Determining if a function is periodic

Posted on 2006-05-13
Last Modified: 2009-07-29
Yup, this is homework. So, here are my answers and thoughts. Am I correct, incorrect or in the wrong ballpark ;)

Determine if the following function is periodic and if so, find the fundamental frequency. (The use of "j" is for imaginary numbers)

1) x(t) = 3cos(4t + pi/3)

  This function is periodic (the amplitude doesn't matter..?..).
  The fundamental frequency is T = pi/2

2) x(t) = exp( j (pi * t - 1))

  From euler form (  exp[x] = cos x - jsin x )

  x(t) = cos(pi t - 1) + j sin(pi t - 1)

  I do not think this is periodic due to the complex portion.

3) x(t) = [ cos(2t - pi/3) ] ^ 2

  Using a Half-Angle identity:

  x(t) = .5 [ 1 + cos( 2 ( 2t - pi/3 ) ) ]
        = .5 [ 1 + cos(4t - 2pi/3) ]

  Period with fundamental period T = pi/2

4) x(t) = Even { cos(4pt*t)) } u(t)

   This is not periodic due to the unit step function (as the amplitude changes at t=0)
Question by:bingie
    LVL 84

    Accepted Solution

    1) correct
    2) with the complex portion, is x(t) = x(t+T) for some T!=0?
    3) correct
    4) I'm not sure how you define Even{} or u(t), but cos(4pt*t) is
    LVL 11

    Author Comment

    2) Not sure....

    I made a table and substituted values for t (starting at 0 and increasing by 90) and the real portion increased by 281 each time and the angle remained constant at 57 degrees.

    Does this indicate a periodic function with fundamental frequency of 2?

    4) Even {} is defined where x(t) = x(-t) and then x(t) = Even{x(t)} + Odd{x(t)} where Odd{x(t)} -> x(t) = -x(-t)
        u(t) is a unit step function ( where x = 0 for t <0 and x = 1 for t>=0)
    LVL 37

    Assisted Solution

    by:Harisha M G
    Hi bingie,

    1) You are right. You know that all trigonometric funcitons are periodic with the period 2*PI (And TAN is periodic with PI)

    You have cos(4t + pi/3) and pi/3 is a constant phase. So, cos(4t). Now, cos(4t+2pi) = cos(4t) = cos(4t'), (say)

    So, 4t' = 4t+2pi => t' = t + pi/2

    So, fundamental frequency is pi/2

    2) x(t) = exp(j(pi*t-1)) = exp(j*pi*t-j) = exp(j*pi*t) * exp(-j) = ( cos (pi*t) - j sin(pi*t) ) * exp(-j)

    which is clearly periodic, with period = 2

    3) You are right, very similar to 1)

    4) You are right.

    But check whether it is given as Even { cos(4pt*t)) } u(t) or Even { cos(4pt*t)) * u(t) }

    In the latter case, it will be periodic

    Even{x(t)} = ( x(t) + x(-t) ) / 2
    Odd{x(t)} = ( x(t) - x(-t) ) / 2
    u(t) = 1 for t >= 0, 0 otherwise

    LVL 26

    Expert Comment

    But more fundamentally, what is your definition of periodic?
    LVL 11

    Author Comment


    >>Even { cos(4pt*t)) } u(t) or Even { cos(4pt*t)) * u(t) }

    What is the difference in these two functions?
    LVL 37

    Expert Comment

    by:Harisha M G
    Only thing is u(t) is outside the even() function
    LVL 11

    Author Comment

    I noticed that, but how does that affect the function?
    LVL 22

    Assisted Solution

    For 1, the _period_ is pi/2, so the _frequency_ is 2/pi.

    2 is periodic.  The complexity doesn't enter into it.  After every period the function generates the same value.  x(t) = x(t+delta) for all t and some particular deltas.

    3 looks right.

    4.  Consider the definition of periodicity, where x(t) = x(t+delta) for all t and some delta.  What is the minimum positive delta for which this definition works?  The fact that there is a unit step function doesn't rule out periodicity.

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