Clarification on the Fourier transformFourier Transform ConvolutionFourier transform with trigonometric and exponential functionsFourier transform of a convolution with upper limit not infinityCompute the following integral given the Fourier transform of $f$Computing Fourier Transformcorrect answer for Fourier Transform of $cos(at+b)$VERY difficult integral (Fourier transform)How to unterstand the condition of absolute integrability for Fourier transform?Fourier transform of a signalInverse Fourier transform of $fracjwLR+jwl$
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Clarification on the Fourier transform
Fourier Transform ConvolutionFourier transform with trigonometric and exponential functionsFourier transform of a convolution with upper limit not infinityCompute the following integral given the Fourier transform of $f$Computing Fourier Transformcorrect answer for Fourier Transform of $cos(at+b)$VERY difficult integral (Fourier transform)How to unterstand the condition of absolute integrability for Fourier transform?Fourier transform of a signalInverse Fourier transform of $fracjwLR+jwl$
$begingroup$
I need to compute the Fourier Transform of $x(t) = cos(t)sin(t)$.
We know that $delta(w) = mathcalF[1](w) = int_-infty^inftye^-jwtdt \$.
And $mathcalF[x(t)](w) = mathcalF[cos(t)sin(t)](w) = frac12pimathcalF[cos(t)](w) circledast mathcalF[sin(t)](w)$.
So I computed : beginsplit
mathcalF[cos(t)](w) &= int_-infty^inftycos(t)e^-jwtdt \
&= int_-infty^inftyfrac12(e^jt+e^-jt)e^-jwtdt \
&= frac12 Big ( int_-infty^inftye^jte^-jwtdt + int_-infty^inftye^-jte^-jwtdt Big ) \
&= frac12 Big ( int_-infty^inftye^-jt(w-1)dt + int_-infty^inftye^-jt(w+1)dt Big ) \
&= frac12 big ( delta(w-1)+delta(w+1) big )
endsplit
and beginsplit
mathcalF[sin(t)](w) &= int_-infty^inftysin(t)e^-jwtdt \
&= int_-infty^inftyfrac12j(e^-jt-e^jt)e^-jwtdt \
&= frac12j Big ( int_-infty^inftye^-jte^-jwtdt - int_-infty^inftye^jte^-jwtdt Big ) \
&= frac12j Big ( int_-infty^inftye^-jt(w+1)dt - int_-infty^inftye^-jt(w-1)dt Big ) \
&= frac12j big ( delta(w+1)-delta(w-1) big )
endsplit
I finally computed the convolution :
$beginsplit
mathcalF[x(t)](w) = mathcalF[cos(t)sin(t)](w) &= frac12pimathcalF[cos(t)](w) circledast mathcalF[sin(t)](w) \
&= frac12pifrac12frac12j big ( delta(w-1)+delta(w+1) big ) circledast big ( delta(w+1)-delta(w-1) big ) \
&= frac18pij big ( delta(w-1)+delta(w+1) big ) circledast big ( delta(w+1)-delta(w-1) big ) \
&= frac18pij big ( delta(w+2)-delta(w-2) big )
endsplit$
Using Matlab and wolframalpha to verify this, I ended up on :
The result of my fourier transform which is the result of the convolution gave me $$frac18pij big ( delta(w+2)-delta(w-2) big )$$
But I don't have any real part anymore.. and my imaginary part is not like on the plot. Using wolframalpha, I should have something like :
$$ ajbig (-delta(w+2)+delta(w-2) big )$$ which is correct according to the imaginary plot.
Where is my mistake ? And how can I compute analytically the phase using $arctan$ if I don't have any real part anymore ? How can I compute analytically the magnitude too ?
Wolframalpha's answer doesn't have a real part too, how can I have one in my plots ?
Thanks for the clarifications !
fourier-transform
asked Mar 13 at 18:46
Romain B.Romain B.
358217
$endgroup$
add a comment |
$begingroup$
I need to compute the Fourier Transform of $x(t) = cos(t)sin(t)$.
We know that $delta(w) = mathcalF[1](w) = int_-infty^inftye^-jwtdt \$.
And $mathcalF[x(t)](w) = mathcalF[cos(t)sin(t)](w) = frac12pimathcalF[cos(t)](w) circledast mathcalF[sin(t)](w)$.
So I computed : beginsplit
mathcalF[cos(t)](w) &= int_-infty^inftycos(t)e^-jwtdt \
&= int_-infty^inftyfrac12(e^jt+e^-jt)e^-jwtdt \
&= frac12 Big ( int_-infty^inftye^jte^-jwtdt + int_-infty^inftye^-jte^-jwtdt Big ) \
&= frac12 Big ( int_-infty^inftye^-jt(w-1)dt + int_-infty^inftye^-jt(w+1)dt Big ) \
&= frac12 big ( delta(w-1)+delta(w+1) big )
endsplit
and beginsplit
mathcalF[sin(t)](w) &= int_-infty^inftysin(t)e^-jwtdt \
&= int_-infty^inftyfrac12j(e^-jt-e^jt)e^-jwtdt \
&= frac12j Big ( int_-infty^inftye^-jte^-jwtdt - int_-infty^inftye^jte^-jwtdt Big ) \
&= frac12j Big ( int_-infty^inftye^-jt(w+1)dt - int_-infty^inftye^-jt(w-1)dt Big ) \
&= frac12j big ( delta(w+1)-delta(w-1) big )
endsplit
I finally computed the convolution :
$beginsplit
mathcalF[x(t)](w) = mathcalF[cos(t)sin(t)](w) &= frac12pimathcalF[cos(t)](w) circledast mathcalF[sin(t)](w) \
&= frac12pifrac12frac12j big ( delta(w-1)+delta(w+1) big ) circledast big ( delta(w+1)-delta(w-1) big ) \
&= frac18pij big ( delta(w-1)+delta(w+1) big ) circledast big ( delta(w+1)-delta(w-1) big ) \
&= frac18pij big ( delta(w+2)-delta(w-2) big )
endsplit$
Using Matlab and wolframalpha to verify this, I ended up on :
The result of my fourier transform which is the result of the convolution gave me $$frac18pij big ( delta(w+2)-delta(w-2) big )$$
But I don't have any real part anymore.. and my imaginary part is not like on the plot. Using wolframalpha, I should have something like :
$$ ajbig (-delta(w+2)+delta(w-2) big )$$ which is correct according to the imaginary plot.
Where is my mistake ? And how can I compute analytically the phase using $arctan$ if I don't have any real part anymore ? How can I compute analytically the magnitude too ?
Wolframalpha's answer doesn't have a real part too, how can I have one in my plots ?
Thanks for the clarifications !
fourier-transform
asked Mar 13 at 18:46
Romain B.Romain B.
358217
$endgroup$
$begingroup$
I am not completely sure where you went wrong, if you did so, but using $sin(t)cos(t)=frac12sin(2t)$ we can directly conclude that, following your notation, that $$mathcal F[sin(t)cos(t)](omega)=mathcal Fleft[frac12sin(2t)right](omega)=frac14j(delta(omega-2)-delta(omega+2))$$ Which agrees with WolframAlpha up to a constant factor.
$endgroup$
– mrtaurho
Mar 13 at 19:04
$begingroup$
@mrtaurho $mathscrF1=2pi delta(omega)$.
$endgroup$
– Mark Viola
Mar 13 at 19:09
$begingroup$
@MarkViola I was not sure which factor to use here. I am getting always kind of confused regarding all these different normalisations of the Fourier Transform which are used out there.
$endgroup$
– mrtaurho
Mar 13 at 19:11
add a comment |
$begingroup$
I need to compute the Fourier Transform of $x(t) = cos(t)sin(t)$.
We know that $delta(w) = mathcalF[1](w) = int_-infty^inftye^-jwtdt \$.
And $mathcalF[x(t)](w) = mathcalF[cos(t)sin(t)](w) = frac12pimathcalF[cos(t)](w) circledast mathcalF[sin(t)](w)$.
So I computed : beginsplit
mathcalF[cos(t)](w) &= int_-infty^inftycos(t)e^-jwtdt \
&= int_-infty^inftyfrac12(e^jt+e^-jt)e^-jwtdt \
&= frac12 Big ( int_-infty^inftye^jte^-jwtdt + int_-infty^inftye^-jte^-jwtdt Big ) \
&= frac12 Big ( int_-infty^inftye^-jt(w-1)dt + int_-infty^inftye^-jt(w+1)dt Big ) \
&= frac12 big ( delta(w-1)+delta(w+1) big )
endsplit
and beginsplit
mathcalF[sin(t)](w) &= int_-infty^inftysin(t)e^-jwtdt \
&= int_-infty^inftyfrac12j(e^-jt-e^jt)e^-jwtdt \
&= frac12j Big ( int_-infty^inftye^-jte^-jwtdt - int_-infty^inftye^jte^-jwtdt Big ) \
&= frac12j Big ( int_-infty^inftye^-jt(w+1)dt - int_-infty^inftye^-jt(w-1)dt Big ) \
&= frac12j big ( delta(w+1)-delta(w-1) big )
endsplit
I finally computed the convolution :
$beginsplit
mathcalF[x(t)](w) = mathcalF[cos(t)sin(t)](w) &= frac12pimathcalF[cos(t)](w) circledast mathcalF[sin(t)](w) \
&= frac12pifrac12frac12j big ( delta(w-1)+delta(w+1) big ) circledast big ( delta(w+1)-delta(w-1) big ) \
&= frac18pij big ( delta(w-1)+delta(w+1) big ) circledast big ( delta(w+1)-delta(w-1) big ) \
&= frac18pij big ( delta(w+2)-delta(w-2) big )
endsplit$
Using Matlab and wolframalpha to verify this, I ended up on :
The result of my fourier transform which is the result of the convolution gave me $$frac18pij big ( delta(w+2)-delta(w-2) big )$$
But I don't have any real part anymore.. and my imaginary part is not like on the plot. Using wolframalpha, I should have something like :
$$ ajbig (-delta(w+2)+delta(w-2) big )$$ which is correct according to the imaginary plot.
Where is my mistake ? And how can I compute analytically the phase using $arctan$ if I don't have any real part anymore ? How can I compute analytically the magnitude too ?
Wolframalpha's answer doesn't have a real part too, how can I have one in my plots ?
Thanks for the clarifications !
fourier-transform
asked Mar 13 at 18:46
Romain B.Romain B.
358217
$endgroup$
I need to compute the Fourier Transform of $x(t) = cos(t)sin(t)$.
We know that $delta(w) = mathcalF[1](w) = int_-infty^inftye^-jwtdt \$.
And $mathcalF[x(t)](w) = mathcalF[cos(t)sin(t)](w) = frac12pimathcalF[cos(t)](w) circledast mathcalF[sin(t)](w)$.
So I computed : beginsplit
mathcalF[cos(t)](w) &= int_-infty^inftycos(t)e^-jwtdt \
&= int_-infty^inftyfrac12(e^jt+e^-jt)e^-jwtdt \
&= frac12 Big ( int_-infty^inftye^jte^-jwtdt + int_-infty^inftye^-jte^-jwtdt Big ) \
&= frac12 Big ( int_-infty^inftye^-jt(w-1)dt + int_-infty^inftye^-jt(w+1)dt Big ) \
&= frac12 big ( delta(w-1)+delta(w+1) big )
endsplit
and beginsplit
mathcalF[sin(t)](w) &= int_-infty^inftysin(t)e^-jwtdt \
&= int_-infty^inftyfrac12j(e^-jt-e^jt)e^-jwtdt \
&= frac12j Big ( int_-infty^inftye^-jte^-jwtdt - int_-infty^inftye^jte^-jwtdt Big ) \
&= frac12j Big ( int_-infty^inftye^-jt(w+1)dt - int_-infty^inftye^-jt(w-1)dt Big ) \
&= frac12j big ( delta(w+1)-delta(w-1) big )
endsplit
I finally computed the convolution :
$beginsplit
mathcalF[x(t)](w) = mathcalF[cos(t)sin(t)](w) &= frac12pimathcalF[cos(t)](w) circledast mathcalF[sin(t)](w) \
&= frac12pifrac12frac12j big ( delta(w-1)+delta(w+1) big ) circledast big ( delta(w+1)-delta(w-1) big ) \
&= frac18pij big ( delta(w-1)+delta(w+1) big ) circledast big ( delta(w+1)-delta(w-1) big ) \
&= frac18pij big ( delta(w+2)-delta(w-2) big )
endsplit$
Using Matlab and wolframalpha to verify this, I ended up on :
The result of my fourier transform which is the result of the convolution gave me $$frac18pij big ( delta(w+2)-delta(w-2) big )$$
But I don't have any real part anymore.. and my imaginary part is not like on the plot. Using wolframalpha, I should have something like :
$$ ajbig (-delta(w+2)+delta(w-2) big )$$ which is correct according to the imaginary plot.
Where is my mistake ? And how can I compute analytically the phase using $arctan$ if I don't have any real part anymore ? How can I compute analytically the magnitude too ?
Wolframalpha's answer doesn't have a real part too, how can I have one in my plots ?
Thanks for the clarifications !
fourier-transform
fourier-transform
asked Mar 13 at 18:46
Romain B.Romain B.
358217
asked Mar 13 at 18:46
Romain B.Romain B.
358217
asked Mar 13 at 18:46
Romain B.Romain B.
358217
asked Mar 13 at 18:46
Romain B.Romain B.
358217
asked Mar 13 at 18:46
Romain B.Romain B.
358217
358217
$begingroup$
I am not completely sure where you went wrong, if you did so, but using $sin(t)cos(t)=frac12sin(2t)$ we can directly conclude that, following your notation, that $$mathcal F[sin(t)cos(t)](omega)=mathcal Fleft[frac12sin(2t)right](omega)=frac14j(delta(omega-2)-delta(omega+2))$$ Which agrees with WolframAlpha up to a constant factor.
$endgroup$
– mrtaurho
Mar 13 at 19:04
$begingroup$
@mrtaurho $mathscrF1=2pi delta(omega)$.
$endgroup$
– Mark Viola
Mar 13 at 19:09
$begingroup$
@MarkViola I was not sure which factor to use here. I am getting always kind of confused regarding all these different normalisations of the Fourier Transform which are used out there.
$endgroup$
– mrtaurho
Mar 13 at 19:11
add a comment |
$begingroup$
I am not completely sure where you went wrong, if you did so, but using $sin(t)cos(t)=frac12sin(2t)$ we can directly conclude that, following your notation, that $$mathcal F[sin(t)cos(t)](omega)=mathcal Fleft[frac12sin(2t)right](omega)=frac14j(delta(omega-2)-delta(omega+2))$$ Which agrees with WolframAlpha up to a constant factor.
$endgroup$
– mrtaurho
Mar 13 at 19:04
$begingroup$
@mrtaurho $mathscrF1=2pi delta(omega)$.
$endgroup$
– Mark Viola
Mar 13 at 19:09
$begingroup$
@MarkViola I was not sure which factor to use here. I am getting always kind of confused regarding all these different normalisations of the Fourier Transform which are used out there.
$endgroup$
– mrtaurho
Mar 13 at 19:11
$begingroup$
I am not completely sure where you went wrong, if you did so, but using $sin(t)cos(t)=frac12sin(2t)$ we can directly conclude that, following your notation, that $$mathcal F[sin(t)cos(t)](omega)=mathcal Fleft[frac12sin(2t)right](omega)=frac14j(delta(omega-2)-delta(omega+2))$$ Which agrees with WolframAlpha up to a constant factor.
$endgroup$
– mrtaurho
Mar 13 at 19:04
$begingroup$
I am not completely sure where you went wrong, if you did so, but using $sin(t)cos(t)=frac12sin(2t)$ we can directly conclude that, following your notation, that $$mathcal F[sin(t)cos(t)](omega)=mathcal Fleft[frac12sin(2t)right](omega)=frac14j(delta(omega-2)-delta(omega+2))$$ Which agrees with WolframAlpha up to a constant factor.
$endgroup$
– mrtaurho
Mar 13 at 19:04
$begingroup$
@mrtaurho $mathscrF1=2pi delta(omega)$.
$endgroup$
– Mark Viola
Mar 13 at 19:09
$begingroup$
@mrtaurho $mathscrF1=2pi delta(omega)$.
$endgroup$
– Mark Viola
Mar 13 at 19:09
$begingroup$
@MarkViola I was not sure which factor to use here. I am getting always kind of confused regarding all these different normalisations of the Fourier Transform which are used out there.
$endgroup$
– mrtaurho
Mar 13 at 19:11
$begingroup$
@MarkViola I was not sure which factor to use here. I am getting always kind of confused regarding all these different normalisations of the Fourier Transform which are used out there.
$endgroup$
– mrtaurho
Mar 13 at 19:11
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
First, $mathscrF1=2pi delta(omega)$.
Next, note that $f(t)=cos(t)sin(t)=frac12sin(2t)=frace^i2t-e^-i2ti4$.
Hence, we see that
$$mathscrFf=frac1i4int_-infty^infty left(e^-i(omega-2)t-e^-i(omega+2)tright),dt=fracpii2left(delta(omega-2)-delta(omega+2)right)tag1$$
One can take the inverse transform, $frac12piint_-infty^infty fracpii2left(delta(omega-2)-delta(omega+2)right),e^iomega t,domega$, of the right-hand side of $(1)$ to confirm that $(1)$ is indeed the Fourier transform of $f$.
answered Mar 13 at 19:06
Mark ViolaMark Viola
133k1278176
$endgroup$
$begingroup$
Thanks ! It was pretty easy finally .. And I don't know about $mathscrF1=2pi delta(omega)$ because I have my $frac12pi$ afterwards, but not really important I guess. How can I compute the magnitude and phase from here ? Where is the real part ?
$endgroup$
– Romain B.
Mar 13 at 19:13
$begingroup$
You're welcome. My pleasure.
$endgroup$
– Mark Viola
Mar 13 at 19:14
$begingroup$
(I edited my comment, I have few more questions)
$endgroup$
– Romain B.
Mar 13 at 19:15
$begingroup$
@RomainB. The Dirac Delta is not a function, it is a distribution (aka, a Generalized Function). As such, it cannot be plotted as a function. You could draw a "pseudo-plot" of the magnitude by drawing (infinite) vertical spikes at $omega=pm2$ with Dirac Delta "amplitudes/weights" of $pi/2$. But what is the point of that?
$endgroup$
– Mark Viola
Mar 13 at 19:18
$begingroup$
Thanks a lot ! Last question, what is the real part of it, if any ? Because I want to compute analytically the phase and magnitude. Thanks again for the clarifications :)
$endgroup$
– Romain B.
Mar 13 at 21:32
|
show 3 more comments
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1 Answer
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1 Answer
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oldest
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oldest
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votes
$begingroup$
First, $mathscrF1=2pi delta(omega)$.
Next, note that $f(t)=cos(t)sin(t)=frac12sin(2t)=frace^i2t-e^-i2ti4$.
Hence, we see that
$$mathscrFf=frac1i4int_-infty^infty left(e^-i(omega-2)t-e^-i(omega+2)tright),dt=fracpii2left(delta(omega-2)-delta(omega+2)right)tag1$$
One can take the inverse transform, $frac12piint_-infty^infty fracpii2left(delta(omega-2)-delta(omega+2)right),e^iomega t,domega$, of the right-hand side of $(1)$ to confirm that $(1)$ is indeed the Fourier transform of $f$.
answered Mar 13 at 19:06
Mark ViolaMark Viola
133k1278176
$endgroup$
$begingroup$
Thanks ! It was pretty easy finally .. And I don't know about $mathscrF1=2pi delta(omega)$ because I have my $frac12pi$ afterwards, but not really important I guess. How can I compute the magnitude and phase from here ? Where is the real part ?
$endgroup$
– Romain B.
Mar 13 at 19:13
$begingroup$
You're welcome. My pleasure.
$endgroup$
– Mark Viola
Mar 13 at 19:14
$begingroup$
(I edited my comment, I have few more questions)
$endgroup$
– Romain B.
Mar 13 at 19:15
$begingroup$
@RomainB. The Dirac Delta is not a function, it is a distribution (aka, a Generalized Function). As such, it cannot be plotted as a function. You could draw a "pseudo-plot" of the magnitude by drawing (infinite) vertical spikes at $omega=pm2$ with Dirac Delta "amplitudes/weights" of $pi/2$. But what is the point of that?
$endgroup$
– Mark Viola
Mar 13 at 19:18
$begingroup$
Thanks a lot ! Last question, what is the real part of it, if any ? Because I want to compute analytically the phase and magnitude. Thanks again for the clarifications :)
$endgroup$
– Romain B.
Mar 13 at 21:32
|
show 3 more comments
$begingroup$
First, $mathscrF1=2pi delta(omega)$.
Next, note that $f(t)=cos(t)sin(t)=frac12sin(2t)=frace^i2t-e^-i2ti4$.
Hence, we see that
$$mathscrFf=frac1i4int_-infty^infty left(e^-i(omega-2)t-e^-i(omega+2)tright),dt=fracpii2left(delta(omega-2)-delta(omega+2)right)tag1$$
One can take the inverse transform, $frac12piint_-infty^infty fracpii2left(delta(omega-2)-delta(omega+2)right),e^iomega t,domega$, of the right-hand side of $(1)$ to confirm that $(1)$ is indeed the Fourier transform of $f$.
answered Mar 13 at 19:06
Mark ViolaMark Viola
133k1278176
$endgroup$
$begingroup$
Thanks ! It was pretty easy finally .. And I don't know about $mathscrF1=2pi delta(omega)$ because I have my $frac12pi$ afterwards, but not really important I guess. How can I compute the magnitude and phase from here ? Where is the real part ?
$endgroup$
– Romain B.
Mar 13 at 19:13
$begingroup$
You're welcome. My pleasure.
$endgroup$
– Mark Viola
Mar 13 at 19:14
$begingroup$
(I edited my comment, I have few more questions)
$endgroup$
– Romain B.
Mar 13 at 19:15
$begingroup$
@RomainB. The Dirac Delta is not a function, it is a distribution (aka, a Generalized Function). As such, it cannot be plotted as a function. You could draw a "pseudo-plot" of the magnitude by drawing (infinite) vertical spikes at $omega=pm2$ with Dirac Delta "amplitudes/weights" of $pi/2$. But what is the point of that?
$endgroup$
– Mark Viola
Mar 13 at 19:18
$begingroup$
Thanks a lot ! Last question, what is the real part of it, if any ? Because I want to compute analytically the phase and magnitude. Thanks again for the clarifications :)
$endgroup$
– Romain B.
Mar 13 at 21:32
|
show 3 more comments
$begingroup$
First, $mathscrF1=2pi delta(omega)$.
Next, note that $f(t)=cos(t)sin(t)=frac12sin(2t)=frace^i2t-e^-i2ti4$.
Hence, we see that
$$mathscrFf=frac1i4int_-infty^infty left(e^-i(omega-2)t-e^-i(omega+2)tright),dt=fracpii2left(delta(omega-2)-delta(omega+2)right)tag1$$
One can take the inverse transform, $frac12piint_-infty^infty fracpii2left(delta(omega-2)-delta(omega+2)right),e^iomega t,domega$, of the right-hand side of $(1)$ to confirm that $(1)$ is indeed the Fourier transform of $f$.
answered Mar 13 at 19:06
Mark ViolaMark Viola
133k1278176
$endgroup$
First, $mathscrF1=2pi delta(omega)$.
Next, note that $f(t)=cos(t)sin(t)=frac12sin(2t)=frace^i2t-e^-i2ti4$.
Hence, we see that
$$mathscrFf=frac1i4int_-infty^infty left(e^-i(omega-2)t-e^-i(omega+2)tright),dt=fracpii2left(delta(omega-2)-delta(omega+2)right)tag1$$
One can take the inverse transform, $frac12piint_-infty^infty fracpii2left(delta(omega-2)-delta(omega+2)right),e^iomega t,domega$, of the right-hand side of $(1)$ to confirm that $(1)$ is indeed the Fourier transform of $f$.
answered Mar 13 at 19:06
Mark ViolaMark Viola
133k1278176
edited Mar 13 at 19:16
answered Mar 13 at 19:06
Mark ViolaMark Viola
133k1278176
answered Mar 13 at 19:06
Mark ViolaMark Viola
133k1278176
answered Mar 13 at 19:06
Mark ViolaMark Viola
133k1278176
133k1278176
$begingroup$
Thanks ! It was pretty easy finally .. And I don't know about $mathscrF1=2pi delta(omega)$ because I have my $frac12pi$ afterwards, but not really important I guess. How can I compute the magnitude and phase from here ? Where is the real part ?
$endgroup$
– Romain B.
Mar 13 at 19:13
$begingroup$
You're welcome. My pleasure.
$endgroup$
– Mark Viola
Mar 13 at 19:14
$begingroup$
(I edited my comment, I have few more questions)
$endgroup$
– Romain B.
Mar 13 at 19:15
$begingroup$
@RomainB. The Dirac Delta is not a function, it is a distribution (aka, a Generalized Function). As such, it cannot be plotted as a function. You could draw a "pseudo-plot" of the magnitude by drawing (infinite) vertical spikes at $omega=pm2$ with Dirac Delta "amplitudes/weights" of $pi/2$. But what is the point of that?
$endgroup$
– Mark Viola
Mar 13 at 19:18
$begingroup$
Thanks a lot ! Last question, what is the real part of it, if any ? Because I want to compute analytically the phase and magnitude. Thanks again for the clarifications :)
$endgroup$
– Romain B.
Mar 13 at 21:32
|
show 3 more comments
$begingroup$
Thanks ! It was pretty easy finally .. And I don't know about $mathscrF1=2pi delta(omega)$ because I have my $frac12pi$ afterwards, but not really important I guess. How can I compute the magnitude and phase from here ? Where is the real part ?
$endgroup$
– Romain B.
Mar 13 at 19:13
$begingroup$
You're welcome. My pleasure.
$endgroup$
– Mark Viola
Mar 13 at 19:14
$begingroup$
(I edited my comment, I have few more questions)
$endgroup$
– Romain B.
Mar 13 at 19:15
$begingroup$
@RomainB. The Dirac Delta is not a function, it is a distribution (aka, a Generalized Function). As such, it cannot be plotted as a function. You could draw a "pseudo-plot" of the magnitude by drawing (infinite) vertical spikes at $omega=pm2$ with Dirac Delta "amplitudes/weights" of $pi/2$. But what is the point of that?
$endgroup$
– Mark Viola
Mar 13 at 19:18
$begingroup$
Thanks a lot ! Last question, what is the real part of it, if any ? Because I want to compute analytically the phase and magnitude. Thanks again for the clarifications :)
$endgroup$
– Romain B.
Mar 13 at 21:32
$begingroup$
Thanks ! It was pretty easy finally .. And I don't know about $mathscrF1=2pi delta(omega)$ because I have my $frac12pi$ afterwards, but not really important I guess. How can I compute the magnitude and phase from here ? Where is the real part ?
$endgroup$
– Romain B.
Mar 13 at 19:13
$begingroup$
Thanks ! It was pretty easy finally .. And I don't know about $mathscrF1=2pi delta(omega)$ because I have my $frac12pi$ afterwards, but not really important I guess. How can I compute the magnitude and phase from here ? Where is the real part ?
$endgroup$
– Romain B.
Mar 13 at 19:13
$begingroup$
You're welcome. My pleasure.
$endgroup$
– Mark Viola
Mar 13 at 19:14
$begingroup$
You're welcome. My pleasure.
$endgroup$
– Mark Viola
Mar 13 at 19:14
$begingroup$
(I edited my comment, I have few more questions)
$endgroup$
– Romain B.
Mar 13 at 19:15
$begingroup$
(I edited my comment, I have few more questions)
$endgroup$
– Romain B.
Mar 13 at 19:15
$begingroup$
@RomainB. The Dirac Delta is not a function, it is a distribution (aka, a Generalized Function). As such, it cannot be plotted as a function. You could draw a "pseudo-plot" of the magnitude by drawing (infinite) vertical spikes at $omega=pm2$ with Dirac Delta "amplitudes/weights" of $pi/2$. But what is the point of that?
$endgroup$
– Mark Viola
Mar 13 at 19:18
$begingroup$
@RomainB. The Dirac Delta is not a function, it is a distribution (aka, a Generalized Function). As such, it cannot be plotted as a function. You could draw a "pseudo-plot" of the magnitude by drawing (infinite) vertical spikes at $omega=pm2$ with Dirac Delta "amplitudes/weights" of $pi/2$. But what is the point of that?
$endgroup$
– Mark Viola
Mar 13 at 19:18
$begingroup$
Thanks a lot ! Last question, what is the real part of it, if any ? Because I want to compute analytically the phase and magnitude. Thanks again for the clarifications :)
$endgroup$
– Romain B.
Mar 13 at 21:32
$begingroup$
Thanks a lot ! Last question, what is the real part of it, if any ? Because I want to compute analytically the phase and magnitude. Thanks again for the clarifications :)
$endgroup$
– Romain B.
Mar 13 at 21:32
|
show 3 more comments
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$begingroup$
I am not completely sure where you went wrong, if you did so, but using $sin(t)cos(t)=frac12sin(2t)$ we can directly conclude that, following your notation, that $$mathcal F[sin(t)cos(t)](omega)=mathcal Fleft[frac12sin(2t)right](omega)=frac14j(delta(omega-2)-delta(omega+2))$$ Which agrees with WolframAlpha up to a constant factor.
$endgroup$
– mrtaurho
Mar 13 at 19:04
$begingroup$
@mrtaurho $mathscrF1=2pi delta(omega)$.
$endgroup$
– Mark Viola
Mar 13 at 19:09
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@MarkViola I was not sure which factor to use here. I am getting always kind of confused regarding all these different normalisations of the Fourier Transform which are used out there.
$endgroup$
– mrtaurho
Mar 13 at 19:11