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anonymous

  • 5 years ago

function x(t) = cos(2*t + pi/4) anyone know whether the signal is periodic, and if it is what's its period. can you show the working out cheers

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  1. anonymous
    • 5 years ago
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    ok it's been a while since i did this. We know it's periodic because all cos functions are periodic, unless there is a discontinuity in there somewhere. (Usually by dividing by 0). Yours is fine. So the way we know its period, is from the angular frequency. The angular frequency is always multiplied by the variable. So in this case the 2 is multiplied by the (t) so the 2 is the angular frequency. The angular frequency is also always equal to 2 * pi* /T. (T= period, t= variable for time). In your case set the 2*pi/T = 2, and then solve for T = pi

  2. anonymous
    • 5 years ago
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    I'm not sure on the diving by 0 and discontinuity can you explain a bit more thanks, i have another example x(t) = u(t)*cos(2*pi*t) apparently this is aperiodic

  3. anonymous
    • 5 years ago
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    Ok so there is formal stuff on discontinuities, you should check with your teacher about them. I wouldn't worry about it unless this is for a calculus course. Also i am not entirely sure if i am correct on the discontinuity bit. Pretty much for a function to be periodic, it must be repeatable on some interval, and then go on forever. So cos(x) is. You can find out what it is from 0 to 2 pi, and then repeat that over and over again. The same can be said for any of the the other trig functions. This is shown by how cos(2 * pi) = 1 = cos(0) = cos (4pi)..... going on forever. I used 0 because it is easy, but in reality the same thing works for any cos(x). ie. cos(48.84*32* x) = cos(48.84*32*x+2*pi) ie. cos(x) = cos(x+2*pi). For your example of x(t) = u(t)*cos(2*pi*t). I am not sure exactly how to answer this because i was taught that, in order for this statement to be true then ALL function of u(t) multiplied by cos(2*pi*t) will make it non periodic. If you make u(t) = t then you get x(t) = t*cos(2*pi*t). This means that x(t) does not equal x(t+2*pi), which means that it is not periodic. The period doesn't have to be 2*pi, but it usually is. If you are stuck you can try graphing it, (either using your calculator or wolfram alpha). On the other hand, if you make u(t) = cos(2*pi*t) you will actually get a periodic function. The function is given by x(t) = cos(2*pi*t)^2, and this has a period of pi. I am not sure how to classify this one, you really need to check with your teacher. I am sure different mathematicians will give you different answers and you just need to regurgitate what yours wants. Hope this helps.

  4. anonymous
    • 5 years ago
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    Ok so there is formal stuff on discontinuities, you should check with your teacher about them. I wouldn't worry about it unless this is for a calculus course. Also i am not entirely sure if i am correct on the discontinuity bit. Pretty much for a function to be periodic, it must be repeatable on some interval, and then go on forever. So cos(x) is. You can find out what it is from 0 to 2 pi, and then repeat that over and over again. The same can be said for any of the the other trig functions. This is shown by how cos(2 * pi) = 1 = cos(0) = cos (4pi)..... going on forever. I used 0 because it is easy, but in reality the same thing works for any cos(x). ie. cos(48.84*32* x) = cos(48.84*32*x+2*pi) ie. cos(x) = cos(x+2*pi). For your example of x(t) = u(t)*cos(2*pi*t). I am not sure exactly how to answer this because i was taught that, in order for this statement to be true then ALL function of u(t) multiplied by cos(2*pi*t) will make it non periodic. If you make u(t) = t then you get x(t) = t*cos(2*pi*t). This means that x(t) does not equal x(t+2*pi), which means that it is not periodic. The period doesn't have to be 2*pi, but it usually is. If you are stuck you can try graphing it, (either using your calculator or wolfram alpha). On the other hand, if you make u(t) = cos(2*pi*t) you will actually get a periodic function. The function is given by x(t) = cos(2*pi*t)^2, and this has a period of pi. I am not sure how to classify this one, you really need to check with your teacher. I am sure different mathematicians will give you different answers and you just need to regurgitate what yours wants. Hope this helps.

  5. anonymous
    • 5 years ago
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    Sorry about the multiple posts, i wasn't logged in and everytime i hit the 'enter' key, it posted it once i logged in.

  6. anonymous
    • 5 years ago
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    No worries thank you for your help

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