Find Period Of Signal

Discussion in 'Homework Help' started by hitmen, Oct 24, 2009.

  1. hitmen

    Thread Starter Active Member

    Sep 21, 2008
    159
    0
    3 sin (2*pi*0.2t) - 2 cos (2*pi*0.3t)
    I need to find the period.
    I know that omega1 = 2*pi / (1/0.2) = 1.2566
    omega 2 = 2*pi / (1/0.3) = 1.884955

    common omega denominator = 2*pi / (1/0.6) = 3.7699

    Now how do I find the period of the equation??
     
  2. someonesdad

    Senior Member

    Jul 7, 2009
    1,585
    141
    It's a sum of two sinusoids -- sin(a) + sin(b). Using a math handbook, you can rewrite that as sin(c)*cos(d) where c and d are related to a and b (look up the formulas, as I'm not going to write down the details). Now you have a sinusoid whose amplitude is modulated by another sinusoid.

    Also use a plotting program to get a feel for what the resultant waveforms will look like to guide your problem solving.
     
  3. Accipiter

    New Member

    Sep 20, 2009
    9
    0
    You're on the right track. Basically you need to find the least common multiple of the periods of the two sinusoids.

    Remember, the period of a signal is the reciprocal of the frequency. (T = 1/f)

    After you calculate the periods of the two sinusoids (T = 1/f), you need to find the least common multiple of the two periods in order to get the period of the overall signal. Suppose, for example, you found that period1 = 4, and period2 = 6. Then, the least common multiple (LCM) of these two numbers (the smallest number that both numbers divide evenly into) is your answer. In this hypothetical case, we would have LCM(4 and 6) = 12. So, the overall period would be 12.

    However, if period1 and/or period2 are fractions, then you should first multiply both periods by their least common denominator (so as to get rid of the denominators), and then find the LCM as was done above with the integers, and then divide your answer by the number that you multipled the two periods by in order to make them integers.

    For example, suppose period1 = 2/3, and period2 = 4/7.

    So, the LCM of the denominators (LCD) is LCM(3 and 7) = 21.

    Remember, we doing this step in order to get rid of the denominators (make the fractions into integers).

    So, 2/3 * 21 = 14, and 4/7 * 21 = 12. So now we find LCM(14 and 12) = 84. Now, divide the answer by 21 (the number we multiplied by above). So, 84/21 = 4. Thus, the overall period would be 4.

    As another example, suppose period1 = 7, and period2 = 5/3.

    So, we have 7*3 = 21
    5/3*3 = 5

    Now that we have converted our two numbers into integers by multiplying them both by 3, we have:

    LCM(21 and 5) = 105.

    So, we have period_overall = 105/3 = 35

    Remember: in order to find the least common multiple of two numbers, you first write out all the prime factors of each number. Then, you choose from these two factor-lists, only the factors necessary so that the factors of each number are included.

    For example, find LCM(25, 100)

    25 = 5*5
    100 = 5*5*2*2
    LCM(25 and 100) = 5*5*2*2 = 100

    Find LCM(24 and 14):
    24 = 2*3*2*2
    14 = 2*7
    LCM(24 and 14) = 2*3*2*2*7 = 168 (notice that I included only three (not four) two's).

    Hope that helps.

    Oh, also, it doesn't really matter that one of the functions is a cosine and the other is a sine. Since cosine and sine only differ by phase, all the above will still work.

    Also, be careful when graphing to pick a large enough window so as to see the the actual fundamental period of the signal. If you only look at a small section of the graph, it is easy to think that the fundamental period of the signal is much smaller than it actually is.

    After you do it, tell us what answer you got.
     
    Last edited: Oct 24, 2009
  4. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    Hello hitmen,

    It's also informative to first plot the waveform (say using excel, Matlab etc) to assist you in determining the period. It's an example of the old saying that "a picture paints a thousand words".
    Keep in mind that the period is effectively the time interval over which the waveform replicates or repeats itself.
    As Accipiter suggests you need to plot the function over a sufficiently long time interval to observe the periodic behaviour.
    Hint: plot for say 0-20 seconds as a starting point
     
  5. hitmen

    Thread Starter Active Member

    Sep 21, 2008
    159
    0
    Thank you for your very very clear explanations! :)
    f= 0.2 f=0.3
    T1 = 5sec T2 = 10/3 sec
    Multiply both side by 3 to get 15 and 10
    LCM of 15 and 10 = 30
    Period = 30/3 =10 :)

    Is this method possible with 2 or more signals?
    Is it possible to use the common omega as well??

    Thanks!
     
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