hi! i make an envelope detector for AM demodulation using Hilbert Transform and complex envelope... but i don't really understand about the advantages of using Hilbert Transform and complex envelope... what is the advantages of using Hilbert Transform and complex envelope? why using Hilbert Transform method is more effective than square-law? and i think i can make an envelope detector just without hilbert transform. i can use the I-phase, then multiply with exp(-jwt), then LPF. and we can multiply the output by 2, because the output will give me only half amplitude... please give me some advice... thanks...

# Advantages of Envelope detector using Hilbert Transform

Started by ●February 11, 2008

Posted by ●February 11, 2008

On Feb 11, 9:06 pm, "c1910" <c_19...@hotmail.com> wrote:> > i make an envelope detector for AM demodulation using Hilbert Transform > and complex envelope... > > but i don't really understand about the advantages of using Hilbert > Transform and complex envelope...i also don't see any advantage, for the purpose of regular AM (no suppressed carrier) detection for computing the complex envelope. square-law might not be what you want, unless you like applying the square-root after filtering out the carrier (actually a frequency that is twice the carrier frequeny). some non-linearity that leave an amplitude proportional to (1+mu*x(t)) where x(t) is your message after unbiasing is what you want. maybe abs(), maybe 1/2 wave rectifier, i dunno. i think that sometimes when the signal is more complex than a modulated sinusoidal carrier (what AM is), such as a broadbanded signal, that the complex envelope might be desired because it gives you an envelope without creating new frequency components. it *does* eliminate the negative frequency components but it does not introduce new frequencies like the square-law does. i dunno when that is advantageous, but i know of audio signal practioners (these guys walk around and fly around to places with expensive analytical test gear and come up with all sorts of numbers describing a room) think that the complex envelope (or the magnitude of the complex "analytic signal") is a super salient measure of stuff. dunno why, and maybe i should, since i'm sorta into audio, but i dunno. r b-j

Posted by ●February 11, 2008

c1910 wrote:> hi! > i make an envelope detector for AM demodulation using Hilbert Transform > and complex envelope... > > but i don't really understand about the advantages of using Hilbert > Transform and complex envelope... > > what is the advantages of using Hilbert Transform and complex envelope? > > why using Hilbert Transform method is more effective than square-law? > > and i think i can make an envelope detector just without hilbert > transform. > i can use the I-phase, then multiply with exp(-jwt), then LPF. and we can > multiply the output by 2, because the output will give me only half > amplitude... > > please give me some advice... > > thanks...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? Jerry -- Engineering is the art of making what you want from things you can get. ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

Posted by ●February 11, 2008

On Feb 11, 10:34 pm, Jerry Avins <j...@ieee.org> wrote:> > 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.yer right, Jerry. that's a good reason and one i didn't think about. r b-j

Posted by ●February 12, 2008

>c1910 wrote:>> hi! >> i make an envelope detector for AM demodulation using HilbertTransform>> and complex envelope... >> >> but i don't really understand about the advantages of using Hilbert >> Transform and complex envelope... >> >> what is the advantages of using Hilbert Transform and complexenvelope?>> >> why using Hilbert Transform method is more effective than square-law? >> >> and i think i can make an envelope detector just without hilbert >> transform. >> i can use the I-phase, then multiply with exp(-jwt), then LPF. and wecan>> multiply the output by 2, because the output will give me only half >> amplitude... >> >> please give me some advice... >> >> thanks... > >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? > >Jerry >-- >Engineering is the art of making what you want from things you can get. >ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ >i use that method because i think there is a lot of distorsion in square law...coz, square law still have the Carrier's Amplitude...

Posted by ●February 12, 2008

On 12 Feb, 03:06, "c1910" <c_19...@hotmail.com> wrote:> what is the advantages of using Hilbert Transform and complex envelope? > why using Hilbert Transform method is more effective than square-law?I don't know much about envelope detectors in AM systems, but generally speaking, when you have more than one way of implementing the same operation, there are a few trade- offs involved: - One method may require fewer components or FLOPS - One method may be more accurate - One method may be faster/simpler to design - One method may be faster/simpler to implement - One method may be better at suppressing artifacts - One method may provide more, more useful internal results and so on. Depending on the constraints on what you try to do, one method may score better on this sort of overview and would thus be the preferred one. It seems to me that processing methods often falls into two groups: Those that are cheap to implement or runs very fast (but which may give inaccurate results), and those that give accurate results but which are expensive to implement and/or cost more time to run. If cost and speed is all that matters, then the 'simpler' (or 'naive' or 'obvious') methods often come out on top in the evaluation. If you need high accuracy or error control, or other processing steps require information that can be squeezed out of the data by more elaborate algorithms, that would be deciding factors what methods to choose. So no methods can be said to be 'best' for any particular purpose. One needs to know one's own priorities (speed/cost or accuracy). Then one needs to know what can be obtained by each method -- and what the cost is. Rune

Posted by ●February 12, 2008

c1910 wrote:>> c1910 wrote: >>> hi! >>> i make an envelope detector for AM demodulation using Hilbert > Transform >>> and complex envelope... >>> >>> but i don't really understand about the advantages of using Hilbert >>> Transform and complex envelope... >>> >>> what is the advantages of using Hilbert Transform and complex > envelope? >>> why using Hilbert Transform method is more effective than square-law? >>> >>> and i think i can make an envelope detector just without hilbert >>> transform. >>> i can use the I-phase, then multiply with exp(-jwt), then LPF. and we > can >>> multiply the output by 2, because the output will give me only half >>> amplitude... >>> >>> please give me some advice... >>> >>> thanks... >> 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 use that method because i think there is a lot of distorsion in square > law...coz, square law still have the Carrier's Amplitude...I don't understand the part the reason for the distortion. Squaring a signal inevitably distorts it. Forget square-law detectors for recovering ordinary AM. Most "simple" AM demodulators are peak detectors. A digital peak detector is not only hard to design, hard even to define what it is. You can approximately extract the envelope by ensuring that the RF or IF signal is zero mean, then taking its absolute value -- not squaring -- and low-pass filtering. That will work fairly well most of the time, but on occasion it can fail horribly. Those failures will be brief enough to go unnoticed, except when you're demonstrating your system. Jerry -- Engineering is the art of making what you want from things you can get. Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯

Posted by ●February 12, 2008

On Feb 11, 9:06 pm, "c1910" <c_19...@hotmail.com> wrote:> hi! > i make an envelope detector for AM demodulation using Hilbert Transform > and complex envelope... > > but i don't really understand about the advantages of using Hilbert > Transform and complex envelope... > > what is the advantages of using Hilbert Transform and complex envelope? > > why using Hilbert Transform method is more effective than square-law? > > and i think i can make an envelope detector just without hilbert > transform. > i can use the I-phase, then multiply with exp(-jwt), then LPF. and we can > multiply the output by 2, because the output will give me only half > amplitude... > > please give me some advice... > > thanks...Basically using a hilbert Xform is a way to find the instantaneous amplitude. As Jerry et al stated this is not needed in standard AM radios where the IF is very high compared to the modulation frequency. In that case a simple diode with lowpass filter does a great job at demodulating (finding the envelope). But as often the case with software radios, the signal is mixed to a very low frequency (to reduced the sampling rate and its attendent overhead). In this case the observed peaks aren't very likely to occur at or near the actuall peaks, so the envelope is not represented very well. So the Hilbert method is a great analytical way to get there. In practice though, all you really need is a pair of filters (flat or nearly flat magnitude in the band of interest) that differ by 90 degrees. Call the outpouts A(t) and B(t) and your amplitude is simply sqrt( A(t)^2 + B(t)^2 ). IHTH, Clay

Posted by ●February 12, 2008

clay@claysturner.com wrote:> On Feb 11, 9:06 pm, "c1910" <c_19...@hotmail.com> wrote: >> hi! >> i make an envelope detector for AM demodulation using Hilbert Transform >> and complex envelope... >> >> but i don't really understand about the advantages of using Hilbert >> Transform and complex envelope... >> >> what is the advantages of using Hilbert Transform and complex envelope? >> >> why using Hilbert Transform method is more effective than square-law? >> >> and i think i can make an envelope detector just without hilbert >> transform. >> i can use the I-phase, then multiply with exp(-jwt), then LPF. and we can >> multiply the output by 2, because the output will give me only half >> amplitude... >> >> please give me some advice... >> >> thanks... > > > Basically using a hilbert Xform is a way to find the instantaneous > amplitude. As Jerry et al stated this is not needed in standard AM > radios where the IF is very high compared to the modulation frequency. > In that case a simple diode with lowpass filter does a great job at > demodulating (finding the envelope). > > But as often the case with software radios, the signal is mixed to a > very low frequency (to reduced the sampling rate and its attendent > overhead). In this case the observed peaks aren't very likely to occur > at or near the actuall peaks, so the envelope is not represented very > well. So the Hilbert method is a great analytical way to get there. In > practice though, all you really need is a pair of filters (flat or > nearly flat magnitude in the band of interest) that differ by 90 > degrees. Call the outpouts A(t) and B(t) and your amplitude is simply > sqrt( A(t)^2 + B(t)^2 ).Or just A(t)^2 + B(t)^2 if you're dead set on a square-law detector. :-) Jerry -- Engineering is the art of making what you want from things you can get. ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

Posted by ●February 13, 2008

"Jerry Avins" <jya@ieee.org> wrote in message news:GLGdnTjnkfQWVyzanZ2dnUVZ_g2dnZ2d@rcn.net...> c1910 wrote: >>> c1910 wrote: >>>> hi! >>>> i make an envelope detector for AM demodulation using Hilbert >> Transform >>>> and complex envelope... >>>> >>>> but i don't really understand about the advantages of using Hilbert >>>> Transform and complex envelope... >>>> >>>> what is the advantages of using Hilbert Transform and complex >> envelope? >>>> why using Hilbert Transform method is more effective than square-law? >>>> >>>> and i think i can make an envelope detector just without hilbert >>>> transform. >>>> i can use the I-phase, then multiply with exp(-jwt), then LPF. and we >> can >>>> multiply the output by 2, because the output will give me only half >>>> amplitude... >>>> >>>> please give me some advice... >>>> >>>> thanks... >>> 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 use that method because i think there is a lot of distorsion in square >> law...coz, square law still have the Carrier's Amplitude... > > I don't understand the part the reason for the distortion. Squaring a > signal inevitably distorts it. Forget square-law detectors for recovering > ordinary AM. Most "simple" AM demodulators are peak detectors. A digital > peak detector is not only hard to design, hard even to define what it is. > You can approximately extract the envelope by ensuring that the RF or IF > signal is zero mean, then taking its absolute value -- not squaring -- and > low-pass filtering. That will work fairly well most of the time, but on > occasion it can fail horribly. Those failures will be brief enough to go > unnoticed, except when you're demonstrating your system. > > Jerry > -- > Engineering is the art of making what you want from things you can get. > ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ Perhaps I'm missing something. I *think* this is a "digital peak detector" that behaves like a diode feeding an RC network y(0) = 0; y(n+1) = x(n) > y(n) ? x(n) : A * y(n); A should be a bit smapper than 1... Best wishes, --Phil Martel

Posted by ●February 26, 2008

>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 peakfor>>> 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, butin>>> 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. >ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ >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...

Posted by ●February 25, 2008

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. ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

Posted by ●February 25, 2008

>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 wantto>> >>> 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 theenvelope...>> >>> For continuous signals, again yes. It doesn't work reliably withsampled>> >>> 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 theywill>>>> 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 foranalog>>>>> linear (distortion-free) demodulation. A square-law devicedistorts.>>>> 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 squarelaw.>>> You calculate sqrt(x^2) where you could simply calculate abs(x) and >>> avoid the square root. A real square-law detector omits thesquare-root>>> operation. There are few processes for which it is appropriate. >>> >>> With either square/square-root or absolute value, you need to selectand>> >>> 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 assurancethat>>> any of yous samples will be near a peak, and even if one is, you needto>> >>> 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 canget.>>> >> >> >> >> 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 andan>> 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 havedifference>> 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 thatok?>> i use this method bcoz when i tried to do abs(), the signal lost theother>> 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?

Posted by ●February 24, 2008

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? > > thanksJust 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.

Posted by ●February 24, 2008

>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 weuse>> 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

Posted by ●February 24, 2008

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?

Posted by ●February 24, 2008

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 lawWhy 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)^2As 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.

Posted by ●February 24, 2008

>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. >ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ > >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...

Posted by ●February 23, 2008

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. ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

Posted by ●February 23, 2008

>c1910 wrote:>> dear Jerry, >>> u said : >> >>> Square-law detectors suffer from distortion (with the rare exceptionof>>> single-sideband with carrier. They have no place in digital designsthat>> >>> I know of. Peak detectors work with continuous signals, but there isno>>> 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