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Fourier-Analysis

Forum|Forum|9 years ago
November 3, 2016
4 replies
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Hello!

What's wrong with the last Fourier-Analysis of a sinus Function?

Thank you for your Help.

Volker

1_Fourier_Analysis.pdf

Statistics_Analysis

Best answer by Werner_E

> The wave is not fully displayed.

And if you would read my various answers her you also would know why.

You don't have to use the Heaviside function to define your signal, but doing so forces all values outside the given intervals to be zero.

In your more clear representation with the if else structure the signal is not defined outside and so you got the error, You could add a simple else at the end to set the function to zero for all other values - which would be good practice anyway.

Your signal is (now) zero for -1 < x < -3/4 but it should be part of sine in this range. Thats the reason your signal is not fully represented by the Fourier-series.

And as already written there are two ways to deal with the problem:

1) Define f.3 in a way so that it also shows the sine for values lower than -3/4

2) Don't integrate from -L to L in FC which is from -1 to 1 in case of your f.3. Use a more flexible function FC as already suggested. In case of your (or F.M.'s) f.3 you would have to integrate at least from -3/4 to 5/4 to get the result you expect.

L

LucMeekes

23-Emerald IV

Can't see. Prime sheet lacks.

From the PDF: What is the difference between the function definitions for FC(f,N,L), FC.1(f,BN,L), FC.2(f,BN,L) andFC.3(f,BN,L)? They appear to be th same !

What if, instead of the offending res.3:=FC.3(f.3,Nt,L)

you call: res.3:=FC(f.3,Nt,L) ?

Luc

V

vlehner

Author

1-Visitor

Luc,

"From the PDF: What is the difference between the function definitions for FC(f,N,L), FC.1(f,BN,L), FC.2(f,BN,L) andFC.3(f,BN,L)? They appear to be th same !"

There are 3 Functions: f, f1, f2 displayed in diagramms and converted into Fourier-functions which are also displayed in diagramms.

"

What if, instead of the offending res.3:=FC.3(f.3,Nt,L)

you call: res.3:=FC(f.3,Nt,L) ?"

I have created a seperat worksheet without the indizes "3" and the error occured again -see attached pdf.

"Can't see. Prime sheet lacks."

I hadn't sent you a mathcad-File

Thanks, Volker

3_Fourier-Analyse_1-mcdx.zip

2_Berechnungen-mcdx.zip

1_Fourier_Analysis_1.pdf

W

Werner_E

25-Diamond I

You should at least have told us what exactly the error message is you get at the definition of res.3

The reason for the error seems to be that you integrate from -1 to 1, but your function f.3 is not defined in the range -1 to -3/4!

It would be more versatile if you rewrite your FC function with two arguments for the start and end of integration interval instead of providing just L. That way you are more flexible in defining the functions you wish to analyze.

Some remarks: There is no need for different functions FC.1, FC.2, FC.3. One fits all.

And in Mathcad you can write something like if -2 < x < 4, so you don't need to write if ( x>-2 ) ^ ( x < 4 ).

L

LucMeekes

23-Emerald IV

"Some remarks: There is no need for different functions FC.1, FC.2, FC.3. One fits all."

That was my point in asking for any differences between them...

Luc

W

Werner_E

25-Diamond I

> That was my point in asking for any differences between them...

Yes, and I found no differences at all, thats making all but one of them obsolete. They are all the same and assume a signal f which is defined in [-L; L] which seems to be the problem as f.3 is not defined at the full range which sure would throw an error.

Therefore I suggested to rewrite FC (one fits all) so that it works with a flexible interval. In case of f.3 it would be avoidable effort to define the parabola additionally i -1 <x< -3/4.

-

-MFra-

21-Topaz II

Hi Volker

f3 is not well defined for mathcad Prime. Using the step function, everything works.

Fourier series.jpg

W

Werner_E

25-Diamond I

> Using the step function, everything works.

No, it doesn't. You set the interval -1 to -3/4 to zero that way, but there should be part of a parabola here!

So you are analyzing a different signal as was intended. There should be a parabola in the range 1 <x< 1.5 but you have just zero there! I am talking about the fourier-row here, not the signal you defined

You see the error even in your coarse approximation with five sines.