# Confusion regarding calculation of fundamental period?

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

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

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

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

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

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:

• Please kindly update 2nd last line of your answer. It is fundamental period that is 0.25 not fundamental frequency – Man Mar 30 '20 at 16:15
• @Man Done, good catch. Sorry about that. – Cedron Dawg Mar 30 '20 at 16:18
• Please kindly make a bit update to your answer to meet the need of updated question – Man Apr 1 '20 at 11:00
• @Man Quit shifting the goal posts. n=3,4,5... can be calculated according to the pattern. the end result is $n4 \pi T=2 \pi$ which is the same as $T=1/(2n)$ – Cedron Dawg Apr 1 '20 at 12:52

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.

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.

• what software are you using? – MBaz Mar 30 '20 at 15:18
• I am using a commercial simulation program named Extend (older version) and ExtendSim (newer versions), from Imagine That, Inc. These are augmented with four libraries of blocks I started developing back in 1990. My libraries, named LightStone, are available for free, with full commented source code. The URL for my libraries is umass.box.com/v/LightStone. I will be updating the libraries by the end of the week so that they work with the latest ExtendSim 10.0.6 version (should be just a recompile). The model above was done with Extend 6.0.8 on an old Mac (I like the way it looks). – Ed V Mar 30 '20 at 15:30
• Thanks, I'll check it out :) – MBaz Mar 30 '20 at 16:40