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 �transform�e� to
describe
the result. Is there such a distinction in English?
--
Jean Castonguay
�lectrocommande Pascal
Dimensional Analysis and the Discrete-Time Fourier Transform(ation?)
Started by ●January 24, 2005
Reply by ●January 24, 20052005-01-24
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 �transform�e� 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
Reply by ●January 24, 20052005-01-24
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
Reply by ●January 25, 20052005-01-25
On Mon, 24 Jan 2005 23:00:19 UTC, Tim Wescott <tim@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 �lectrocommande Pascal
Reply by ●January 25, 20052005-01-25
On Mon, 24 Jan 2005 23:21:26 UTC, glen herrmannsfeldt <gah@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 �lectrocommande Pascal
Reply by ●January 25, 20052005-01-25
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. �����������������������������������������������������������������������
Reply by ●January 25, 20052005-01-25
Jean Castonguay wrote:> On Mon, 24 Jan 2005 23:21:26 UTC, glen herrmannsfeldt > <gah@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
Reply by ●January 25, 20052005-01-25
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. �����������������������������������������������������������������������
Reply by ●January 25, 20052005-01-25
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
Reply by ●January 25, 20052005-01-25
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. �����������������������������������������������������������������������
