Confusion regarding calculation of fundamental period?

Question

I am reading signal processing first and in chapter 3 ex3.8 i came across an example of fundamental period as shown in attached photo

It apparently shows that signal $$x(t)=\cos^2(4\pi t)$$ has period 0.5 but then it also writes that fundamental period is 0.25

How is he doing that?

Also will be fundamental period if $$x(t)=\cos^n(4\pi t)$$ where n can be 3 or 4 or 5

Answer 1

The trigonometric functions are essentially exponential. Thus, a doubling of the argument corresponds to a squaring of the function (in a sense). In this case, it can be seen by applying the angle addition formula:

$$ \begin{aligned} \cos( 2\theta ) &= \cos( \theta + \theta ) \\ &= \cos(\theta)\cos(\theta) - \sin(\theta)\sin(\theta) \\ &= \cos^2(\theta) - ( 1- \cos^2(\theta) ) \\ &= 2 \cos^2(\theta) - 1 \end{aligned} $$

Making

$$ \cos^2(\theta) = \frac{\cos( 2\theta ) + 1}{2} $$

Applying it to your equation:

$$ x(t)=\cos^2(4\pi t) = \frac{\cos( 8 \pi t ) + 1}{2} $$

From this it is pretty clear the fundamental period is 0.25 as that makes $8 \pi t = 2\pi$.

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Upon request:

$$ \begin{aligned} x(t) &= \cos^3(4\pi t) \\ &= \left( \frac{ e^{i 4\pi t} + e^{-i 4\pi t} }{2} \right)^3 \\ &= \frac{1}{8}\left( e^{i 12\pi t} + 3 e^{i 4\pi t} + 3 e^{-i 4\pi t} + e^{-i 12\pi t} \right) \\ &= \frac{1}{4}\left[ \cos(12\pi t) + 3 \cos( 4\pi t) \right] \\ \end{aligned} $$

You should be able to figure from there. Note, the squared case could have been handled the same way.

I use this technique extensively for these formulas:

• Exact Near Instantaneous Frequency Formulas Best at Peaks (Part 1)
• Exact Near Instantaneous Frequency Formulas Best at Peaks (Part 2)
• Exact Near Instantaneous Frequency Formulas Best at Zero Crossings

Answer 2

This seems like more of a semantics problem.

A signal is periodic with time $T$ if

$$x(t+n\cdot T) = x(t), n \in \mathbb{Z}$$

So the signal is periodic in $0.5$ since the for $T = 0.5 \cdot n$ the argument of the cosine is an integer multiple of $2 \pi$. Since it's periodic in $0.5$ it's also periodic in all integer multiples of $0.5$, i.e $1$, $1.5$, $2$ etc.

In this case it's also periodic in $0.25$ since $$ \cos^2(4 \cdot \pi \cdot t ) = 0.5 \cdot (1+\cos(8 \cdot \pi \cdot t))$$

So any periodic signal has an infinite number of periods, the fundamental one is the smallest one and all the others are integer multiples of the fundamental.

Answer 3

If it helps any, generate a unit amplitude sinewave at 1 Hz and its square:

Then the sinewave and its square look like this:

You can see the DC component: the averaged value of the squared sinewave (averaged over an integer number of periods) is 1/2. And the red sinewave frequency is exactly doubled, so the period is halved. The DC and doubled frequency are the 'beat frequencies' obtained by multiplying the sinewave by itself.