Are all discrete sinusoids distinct?

  • #1
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Main Question or Discussion Point

For the discrete complex sinusoid with period N, how many distnict sinusoids are there? And why?
 

Answers and Replies

  • #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?
 
  • #3
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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?
Nope that wasnt my point.. I meant in terms of the frequency how many distinct sinusoids are there for discrete
 
  • #4
jim hardy
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I meant in terms of the frequency how many distinct sinusoids are there for discrete
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

[URL said:
http://www.mathworks.com/help/dsp/ref/sinewave.html][/url]
Description
The SineWave object generates a discrete-time sinusoid. The sine wave object generates a real–valued sinusoid or a complex exponential. A real-valued, discrete-time sinusoid is defined as:

y(n)=Asin(2πfn+ϕ)

where A is the amplitude, f is the frequency in hertz, and φ is the initial phase, or phase offset, in radians.
A complex exponential is defined as:

y(n)=Aej(2πfn+ϕ)

For both real and complex sinusoids, the amplitude, frequency, and phase offsets can be scalars or length-N vectors, where N is the desired number of channels in the output. When you specify at least one of these properties as a length-N vector, scalar values specified for the other properties are applied to each of the N channels.
and
http://www.mathworks.com/help/dsp/ref/sinewave.html said:
Generating Multichannel Outputs
For both real and complex sinusoids, the Amplitude, Frequency, and Phase offset parameter values (A, f, and ϕ) can be scalars or length-N vectors, where N is the desired number of channels in the output. When you specify at least one of these parameters as a length-N vector, scalar values specified for the other parameters are applied to every channel.

For example, to generate the three-channel output containing the real sinusoids below, set Output complexity to Real and the other parameters as follows:

  • Amplitude = [1 2 3]

  • Frequency = [1000 500 250]

  • Phase offset = [0 0 pi/2]
y1 = sin(2000πt) ; (channel 1)
y2 = 2sin(1000πt) ; (channel 2)
y3 = 3sin(500πt+π2) ; (channel 3)
one per amplitude.

But i am likely way off target , so will await a better informed response.
 
  • #5
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For the discrete complex sinusoid with period N, how many distnict sinusoids are there? And why?
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)
:)
 
  • #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
 
  • #7
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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
Well, I'm not certain what he is asking. If he really means complex sine, that just complicates things ;)
 
  • #8
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Well, I'm not certain what he is asking. If he really means complex sine, that just complicates things ;)
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)
 
  • #9
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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.
Thank you for the effort you spent here, but its my fault for not expressing the question unambiguously
 
  • #10
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For the discrete complex sinusoid with period N, how many distnict sinusoids are there? And why?
Where e^(jwn) referes to the discrete complex sinusoid with period N and angular frequency w
 
  • #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
 
  • #12
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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.
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

For the discrete complex sinusoid with period N, how many distnict sinusoids are there? And why?
Sorry, I don't understand what you're trying to ask
 
  • #13
jim hardy
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Superposition of two or more sine functions gives another sine function only if their frequencies are equal
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.
 
  • #14
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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.
I know that you know/understand this, but I'm not sure for OP. Maybe this remark could help him.
 
  • #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
 
  • #16
Baluncore
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For the discrete complex sinusoid with period N, how many distnict sinusoids are there? And why?
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.
 
  • #17
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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?
 
  • #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
 
  • #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).”
 
  • #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.
 
  • #21
Baluncore
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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.
What do you mean by "point"?
If there are 128 samples in your time record then there will be 64 discrete phasors in your frequency domain. Each phasor is represented by a Cos() and a Sin() component, the combination of which is a single sinusoid specified by amplitude and phase.
 
  • #22
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Thank you all but i am sorry that none of you seem to have solved my problem.

If i can simply restate the question... Why does the discrete fourier series summation go from zero to N-1 wheras for continuous fourier series, the summation goes from negative infiniti to infiniti...
 
  • #23
Baluncore
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How would you place N discrete points between +infinity and –infinity?
 
  • #24
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How would you place N discrete points between +infinity and –infinity?
Why N
 
  • #25
Baluncore
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from zero to N-1
That is N discrete samples in time, giving N/2 discrete frequencies.
 

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