Fourier Transform question (kind of Urgent)

  • Thread starter toneboy1
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  • #1
toneboy1
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This is for an assignment, (not sure if its in the right section) but anyway I'm considering the system response to H(w) = 10/(jw + 10)

when the input is x(t) = 2 + 2*cos(50*t + pi/2)

so I know that Y(w) = X(w).H(w) but I'm not sure what to do about the '2 + ' in the input.

I know that '1' transforms into 2piδ(w) frequency domain.

so does this mean that 2 becomes 4piδ(w) ? times by H(w)...?

And if so, does this mean it in x(t) is only 4pi at w = 0 ? Because the w = 50 from that input.

Alternatively I concidered 2*H(0) or 2*H(50), but I see no reason for them.

My calculation would be as follows:

% y(t) = 2*something??* + |H(50)|*2.*cos(50*t + pi/2 + <H(50) )
%
% where <H(50) = ( 10<0 / 50.99<78.69 )
% and 360 - 78.69 = 281.31 and plus pi/2 that equals 11.31 deg
% in rad that is 0.197
%
% y(t) = 2 + 0.392*cos(50*t + 0.197)


Please help asap!

Thanks
 

Answers and Replies

  • #2
rude man
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You shold stay in the Fourier domain until the end.

You have X(w) = 4πδ(w) + the Fourier transform of 2*cos(50*t + pi/2)
You might wanty to exploit the fact that cos(x+π/2) = -sin(x)
You have H(w) = 10/(jw + 10)
So you have Y(w) = X(w)H(w)
and finally you have y(t) = F-1(Y(w).
 
  • #3
toneboy1
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You might wanty to exploit the fact that cos(x+π/2) = -sin(x)

What you said makes perfect sense, except that we were told that
x(t) = Acos(w0t + theta) = Re[Ae^(j.theta).e^(j.w.t)] (I'm assuming from Euler identity)

and so the response is y(t) = A|H(w0)|cos(w0t + theta + <H(w0) )

So I don't have to inverse that bit?

Ah, I thought it might be 4πδ(w), so by what you've said I'd need to inverse transform:
4πδ(w)* 10/(jw + 10) = Y(w) of the response to '2 +'.

What would that be? (is w = 50 or 0?)

Thanks!
 
  • #4
rude man
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Ah, I thought it might be 4πδ(w), so by what you've said I'd need to inverse transform:
4πδ(w)* 10/(jw + 10) = Y(w) of the response to '2 +'.

What would that be? (is w = 50 or 0?)

Thanks!

w is not a constant.

What is the inverse transform integral? I.e. going from Y(w) to y(t)?
 
  • #5
toneboy1
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w is not a constant.

What is the inverse transform integral? I.e. going from Y(w) to y(t)?

Good point!

U'mm, ok so Y(w) being 40piδ(w)

y(t) = 1/2pi∫-∞→∞ [H(w).Y(w).e^(j.w.t)] dt for -∞=< t =< ∞

y(t) = 1/2pi∫-∞→∞ [(4piδ(w).e^(j.w.t))*10/(jw + 10)] dt for -∞=< t =< ∞

so y(t) = 1/2pi∫-∞→∞ [(40piδ(w).e^(j.w.t))/(jw + 10)] dt for -∞=< t =< ∞

therefore y(t) = (20.e^(jwt))/(10j.w - w^2) ? that doesn't look right
 
  • #6
rude man
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Good point!

U'mm, ok so Y(w) being 40piδ(w)

y(t) = 1/2pi∫-∞→∞ [H(w).Y(w).e^(j.w.t)] dt for -∞=< t =< ∞

y(t) = 1/2pi∫-∞→∞ [(4piδ(w).e^(j.w.t))*10/(jw + 10)] dt for -∞=< t =< ∞

so y(t) = 1/2pi∫-∞→∞ [(40piδ(w).e^(j.w.t))/(jw + 10)] dt for -∞=< t =< ∞

therefore y(t) = (20.e^(jwt))/(10j.w - w^2) ? that doesn't look right

No, we said X(w) = 4πδ(w), right? So Y(w) = X(w)H(w) with H(w) = 10/(10 + jw).

So since y(t) = F-1{Y(w)}, the inverse integral had better be (1/2π)∫-∞+∞X(w)H(w)ejwtdw, don't you think?

BTW yes you could compute the cosine part of your output the way you said. That is the phasor method which is normally used with continuous sinusoidal inputs. For that matter, you could have handled the "2" input the same way: |H(0)| = 20/(10 + jw) with w = 0 so output = 2. But I thought you were supposed to use the Fourier transform.

EDIT: I noticed later your integral is almost right. But the integrating variable is w, not t.

Make the correction to that integral the way I said and now ask yourself the $64,000 question: what is
-∞+∞f(x)δ(x)dx for any f(x)? Hint: it's a really easy integration!
Check with the sampling theorem if you don't know.
 
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  • #7
toneboy1
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Sorry, I ment X(w)!
Umm, its just finding the system response to the input, I was using the transform because I thought I had to.

That cosine part "phasor method" was just in my lecture notes, I'm still not 100% on the basic principle of how it works, hence my confusion in using it here with the '2 +' part of the input.

So X(w) being 40piδ(w) was what I did correct?

(I really am struggling with applying the δ function, I'm also trying to find the exponential Fourier series of: x(t) = Ʃ δ(t - kT) for k = -∞ → ∞ But I can't understand how to apply it.


Thanks again!

EDIT: I just saw your EDIT and disregard this, I haven't read it enough to properly reply
 
  • #8
toneboy1
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Ok, *composed reply:*
...
(I really am struggling with applying the δ function, I'm also trying to find the exponential Fourier series of: x(t) = Ʃ δ(t - kT) for k = -∞ → ∞ But I can't understand how to apply it.


EDIT: I just saw your EDIT and disregard this, I haven't read it enough to properly reply

EDIT: I noticed later your integral is almost right. But the integrating variable is w, not t.

Make the correction to that integral the way I said and now ask yourself the $64,000 question: what is
-∞+∞f(x)δ(x)dx for any f(x)? Hint: it's a really easy integration!
Check with the sampling theorem if you don't know.

Well from what I do understand about δ
-∞+∞f(x)δ(x)dx = f(0)

EDIT: [ I lost my train of thought and forgot to include the δ amongst my messy working out,
f(x) = 40pi, δ(x) = δ.exp(jwt)
so 1/2pi * Int [ f(x)δ(x) ] dx * H(w) = f(0)*H(w) = 40pi/(2pi(jw+10)) = 2 at w = 0

1/(jw+10) * [20*1] (the '1' being f(0)) I hope you don't see what I embarrassingly originally wrote]

So with 'w' as the variable and '2' as the input signal:

y(t) = 20/(10j.w + 10) = 2 at w = 0 because the input of '2' is DC, so to speak?


Appreciate your help so much!
 
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  • #9
toneboy1
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As I said, I hope you didn't see my un-edited last reply before I changed it, it would be enough to demoralise anyone from giving further advice ever again.
∫-∞+∞f(x)δ(x)dx = f(0) was a helpful relation, though I was wondering if: f(x) = 40pi, δ(x) = δ.exp(jwt) is the right way of 'putting it'?

Could I lastly ask, the Fourier series of:
x(t) = Ʃ δ(t - kT) for k = -∞ → ∞

I'm assuming they mean that 'T' is the period, so would x(t) be like: a δ on the multiples of the period, like -T, 0, T, 2T... → ∞ ?

so integrating that Fourier Series would Cn = 1/T ∫ 0→T [1* exp(-jKwt)] dt ?

Cheers
 
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  • #10
rude man
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∫-∞+∞f(x)δ(x)dx = f(0) was a helpful relation, though I was wondering if: f(x) = 40pi, δ(x) = δ.exp(jwt) is the right way of 'putting it'?

I don't understand this question, toneboy.

You did get the right answer for y(t) = 2 going the Fourier route and yes, that's a dc input (w = 0 → zero frequency = dc!).

If you want to practice your Fourier transform, doing the cosine input is a really good exercise! I just went thru it myself with another op & it had me in a spin for a bit. But, as I said, the conventional way is to use phasors. BTW the term 'phasor' is really not applicable for dc voltages. You just set w = 0 everywhere and get your answer pronto.


Could I lastly ask, the Fourier series of:
x(t) = Ʃ δ(t - kT) for k = -∞ → ∞

I'm assuming they mean that 'T' is the period, so would x(t) be like: a δ on the multiples of the period, like -T, 0, T, 2T... → ∞ ?

so integrating that Fourier Series would Cn = 1/T ∫ 0→T [1* exp(-jKwt)] dt ?

Cheers

Yes, T is the time spacing between delta functions. This is a very important function in sample-data theory. The idea is every delta function δ(t - kT), k = 1,2, ... with the appropriate coefficient, represents one sample of a sampled continuous ("analog") function, like think voltage from a microphone. The series thus represents discretization, and suggests digitization, of the analog signal.

Finding the Fourier series is quite straight-forward. But why do you have "1" in your integrand for finding the complex Cn coefficients? (I assume you're referring to the exponential form of the series). Your function is not "1", it's δ!

Use the exponential form of the series and compute the Cn complex coefficients. The period is indeed T.

Cheers also.
 
  • #12
toneboy1
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I don't understand this question, toneboy.

Your function is not "1", it's δ!

I thought '1' because I thought that was the magnitude of δ at T was '1'.
would it be fair to say:

Cn = 1/T ∫ 0→T [δ(t-kT)* exp(-jKwt)] dt

= 1/-jkwT*[δ(T-kT)*exp(-jkwT) - δ(0-kT)*1]

=1/jkwT * δ(0-kT) ?


(thanks for the Fourier link btw)
 
  • #13
rude man
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I thought '1' because I thought that was the magnitude of δ at T was '1'.
The value of δ(t-T) at t = T is infinity!

It's the integral of δ(t-T) from -∞ to +∞ that's equal to 1.

would it be fair to say:

Cn = 1/T ∫ 0→T [δ(t-kT)* exp(-jKwt)] dt

= 1/-jkwT*[δ(T-kT)*exp(-jkwT) - δ(0-kT)*1]

=1/jkwT * δ(0-kT) ?

Well, first, you don't want to integrate from 0 to T because T is where the function is infinite & you need to include your entire function x(t) when integrating over one period T. You might be missing half of infinity! Let's say instead you integrate from t = T-ε to T+ε where ε is an arbitrarily small number; that would include your entire function over one period T, right? Then, realize that you can integrate δ(t-T) from -∞ to +∞ and get the same answer!

In what follows I assume Cn are the coefficients of the exponential form of the Fourier series.

I don't think your formula for Cn is correct. For one thing, n is missing on the right-hand side. One correct form for Cn is:

Cn = (1/T)∫ over one period T of δ(t - kT)e-j2nt/Tdt

By that convention then,

x(t) = Ʃ from m = -∞ to +∞ of Cmejmw0t
where w0 = 2π/T.

Now, think about solving that integral, bearing in mind what we said about solving an integral with δ(t-T) in it - how simple that is!
 
  • #14
rude man
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I notice now that you've confused n and k in your Cn expression, otherwise it's actually quite spot-on.
So it's δ(t - kT) but it's exp(-jnwt) but realize that your w is really 2π/T.
Also of course your evaluation of the integral is wrong. What did we say about ∫f(x)δ(x-x0)dx?

Bed time for me, be back in about 9 hrs. I'm at GMT-7.
 
  • #15
toneboy1
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Do you have like 1 more min! ? please!
 
  • #16
toneboy1
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Ah, I thought you'd be in a different time zone. I had no idea it was so late. (locally)

So Cn = 1/T ∫ T-ε→T+ε [δ(t-kT)* exp(-jn2pi.t/T)] dt

= 1/T*[exp(-jn.k.2pi)] (that is, only f(x0) remain at t = Tk and the 'T' in the exponent cancel out)

IF you're still there, is that right?

EDIT: forget it, go to bed lol, I don't want to keep you up, I need too long to think about it, thanks
 
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  • #17
rude man
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Ah, I thought you'd be in a different time zone. I had no idea it was so late. (locally)

So Cn = 1/T ∫ T-ε→T+ε [δ(t-kT)* exp(-jn2pi.t/T)] dt

= 1/T*[exp(-jn.k.2pi)] (that is, only f(x0) remain at t = Tk and the 'T' in the exponent cancel out)

IF you're still there, is that right?

EDIT: forget it, go to bed lol, I don't want to keep you up, I need too long to think about it, thanks

Right you are!
 
  • #18
toneboy1
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Right you are!

You're a genius, a legend and a saviour...not to mention a gentlemen and a scholor.

Thanks for your help, rest well.
 
  • #19
rude man
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BTW note that exp(j2πk) = 1 for integer k!
 
  • #20
toneboy1
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H'mm, I'm yet to reason why that is...

EDIT: ofcourse, Euler Identity!
 
  • #21
toneboy1
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So exp(-j2πk) = 1 too...

but does that work here with the 'n' in the exp?

EDIT:
I guess it doesn't matter because the 'n' is practically the coefficient of the 'k', so it would I suppose still = 1.
 
  • #22
rude man
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So exp(-j2πk) = 1 too...

but does that work here with the 'n' in the exp?

EDIT:
I guess it doesn't matter because the 'n' is practically the coefficient of the 'k', so it would I suppose still = 1.

You should wind up with Cn = (1/T)Ʃm = -∞ to +∞ exp(jmw0t). The "m" is a dummy variable, just for the summation. It's not the same as "n" which represents the harmonic number. And it's not the same as "k" which gives the time-displacement of each individual delta function.

What you may notice is the Fourier series for this function (sometimes called a "comb" function, for obvious reasons) has equal-amplitude components from dc all the way to infinity.

Suspecting you are in the UK somewhere: Cheerio!

(What would I do without your fabulous TV mysteries like Midsomer Murders , The Last Detective, and Morse, or your mystery writers, starting with Aggie C. herself!)

EDIT: actually, m is effectively the harmonic number also, really same as n. But my textbook separates them out of mathematical rigor (OK, 'rigour' to you! :-)
 
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  • #23
toneboy1
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I see...are you saying its just like: -1/T+ 0 + 1/T + 2/T ...

I had: say, summation from n = -1 to 1:exp(-j2*pi/T * t) / T + 1/T + exp(j2*pi/T * t) / T

Where have I gone wrong?



I'm reluctant to say Australia actually (because my questions have been so rudementary it's enbarrasing). But...

I Don't really like Poirot or miss Marple, or Jonothan Creek or Midsomer Murders for that matter, too slow, but my girl friend can't get enough of them!
I say if you're after mystery, you CAN'T beat Jeremy Brett as sherlock holmes, best series of the genre ever (ran for years too, so there's a lot to enjoy).
I've never seen 'Morse' but Mark from 'Peep show' seems to admire him, so that says something lol. (If you haven't seen Peep Show its definitely worth checking out, one of the best comedies in history.



EDIT:
I'm going to look into 'The Last Detective' I haven't heard of that.

Also, I remembered 'Dirk Gently', that's another great detective show!
 
  • #24
rude man
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Thanks for your tv recommendations! Ooh, Australia! Not that different from Arizona where I hang out. Is that the 'new' sherlock holmes? He talks so fast my aged brain can't follow up.

No, you're not expanding the summation correctly. The exponential series looks like
x(t) = (1/T){ ... C-2exp(-j2w0t) + C-1exp(-jw0t) + C0 + C1exp(jw0t + C2exp(j2w0t) + ... }.

For each harmonic you combine the -m and +m terms, which, after using the Euler identity, gives a real number for each harmonic component. Better review your exponential series stuff. If your course hasn't covered them I suggest doing it on your own. The exp. series usually simplifies the algebra.
 
  • #25
rude man
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I had: say, summation from n = -1 to 1:exp(-j2*pi/T * t) / T + 1/T + exp(j2*pi/T * t) / T

Where have I gone wrong?

Sorry, I missed that line, just saw the first one.
That's close, but why did you omit for m=+/-1? You only had m=+/.2 and m=0. Otherwise, perfect.
 
  • #26
toneboy1
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Sorry, it's twenty past 4 in the morning here and I'm not following, what did I omit? Are you talking about in: e^(...) like e^(j*m*2pi/T * t) because I put in -1, 0, 1 there.



I heard they legalised gold and silver as legal tender in Arizona, hopefully one day they will federally.
Yeah it gets pretty hot and dry here depending on the 5-10 yr weather cycle, but I'm from the mountains where its a bit more mild.


No, not that the new series of SH isn't good too but it's the series that ran from 1984 - 1994.
Dirk Gently on the other hand is more satirical and canceled due to budget cutbacks after about 6 episodes :P
 
  • #27
rude man
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Sorry, it's twenty past 4 in the morning here and I'm not following, what did I omit? Are you talking about in: e^(...) like e^(j*m*2pi/T * t) because I put in -1, 0, 1 there.

Sorry again, you did it right. I forgot that you're using 2π/T instead of w0.

I heard they legalised gold and silver as legal tender in Arizona, hopefully one day they will federally.
Yeah it gets pretty hot and dry here depending on the 5-10 yr weather cycle, but I'm from the mountains where its a bit more mild.
Nope, Au and Ag not legal tender here. But you can get what you want with it anyway, just like other places.

No, not that the new series of SH isn't good too but it's the series that ran from 1984 - 1994.:P

I agree, I like the 84-94 series. Just watched one yesterday in fact about a bloke who murdered wife #1, had wife #2 cooped up like an animal in his castle, and was working on #3, a real doll BTW.

Best thing about 'new detectives' is the humor. Too few writers do that.
 
  • #28
toneboy1
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Great! I could use some good news.
At the moment I'm Trying to understand what the question means:

[Edit: there was another question here but I think I was reading into it a bit too much]

So far I'm going around in circles of confusion and this thing's due in, in a few hours.
I'm using the ifourier() function in MATLAB for the first time tonight/this morrow and I had the 'syms a real' in the wrong spot and I had been trying to figure out why MATLAB S*#t itself.

If I:

w = -45:0.1:45;
H_amp = 10./(sqrt( (w).^2 + 100) ); % OR = 10./(j*w + 10);
X = 1./(j*(w+4)+2) + 1./(j*(w-4)+2) + 0.5./(j*(w+20)+2) + 0.5./(j*(w-20)+2);
x_small = X.*H_amp;
n = 0:.01:2.5;
x1 = 2*exp(-2*n).*cos(4*n).*u1(n) + exp(-2*n).*cos(20*n).*u1(n);
figure;
y = ifft(x_small);
plot(n,x1);
figure;
plot(y)


The graph, rather than being a filtered virsion of X, comes out as a bug splattered on a wind-screen. It has heaps of elements and the magnitudes of them are very small.

I'm impressed that you also enjoy that SH, though I can't say I remember that episode, I haven't seen all of The Casebook or Memiors of SH, eras.
 
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  • #29
rude man
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Great! I could use some good news.
At the moment I'm Trying to understand what the question means:

[Edit: there was another question here but I think I was reading into it a bit too much]

So far I'm going around in circles of confusion and this thing's due in, in a few hours.
I'm using the ifourier() function in MATLAB for the first time tonight/this morrow and I had the 'syms a real' in the wrong spot and I had been trying to figure out why MATLAB S*#t itself.

If I:

w = -45:0.1:45;
H_amp = 10./(sqrt( (w).^2 + 100) ); % OR = 10./(j*w + 10);
X = 1./(j*(w+4)+2) + 1./(j*(w-4)+2) + 0.5./(j*(w+20)+2) + 0.5./(j*(w-20)+2);
x_small = X.*H_amp;
y = ifft(x_small,w);
plot(y)


The graph, rather than being a filtered virsion of X, comes out as a bug splattered on a wind-screen. It has heaps of elements and the magnitudes of them are very small.

I'm impressed that you also enjoy that SH, though I can't say I remember that episode, I haven't seen all of The Casebook or Memiors of SH, eras.

I'm embarrassed to say I'm not familiar with matlab. This and similar math packages came along when this dog was too old to learn new tricks (for math I still use Derive occasionally but it's nothing like Matlab, Wolframm, mathCad etc.)

All I can offer is you know what the response to x(t) of 10/(jw1 + 10) should look like, right? Close to flat response for n(2π/T) < w1/10, then a 20 dB/decade rolloff. Those are the asymptotes, anyway. What are your w1 and T values?
 
  • #30
toneboy1
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Ah, precisely what I expected, which frustratingly means...'I'm all out of ideas'. Too late anyway I'm going to go in and hand this assignment in soon.

I'm sort of new to MATLAB but I am impressed by it's power, still if you've got no reason to...than you've no reason to.

Since I've just been up all night here, when I stumbled across sin(w) and I was trying to inverse Fourier transform it, I thought I'd done it before, it should be "elementary!", but now looking at it I tried and found myself integrating (from Euler's Identity):
1/2pi * [(exp(2jwt)-1)/2j] ∞→-∞

which obviously doesn't work or trying to use Duality on the table with no success in getting the: j√(pi/2)*[del(t+1)-del(t-1)] that wolfram generated as the answer.

Anyway, a question for another time perhaps.

Thanks for your assistance!
 
  • #31
rude man
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Since I've just been up all night here, when I stumbled across sin(w) and I was trying to inverse Fourier transform it, I thought I'd done it before, it should be "elementary!", but now looking at it I tried and found myself integrating (from Euler's Identity):
1/2pi * [(exp(2jwt)-1)/2j] ∞→-∞

which obviously doesn't work or trying to use Duality on the table with no success in getting the: j√(pi/2)*[del(t+1)-del(t-1)] that wolfram generated as the answer.

Anyway, a question for another time perhaps.

Thanks for your assistance!

Look at the link I sent you. It tells you how F{sin(wt)} is derived. They use the time-displaced delta function in the w domain in a clever way.

I don't know what field you hope to go into, but if it's electrical engineering I would de-emphasize the Fourier transform and concentrate on the LaPlace. The Fourier comes into its own in optics and statistical control system design (for which the two-sided LaPlace is equally suitable).
 
  • #32
toneboy1
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BTW I like the following site for the Fourier transform:

http://www.thefouriertransform.com/transform/fourier.php

Look at the link I sent you. It tells you how F{sin(wt)} is derived. They use the time-displaced delta function in the w domain in a clever way.

I don't know what field you hope to go into, but if it's electrical engineering I would de-emphasize the Fourier transform and concentrate on the LaPlace. The Fourier comes into its own in optics and statistical control system design (for which the two-sided LaPlace is equally suitable).

Right'o, I will concentrate on LaPlace because I am in Electrical Eng; I have done more LaPlace a couple of years ago, than I ever have Fourier.

In that link you sent me, I assume the Sinc and the Sin in which you refer are one and the same. I did notice that using L'Hopitals rule sinc(0) = 1 for division by 0 which makes sense.

[ On what might be a side note: I'm confused because my Transform table says (using κ for tau) Pof width κ(t) is transformed to κSinc(κω/2pi) but in examples I can see Pκ(t) is transformed into (2/ω)*sin(κω/2) Which would mean that Pκ(t) is transformed into κSinc(κω/2), so where has the pi in the denominator gone? ]


To be clear, we're talking about sin(w) FROM frequency domain to time domain aren't we?
(I was trying to use Duality to multiply Sin(w) by 2.pi to get the: j√(pi/2)*[δ(t+1)-δ(t-1)]
Wolfram generated.)

Cheers
 
  • #33
rude man
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Right'o, To be clear, we're talking about sin(w) FROM frequency domain to time domain aren't we?
(I was trying to use Duality to multiply Sin(w) by 2.pi to get the: j√(pi/2)*[δ(t+1)-δ(t-1)]
Wolfram generated.)

Cheers

I goofed again. I was thinking F{sin(wt)}, not F-1{sin(w)}. Sorry.

But the technique is similar: think the time-delayed delta function.
 
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  • #34
rude man
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Right'o, [ On what might be a side note: I'm confused because my Transform table says (using κ for tau) Pof width κ(t) is transformed to κSinc(κω/2pi) but in examples I can see Pκ(t) is transformed into (2/ω)*sin(κω/2) Which would mean that Pκ(t) is transformed into κSinc(κω/2), so where has the pi in the denominator gone? ]

The Fourier transform of a pulse of height 1 and width 2T is F(w) = (T/π)(sin(wT)/wT.
Some folks call sin(x)/x = sinc(x) but I've also seen different definitions of sinc so I avoid that designation in general.

As an aside: you can't let T → ∞ to derive F(1). You have to invoke the delta function.
 
  • #35
toneboy1
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The Fourier transform of a pulse of height 1 and width 2T is F(w) = (T/π)(sin(wT)/wT.
Some folks call sin(x)/x = sinc(x) but I've also seen different definitions of sinc so I avoid that designation in general.

Just experimenting, I noticed if w = 3, width = 2 then
(2/w)*sin(2*w/2) is ALMOST the same as (2.pi/w)*sin(2*w/2.pi),
so you seem right about it being a notational thing, but what I can't understand is that the one with the pi in it seems more accurate because in one instance the number was finite and the other kept going. Can you see why this would be?

I goofed again. I was thinking F{sin(wt)}, not F-1{sin(w)}. Sorry.

But the technique is similar: think the time-delayed delta function.


H'mm, I'm still not clear, I'll elaborate on where I got stuck:

x(t) = (1/2pi)* ∫ (exp(jwt)-exp(-jwt) /2j)*exp(jwt) dw -∞ to ∞

∴ (1/2pi)* ∫(exp(2jwt)-exp(0) / 2j) dw -∞ to ∞

∴ (1/j4pi)*[exp(2jwt)/2jw - w ] -∞ to ∞

And I'm not sure where to go from here. (especially with your time delay)

Alternatively using duality and linearity I attempted to make

sin(ω) = 2pi*sin(-w)/2pi

inverse transformed into: j.pi/2pi * [δ(1-t)-δ(t-1)] which is wrong according to wolfram.

Cheers for any input!
 
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