Are all discrete sinusoids distinct?

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?

 

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.

 

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)
:)

 

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 ;)

 

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

 

Time depedent sine function f_n(t)=A_n⋅sin(ω_n⋅t+φ_n) can be represented by a time depedent complex number in a complex plane, the rotating phasor z_n(t):
f_n=Im {z_n(t)}=Im {A_n⋅e^jφ_n⋅e^jω_nt}
In t=0 its' position is determined by modulus and initial phase angle φ_n:
z_n(0)=A_n⋅e^φ_n
After time t, this phasor gets rotated by phase angle (ω_n⋅t+φ_n) 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 z_n(0)=A_n⋅e^jφ_n, 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

 

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.

 

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

 

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.

 

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

 

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).”

 

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.

 

How would you place N discrete points between +infinity and –infinity?

 

That is N discrete samples in time, giving N/2 discrete frequencies.