Chapter 2 - Signal Properties

Periodicity

Worked Example 2:

Find the fundamental period of the following signal:

\[ f(t)=\cos{(2\pi t)} + 3\sin{(6\pi t)} - 5\cos{\left(10\pi t - \tfrac{\pi}{4}\right)} \]

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\[ T = 1 \]

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Let's determine the periods of each component, since we know that sines and cosines are periodic.

\[ \exists\,T_{1},\quad \cos{(2\pi t)} = \cos{\left(2\pi \left(t+T_{1}\right)\right)} \]

Expanding the argument, we have:

\[ \cos{\left(2\pi (t+T_{1})\right)} = \cos{\left(2\pi t + 2\pi T_{1}\right)} \]

We know that:

\[ \cos{(x + 2\pi n)} = \cos{(x)},\quad n\in\mathbb{Z} \]

So we have:

\[ 2\pi T_{1} = 2\pi k_{1} \Rightarrow T_{1} = k_{1} \]

We can pick the smallest positive number \(k_{1}=1\), which gives us:

\[ T_{1} = 1 \]

Repeating this process for the second term:

\[ 3\sin{(6\pi t)} = 3\sin{\left(6\pi\left(t + T_{2}\right)\right)} = 3\sin{\left(6\pi t + 6\pi T_{2}\right)} \]

\[ \Rightarrow 6\pi T_{2} = 2\pi k \Rightarrow T_{2} = \frac{k_{2}}{3} \]

Then picking \(k_{2}=1\) gives us:

\[ T_{2} = \frac{1}{3} \]

Finally, for the third term, we have: $$ T_{3} = \frac{k_{3}}{5} $$

\[ k_{3}=1\Rightarrow T_{3} = \frac{1}{5} \]

To work out the periodicity of the whole signal, we need to consider when the periods 'line-up'.

Since the individual components are periodic with integer multiples of:

\[ T_{1} = 1,\quad T_{2} = \frac{1}{3},\quad T_{3}=\frac{1}{5} \]

The point where they line up can be computer by finding the lowest common multiple, equivalent to finding the smallest \(T\) that satisfies:

\[ T = k_{1}T_{1} = k_{2}T_{2} = k_{3}T_{3} \]

for some \(k_{1},k_{2},k_{3}\in\mathbb{Z}^{+}\).

Note

Lowest-common-multiple is typically used for integers only, but we can extend this to fractions using:

\[ \mathrm{LCM}\left(\tfrac{n_{1}}{d_{1}},\dots, \tfrac{n_{k}}{d_{k}}\right) = \frac{\mathrm{LCM}\left(n_{1},\dots,n_{k}\right)}{\mathrm{GCD}(d_{1},\dots, d_{k})} \]

\[ T = \mathrm{LCM}\left(1,\tfrac{1}{3},\tfrac{1}{5}\right) = \frac{\mathrm{LCM}\left(1,1,1\right)}{\mathrm{GCD}(1,3,5)} = \frac{1}{1} = 1 \]

So our fundamental period would be \(T=1\), which makes sense, since you need \(3\times\) the second component's period and \(5\times\) the third component's period to get the first components period.

Or for a more complicated example, if we had periods of:

\[ T_{1} = \frac{2}{3},\quad T_{2} = \frac{5}{6},\quad T_{3} = \frac{1}{20} \]

We'd get:

\[ T = \mathrm{LCM}\left(\tfrac{2}{3},\tfrac{5}{6},\tfrac{1}{24}\right) = \frac{\mathrm{LCM}\left(2,5,1\right)}{\mathrm{GCD}(3,6,24)} = \frac{10}{3} \]