fundamental period

Thread Starter

curw55

Joined Aug 27, 2011
2
How would one find the fundamental period of the function cos^2 t, cos((pi*n^2)/8), and e^(j((t/4)-pi)). We can't use graphing calculators and I know to find the fundamental period for Acos(wt+theta) would be 2*pi/w but I don't know how to apply that to these problems.
 

t_n_k

Joined Mar 6, 2009
5,455
You can generally deal with the first and last functions keeping in mind the following relationships ....

\(cos^2(\omega t)=\frac{1+2cos(2 \omega t)}{2}\)

and using Euler's relationship in conventional sinusoidal function notation

\(R_e [ e^{j \omega t}e^{j \theta}]=R_e[e^{j(\omega t + \theta)}]=cos(\omega t +\theta)\)

Where Re refers to the "real part".

The second function is not defined in terms of time [t] so I'm not sure how one defines the periodicity.
 

Thread Starter

curw55

Joined Aug 27, 2011
2
I'm not understanding why those relationships are true, so I don't know how to make sense out of them and remember them in the future.
 

someonesdad

Joined Jul 7, 2009
1,583
They're proven in basic trigonometry classes. They're also good for exercising your math manipulation skills by figuring out how to prove them yourself. You can find good lists of these types of identities (the one given is of a type called "multiple angle identities") in the math handbooks; also check here.
 

djsfantasi

Joined Apr 11, 2010
6,791
Googling for the second equation provides several solutions. Apparently it is a common problem in audio signal processing. I searched for:
"fundamental period cos(pi*(n^2)/8)"
 

t_n_k

Joined Mar 6, 2009
5,455
I'm not understanding why those relationships are true, so I don't know how to make sense out of them and remember them in the future.
It can be confusing at first.

OK - consider the function

\(f(t)=cos^2(t)\)

Using the relationship

\(cos^2(\omega t)=\frac{1+cos(2 \omega t)}{2}\)

One can therefore write that

\(f(t)=cos^2(t)=\frac{1+cos(2t)}{2}\)

Note that in this case, we can conclude ω=1 radians per second and the function f(t) is therefore periodic at angular frequency 2ω or 2 radians per second.

One could also recast f(t) as

\(f(t)=\frac{1}{2}+\frac{1}{2}cos(2t)\)

In other words this is a level shifted (by 0.5) sinusoid of amplitude 0.5 with a period of

\(T=\frac{2 \pi}{2}=\pi \ seconds\)
 
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