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Are all discrete sinusoids distinct?

  1. Dec 15, 2014 #1
    For the discrete complex sinusoid with period N, how many distnict sinusoids are there? And why?
     
  2. jcsd
  3. Dec 15, 2014 #2

    jim hardy

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    Let ω = 2pi/N,

    How many Asin(ωt + Φ) 's are there? I suppose A can only be positive but Φ can have any value?

    Did i address the right question?
     
  4. Dec 15, 2014 #3
    Nope that wasnt my point.. I meant in terms of the frequency how many distinct sinusoids are there for discrete
     
  5. Dec 15, 2014 #4

    jim hardy

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    I guess "discrete" has some meaning in DSP that i don't know.

    Are you asking in reference to a computer program of some sort ?
    I'm guessing it's user defined

    You defined frequency as 1/N

    and
    one per amplitude.

    But i am likely way off target , so will await a better informed response.
     
  6. Dec 15, 2014 #5
    Complex sine? This is a complex sine:
    sin z = sin(x+iy) = sin(x)⋅cos(iy) + cos(x)⋅sin(iy) = sin(x)⋅cosh(y) + i⋅cos(x)⋅sinh(y)
    :)
     
  7. Dec 15, 2014 #6

    jim hardy

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    Aha ! And THANKS !
    A question well stated is half answered
    and as i said i'll await an answer from somebody who's more fresh on the subject than me.

    old jim
     
  8. Dec 15, 2014 #7
    Well, I'm not certain what he is asking. If he really means complex sine, that just complicates things ;)
     
  9. Dec 16, 2014 #8
    I was referring to e^jwn (where w is the angular frequency) as the complex sinusoid (thats how my book refers to it, i honestly thought it was not an ambiguous term)
     
  10. Dec 16, 2014 #9
    Thank you for the effort you spent here, but its my fault for not expressing the question unambiguously
     
  11. Dec 16, 2014 #10
    Where e^(jwn) referes to the discrete complex sinusoid with period N and angular frequency w
     
  12. Dec 16, 2014 #11

    jim hardy

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    Hmmmm

    Pure math is not my strength.
    I was taught a long time ago in high school electronics class how to use rotating phasors .
    But that was 1962 and just high school so we didn't study Euler's equation.
    And i observed with an oscilloscope the truism that when you add two sinewaves you get just another sinewave.
    So in my simple world, any sinewave can be broken into its real and imaginary parts but it's still just one sinewave.


    This tutorial agrees in the second slide, page 2 of 22, with my understanding ,of course in terms of t not n,
    http://web.mit.edu/6.02/www/s2007/lec3.pdf
    so e^(jwn) to me represents one sine function, represented by that rotating phasor.
    To me they rotate in time; what would rotating in 'n' mean physically i can't say.

    so i would answer your question: one.
    But i'm no Euler.

    hoping a math guru can help you further.

    old jim
     
  13. Dec 17, 2014 #12
    Time depedent sine function fn(t)=An⋅sin(ωn⋅t+φn) can be represented by a time depedent complex number in a complex plane, the rotating phasor zn(t):
    fn=Im {zn(t)}=Im {An⋅en⋅ent}
    In t=0 its' position is determined by modulus and initial phase angle φn:
    zn(0)=An⋅eφn
    After time t, this phasor gets rotated by phase angle (ωn⋅t+φn) in a complex plane.
    IF all the rotating phasors have the same angular frequency ω, corresponding sine functions can be represented in the same picture by still phasors zn(0)=An⋅en, where coordinate system rotates in the opposite direction at constant frequency ω.
    Superposition of two or more sine functions gives another sine function only if their frequencies are equal

    Sorry, I don't understand what you're trying to ask
     
  14. Dec 17, 2014 #13

    jim hardy

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    Of course - i neglected to add that condition.
    adding different frequency sines gives a complex complicated wave
    I've done that both with oscilloscope in high school and decades later with Fourier in Basic.

    Thanks, zoki that helps me a lot.
     
  15. Dec 17, 2014 #14
    I know that you know/understand this, but I'm not sure for OP. Maybe this remark could help him.
     
  16. Dec 17, 2014 #15

    jim hardy

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    You overestimate me, zoki.

    I never understood Laplace transforms just learned to use them .
    I have a hard time imagining √-1 , so just think of j as making a quarter turn because if you do that twice you've reversed direction...or negated a vector.
    One needn't understand impulse and momentum to use a hammer.....

    I envy you fellows who are fluent in higher math. Perhaps at my advanced age i should have another go at it - i have more patience now than in my twenties.

    thanks again,

    old jim
     
  17. Dec 17, 2014 #16

    Baluncore

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    In my opinion the answer to the OP can be either 1 or 2, depending on the interpretation of counting “how many” and the meaning of “distinct sinusoid”. The term “discrete complex sinusoid” is singular.

    Consider Z = a + bi.
    1. The “complex number” Z is singular. It represents only one “point” on the complex plane.
    2. But Z can be resolved into the two separate real numbers a and b, kept orthogonal by the “turn left” operator i.
     
  18. Dec 18, 2014 #17
    Okay i understand this, but perhaps its easier to just memtion where i came across my question... I
    Expressing a discrete function as fourier series means that you take the summation of k from 0 to N-1 of A[k]e^(-jkwn).. What i really wanted to know is why is the summation only from 0 to N-1 for discrete but from minus infiniti to infiniti for continuous? My book for signals and systems (by simon haykin) says that there are only N discrete complex sinusoids (e^jkwn) where N is the fundamental period of e^jwn, but how and why?
     
  19. Dec 18, 2014 #18

    donpacino

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    I dont know how to derive all of it, but this may be a start

    You can think of N as the number of discrete points making up the sinusoids.
    In order for the discrete points to represent a sinusoid, there must be at least one discrete point per period and N must be divisible by pi (which will mean that the discrete points will show up at the same amplitude each period and will show up each period).

    This also means that the smallest number of discrete points that can represent one cycle of a sinusoid is 1.

    I know that information is part of the proof, but im not sure where to go from there
     
  20. Dec 18, 2014 #19

    Baluncore

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    The discrete transform between time and frequency is numerically reversible. When reversed it will return the original dataset. It is important therefore that information be stored in both the time and frequency domains with the same “resolution".

    For N points in time, complex time data from 0 to N-1 is transformed to complex frequencies 0 to N-1. The frequency 0 must be present as it represents the DC offset. The complex time data is actually constructed from 2*N real samples. This satisfies the requirement that only N complex frequencies can be transformed from 2*N real data points.

    See; http://en.wikipedia.org/wiki/Nyquist–Shannon_sampling_theorem “In the field of digital signal processing, the sampling theorem is a fundamental bridge between continuous signals (analog domain) and discrete signals (digital domain).”
     
  21. Dec 20, 2014 #20

    meBigGuy

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    Basically, if there are 128 points in your original discrete time waveform, there will be 128 points in your discrete fourier transform.
     
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