3:

$$x[n]=sin(\frac{5\pi}{6}n)+cos(\frac{3\pi}{4}n)+sin(\frac{\pi}{3}n)$$

The least common multiple of the denominators is 12, therefore

$$x[n]=sin(\frac{10\pi}{12}n)+cos(\frac{9\pi}{12}n)+sin(\frac{4\pi}{12}n)$$

i.e., the fundamental frequency is $\omega_0=\pi/12$, the fundamental period is $T=2\pi/\omega_0=24$ and the three terms are the 4th, 9th and 10th harmonic of $\omega_0$, respectively.