>c1910 wrote:
>
>   ...
>
>>>>>> using Hilbert Trans
>>>>>> if : A = amplitude of carrier
>>>>>>
>>>>>> w = freq of carrier
>>>>>>
>>>>>> m(t)= information signal
>>>>>>
>>>>>> I = A*cos(wt)*m(t)
>>>>>> Q = A*sin(wt)*m(t)
>>>>>>
>>>>>>    I^2 + Q^2 = A^2*cos^2(wt)*m(t)^2 + A^2*sin^2(wt)*m(t)^2
>>>>>> = A^2*m(t)^2*[cos^2(wt) + sin^2(wt)] = A^2*m(t)^2*1 = A^2*m(t)^2
>>>>>>
>>>>>> sqrt(A^2*m(t)^2) = A*m(t)...this is the envelope right?
>>>>> Yes.
>
>   ...
>
>>> If the sample rate is high relative to the carrier frequency -- What a

>>> waste! -- then some of the samples will be close enough to the peak
for 
>>> a detector to work is you just average all the absolute values.  In a

>>> simulation, you can adjust all these variables to suit the demo, but
in 
>>> real live, you don't have that luxury.
>
>   ...
>
>> ooo
>> 
>> ic...
>
>How do integrated circuits come into this?
>
>> it's too late to change my method...
>> but it's work...
>> hehe...thanks for the input...
>
>So in your school, it's better to do it wrong than do it late? What 
>school is that? I'm sure many would like to know.
>
>> so, there is no explanation in math about hilbert T. if we use it for
>> discrete signal?
>
>I explained it to you. You repeated the proof above. Did you forget it, 
>or did you not understand what you were writing?
>
>Jerry
>-- 
>Engineering is the art of making what you want from things you can get.
>�����������������������������������������������������������������������
>




i think my method doesn't fully wrong...
it's just different from the original one...
as long as the hilbert T. is cover the uncover peak of the signal...
and it does work well...

the proof, i have is in continues signal...
i don't have the proof in discrete signal...
and the proof in continues signal, i think it is the same when we use
square law...

c1910 wrote:

   ...

>>>>> using Hilbert Trans
>>>>> if : A = amplitude of carrier
>>>>>
>>>>> w = freq of carrier
>>>>>
>>>>> m(t)= information signal
>>>>>
>>>>> I = A*cos(wt)*m(t)
>>>>> Q = A*sin(wt)*m(t)
>>>>>
>>>>>    I^2 + Q^2 = A^2*cos^2(wt)*m(t)^2 + A^2*sin^2(wt)*m(t)^2
>>>>> = A^2*m(t)^2*[cos^2(wt) + sin^2(wt)] = A^2*m(t)^2*1 = A^2*m(t)^2
>>>>>
>>>>> sqrt(A^2*m(t)^2) = A*m(t)...this is the envelope right?
>>>> Yes.

   ...

>> If the sample rate is high relative to the carrier frequency -- What a 
>> waste! -- then some of the samples will be close enough to the peak for 
>> a detector to work is you just average all the absolute values.  In a 
>> simulation, you can adjust all these variables to suit the demo, but in 
>> real live, you don't have that luxury.

   ...

> ooo
> 
> ic...

How do integrated circuits come into this?

> it's too late to change my method...
> but it's work...
> hehe...thanks for the input...

So in your school, it's better to do it wrong than do it late? What 
school is that? I'm sure many would like to know.

> so, there is no explanation in math about hilbert T. if we use it for
> discrete signal?

I explained it to you. You repeated the proof above. Did you forget it, 
or did you not understand what you were writing?

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

>c1910 wrote:
>>> c1910 wrote:
>>>> ok i understand that...
>>>>
>>>> using Hilbert Trans
>>>> if : A = amplitude of carrier
>>>>
>>>> w = freq of carrier
>>>>
>>>> m(t)= information signal
>>>>
>>>> I = A*cos(wt)*m(t)
>>>> Q = A*sin(wt)*m(t)
>>>>
>>>>    I^2 + Q^2 = A^2*cos^2(wt)*m(t)^2 + A^2*sin^2(wt)*m(t)^2
>>>> = A^2*m(t)^2*[cos^2(wt) + sin^2(wt)] = A^2*m(t)^2*1 = A^2*m(t)^2
>>>>
>>>> sqrt(A^2*m(t)^2) = A*m(t)...this is the envelope right?
>>> Yes.
>>>
>>>> using square law
>>> Why do you insist on square law rather than abs()? do you really want
to
>> 
>>> represent the carrier nonlinearly?
>>>
>>>> if : A = amplitude of carrier
>>>>
>>>> w = freq of carrier
>>>>
>>>> m(t)= information signal
>>>>
>>>> eq. of AM = A*cos(wt)*m(t)
>>> That's the continuous-time equation of the carrier, yes.
>>>
>>>> A^2*cos^2(wt)*m(t)^2  
>>>>
>>>> then low pass filter, become :
>>>>
>>>> A^2*m(t)^2
>>> As an analog signal, yes.
>>>
>>>> then :
>>>> sqrt(A^2*m(t)^2) = A*m(t)...same with Hilbert T., show the
envelope...
>> 
>>> For continuous signals, again yes. It doesn't work reliably with
sampled
>> 
>>> signals because there is no control over the part of the carrier 
>>> waveform at which the samples are taken.
>>>
>>>> so my conclusion is in math, both of the method show us that they
will
>>>> recover all the envelope, if we use continues signal...
>>>> and as u said too, that square law won't recover the envelope if we
>> use
>>>> discrete signal, b'coz the sampling freq problem...
>>>> how to proof it, if we use discrete signal?
>>> Logic. Your square-law method depends on sample instants occurring at

>>> the peak of the analog waveform. There is no way to guarantee that.
>>>
>>>> oya...
>>>> u said,
>>>>> Square-law detectors are suitable neither for digital nor for
analog
>>>>> linear (distortion-free) demodulation. A square-law device
distorts.
>>>> what kind of distortion? and what kind of device?
>>>>
>>>> i need this to defend my thesis...
>>> Observe that your proposed demodulation system isn't really square
law. 
>>> You calculate sqrt(x^2) where you could simply calculate abs(x) and 
>>> avoid the square root. A real square-law detector omits the
square-root 
>>> operation. There are few processes for which it is appropriate.
>>>
>>> With either square/square-root or absolute value, you need to select
and
>> 
>>> use only those samples that fall at the peak if the carrier. For the 
>>> analog case too, you need a peak detector. With I and Q, you don't.
With
>> 
>>> analog, you have every peak available and a peak detector is simple 
>>> (diode and capacitor). With digital systems, you have no assurance
that 
>>> any of yous samples will be near a peak, and even if one is, you need
to
>> 
>>> discard the others. The I/Q demodulation lets you infer peaks that 
>>> haven't been sampled. If you can find the peaks some other way, that 
>>> would indeed be material for a thesis.
>>>
>>> Jerry
>>> -- 
>>> Engineering is the art of making what you want from things you can
get.
>>>
>> 
>> 
>> 
>> ok, i understand that, thanks...
>> 
>> i already made my demodulation system, but a little bit different
>> 
>> i use Hilbert T., i use multiplyer with cos and sine wave, and LPF and
an
>> adder...
>> what i do is:
>> 1. separate the signal to I and Q channel
>> 2. Q channel, using Hilbert T., so between I and Q channel have
difference
>> phase of pi/2 
>> 3. multiply the I-phase with Cosine wave with carrier freq
>> 4. multiply the Q-phase with sine wave with carrier freq
>> 5. pass it thru LPF on each channel
>> 6. sum the output from LPF from eanch channel
>> 7. then i got the envelope
>> 
>> you said, if we use sqrt(), will give non-linearity...
>> as the output, i found my signal have a lot of sampling quite look the
>> same with the input...just don't reach the peak, very well...is that
ok?
>> i use this method bcoz when i tried to do abs(), the signal lost the
other
>> sample, but it reach the peak...
>> with this method, the signal looks better...is my method ok?
>> 
>> thanks
>
>Just square each of I and Q, add then and take the square root of the 
>sum. There's no need to filter. A*m = sqrt(I^2 + Q^2). There are square 
>root approximations that may be good enough depending on the fidelity 
>you need.
>
>Sqrt() will give non-linearity is you don't square first, and square 
>will give non-linearity is you don't square-root first. What don't you 
>understand?
>
>If the sample rate is high relative to the carrier frequency -- What a 
>waste! -- then some of the samples will be close enough to the peak for 
>a detector to work is you just average all the absolute values.  In a 
>simulation, you can adjust all these variables to suit the demo, but in 
>real live, you don't have that luxury.
>
>Jerry
>-- 
>Engineering is the art of making what you want from things you can get.
>



ooo

ic...

it's too late to change my method...
but it's work...
hehe...thanks for the input...

so, there is no explanation in math about hilbert T. if we use it for
discrete signal?

c1910 wrote:
>> c1910 wrote:
>>> ok i understand that...
>>>
>>> using Hilbert Trans
>>> if : A = amplitude of carrier
>>>
>>> w = freq of carrier
>>>
>>> m(t)= information signal
>>>
>>> I = A*cos(wt)*m(t)
>>> Q = A*sin(wt)*m(t)
>>>
>>>    I^2 + Q^2 = A^2*cos^2(wt)*m(t)^2 + A^2*sin^2(wt)*m(t)^2
>>> = A^2*m(t)^2*[cos^2(wt) + sin^2(wt)] = A^2*m(t)^2*1 = A^2*m(t)^2
>>>
>>> sqrt(A^2*m(t)^2) = A*m(t)...this is the envelope right?
>> Yes.
>>
>>> using square law
>> Why do you insist on square law rather than abs()? do you really want to
> 
>> represent the carrier nonlinearly?
>>
>>> if : A = amplitude of carrier
>>>
>>> w = freq of carrier
>>>
>>> m(t)= information signal
>>>
>>> eq. of AM = A*cos(wt)*m(t)
>> That's the continuous-time equation of the carrier, yes.
>>
>>> A^2*cos^2(wt)*m(t)^2  
>>>
>>> then low pass filter, become :
>>>
>>> A^2*m(t)^2
>> As an analog signal, yes.
>>
>>> then :
>>> sqrt(A^2*m(t)^2) = A*m(t)...same with Hilbert T., show the envelope...
> 
>> For continuous signals, again yes. It doesn't work reliably with sampled
> 
>> signals because there is no control over the part of the carrier 
>> waveform at which the samples are taken.
>>
>>> so my conclusion is in math, both of the method show us that they will
>>> recover all the envelope, if we use continues signal...
>>> and as u said too, that square law won't recover the envelope if we
> use
>>> discrete signal, b'coz the sampling freq problem...
>>> how to proof it, if we use discrete signal?
>> Logic. Your square-law method depends on sample instants occurring at 
>> the peak of the analog waveform. There is no way to guarantee that.
>>
>>> oya...
>>> u said,
>>>> Square-law detectors are suitable neither for digital nor for analog
>>>> linear (distortion-free) demodulation. A square-law device distorts.
>>> what kind of distortion? and what kind of device?
>>>
>>> i need this to defend my thesis...
>> Observe that your proposed demodulation system isn't really square law. 
>> You calculate sqrt(x^2) where you could simply calculate abs(x) and 
>> avoid the square root. A real square-law detector omits the square-root 
>> operation. There are few processes for which it is appropriate.
>>
>> With either square/square-root or absolute value, you need to select and
> 
>> use only those samples that fall at the peak if the carrier. For the 
>> analog case too, you need a peak detector. With I and Q, you don't. With
> 
>> analog, you have every peak available and a peak detector is simple 
>> (diode and capacitor). With digital systems, you have no assurance that 
>> any of yous samples will be near a peak, and even if one is, you need to
> 
>> discard the others. The I/Q demodulation lets you infer peaks that 
>> haven't been sampled. If you can find the peaks some other way, that 
>> would indeed be material for a thesis.
>>
>> Jerry
>> -- 
>> Engineering is the art of making what you want from things you can get.
>>
> 
> 
> 
> ok, i understand that, thanks...
> 
> i already made my demodulation system, but a little bit different
> 
> i use Hilbert T., i use multiplyer with cos and sine wave, and LPF and an
> adder...
> what i do is:
> 1. separate the signal to I and Q channel
> 2. Q channel, using Hilbert T., so between I and Q channel have difference
> phase of pi/2 
> 3. multiply the I-phase with Cosine wave with carrier freq
> 4. multiply the Q-phase with sine wave with carrier freq
> 5. pass it thru LPF on each channel
> 6. sum the output from LPF from eanch channel
> 7. then i got the envelope
> 
> you said, if we use sqrt(), will give non-linearity...
> as the output, i found my signal have a lot of sampling quite look the
> same with the input...just don't reach the peak, very well...is that ok?
> i use this method bcoz when i tried to do abs(), the signal lost the other
> sample, but it reach the peak...
> with this method, the signal looks better...is my method ok?
> 
> thanks

Just square each of I and Q, add then and take the square root of the 
sum. There's no need to filter. A*m = sqrt(I^2 + Q^2). There are square 
root approximations that may be good enough depending on the fidelity 
you need.

Sqrt() will give non-linearity is you don't square first, and square 
will give non-linearity is you don't square-root first. What don't you 
understand?

If the sample rate is high relative to the carrier frequency -- What a 
waste! -- then some of the samples will be close enough to the peak for 
a detector to work is you just average all the absolute values.  In a 
simulation, you can adjust all these variables to suit the demo, but in 
real live, you don't have that luxury.

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

>c1910 wrote:
>> ok i understand that...
>> 
>> using Hilbert Trans
>> if : A = amplitude of carrier
>> 
>> w = freq of carrier
>> 
>> m(t)= information signal
>> 
>> I = A*cos(wt)*m(t)
>> Q = A*sin(wt)*m(t)
>> 
>>    I^2 + Q^2 = A^2*cos^2(wt)*m(t)^2 + A^2*sin^2(wt)*m(t)^2
>> = A^2*m(t)^2*[cos^2(wt) + sin^2(wt)] = A^2*m(t)^2*1 = A^2*m(t)^2
>> 
>> sqrt(A^2*m(t)^2) = A*m(t)...this is the envelope right?
>
>Yes.
>
>> using square law
>
>Why do you insist on square law rather than abs()? do you really want to

>represent the carrier nonlinearly?
>
>> if : A = amplitude of carrier
>> 
>> w = freq of carrier
>> 
>> m(t)= information signal
>> 
>> eq. of AM = A*cos(wt)*m(t)
>
>That's the continuous-time equation of the carrier, yes.
>
>> A^2*cos^2(wt)*m(t)^2  
>> 
>> then low pass filter, become :
>> 
>> A^2*m(t)^2
>
>As an analog signal, yes.
>
>> then :
>> sqrt(A^2*m(t)^2) = A*m(t)...same with Hilbert T., show the envelope...

>
>For continuous signals, again yes. It doesn't work reliably with sampled

>signals because there is no control over the part of the carrier 
>waveform at which the samples are taken.
>
>> so my conclusion is in math, both of the method show us that they will
>> recover all the envelope, if we use continues signal...
>> and as u said too, that square law won't recover the envelope if we
use
>> discrete signal, b'coz the sampling freq problem...
>> how to proof it, if we use discrete signal?
>
>Logic. Your square-law method depends on sample instants occurring at 
>the peak of the analog waveform. There is no way to guarantee that.
>
>> oya...
>> u said,
>>> Square-law detectors are suitable neither for digital nor for analog
>>> linear (distortion-free) demodulation. A square-law device distorts.
>> 
>> what kind of distortion? and what kind of device?
>> 
>> i need this to defend my thesis...
>
>Observe that your proposed demodulation system isn't really square law. 
>You calculate sqrt(x^2) where you could simply calculate abs(x) and 
>avoid the square root. A real square-law detector omits the square-root 
>operation. There are few processes for which it is appropriate.
>
>With either square/square-root or absolute value, you need to select and

>use only those samples that fall at the peak if the carrier. For the 
>analog case too, you need a peak detector. With I and Q, you don't. With

>analog, you have every peak available and a peak detector is simple 
>(diode and capacitor). With digital systems, you have no assurance that 
>any of yous samples will be near a peak, and even if one is, you need to

>discard the others. The I/Q demodulation lets you infer peaks that 
>haven't been sampled. If you can find the peaks some other way, that 
>would indeed be material for a thesis.
>
>Jerry
>-- 
>Engineering is the art of making what you want from things you can get.
>



ok, i understand that, thanks...

i already made my demodulation system, but a little bit different

i use Hilbert T., i use multiplyer with cos and sine wave, and LPF and an
adder...
what i do is:
1. separate the signal to I and Q channel
2. Q channel, using Hilbert T., so between I and Q channel have difference
phase of pi/2 
3. multiply the I-phase with Cosine wave with carrier freq
4. multiply the Q-phase with sine wave with carrier freq
5. pass it thru LPF on each channel
6. sum the output from LPF from eanch channel
7. then i got the envelope

you said, if we use sqrt(), will give non-linearity...
as the output, i found my signal have a lot of sampling quite look the
same with the input...just don't reach the peak, very well...is that ok?
i use this method bcoz when i tried to do abs(), the signal lost the other
sample, but it reach the peak...
with this method, the signal looks better...is my method ok?

thanks

On Feb 23, 10:08 pm, "c1910" <c_19...@hotmail.com> wrote:
> so my conclusion is in math, both of the method show us
> that they will
> recover all the envelope, if we use continues signal...
> and as u said too, that square law won't recover the
> envelope if we use
> discrete signal, b'coz the sampling freq problem...
> how to proof it, if we use discrete signal?

What you might want to do in look into the math of a generic
sampling problem.  Does the measured amplitude of samples
of a sine wave depend on the relationship to the sampling
frequency and phase?  If so, what is that relationship?

c1910 wrote:
> ok i understand that...
> 
> using Hilbert Trans
> if : A = amplitude of carrier
> 
> w = freq of carrier
> 
> m(t)= information signal
> 
> I = A*cos(wt)*m(t)
> Q = A*sin(wt)*m(t)
> 
>    I^2 + Q^2 = A^2*cos^2(wt)*m(t)^2 + A^2*sin^2(wt)*m(t)^2
> = A^2*m(t)^2*[cos^2(wt) + sin^2(wt)] = A^2*m(t)^2*1 = A^2*m(t)^2
> 
> sqrt(A^2*m(t)^2) = A*m(t)...this is the envelope right?

Yes.

> using square law

Why do you insist on square law rather than abs()? do you really want to 
represent the carrier nonlinearly?

> if : A = amplitude of carrier
> 
> w = freq of carrier
> 
> m(t)= information signal
> 
> eq. of AM = A*cos(wt)*m(t)

That's the continuous-time equation of the carrier, yes.

> A^2*cos^2(wt)*m(t)^2  
> 
> then low pass filter, become :
> 
> A^2*m(t)^2

As an analog signal, yes.

> then :
> sqrt(A^2*m(t)^2) = A*m(t)...same with Hilbert T., show the envelope... 

For continuous signals, again yes. It doesn't work reliably with sampled 
signals because there is no control over the part of the carrier 
waveform at which the samples are taken.

> so my conclusion is in math, both of the method show us that they will
> recover all the envelope, if we use continues signal...
> and as u said too, that square law won't recover the envelope if we use
> discrete signal, b'coz the sampling freq problem...
> how to proof it, if we use discrete signal?

Logic. Your square-law method depends on sample instants occurring at 
the peak of the analog waveform. There is no way to guarantee that.

> oya...
> u said,
>> Square-law detectors are suitable neither for digital nor for analog
>> linear (distortion-free) demodulation. A square-law device distorts.
> 
> what kind of distortion? and what kind of device?
> 
> i need this to defend my thesis...

Observe that your proposed demodulation system isn't really square law. 
You calculate sqrt(x^2) where you could simply calculate abs(x) and 
avoid the square root. A real square-law detector omits the square-root 
operation. There are few processes for which it is appropriate.

With either square/square-root or absolute value, you need to select and 
use only those samples that fall at the peak if the carrier. For the 
analog case too, you need a peak detector. With I and Q, you don't. With 
analog, you have every peak available and a peak detector is simple 
(diode and capacitor). With digital systems, you have no assurance that 
any of yous samples will be near a peak, and even if one is, you need to 
discard the others. The I/Q demodulation lets you infer peaks that 
haven't been sampled. If you can find the peaks some other way, that 
would indeed be material for a thesis.

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

>Yes. Al Clark demonstrated it above. I will paraphrase his proof here.
>
>A is the amplitude of the carrier envelope at instant t.
>w is the frequency of the carrier.
>
>I = A*cos(wt)
>Q = A*sin(wt)
>
>    I^2 + Q^2 = A^2*cos^2(wt) + A^2*sin^2(wt)
>  = A^2*[cos^2(wt) + sin^2(wt)] = A^2*1 = A^2
>
>sqrt(A^2) = A = (by definition) the magnitude of the envelope.
>
>Therefore sqrt(I^2 + Q^2) recovers the envelope.
>
>Jerry
>-- 
>Engineering is the art of making what you want from things you can get.
>�����������������������������������������������������������������������
>
>


ok i understand that...

using Hilbert Trans
if : A = amplitude of carrier

w = freq of carrier

m(t)= information signal

I = A*cos(wt)*m(t)
Q = A*sin(wt)*m(t)

   I^2 + Q^2 = A^2*cos^2(wt)*m(t)^2 + A^2*sin^2(wt)*m(t)^2
= A^2*m(t)^2*[cos^2(wt) + sin^2(wt)] = A^2*m(t)^2*1 = A^2*m(t)^2

sqrt(A^2*m(t)^2) = A*m(t)...this is the envelope right?

using square law
if : A = amplitude of carrier

w = freq of carrier

m(t)= information signal

eq. of AM = A*cos(wt)*m(t)


A^2*cos^2(wt)*m(t)^2  

then low pass filter, become :

A^2*m(t)^2

then :
sqrt(A^2*m(t)^2) = A*m(t)...same with Hilbert T., show the envelope... 



so my conclusion is in math, both of the method show us that they will
recover all the envelope, if we use continues signal...
and as u said too, that square law won't recover the envelope if we use
discrete signal, b'coz the sampling freq problem...
how to proof it, if we use discrete signal?

oya...
u said,
>Square-law detectors are suitable neither for digital nor for analog
>linear (distortion-free) demodulation. A square-law device distorts.

what kind of distortion? and what kind of device?

i need this to defend my thesis...

thanks...

Second try. The first didn't show up after a couple of hours.

c1910 wrote:

  ...

 >>> i need a mathematical calculation or proof in math...
 >>> is there any math explanation for this?
 >> Proof of what? The problems of peak detectors, or the identity
 >> showing that sin^2(x) + cos^2(x) = 1?

   ...

 > problem of peak detectors...
 >
 > from what u've said, my conclusion is that square-law detector doesn't
 > suit with digital processing because the sampling freq of a signal
 > mostly never near the peak...

Square-law detectors are suitable neither for digital nor for analog
linear (distortion-free) demodulation. A square-law device distorts.

 > and by using hilbert transform, we can solve the sampling problem...i
 > that right?

Yes.

 > i don't understand how to proof that thing in math?!
 > can it be proof?

Yes. Al Clark demonstrated it above. I will paraphrase his proof here.

A is the amplitude of the carrier envelope at instant t.
w is the frequency of the carrier.

I = A*cos(wt)
Q = A*sin(wt)

    I^2 + Q^2 = A^2*cos^2(wt) + A^2*sin^2(wt)
  = A^2*[cos^2(wt) + sin^2(wt)] = A^2*1 = A^2

sqrt(A^2) = A = (by definition) the magnitude of the envelope.

Therefore sqrt(I^2 + Q^2) recovers the envelope.

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

>c1910 wrote:
>> dear Jerry,
>>> u said :
>> 
>>> Square-law detectors suffer from distortion (with the rare exception
of 
>>> single-sideband with carrier. They have no place in digital designs
that
>> 
>>> I know of. Peak detectors work with continuous signals, but there is
no 
>>> reason to think that most samples will be near the carrier peak in a 
>>> sampled system unless the oversampling ratio is quite high relative to

>>> the carrier or IF frequency. I-Q demodulation allows you to get the 
>>> magnitude at much lower sample rates.
>>>
>>> If you didn't know that, what led you to that method?
>
>   ...
>
>> i need a mathematical calculation or proof in math...
>> is there any math explanation for this?
>
>Proof of what? The problems of peak detectors, or the identity showing 
>that sin^2(x) + cos^2(x) = 1?
>
>Jerry
>-- 
>Engineering is the art of making what you want from things you can get.
>


problem of peak detectors...

from what u've said, my conclusion is that square-law detector doesn't
suit with digital processing because the sampling freq of a signal mostly
never near the peak...
and by using hilbert transform, we can solve the sampling problem...i that
right?

i don't understand how to proof that thing in math?!
can it be proof?

thanks