fundamental period

Discussion in 'Homework Help' started by curw55, Aug 27, 2011.

  1. curw55

    Thread Starter New Member

    Aug 27, 2011
    2
    0
    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.
     
  2. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    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.
     
    curw55 likes this.
  3. curw55

    Thread Starter New Member

    Aug 27, 2011
    2
    0
    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.
     
  4. someonesdad

    Senior Member

    Jul 7, 2009
    1,585
    141
    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.
     
  5. djsfantasi

    AAC Fanatic!

    Apr 11, 2010
    2,809
    834
    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)"
     
  6. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    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
     
Loading...