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Fundamental Frequency of Continuous Signals

To identify the period $T$, the frequency $f=1/T$, or the angular frequency $\omega=2\pi f=2\pi/T$ of a given sinusoidal or complex exponential signal, it is always helpful to write it in any of the following forms:

\begin{displaymath}sin(\omega t)=sin(2\pi ft)=sin(2\pi t/T) \end{displaymath}

The fundamental frequency of a signal is the greatest common divisor (GCD) of all the frequency components contained in a signal, and, equivalently, the fundamental period is the least common multiple (LCM) of all individual periods of the components.

Example 1: Find the fundamental frequency of the following continuous signal:

\begin{displaymath}x(t)=cos(\frac{10\pi}{3}t)+sin(\frac{5\pi}{4}t) \end{displaymath}

The frequencies and periods of the two terms are, respectively,

\begin{displaymath}\omega_1=\frac{10\pi}{3},\;f_1=\frac{5}{3},\;T_1=\frac{3}{5},...
...; \omega_2=\frac{5\pi}{4},\;f_2=\frac{5}{8},\; T_2=\frac{8}{5}
\end{displaymath}

The fundamental frequency $f_0$ is the GCD of $f_1=5/3$ and $f_2=5/8$:

\begin{displaymath}f_0=GCD(\frac{5}{3},\frac{5}{8})=GCD(\frac{40}{24},\frac{15}{24})=\frac{5}{24} \end{displaymath}

Alternatively, the period of the fundamental $T_0$ is the LCM of $T_1=3/5$ and $T_2=8/5$:

\begin{displaymath}T_0=LCM(\frac{3}{5},\frac{8}{5})=\frac{24}{5} \end{displaymath}

Now we get $\omega_0=2\pi f_0=2\pi/T_0=5\pi/12$ and the signal can be written as

\begin{displaymath}x(t)=cos(8\frac{5\pi}{12}t)+sin(3\frac{5\pi}{12}t)
=cos(8\omega_0t)+sin(3\omega_0t)
\end{displaymath}

i.e., the two terms are the 3th and 8th harmonic of the fundamental frequency $\omega_0$, respectively.

Example 2:

\begin{displaymath}x(t)=sin(\frac{5\pi}{6}t)+cos(\frac{3\pi}{4}t)+sin(\frac{\pi}{3}t) \end{displaymath}

The frequencies and periods of the three terms are, respectively,

\begin{displaymath}\omega_1=\frac{5\pi}{6},\;f_1=\frac{5}{12},\;T_1=\frac{12}{5}...
...c{8}{3},\;\;
\omega_3=\frac{\pi}{3},\;f_3=\frac{1}{6},\;T_3=6 \end{displaymath}

The fundamental frequency $f_0$ is the GCD of $f_1$, $f_2$ and $f_3$:

\begin{displaymath}f_0=GCD(\frac{5}{12},\frac{3}{8},\frac{1}{6})
=GCD(\frac{10}{24},\frac{9}{24},\frac{4}{24})=\frac{1}{24} \end{displaymath}

Alternatively, the period of the fundamental $T_0$ is the LCM of $T_1$, $T_2$ and $T_3$:

\begin{displaymath}T_0=LCM(\frac{12}{5},\frac{8}{3},6)=LCM(\frac{36}{15},\frac{40}{15},\frac{90}{15})
\frac{5}{24} \end{displaymath}

The signal can be written as

\begin{displaymath}x(t)=sin(\frac{10\pi}{12}t)+cos(\frac{9\pi}{12}t)+sin(\frac{4\pi}{12}t) \end{displaymath}

i.e., the fundamental frequency is $\omega_0=\pi/12$, the fundamental period is $T_0=2\pi/\omega)=24$, and the three terms are the 4th, 9th and 10th harmonic of $\omega_0$, respectively.

Example 3: Find the fundamental frequency of the following continuous signal:

\begin{displaymath}x(t)=cos(\frac{10}{3}t)+sin(\frac{5\pi}{4}t) \end{displaymath}

Here the angular frequencies of the two terms are, respectively,

\begin{displaymath}\omega_1=\frac{10}{3}, \;\;\;\;\; \omega_2=\frac{5\pi}{4} \end{displaymath}

The fundamental frequency $\omega_0$ should be the GCD of $\omega_1$ and $\omega_2$:

\begin{displaymath}\omega_0=GCD(\frac{10}{3},\frac{5\pi}{4}) \end{displaymath}

which does not exist as $\pi$ is an irrational number which cannot be expressed as a ratio of two integers, therefore the two frequencies can not be multiples of the same fundamental frequency. In other words, the signal as the sum of the two terms is not a periodic signal.


next up previous
Next: Fundamental Frequency of Discrete Up: Fundamental_Frequency Previous: Fundamental_Frequency
Ruye Wang 2009-02-02