# Fourier transform basic

Discussion in 'Mathematics and Physics' started by neptune, Dec 17, 2011.

1. ### RatchitWell-Known Member

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neptune,

What it seems and what it is are two different things. Finding the sinusoidal equivalent is what Fourier analysis does.

It depends on how you define a dimension.

There are components in a signal that are orthogonal to each other with repect to phase. It is not a matter of what they have to be, it is a matter of what they are.

Ratch

2. ### neptuneMember

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ok let me ask some basic first, can you plot graph of i.sinx ? where i=imaginary

3. ### MrAlWell-Known MemberMost Helpful Member

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Hi neptune,

I got an idea, why do you plot that, or better yet, why dont you plot this:

y=cos(x)+sin(x)

and then plot this one:

Y=cos(TH)+i*sin(TH), TH=Theta which is angle.

and then compare the two. You'll see a very marked difference because one is in one dimension and the other is in two dimensions (upper case Y for complex). Note what the plot of y looks like and what the plot of Y looks like.

Last edited: Dec 29, 2011

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5. ### MrAlWell-Known MemberMost Helpful Member

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Hi again,

I wasnt sure if this was clear enough so i went ahead and graphed both of those functions for you. See the attachment.

Note the red plot is not closed because i only ran Theta (TH) from 0 to 6.2 instead of from 0 to 2*pi (6.283...) to show where the start of the plot is and where the end of the plot is.

Clearly the top plot cos(x)+sin(x) is just a phase shifted sinusoidal wave. To produce this plot each value of cos(x) is simply added to each value of sin(x) for every x from 0 to 6.2 .

Clearly the bottom plot cos(TH)+i*sin(TH) is a circle, where the x coordinate is obtained from cos(TH) and the y coordinate is obtained from sin(TH). Note this is not a simple addition of the two terms as the above plot, but a two dimensional parametric representation.

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6. ### neptuneMember

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Hello,
sorry was busy in new year Celebration
You have ploted Cosx , iSinx in complex plain. but you have not added them , but lets leave it i asked iSinx graph in X,Y plane , obviously it cannot be ploted because it is imaginary.
and also Cosx + Sinx can be plotted on Complex Graph but it would be one dimension, There value would only be on Real axis.
so when a Simple time domain signal is coming it can be easily plotted in both X,Y plain, But would become 1 dimensional signal in Complex plain , it would keep on Varying on real Axis. So what it all has to do wiyh phase ?

7. ### MrAlWell-Known MemberMost Helpful Member

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Hi,

cos x and i sin x do not add directly. How many times do you have to hear this before you actually sit down and try an example? That's how you start to understand why and how this works. You can add cos x and sin x, but you cant add cos x and i sin x. That's because cos x is the real part and sin x is the imaginary part, and you do not add a+bj directly because a is real and b is imaginary. This is where the phase shift comes into play. Because we have two numbers rather than one, we can say that there is a phase shift and an amplitude rather than saying that there is a real and imag part.

If you sit down and do the pulse example i was talking about you'll see how this all fits together. The ck has a real and imaginary part which can be converted into a phase and amplitude. Once converted into phase and amplitude we can use that information to show the signal as a single sin or cos wave with a phase shift rather than both a sin and cos wave added together.

Last edited: Jan 2, 2012
8. ### neptuneMember

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Hello ,
well i tried to find Ck by hand for this Ck = 1/2 ∫ 1.e^(-j.k.2.pie.t) dt limits 0.50 to 0.75
But this is giving me personnal nightmares to find solution.
So can i use another limits for this pulse function. like 1 to 2

9. ### MrAlWell-Known MemberMost Helpful Member

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Hi again,

Let me see if this helps.

The integral of e^(-i*k*w0*t) dt from t0 to t1 is:
i*e^(-i*k*t1*w0)/(k*w0)-i*e^(-i*k*t0*w0)/(k*w0)

(You can probably see all we did was a simple integration of the form e^nt and then applied t0 and t1)

Now that solution allows the selection of any starting and ending point for the pulse, start at t0 and end at t1 as long as t0 and t1 are less than the repetition period.

To proceed you could turn that expression into sin and cos terms, where there will be a real part and an imaginary part. You then do the solution using that and you end up with a real and imaginary part of the ck being done. There may be two sinusoidal terms for each part for a total of four sinusoidal terms, two of which are imaginary and so form the imaginary part.
For example if you end up with something in this form:
A*sin(xt)+B*sin(xt)+i*C*cos(xt)+i*D*cos(xt)

then the real part is:
A*sin(xt)+B*sin(xt)

and the imag part is:
C*cos(xt)+D*cos(xt)

You'll see how easy this really is once you get going.

After we do this we then turn it into a sinusoidal signal with a combined amplitude and a phase shift instead of two terms for each component.

Last edited: Jan 5, 2012
10. ### neptuneMember

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Taking my math from here ... what should be the solution of ??? in pic

There is no math editor on Math site ! Where are all old symbols gone >:-{

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11. ### MrAlWell-Known MemberMost Helpful Member

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Hi,

Instead of doing that use Euler identities to convert the form i gave you for the integral into sine and cosine terms. Then go from there.

Oh yeah, the other constraint is t1>t0 but that's obvious from the pulse shape we want to deal with because the pulse starts at t0 and ends at t1.

Last edited: Jan 6, 2012
12. ### neptuneMember

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Euler's identity is defined for +ve power of e as e^i.pie = -1, there is no +ve power of e here !

13. ### MrAlWell-Known MemberMost Helpful Member

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Hi,

Here are some identities you should check out:
e^(i*x)=cos(x)+i*sin(x)
e^(-i*x)=cos(x)-i*sin(x)

Do you know how to find the identity for:
i*e^(-i*x)

???

14. ### neptuneMember

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hey That what i have done.. in Pic attached , i dont know what to do with this function CosK.3π/2 - j.sink.3π/2 after opening its identity..

15. ### MrAlWell-Known MemberMost Helpful Member

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Hi again,

I dont see any pic attached, but lets more forward a little here anyway...

After we integrate we get two exponential terms, then we convert them to sin and cos terms and we get:
sin(k*t1*w0)/(k*w0)-sin(k*t0*w0)/(k*w0)+i*cos(k*t1*w0)/(k*w0)-i*cos(k*t0*w0)/(k*w0)

The above contains two terms that are real and two terms that are imaginary. The two terms that are imaginary are the ones with the multiplication by the operator "i". We separate them into the real and imag components by simply regrouping them:
real=sin(k*t1*w0)/(k*w0)-sin(k*t0*w0)/(k*w0)
imag=cos(k*t1*w0)/(k*w0)-cos(k*t0*w0)/(k*w0)

So now for each k we will have one complete real part and one complete imag part, which forms a single complex number:
real+i*imag

t0 is the start of the pulse, t1 is the end of the pulse, w0 is the angular frequency of the repetition period Tp. The frequency of the pulse is F=w0/2pi, so the repetition period is 1/F=2pi/w0. So with a period of 1 second we have F=1Hz and so w0=2*pi*F=2*pi.

To proceed we use the two equations above and replace w0 and t0 and t1 with their values and calculate a real and imag part for each k. We thus end up with one complex number real+i*imag for each k.

16. ### neptuneMember

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The Ck = 1/2 ∫ 1.e-jkωt dt limits 0.5 to 1.5 <-- This is shlifted version
is coming = -1/jkπ{e^-jk3π/2 - e^-jkπ/2} when i open by Eulers identity..
1> CosK.3π/2 - j.sink.3π/2 when we open this with K=1,2,3... we get
SINE 0,-1,0,1,0,-1,0....
COSINE1,0,-1,0,1,0,-1....
And
2> CosK.π/2 - j.sink.π/2
SINE 0,1,0,-1,0,1,0....
COSINE1,0,-1,0,1,0,-1....

So resultant of -1/jkπ{e^-jk3π/2 - e^-jkπ/2} = -2/jkπ , 2/jkπ , -2/jkπ ...For K=1,2,3....
So it is oscillating series -,+,-,+,-... at 0.5 to 1.5 time
What should i take resultant as ?

17. ### MrAlWell-Known MemberMost Helpful Member

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Hi again,

Evaluating the real part for t0=0.5 and t1=0.75 we get:
sin(1.5*pi*k)/(2*pi*k)-sin(1.0*pi*k)/(2*pi*k)
Evaluating this at k=1 we get:
-1/(2*pi)=-0.159155 approximately

Note that to evaluate this for k=0 we have to take the limit of that expression as k approaches zero because it can not be evaluated directly as it can with k=1 or greater.

18. ### neptuneMember

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Ok i have managed to know Fourier series from all this discussion.
and i posted a little pic for you to see my inference from it.
(Please message me if you cant see image rathere then posting here)

But i still dont understand Fourier Transform, why is it different from fourier series.
Wikipidea is not helping me in it..
(If possible please explain me with drawings)

what i understand from Fourier series is there is nothing as phase of signal.
signals are defined by Cos wave or Sine wave, you cant add both of them therefore they write both of these in Complex form.

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19. ### MrAlWell-Known MemberMost Helpful Member

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Hi,

I can see your drawing now. Im not sure what you mean by "Sine wave can not define it".

I gave you the solution to one part of the expression, cant you take it from there and do the others the same way? If not i'll give more info.

If you can not tell the difference between the Fourier Series and Fourier Transform that's probably because you've so far only looked at periodic signals. To get more insight into this, try to find the Fourier Series of the SINGLE unit impulse located at t=t0 (this is a one time only impulse that does NOT repeat). This signal looks like this:
______|____ (and that impulse is located at t=t0 and never repeats)

To see how the phase comes into play, i'll post more if you still dont see it yet.

Last edited: Jan 12, 2012
20. ### neptuneMember

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FT of impulse signal is 1 all over spectrum

21. ### MrAlWell-Known MemberMost Helpful Member

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Hi again,

That's nice, but i didnt say, "try to find the Fourier Transform of the single unit impluse", i had said, "try to find the Fourier Series of the single unit impulse". Think about this for a minute