All the textbooks I have define the Discrete-Time Fourier 
Transform(ation?) as:

                 +inf
	X(f) =  Sum  x[k] . exp(-j.2.pi.f.n)
               n=-inf
 
From a dimensional analysis viewpoint, it seems incorrect:the phasor 
exponent is not dimensionless, it is a frequency.

The Discrete-Time Fourier Transform(ation?) of a signal is, hopefully 
so, the Continuous-Time Fourier Transform(ation?) of that sampled 
signal:

	x[n] = xs(t) = xs(n.Ts) where Ts is the sampling interval

                 +inf
	X(f) = Integral[xs(t) . exp(-j.2.pi.f.t) . dt]
                 -inf

                 +inf            +inf
	     = Integral[xs(t) . Sum[d(t-n.Ts)] . exp(-j.2.pi.f.t) . dt]
                 -inf          n=-inf

Since the integrand is defined only when t = n.Ts, we can replace t by
n.Ts.

                 +inf               +inf
	     = Integral[xs(n.Ts) . Sum[d(t-n.Ts)] . exp(-j.2.pi.f.Ts.n) . dt]
                 -inf             n=-inf

With a little rearranging, we get:

                 +inf            +inf
	     = Sum[xs(n.Ts)] . Integral[d(t-n.Ts) . exp(-j.2.pi.f.Ts.n) . dt]
              n=-inf             -inf

The Dirac impulse integrating to 1, the result of the integral is: 
exp(-j.2.pi.f.Ts.n); so we can write finally:

                 +inf
	X(f) = Sum[x[n]] . exp(-j.2.pi.f.Ts.n)
              n=-inf

In this expression, the phasor exponent is dimensionless. X(f) is 
periodic but its periodicity is not 1, it is 1/Ts, the sampling 
frequency.  X(f.Ts) has a periodicity of 1: f.Ts is dimensionless.

Do you think it helps using dimensional analysis?

Thank you for your comments.


P.S.: In French, we use the word «transformation» to describe the 
      Transform mathematical process and the word «transformée» to 
describe 
      the result.  Is there such a distinction in English?
-- 
Jean Castonguay
Électrocommande Pascal

Jean Castonguay wrote:

> All the textbooks I have define the Discrete-Time Fourier 
> Transform(ation?) as:
> 
>                  +inf
> 	X(f) =  Sum  x[k] . exp(-j.2.pi.f.n)
>                n=-inf
>  
> From a dimensional analysis viewpoint, it seems incorrect:the phasor 
> exponent is not dimensionless, it is a frequency.

In this case 'f' is the frequency in cycles/sample -- since both cycles 
and samples are dimensionless f is dimensionless.
> 
> The Discrete-Time Fourier Transform(ation?) of a signal is, hopefully 
> so, the Continuous-Time Fourier Transform(ation?) of that sampled 
> signal:
> 
> 	x[n] = xs(t) = xs(n.Ts) where Ts is the sampling interval
> 
>                  +inf
> 	X(f) = Integral[xs(t) . exp(-j.2.pi.f.t) . dt]
>                  -inf
> 
>                  +inf            +inf
> 	     = Integral[xs(t) . Sum[d(t-n.Ts)] . exp(-j.2.pi.f.t) . dt]
>                  -inf          n=-inf
> 
> Since the integrand is defined only when t = n.Ts, we can replace t by
> n.Ts.
> 
>                  +inf               +inf
> 	     = Integral[xs(n.Ts) . Sum[d(t-n.Ts)] . exp(-j.2.pi.f.Ts.n) . dt]
>                  -inf             n=-inf
> 
> With a little rearranging, we get:
> 
>                  +inf            +inf
> 	     = Sum[xs(n.Ts)] . Integral[d(t-n.Ts) . exp(-j.2.pi.f.Ts.n) . dt]
>               n=-inf             -inf
> 
> The Dirac impulse integrating to 1, the result of the integral is: 
> exp(-j.2.pi.f.Ts.n); so we can write finally:
> 
>                  +inf
> 	X(f) = Sum[x[n]] . exp(-j.2.pi.f.Ts.n)
>               n=-inf
> 
> In this expression, the phasor exponent is dimensionless. X(f) is 
> periodic but its periodicity is not 1, it is 1/Ts, the sampling 
> frequency.  X(f.Ts) has a periodicity of 1: f.Ts is dimensionless.
> 
> Do you think it helps using dimensional analysis?

Yes, and it points out that the 'f' in your first equation is equal to 
f.Ts in your last.  I prefer to use ? = 2?f (theta equals two pi f, if 
your reader doesn't render Latin) for the sampled-time "frequency", 
since it is the angular advance each sample, it reminds you that you 
have to take sampling into account, and it doesn't boggle the mind with 
questions of dimensionality!
> 
> Thank you for your comments.
> 
> 
> P.S.: In French, we use the word «transformation» to describe the 
>       Transform mathematical process and the word «transformée» to 
> describe 
>       the result.  Is there such a distinction in English?

In American usage one uses "transform" for just about all of it.  I 
don't know why, because the word "transformation" is a valid English 
word that is used in other contexts.

So if x(t) is the time-domain signal then X(omega) is _its_ Fourier 
transform, which was derived using _the_ Fourier transform. 
Understanding this is one of your first steps in the transformation from 
an ordinary person to a mathematician.

-- 

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com

Jean Castonguay wrote:

> All the textbooks I have define the Discrete-Time Fourier 
> Transform(ation?) as:
> 
>                  +inf
> 	X(f) =  Sum  x[k] . exp(-j.2.pi.f.n)
>                n=-inf
>  
> From a dimensional analysis viewpoint, it seems incorrect:the phasor 
> exponent is not dimensionless, it is a frequency.

 From a dimensional point of view for a discrete time transform,
it doesn't make sense that one variable is dimensionless and
the other isn't.

For continuous time the variables have dimensions time and
either cycles/time or radians/time.

For discrete time they are sample number and cycles/sample.
In the case where the transform is exp(-j 2 pi n m/N),
N is the length of the periodic transform in samples,
then m/N is in cycles/sample.   It seems then in your
case the f=m/N should also be in cycles/sample, even
if lim N --> infinity, as in your case.

Note that there is no requirement that frequency be cycles
per time, as in optics problems it is often cycles/length,
and here cycles/sample.  That is, cycles per whatever unit
is being transformed.   The phase is exp(-j 2 pi cycles).

Personally I prefer the radian units, omega of radian/second,
k for radians/length, and here radians/sample, and
exp( i k.x - i w t) where the . is a dot product between
vector x (position) and vector k (wave number or wave vector).

In the discrete time discrete frequency case, with
exp(-j 2 pi n m/N) they would be sample number and cycles per 
period of the periodic transform, again dimensionless.

-- glen

On Mon, 24 Jan 2005 23:00:19 UTC, Tim Wescott 
<t...@wescottnospamdesign.com> wrote:

> Yes, and it points out that the 'f' in your first equation is equal to 
> f.Ts in your last.

I wish authors would explain this: it would be much easier on 
self-learners like me.  

> I prefer to use ? = 2?f (theta equals two pi f,

My computer uses the 850 code page.  My news reader ProNews/2 (under 
IBM OS/2) tranlates my message to  charset=ISO-8859-1.

> if 
> your reader doesn't render Latin) for the sampled-time "frequency", 
> since it is the angular advance each sample, ...
> 

Unfortunately, the 850 code page does not have the Greek letters.  I 
miss them.
-- 
Jean Castonguay
Électrocommande Pascal

On Mon, 24 Jan 2005 23:21:26 UTC, glen herrmannsfeldt 
<g...@ugcs.caltech.edu> wrote:

> Note that there is no requirement that frequency be cycles
> per time, as in optics problems it is often cycles/length,
> and here cycles/sample.  That is, cycles per whatever unit
> is being transformed.   The phase is exp(-j 2 pi cycles).
>
Could you suggest a few titles where the Fourier Transform is used 
outside of DSP: optics, antennas, etc.?
  
> Personally I prefer the radian units, omega of radian/second,
> k for radians/length, and here radians/sample, and
> exp( i k.x - i w t) where the . is a dot product between
> vector x (position) and vector k (wave number or wave vector).
>  
I do too.  At times though, it is easier to make proof using the 
frequency: you can eschew the 2.pi «bad penny».

-- 
Jean Castonguay
Électrocommande Pascal

Jean Castonguay wrote:

  ...

>>I prefer to use ? = 2?f (theta equals two pi f,

It's evident why that is not good practice.

Jerry
-- 
Engineering is the art of making what you want from things you can get.
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

Jean Castonguay wrote:

> On Mon, 24 Jan 2005 23:21:26 UTC, glen herrmannsfeldt 
> <g...@ugcs.caltech.edu> wrote:
> 
> 
>>Note that there is no requirement that frequency be cycles
>>per time, as in optics problems it is often cycles/length,
>>and here cycles/sample.  That is, cycles per whatever unit
>>is being transformed.   The phase is exp(-j 2 pi cycles).
>>
> 
> Could you suggest a few titles where the Fourier Transform is used 
> outside of DSP: optics, antennas, etc.?
>   
> 
>>Personally I prefer the radian units, omega of radian/second,
>>k for radians/length, and here radians/sample, and
>>exp( i k.x - i w t) where the . is a dot product between
>>vector x (position) and vector k (wave number or wave vector).
>> 
> 
> I do too.  At times though, it is easier to make proof using the 
> frequency: you can eschew the 2.pi «bad penny».
> 
The 2D Fourier transform is used in analyzing optical systems.  I don't 
know all of it by any means, but apparently you can express the action 
of each lens using the Fourier transform in a way that takes diffraction 
into account.  I _do_ know that the diffraction pattern of a telescope 
focused as best as it can be is the 2D Fourier transform of the 
effective aperture shape.  Normally this is just the "Airy ring", but 
it's a snap to predict optical resolution for a given aperture.

Generally anywhere that you need to do analysis by solving linear 
time-invariant differential equations you can use the Fourier transform. 
  There are some time-varying problems that will reluctantly yield to 
Fourier analysis -- and sampling theory is a special case that will 
enthusiastically yield to Fourier analysis, thank goodness!

-- 

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com

Tim Wescott wrote:

  ...

>          I _do_ know that the diffraction pattern of a telescope
> focused as best as it can be is the 2D Fourier transform of the
> effective aperture shape.  Normally this is just the "Airy ring", but
> it's a snap to predict optical resolution for a given aperture.

Point object is implicit here. (What is being imaged has to matter.)

 ...

Jerry
-- 
Engineering is the art of making what you want from things you can get.
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

Jerry Avins wrote:

> Tim Wescott wrote:
> 
>   ...
> 
> 
>>         I _do_ know that the diffraction pattern of a telescope
>>focused as best as it can be is the 2D Fourier transform of the
>>effective aperture shape.  Normally this is just the "Airy ring", but
>>it's a snap to predict optical resolution for a given aperture.
> 
> 
> Point object is implicit here. (What is being imaged has to matter.)
> 
Well, the diffraction pattern itself assumes a point object (plane 
waves).  Any image will convolved by the diffraction pattern (or have 
it's FT multiplied, yadda yadda), keeping in mind that what gets 
detected is the magnitude squared, not the quantity that's having it's 
FT taken.

-- 

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com

Tim Wescott wrote:

> Jerry Avins wrote:
> 
>> Tim Wescott wrote:
>>
>>   ...
>>
>>
>>>         I _do_ know that the diffraction pattern of a telescope
>>> focused as best as it can be is the 2D Fourier transform of the
>>> effective aperture shape.  Normally this is just the "Airy ring", but
>>> it's a snap to predict optical resolution for a given aperture.
>>
>>
>>
>> Point object is implicit here. (What is being imaged has to matter.)
>>
> Well, the diffraction pattern itself assumes a point object (plane
> waves).  Any image will convolved by the diffraction pattern (or have
> it's FT multiplied, yadda yadda), keeping in mind that what gets
> detected is the magnitude squared, not the quantity that's having it's
> FT taken.

@-D Fourier transforms, that is to say, diffraction optics, had a direct
impact on Abbe's developing a rational theory of microscope optics which
led to strikingly better objectives. The diffraction pattern of a line
is a grating (naturally!). When only one grating line falls within the
scope of the ocular, the line can't be resolved. Simple, ain't it? Hah!

http://micro.magnet.fsu.edu/optics/timeline/people/abbe.html

Jerry
-- 
Engineering is the art of making what you want from things you can get.
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯