Re: [music-dsp] Fourier and its negative exponent

[Stijn Frishert]

I think it's all starting to sink in.

(This may seem obvious to some, but as a way to chronicle this and for people 
reading along and struggling with this as well.)

I've always seen the correlation as the original signal doing something to/with 
the sinusoid (like scaling or projecting on it). If you look at it the other 
way around, essentially each sample in your original signal is multiplied by an 
e^±i. In other words, you're rotating it in the complex plane. Sure, each 
sample is rotated by a different angle (except for a sinusoid of frequency 0 -> 
DC), but it's a rotation.

Now if all samples after rotation line up, they evaluate to a huge magnitude. 
To do that, you need to rotate them all by the negative of the angle that got 
them there in the first place, hence the minus. If they all stack up, but not 
on the real, positive axis, you've got your angle.

All the samples that don't stack up will be scattered on the imaginary plane, 
meaning the frequency is not fully present.

2015/10/07 21:02、Theo Verelst  のメッセージ:
> Depending on how deep you want to study/understand the subject, get a good 
> textbook on the subject, like "The Fourier Transform and its Applications" 
> from the Stanford "see" courses...

That one seems really interesting :) Thank you for that link.

Stijn
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Re: [music-dsp] Fourier and its negative exponent

[robert bristow-johnson]

On 10/7/15 3:02 PM, Theo Verelst wrote:

Stijn Frishert wrote:

Hey all,

In trying to get to grips with the discrete Fourier transform,


Depending on how deep you want to study/understand the subject, get a 
good textbook on the subject, like "The Fourier Transform and its 
Applications" from the Stanford "see" courses, or this one which 
caught my eye because it quotes a book from  a course I did (Circuits 
and Systems, Papoulis) :



https://books.google.nl/books?id=Sp7O4bocjPAC=PA53=PA53=the+DFT+is+a+special+case+of+the+sampled+DTFT=bl=dqfZLS8G9N=rWuIFQikiPPr_f3snYb0FHYZpFI=en=X=0CCsQ6AEwAmoVChMI3vKC4oKxyAIVy2sUCh178gpd#v=onepage=the%20DFT%20is%20a%20special%20case%20of%20the%20sampled%20DTFT=false 



which contains a good quote at the top of page 53

"the DFT is a special case of the sampled DTFT"



the quote is "good" and it's salient, but it only one way to look at it.

the other is "the DFT is synonymous with the DFS which bijectively maps 
a discrete-time sequence of period N to a discrete-frequency sequence 
with the same period N."


(besides the commonly-considered zero-padded x[n] that goes into the 
DTFT and is sampled for the DFT, there are many other aliases of x[n] 
that can be DTFT and sampled to get the very same DFT.)



--

r b-j  r...@audioimagination.com

"Imagination is more important than knowledge."



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Re: [music-dsp] Fourier and its negative exponent

[Earl Vickers]

Ethan Duni wrote:

> Also there are different conventions about where to put the normalization
> constants (on the analysis side, or on the synthesis side, or take the
> square root and include it on both).

I remember it blew my mind when I first learned there was a symmetric form of 
the Fourier transform that gave frequency-domain coefficients in response to 
time-domain data and vice versa, with no flag telling it which domain the input 
represented. Like, how does it know?? 

I now “understand” that it’s just a rotation, but that’s like saying I 
understand gravity. I’ve gotten used to it, or maybe just repressed how weird 
it is.

Earl
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Re: [music-dsp] Fourier and its negative exponent

[Theo Verelst]

Think of the Fast Fourier Transform as computing the inner product of a piece of signal 
(the length of the transform) with all kinds of basis functions: the various frequencies 
that can "fit" in the interval. Without going into engineering basics, you can take a sine 
and a cosine as a basis function got each frequency that "fits" in the chosen interval (so 
a sine the exact length of the interval, two sines that exactly fit the chosen interval, 3 
full sines, etc., until you have a sine that isn't distinguishable anymore because it's 
frequency is too high because it's peaks appear shorter together than the elements of you 
time vector. As it appears, it is a good idea to take sine and cosines, because than you 
can proof/make credible that there is one precise and only one FFT transformed signal that 
can be added to your time "sample" or signal data point vector, which given certain 
accuracies is also a bi-jective mapping. The idea of measuring the sine and cosine 
correlation is as an ancient EE trick (probably in mechanics and physics before that) 
connected with the idea that each signal (given certain conditions) can be seen as an 
addition of sine waves, where there's phase associated with each component. Writing that 
as a complex number is a trick, it has in this case not really to do with the transform, 
it's a short way of writing things. I assure you there's a lot of mathematical hassle with 
complex numbers and matrices possible you might not want to get into, because it is often 
not very insightful.


T.
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Re: [music-dsp] Fourier and its negative exponent

[Esteban Maestre]

By the way: complex-conjugate does not mean it rotates in opposite 
direction; check out this picture:


http://www.eetasia.com/STATIC/ARTICLE_IMAGES/200902/EEOL_2009FEB04_DSP_RFD_NT_01c.gif

Rotation in opposite direction happens with negative frequencies.

Cheers,
Esteban

On 10/5/2015 8:06 PM, Stijn Frishert wrote:

Thanks Allen, Esteban and Sebastian.

My main thought error was thinking that negating the exponent was the 
complex equivalent of flipping the sign of a non-complex sinusoid (sin 
and -sin). Of course it isn’t. e^-a isn’t the same as -e^a. The real 
part of a complex sinusoid and its complex conjugate are the same, 
they only rotate in different directions.


And so the minus is to negate that rotation in the complex plane. 
Correct me if I’m wrong, of course.


Stijn

On 5 Oct 2015, at 15:51, Allen Downey > wrote:


In Chapter 7 of Think DSP, I develop the DFT in a way that might help 
with this:


http://greenteapress.com/thinkdsp/html/thinkdsp008.html

If you think of the inverse DFT as matrix multiplication where the 
matrix, M, contains complex exponentials as basis vectors, the 
(forward) DFT is the multiplication by the inverse of M.  Since M is 
unitary, its inverse is its conjugate transpose.  The conjugation is 
the source of the negative sign, when you write the DFT in summation 
form.


Allen



On Mon, Oct 5, 2015 at 9:28 AM, Stijn Frishert 
> wrote:


Hey all,

In trying to get to grips with the discrete Fourier transform, I
have a question about the minus sign in the exponent of the
complex sinusoids you correlate with doing the transform.

The inverse transform doesn’t contain this negation and a quick
search on the internet tells me Fourier analysis and synthesis
work as long as one of the formulas contains that minus and the
other one doesn’t.

So: why? If the bins in the resulting spectrum represent how much
of a sinusoid was present in the original signal
(cross-correlation), I would expect synthesis to use these exact
same sinusoids to get back to the original signal. Instead it
uses their inverse! How can the resulting signal not be 180 phase
shifted?

This may be text-book dsp theory, but I’ve looked and searched
and everywhere seems to skip over it as if it’s self-evident.

Stijn Frishert
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CIRMMT/CAML - McGill Univ
MTG - Univ Pompeu Fabra
http://ccrma.stanford.edu/~esteban

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Re: [music-dsp] Fourier and its negative exponent

[Esteban Maestre]

"does not mean" > "does mean"

Esteban

On 10/5/2015 8:47 PM, Esteban Maestre wrote:
By the way: complex-conjugate does not mean it rotates in opposite 
direction; check out this picture:


http://www.eetasia.com/STATIC/ARTICLE_IMAGES/200902/EEOL_2009FEB04_DSP_RFD_NT_01c.gif

Rotation in opposite direction happens with negative frequencies.

Cheers,
Esteban

On 10/5/2015 8:06 PM, Stijn Frishert wrote:

Thanks Allen, Esteban and Sebastian.

My main thought error was thinking that negating the exponent was the 
complex equivalent of flipping the sign of a non-complex sinusoid 
(sin and -sin). Of course it isn’t. e^-a isn’t the same as -e^a. The 
real part of a complex sinusoid and its complex conjugate are the 
same, they only rotate in different directions.


And so the minus is to negate that rotation in the complex plane. 
Correct me if I’m wrong, of course.


Stijn


On 5 Oct 2015, at 15:51, Allen Downey  wrote:

In Chapter 7 of Think DSP, I develop the DFT in a way that might 
help with this:


http://greenteapress.com/thinkdsp/html/thinkdsp008.html

If you think of the inverse DFT as matrix multiplication where the 
matrix, M, contains complex exponentials as basis vectors, the 
(forward) DFT is the multiplication by the inverse of M.  Since M is 
unitary, its inverse is its conjugate transpose.  The conjugation is 
the source of the negative sign, when you write the DFT in summation 
form.


Allen



On Mon, Oct 5, 2015 at 9:28 AM, Stijn Frishert 
 wrote:


Hey all,

In trying to get to grips with the discrete Fourier transform, I
have a question about the minus sign in the exponent of the
complex sinusoids you correlate with doing the transform.

The inverse transform doesn’t contain this negation and a quick
search on the internet tells me Fourier analysis and synthesis
work as long as one of the formulas contains that minus and the
other one doesn’t.

So: why? If the bins in the resulting spectrum represent how
much of a sinusoid was present in the original signal
(cross-correlation), I would expect synthesis to use these exact
same sinusoids to get back to the original signal. Instead it
uses their inverse! How can the resulting signal not be 180
phase shifted?

This may be text-book dsp theory, but I’ve looked and searched
and everywhere seems to skip over it as if it’s self-evident.

Stijn Frishert
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CIRMMT/CAML - McGill Univ
MTG - Univ Pompeu Fabra
http://ccrma.stanford.edu/~esteban  


--

Esteban Maestre
CIRMMT/CAML - McGill Univ
MTG - Univ Pompeu Fabra
http://ccrma.stanford.edu/~esteban

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Re: [music-dsp] Fourier and its negative exponent

[robert bristow-johnson]

On 10/5/15 9:28 AM, Stijn Frishert wrote:

In trying to get to grips with the discrete Fourier transform, I have a 
question about the minus sign in the exponent of the complex sinusoids you 
correlate with doing the transform.

The inverse transform doesn’t contain this negation and a quick search on the 
internet tells me Fourier analysis and synthesis work as long as one of the 
formulas contains that minus and the other one doesn’t.

So: why? If the bins in the resulting spectrum represent how much of a sinusoid 
was present in the original signal (cross-correlation), I would expect 
synthesis to use these exact same sinusoids to get back to the original signal. 
Instead it uses their inverse! How can the resulting signal not be 180 phase 
shifted?

This may be text-book dsp theory, but I’ve looked and searched and everywhere 
seems to skip over it as if it’s self-evident.



hi Stijn,

so just to confuse things further, i'll add my 2 cents that i had always 
thought made it less confusing.  (but people have disabused me of that 
notion.)


first of all, it's a question oft asked in DSP circles, like the USENET 
comp.dsp or, more recently at Stack Exchange (not a bad thing to sign up 
and participate in):


 
http://dsp.stackexchange.com/questions/19004/why-is-a-negative-exponent-present-in-fourier-and-laplace-transform



in my opinion, the answer to your question is one word: "convention".

the reason why it's merely convention is that if the minus sign was 
swapped between the forward and inverse Fourier transform in all of the 
literature and practice, all of the theorems would work the same as they 
do now.


the reason for that is that the two imaginary numbers +j and -j are, 
qualitatively, *exactly* the same even though they are negatives of each 
other and are not zero.  (the same cannot be said for +1 and -1, which 
are qualitatively different.) both +j and -j are purely imaginary and 
have equal claim to squaring to become -1.


so, by convention, they chose +j in the inverse Fourier Transform and -j 
had to come out in the forward Fourier transform.  they could have 
chosen -j for the inverse F.T., but then they would need +j in the 
forward F.T.


so why did they do that?  in signal processing, where we are as 
comfortable with negative frequency as we are with positive frequency 
it's because if you want to represent a single (complex) sinusoid at an 
angular frequency of omega_0 with an amplitude of 1 and phase offset of 
zero, it is:



   e^(j*omega_0*t)

so, when we represent a periodic signal with fundamental frequency of 
omega_0>0 (that is, the period is 2*pi/omega_0), it is:


 +inf
   x(t)  =   SUMX[k] * e^(j*k*omega_0*t)
k=-inf


each frequency component is at frequency k*omega_0.  for positive 
frequencies, k>0, for negative, k<0.



to extract the coefficient X[m], we must multiply x(t) by 
e^(-j*m*omega_0*t) to cancel the factor e^(j*m*omega_0*t) in that term 
(when k=m) in that summation, and then we average.  the m-th term is now 
DC and averaging will get X[m].  all of the other terms are AC and 
averaging will eventually make those terms go to zero.  so only X[m] is 
left.


that is conceptually the basic way in which Fourier series or Fourier 
transform works.  (discrete or continuous.)



but, we could do the same thing all over again, this time replace every 
occurrence of +j with -j and every -j with +j, and the same results will 
come out.  the choice of +j in the above two expressions is one of 
convention.




--

r b-j  r...@audioimagination.com

"Imagination is more important than knowledge."



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Re: [music-dsp] Fourier and its negative exponent

[robert bristow-johnson]

On 10/5/15 5:40 PM, robert bristow-johnson wrote:


about an hour ago i posted to this list and it hasn't shown up on my end.



okay, something got lost in the aether. i am reposting this:


On 10/5/15 9:28 AM, Stijn Frishert wrote:
In trying to get to grips with the discrete Fourier transform, I have 
a question about the minus sign in the exponent of the complex 
sinusoids you correlate with doing the transform.


The inverse transform doesn’t contain this negation and a quick search 
on the internet tells me Fourier analysis and synthesis work as long 
as one of the formulas contains that minus and the other one doesn’t.


So: why? If the bins in the resulting spectrum represent how much of a 
sinusoid was present in the original signal (cross-correlation), I 
would expect synthesis to use these exact same sinusoids to get back 
to the original signal. Instead it uses their inverse! How can the 
resulting signal not be 180 phase shifted?


This may be text-book dsp theory, but I’ve looked and searched and 
everywhere seems to skip over it as if it’s self-evident.



hi Stijn,

so just to confuse things further, i'll add my 2 cents that i had always 
thought made it less confusing. (but people have disabused me of that 
notion.)


first of all, it's a question oft asked in DSP circles, like the USENET 
comp.dsp or, more recently at Stack Exchange (not a bad thing to sign up 
and participate in):


http://dsp.stackexchange.com/questions/19004/why-is-a-negative-exponent-present-in-fourier-and-laplace-transform 





in my opinion, the answer to your question is one word: "convention".

the reason why it's merely convention is that if the minus sign was 
swapped between the forward and inverse Fourier transform in all of the 
literature and practice, all of the theorems would work the same as they 
do now.


the reason for that is that the two imaginary numbers +j and -j are, 
qualitatively, *exactly* the same even though they are negatives of each 
other and are not zero. (the same cannot be said for +1 and -1, which 
are qualitatively different.) both +j and -j are purely imaginary and 
have equal claim to squaring to become -1.


so, by convention, they chose +j in the inverse Fourier Transform and -j 
had to come out in the forward Fourier transform. they could have chosen 
-j for the inverse F.T., but then they would need +j in the forward F.T.


so why did they do that? in signal processing, where we are as 
comfortable with negative frequency as we are with positive frequency 
it's because if you want to represent a single (complex) sinusoid at an 
angular frequency of omega_0 with an amplitude of 1 and phase offset of 
zero, it is:



e^(j*omega_0*t)

so, when we represent a periodic signal with fundamental frequency of 
omega_0>0 (that is, the period is 2*pi/omega_0), it is:


+inf
x(t) = SUM X[k] * e^(j*k*omega_0*t)
k=-inf


each frequency component is at frequency k*omega_0. for positive 
frequencies, k>0, for negative, k<0.



to extract the coefficient X[m], we must multiply x(t) by 
e^(-j*m*omega_0*t) to cancel the factor e^(j*m*omega_0*t) in that term 
(when k=m) in that summation, and then we average. the m-th term is now 
DC and averaging will get X[m]. all of the other terms are AC and 
averaging will eventually make those terms go to zero. so only X[m] is 
left.


that is conceptually the basic way in which Fourier series or Fourier 
transform works. (discrete or continuous.)



but, we could do the same thing all over again, this time replace every 
occurrence of +j with -j and every -j with +j, and the same results will 
come out. the choice of +j in the above two expressions is one of 
convention.






--

r b-j  r...@audioimagination.com

"Imagination is more important than knowledge."



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Re: [music-dsp] Fourier and its negative exponent

[Ethan Duni]

>the reason why it's merely convention is that if the minus sign was
swapped
>between the forward and inverse Fourier transform in all of the literature
and
>practice, all of the theorems would work the same as they do now.

Note that in some other areas they do actually use other conventions. It's
been a while since I've looked at it but IIRC in areas like geophysics they
have the signs swapped around.

Also there are different conventions about where to put the normalization
constants (on the analysis side, or on the synthesis side, or take the
square root and include it on both). Those make a bit more difference for
some of the theorems like Parseval, but again it all works the same you
just gotta be careful to be consistent.

E

On Mon, Oct 5, 2015 at 2:52 PM, robert bristow-johnson <
r...@audioimagination.com> wrote:

> On 10/5/15 5:40 PM, robert bristow-johnson wrote:
>
>>
>> about an hour ago i posted to this list and it hasn't shown up on my end.
>>
>>
> okay, something got lost in the aether. i am reposting this:
>
>
> On 10/5/15 9:28 AM, Stijn Frishert wrote:
>
>> In trying to get to grips with the discrete Fourier transform, I have a
>> question about the minus sign in the exponent of the complex sinusoids you
>> correlate with doing the transform.
>>
>> The inverse transform doesn’t contain this negation and a quick search on
>> the internet tells me Fourier analysis and synthesis work as long as one of
>> the formulas contains that minus and the other one doesn’t.
>>
>> So: why? If the bins in the resulting spectrum represent how much of a
>> sinusoid was present in the original signal (cross-correlation), I would
>> expect synthesis to use these exact same sinusoids to get back to the
>> original signal. Instead it uses their inverse! How can the resulting
>> signal not be 180 phase shifted?
>>
>> This may be text-book dsp theory, but I’ve looked and searched and
>> everywhere seems to skip over it as if it’s self-evident.
>>
>
>
> hi Stijn,
>
> so just to confuse things further, i'll add my 2 cents that i had always
> thought made it less confusing. (but people have disabused me of that
> notion.)
>
> first of all, it's a question oft asked in DSP circles, like the USENET
> comp.dsp or, more recently at Stack Exchange (not a bad thing to sign up
> and participate in):
>
>
> http://dsp.stackexchange.com/questions/19004/why-is-a-negative-exponent-present-in-fourier-and-laplace-transform
>
>
>
> in my opinion, the answer to your question is one word: "convention".
>
> the reason why it's merely convention is that if the minus sign was
> swapped between the forward and inverse Fourier transform in all of the
> literature and practice, all of the theorems would work the same as they do
> now.
>
> the reason for that is that the two imaginary numbers +j and -j are,
> qualitatively, *exactly* the same even though they are negatives of each
> other and are not zero. (the same cannot be said for +1 and -1, which are
> qualitatively different.) both +j and -j are purely imaginary and have
> equal claim to squaring to become -1.
>
> so, by convention, they chose +j in the inverse Fourier Transform and -j
> had to come out in the forward Fourier transform. they could have chosen -j
> for the inverse F.T., but then they would need +j in the forward F.T.
>
> so why did they do that? in signal processing, where we are as comfortable
> with negative frequency as we are with positive frequency it's because if
> you want to represent a single (complex) sinusoid at an angular frequency
> of omega_0 with an amplitude of 1 and phase offset of zero, it is:
>
>
> e^(j*omega_0*t)
>
> so, when we represent a periodic signal with fundamental frequency of
> omega_0>0 (that is, the period is 2*pi/omega_0), it is:
>
> +inf
> x(t) = SUM X[k] * e^(j*k*omega_0*t)
> k=-inf
>
>
> each frequency component is at frequency k*omega_0. for positive
> frequencies, k>0, for negative, k<0.
>
>
> to extract the coefficient X[m], we must multiply x(t) by
> e^(-j*m*omega_0*t) to cancel the factor e^(j*m*omega_0*t) in that term
> (when k=m) in that summation, and then we average. the m-th term is now DC
> and averaging will get X[m]. all of the other terms are AC and averaging
> will eventually make those terms go to zero. so only X[m] is left.
>
> that is conceptually the basic way in which Fourier series or Fourier
> transform works. (discrete or continuous.)
>
>
> but, we could do the same thing all over again, this time replace every
> occurrence of +j with -j and every -j with +j, and the same results will
> come out. the choice of +j in the above two expressions is one of
> convention.
>
>
>
>
>
> --
>
> r b-j  r...@audioimagination.com
>
> "Imagination is more important than knowledge."
>
>
>
> ___
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>
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Re: [music-dsp] Fourier and its negative exponent

[Esteban Maestre]

HI Stijn,

That "minus" comes from complex-conjugate (of Euler's formula). To find 
the projection coefficients (Fourier Transform), in each of the terms in 
the summation one computes the inner product of two complex vectors: the 
complex sinusoid you are "testing", and its complex-conjugate. The 
resulting complex number (each bin is a complex number) will not only 
tell you "how much of a sinusoid was present in the original signal", 
but also its relative phase.


This is an excellent read:

https://ccrma.stanford.edu/~jos/st/

Cheers,
Esteban





On 10/5/2015 4:28 PM, Stijn Frishert wrote:

Hey all,

In trying to get to grips with the discrete Fourier transform, I have a 
question about the minus sign in the exponent of the complex sinusoids you 
correlate with doing the transform.

The inverse transform doesn’t contain this negation and a quick search on the 
internet tells me Fourier analysis and synthesis work as long as one of the 
formulas contains that minus and the other one doesn’t.

So: why? If the bins in the resulting spectrum represent how much of a sinusoid 
was present in the original signal (cross-correlation), I would expect 
synthesis to use these exact same sinusoids to get back to the original signal. 
Instead it uses their inverse! How can the resulting signal not be 180 phase 
shifted?

This may be text-book dsp theory, but I’ve looked and searched and everywhere 
seems to skip over it as if it’s self-evident.

Stijn Frishert
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CIRMMT/CAML - McGill Univ
MTG - Univ Pompeu Fabra
http://ccrma.stanford.edu/~esteban

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[music-dsp] Fourier and its negative exponent

[Stijn Frishert]

Hey all,

In trying to get to grips with the discrete Fourier transform, I have a 
question about the minus sign in the exponent of the complex sinusoids you 
correlate with doing the transform.

The inverse transform doesn’t contain this negation and a quick search on the 
internet tells me Fourier analysis and synthesis work as long as one of the 
formulas contains that minus and the other one doesn’t.

So: why? If the bins in the resulting spectrum represent how much of a sinusoid 
was present in the original signal (cross-correlation), I would expect 
synthesis to use these exact same sinusoids to get back to the original signal. 
Instead it uses their inverse! How can the resulting signal not be 180 phase 
shifted?

This may be text-book dsp theory, but I’ve looked and searched and everywhere 
seems to skip over it as if it’s self-evident.

Stijn Frishert
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Re: [music-dsp] Fourier and its negative exponent

[Allen Downey]

In Chapter 7 of Think DSP, I develop the DFT in a way that might help with
this:

http://greenteapress.com/thinkdsp/html/thinkdsp008.html

If you think of the inverse DFT as matrix multiplication where the matrix,
M, contains complex exponentials as basis vectors, the (forward) DFT is the
multiplication by the inverse of M.  Since M is unitary, its inverse is its
conjugate transpose.  The conjugation is the source of the negative sign,
when you write the DFT in summation form.

Allen



On Mon, Oct 5, 2015 at 9:28 AM, Stijn Frishert 
wrote:

> Hey all,
>
> In trying to get to grips with the discrete Fourier transform, I have a
> question about the minus sign in the exponent of the complex sinusoids you
> correlate with doing the transform.
>
> The inverse transform doesn’t contain this negation and a quick search on
> the internet tells me Fourier analysis and synthesis work as long as one of
> the formulas contains that minus and the other one doesn’t.
>
> So: why? If the bins in the resulting spectrum represent how much of a
> sinusoid was present in the original signal (cross-correlation), I would
> expect synthesis to use these exact same sinusoids to get back to the
> original signal. Instead it uses their inverse! How can the resulting
> signal not be 180 phase shifted?
>
> This may be text-book dsp theory, but I’ve looked and searched and
> everywhere seems to skip over it as if it’s self-evident.
>
> Stijn Frishert
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> music-dsp@music.columbia.edu
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Re: [music-dsp] Fourier and its negative exponent

[Esteban Maestre]

On 10/5/2015 6:15 PM, Esteban Maestre wrote:

the complex sinusoid you are "testing", and its complex-conjugate

Sorry:

I mean "your signal and the complex sinusoid your are testing".

Esteban

--

Esteban Maestre
CIRMMT/CAML - McGill Univ
MTG - Univ Pompeu Fabra
http://ccrma.stanford.edu/~esteban

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