DFT spectrum and coefficient insight

Hi,

I have created a new flash program that gives further insight into the DFT. 
it is located here.

http://www.fourier-series.com/fourierseries2/DFT_tutorial.html

I have the new program highlighted on the page.  This program explains the 
DFT coeffiecients and also gives insight into why windowing is needed.

I intended to do a program on windowing, but got sidetracked and did this 
first , as a prcurser to windowing. 

"Blocher's spokesman" <no@nono.com> wrote in message 
news:48028b7c$0$5720$4c368faf@roadrunner.com...
> Hi,
>
> I have created a new flash program that gives further insight into the 
> DFT. it is located here.
>
> http://www.fourier-series.com/fourierseries2/DFT_tutorial.html
>
> I have the new program highlighted on the page.  This program explains the 
> DFT coeffiecients and also gives insight into why windowing is needed.
>
> I intended to do a program on windowing, but got sidetracked and did this 
> first , as a prcurser to windowing.
>

Very nice

Fred 

On Apr 14, 10:23&#4294967295;am, "Fred Marshall" <fmarshallx@remove_the_x.acm.org>
wrote:
> "Blocher's spokesman" <n...@nono.com> wrote in message
>
> news:48028b7c$0$5720$4c368faf@roadrunner.com...
>
> > Hi,
>
> > I have created a new flash program that gives further insight into the
> > DFT. it is located here.
>
> >http://www.fourier-series.com/fourierseries2/DFT_tutorial.html
>
> > I have the new program highlighted on the page. &#4294967295;This program explains the
> > DFT coeffiecients and also gives insight into why windowing is needed.
>
> > I intended to do a program on windowing, but got sidetracked and did this
> > first , as a prcurser to windowing.
>
> Very nice
>
> Fred

Thanks,

I am starting on windows, but I want to first explain how The DFT
coefficients are converted into magnitude and go over the log
conversion,
I think the books switch to magnitude representation too lightly.

I also think that a good explanation is lacking for when to use 20log
and 10 log for doing dB's.  For instance, To convert the DSP to
magnitude you need to  square both real and imaginary coefficients and
then take the square root.  Then to do a logrithnic presentation you
need to square the magnitude again.  I am not sure its clear that the
re-squaring of the magnitude is becuase the DFT coefficients are like
voltages and thats why it gets squared again.

Anyhow,  I hope to make all the log stuff clear (dB's , dBm's , dBW's
dBc's etc- and when to do 10 log and 20 log).

On Apr 15, 9:54 am, buleg...@columbus.rr.com wrote:
> On Apr 14, 10:23 am, "Fred Marshall" <fmarshallx@remove_the_x.acm.org>
> wrote:
>
>
>
> > "Blocher's spokesman" <n...@nono.com> wrote in message
>
> >news:48028b7c$0$5720$4c368faf@roadrunner.com...
>
> > > Hi,
>
> > > I have created a new flash program that gives further insight into the
> > > DFT. it is located here.
>
> > >http://www.fourier-series.com/fourierseries2/DFT_tutorial.html
>
> > > I have the new program highlighted on the page.  This program explains the
> > > DFT coeffiecients and also gives insight into why windowing is needed.
>
> > > I intended to do a program on windowing, but got sidetracked and did this
> > > first , as a prcurser to windowing.
>
> > Very nice
>
> > Fred
>
> Thanks,
>
> I am starting on windows, but I want to first explain how The DFT
> coefficients are converted into magnitude and go over the log
> conversion,
> I think the books switch to magnitude representation too lightly.
>
> I also think that a good explanation is lacking for when to use 20log
> and 10 log for doing dB's.  For instance, To convert the DSP to
> magnitude you need to  square both real and imaginary coefficients and
> then take the square root.  Then to do a logrithnic presentation you
> need to square the magnitude again.  I am not sure its clear that the
> re-squaring of the magnitude is becuase the DFT coefficients are like
> voltages and thats why it gets squared again.

.> Anyhow,  I hope to make all the log stuff clear (dB's , dBm's ,
dBW's
.> dBc's etc- and when to do 10 log and 20 log).

Another way to look at this might be that if you are going to take the
logarithm you don't need to take the square root (or square it again).
That's one of the reasons to use '10 log'.

While the power spectrum can be -calculated- as the sum of the squares
of the real and imaginary components, the formalists usually -define-
it as the product of the complex coefficient and its complex
conjugate. It might be useful to understand why, and explain it at
this point.

Dale B. Dalrymple
http://dbdimages.com

"dbd" <dbd@ieee.org> wrote in message

> Another way to look at this might be that if you are going to take the
> logarithm you don't need to take the square root (or square it again).
> That's one of the reasons to use '10 log'.

Agreed, but it glosses over the fact that the magnitude of the DFT is the 
square root of the square of the
Imag and real parts.

>
> While the power spectrum can be -calculated- as the sum of the squares
> of the real and imaginary components, the formalists usually -define-
> it as the product of the complex coefficient and its complex
> conjugate. It might be useful to understand why, and explain it at
> this point.
>
I spent all afternoon thinking about this.  Power *seems* so straight 
forward, but it is not in my opinion.
I finally got it I think.  I really had to look at Parsevals theorom for the 
half frequency point.  Parsevals theorom works at the half freq coefficient, 
but that does not mean that the DFT coefficient properly reflects the 
correct power of the signal there.

Any how,  I started with an amplitude 1V cosine signal on an exact harmonic 
of the windowing interval. After reviewing my basic EE theory (that took 
longer than I care to admit to remember that this 1V-peak signal (across 1 
ohm) gives 0.5 Watt).  I took the DFT coefficients , which had a magnitude 
of 0.5 when normalized.  Then resquare the magnitudes to get power , and 
wholla, 0.25 + 0.25 for the conjugate = 0.5 V.

And it works beautiful at DC, because the power of a 1vdc signal is 1 watt, 
and the normalized DFT coefficient is 1 , instead of 0.5, so the power works 
out correctly there too.

Any how your post got me thinking to pull the, so called, straight forward 
power calculations together.

Brent

 

On Apr 15, 6:59 pm, "Blocher's spokesman" <n...@nono.com> wrote:
> "dbd" <d...@ieee.org> wrote in message
> > Another way to look at this might be that if you are going to take the
> > logarithm you don't need to take the square root (or square it again).
> > That's one of the reasons to use '10 log'.
>
> Agreed, but it glosses over the fact that the magnitude of the DFT is the
> square root of the square of the
> Imag and real parts.
>
>
>
> > While the power spectrum can be -calculated- as the sum of the squares
> > of the real and imaginary components, the formalists usually -define-
> > it as the product of the complex coefficient and its complex
> > conjugate. It might be useful to understand why, and explain it at
> > this point.
>
> I spent all afternoon thinking about this.  Power *seems* so straight
> forward, but it is not in my opinion.
> I finally got it I think.  I really had to look at Parsevals theorom for the
> half frequency point.  Parsevals theorom works at the half freq coefficient,
> but that does not mean that the DFT coefficient properly reflects the
> correct power of the signal there.
>
> Any how,  I started with an amplitude 1V cosine signal on an exact harmonic
> of the windowing interval. After reviewing my basic EE theory (that took
> longer than I care to admit to remember that this 1V-peak signal (across 1
> ohm) gives 0.5 Watt).  I took the DFT coefficients , which had a magnitude
> of 0.5 when normalized.  Then resquare the magnitudes to get power , and
> wholla, 0.25 + 0.25 for the conjugate = 0.5 V.
>
> And it works beautiful at DC, because the power of a 1vdc signal is 1 watt,
> and the normalized DFT coefficient is 1 , instead of 0.5, so the power works
> out correctly there too.
>
> Any how your post got me thinking to pull the, so called, straight forward
> power calculations together.
>
> Brent

If it is your goal to understand DSP and spread knowledge of DSP you
would do better to dispense with the association of DFT coefficients
and concepts like voltage, impedance and any correspondence to an
analog concept of power until you are able appreciate and explain the
impact of deterministic, random and transient signals on the meaning
of the values calculated from DFT coefficients.

Dale B. Dalrymple

"dbd" <dbd@ieee.org> wrote in message 
news:5e50ccef-fd55-45d9-94b3-e6c211a4e7f7@w5g2000prd.googlegroups.com...

> If it is your goal to understand DSP and spread knowledge of DSP you
> would do better to dispense with the association of DFT coefficients
> and concepts like voltage, impedance and any correspondence to an
> analog concept of power until you are able appreciate and explain the
> impact of deterministic, random and transient signals on the meaning
> of the values calculated from DFT coefficients.
>
> Dale B. Dalrymple

If I took that advice I may have to wait 10 more years to make my next move. 
I have perused the archives here and have seen other comments about not 
using analog models to explain DSP.  However, it is my major frame of 
reference, and if I cannot translate how an A/D converter which takes in a 1 
volts sinusoidal signal produces  appropriate coefficients to match that 
incoming signal... Well, at this point it is the only way I know to complete 
the loop.

So I guess I am working on an explanation of a deterministic signal.  Random 
signals seem like a whole new level.  Actually, noise and probability is 
another area I would like to explore with these flash programs. With 
transient signals, heck, from what I remember you need to be an expert at 
eigenvalues, initial conditions,  and Laplace transform. Well actually, 
impulse response and convolution seem much easier in the digital realm than 
in the analog realm, fortunately.

But for now, I am focusing on explaining dB's and how a coefficient plot is 
converted to magnitude and then to dBs.

Thanks for your thoughtful inputs, they really do get me to think.

Brent 

On Apr 15, 10:15 pm, "Blocher's spokesman" <n...@nono.com> wrote:
> "dbd" <d...@ieee.org> wrote in message
>
> news:5e50ccef-fd55-45d9-94b3-e6c211a4e7f7@w5g2000prd.googlegroups.com...
>
> > If it is your goal to understand DSP and spread knowledge of DSP you
> > would do better to dispense with the association of DFT coefficients
> > and concepts like voltage, impedance and any correspondence to an
> > analog concept of power until you are able appreciate and explain the
> > impact of deterministic, random and transient signals on the meaning
> > of the values calculated from DFT coefficients.
>
> > Dale B. Dalrymple
>
> If I took that advice I may have to wait 10 more years to make my next move.
> I have perused the archives here and have seen other comments about not
> using analog models to explain DSP.  However, it is my major frame of
> reference, and if I cannot translate how an A/D converter which takes in a 1
> volts sinusoidal signal produces  appropriate coefficients to match that
> incoming signal... Well, at this point it is the only way I know to complete
> the loop.
>
> So I guess I am working on an explanation of a deterministic signal.  Random
> signals seem like a whole new level.  Actually, noise and probability is
> another area I would like to explore with these flash programs. With
> transient signals, heck, from what I remember you need to be an expert at
> eigenvalues, initial conditions,  and Laplace transform. Well actually,
> impulse response and convolution seem much easier in the digital realm than
> in the analog realm, fortunately.
>
> But for now, I am focusing on explaining dB's and how a coefficient plot is
> converted to magnitude and then to dBs.
>
> Thanks for your thoughtful inputs, they really do get me to think.
>
> Brent

I'm glad you are thinking and are not discouraged. The point of
avoiding the use of analog concepts like impedance and power flow on
the digital region between A/Ds and D/As is to try to get you into DSP
in a shorter time.

I've quoted Will Rogers on this kind of thing here before:
"The problem isn't so much what we don't know, as what we know that
ain't so."

Dale B. Dalrymple

On Apr 14, 3:38&#4294967295;am, "Blocher's spokesman" <n...@nono.com> wrote:
> Hi,
>
> I have created a new flash program that gives further insight into the DFT.
> it is located here.
>
> http://www.fourier-series.com/fourierseries2/DFT_tutorial.html
>
> I have the new program highlighted on the page. &#4294967295;This program explains the
> DFT coeffiecients and also gives insight into why windowing is needed.
>
> I intended to do a program on windowing, but got sidetracked and did this
> first , as a prcurser to windowing.

nice work!!!! looking forward for your windowing flash tutorial.

Hari.

bulegoge@columbus.rr.com wrote:

   ...

> I also think that a good explanation is lacking for when to use 20log
> and 10 log for doing dB's.  For instance, To convert the DSP to
> magnitude you need to  square both real and imaginary coefficients and
> then take the square root.  Then to do a logrithnic presentation you
> need to square the magnitude again.  I am not sure its clear that the
> re-squaring of the magnitude is becuase the DFT coefficients are like
> voltages and thats why it gets squared again.
> 
> Anyhow,  I hope to make all the log stuff clear (dB's , dBm's , dBW's
> dBc's etc- and when to do 10 log and 20 log).

It's not an explanation of details that's needed, but an understanding 
of the fundamentals. I'll try to help.

A decibel is a ratio of powers. It is a tenth of a bel; one bel 
represents a power ratio of ten, the familiar "order of magnitude".

Ten decibels also represents a power ratio 10, so 20 dB is a ratio of 
100, and so on. A table of logs will show that a ratio of 4 is about 6 
dB. The formula is dB = 10*log10(power ratio).

In linear systems, power at any given node is proportional to the square 
of voltage, so (power ratio) = (voltage ratio)^2. Observe that if
10*log10(power ratio) = 10*log10((voltage ratio)^2), then it also equals 
20*log10(voltage ratio).

The various subscripted dB types are named for different references. A 
power of 1 milliwatt = 0 dBm. On a 600-ohm line, that represents a 
voltage of .7746; sqrt(.6). .7746 volts on any impedance = 0 dBu. Note 
that when computing dB from voltage, the multiplier is 20. 0 dBm = 0 dBu 
in a 600-ohm circuit.

I'd be happy to try to clarify any of that if is wasn't as helpful as I 
would have liked.

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