Continue to Site

Welcome to EDAboard.com

Welcome to our site! EDAboard.com is an international Electronics Discussion Forum focused on EDA software, circuits, schematics, books, theory, papers, asic, pld, 8051, DSP, Network, RF, Analog Design, PCB, Service Manuals... and a whole lot more! To participate you need to register. Registration is free. Click here to register now.

Register Log in

Help me find the proof for a Fourier Transform equation

Status
Not open for further replies.
safwatonline

safwatonline

Advanced Member level 4
Joined
Nov 19, 2005
Messages
1,341
Helped
219
Reputation
438
Reaction score
48
Trophy points
1,328
Location
EGYPT
Activity points
9,038
Hello,
i was wondering if someone can provide a proof for the FT of u(t) which is equal to (1/jω)+(πδ(ω))


as when i solve it using the direct formula it gave me the first term only, also using some properties along with u'(t)= δ(t) also gives the first term only, so if someone can provide the proof and tell me whether the methods i used are wrong.

regards,
Safwat
 

Re: Fourier Transform

Assume u(t) = 1 when t > 0.
Then, step function can be considered as u(t) = ½ + ½*sgn(t)
Let F denotes Fourier transform.
F(1/2) = π*δ(ω)
F(sgn(t)) = 2/jω (It can be derived from F(1/t) = k*sgn(ω), where k is a constant I forgot..
So, F(u(t)) = F(½ + ½*sgn(t)) = π*δ(ω) + 1/jω
 

    safwatonline

    safwatonline

    Points: 2
    Helpful Answer Positive Rating
Fourier Transform

thnx johnchau,
ur solution seems working, but i still cannot get why using the direct formula or u'(t)=delta(t) fails and gives wrong answer, this is really confusing me.
 

Fourier Transform

Hi safwatonline,

The integrals you found applying the direct formula are improper. They do not converge and need special consideration for w=0.
Regards

Z
 

    safwatonline

    safwatonline

    Points: 2
    Helpful Answer Positive Rating
Fourier Transform

zorro said:
Hi safwatonline,

The integrals you found applying the direct formula are improper. They do not converge and need special consideration for w=0.
Regards

Z
Click to expand...

Code: 
what i did is F{u(t)}=∫u(t)*(e^-iwt)dt     (-∞,∞)
                            =∫(e^-iwt)dt            (0,∞)
                            =(e^-iwt)/(-iw)         (t=0,t=∞)
                            =1/jw

i don't get why it is improper.

Added after 1 minutes:

beside i also used the differentiation property along with u'(t)=δ(t) and it gave the same value as the integration
 

Re: Fourier Transform

hi,

when thinking of the fourier transform of u(t) away from the direct formula it makes sense to have the term πδ(ω) since u(t) has a dc component of 1/2, that is its average value is 1/2, and so it must have an impulse at w=0.

but when i applied the direct formula and the properties of the forier transform on the derivative which is δ(t) , as you have explained, i became confused my self.

so i searched through the net and this is the closest answer that i found at **broken link removed**

For a number of signals of interest, the Fourier transform integral does not converge in the usual sense of elementary calculus. Some of these signals can be treated in a consistent fashion by admitting Fourier transforms that contain impulses. For example, if , the unit-step signal, then

x(w) = (1/jω)+(πδ(ω))

For such a Fourier transform, we treat impulse components as separate in computing the magnitude spectrum since an impulse is zero at all values of but one, though admittedly something very special happens at that one point. Thus

|x(w)| = (1/ω)+(πδ(ω))
Click to expand...
 

    safwatonline

    safwatonline

    Points: 2
    Helpful Answer Positive Rating
Fourier Transform

well, i do agree that the known solution is the right solution but if this is a special case then the integration should be wrong and there should be like a rule when to use the integration.
what i mean is that ok i know the problem in u(t) but is there some other functions with the same issue.
 

Re: Fourier Transform

safwatonline said:
zorro said:
Hi safwatonline,

The integrals you found applying the direct formula are improper. They do not converge and need special consideration for w=0.
Regards

Z
Click to expand...

Code: 
what i did is F{u(t)}=∫u(t)*(e^-iwt)dt     (-∞,∞)
                            =∫(e^-iwt)dt            (0,∞)
                            =(e^-iwt)/(-iw)         (t=0,t=∞)
                            =1/jw

i don't get why it is improper.
Click to expand...

I mean "improper" in the mathematical sense (https://en.wikipedia.org/wiki/Improper_integral , although i would prefer a calculus book rather than this wikipedia article).
For the value w=0, the integral does not converge: it becomes ∫dt (0,∞) ; the result is not finite and it is needed to consider that special case (w=0) separately.

Regards

Z
 

    safwatonline

    safwatonline

    Points: 2
    Helpful Answer Positive Rating
Fourier Transform

well, i know that the result is not finite but in the correct answer (1/jw)+pi*delta(w) the answer doesn't converge also at w=0
 

Re: Fourier Transform

this means that at all w except w=0 the fourier transform is 1/jw, and at w=0 it is pi*δ(ω). hence, as zorro mensioned, considering the special case of nonconvergence at w=0.

i think that the problem is with the way of writing the fourier transform, which is using the addition between the two terms, which doesn't exclude w=0 from being applied to 1/jw.
 

    safwatonline

    safwatonline

    Points: 2
    Helpful Answer Positive Rating
Re: Fourier Transform

ok,what i mean is that lets for example try to get the Fourier transform of sgn(t) , the method normally used is by getting the F.T. of [e^-|t|/n)]*sgn(t) using the direct formula
please see attached file
u will notice that the result is F(sgn(t))=2/jw , which is exactly the same as in u(t) but here we didn't say there is something wrong in the result.
 

Re: Fourier Transform

but does the fourier transform of sgn(t) have a value at w=0?

it doesn't.

in u(t) we have some DC component (=1/2) while sgn(t) doesn't have DC component (=0). and so in u(t) we have the impulse while in sgn(t) we don't.

well, this is my personal understanding of the problem.

can't deny that it is confusing, since i have never seen any statement that excludes w=0 neither in u(t) nor in sgn(t) .
 

Re: Fourier Transform

safwatonline said:
well, i do agree that the known solution is the right solution but if this is a special case then the integration should be wrong and there should be like a rule when to use the integration.
Click to expand...

In the formal definition of the Fourier Transform, there is a rule as to when the integration is valid, and it has to do with L1- and L2-integrable functions. The same applies for the inverse Fourier Transform. Things like the unit step, the signum function, and the delta impulse aren't, in the strictest sense, treated as functions in the Fourier Transform, but rather distributions, and are obtained by taking limits of valid Fourier integral results. However, integrating the delta function or applying the Fourier Transform formula to it doesn't actually jive mathematically.

If you're having trouble thinking about this, and don't like to just blindly accept that those Fourier Transforms are satisfied, you can check out Section 5.11 in Lokenath Debnath's "Introduction to Hilbert Spaces" to get a formal understanding of where the Fourier Transform comes from, but this will probably require that you read a good portion of the book to get to that point.
 

Status
Not open for further replies.

Similar threads

Part and Inventory Search

Welcome to EDABoard.com

Sponsor

Back
Top