I have a question about envelope detection. I am an "analog" guy trying to understand the advantage of using the Hilbert transformer (HT) method for obtaining the envelope of a carrier signal. I have read that the standard DSP approach is to use a Hilbert transformer (HT) to create a 90 deg shifted version of the carrier then: envelope = sqrt (I^2+Q^2) Does the output of this contain ripple at 4x the carrier frequency? I know that "simply" taking the absolute value or squaring the carrier results in an envelope signal that contains ripple at 2x (and above) the carrier frequency. I know that "simple" half wave rectification contains ripple at 1X (and above) the carrier frequency. So is it true that the HT method still requires low pass filtering to remove the 4x carrier component but that the HT method is considered better because the 4x carrier component is easier to filter (remove) than a 2x or 1x component? thanks Mark

# Evelope detection using Hilbert transformer and carrier ripple

Started by ●April 7, 2005

Reply by ●April 7, 20052005-04-07

Reply by ●April 7, 20052005-04-07

Mark wrote:> I have a question about envelope detection. I am an "analog" guy > trying to understand the advantage of using the Hilbert transformer > (HT) method for obtaining the envelope of a carrier signal. > > I have read that the standard DSP approach is to use a Hilbert > transformer (HT) to create a 90 deg shifted version of the carrier > then: > > envelope = sqrt (I^2+Q^2) > > Does the output of this contain ripple at 4x the carrier frequency? > > I know that "simply" taking the absolute value or squaring thecarrier> results in an envelope signal that contains ripple at 2x (and above) > the carrier frequency. > > I know that "simple" half wave rectification contains ripple at 1X(and> above) the carrier frequency. > > So is it true that the HT method still requires low pass filtering to > remove the 4x carrier component but that the HT method is considered > better because the 4x carrier component is easier to filter (remove) > than a 2x or 1x component? > > thanks > MarkThis method will not produce harmonics. One thing that is often done is to multiply the carrier wave with exp(-j*2*pi*Fc*t), then lowpass filter and decimate to reduce the number of magnitude calculations per second and limit the noise bandwidth. Next a smoothing filter can be applied to the magnitude result. John

Reply by ●April 8, 20052005-04-08

Hi Mark, this type of amplitude demodulation or amplitude detection is one of the very few methods that needs NO lowpass fltering because it generates NO ripple on whatever frequency. And it is very easy to see why: Consider your carrier (the In-phase signal) being A*Sin(w*t). Then your Q(uadrature) signal is A*Cos(w*t). Now compute sqrt (I^2+Q^2): sqrt(A^2*Sin^2(w*t)+A^2*Cos^2(w*t)) = A*sqrt(Sin^2(w*t)+Cos^2(w*t)) The next step is to see that Sin^2(Something)+Cos^2(Something) can be reduced to "1" if you consider Phytagoras's law A^2+B^2=C^2 being applied to a rectangle triangle with a hypotenusis of length "1". Therefore A*sqrt(Sin^2(w*t)+Cos^2(w*t)) = A, no matter what w*t is. Regards Ulrich "Mark" <makolber@yahoo.com> schrieb im Newsbeitrag news:1112912339.014082.273280@g14g2000cwa.googlegroups.com...> I have a question about envelope detection. I am an "analog" guy > trying to understand the advantage of using the Hilbert transformer > (HT) method for obtaining the envelope of a carrier signal. > > I have read that the standard DSP approach is to use a Hilbert > transformer (HT) to create a 90 deg shifted version of the carrier > then: > > envelope = sqrt (I^2+Q^2) > > Does the output of this contain ripple at 4x the carrier frequency? > > I know that "simply" taking the absolute value or squaring the carrier > results in an envelope signal that contains ripple at 2x (and above) > the carrier frequency. > > I know that "simple" half wave rectification contains ripple at 1X (and > above) the carrier frequency. > > So is it true that the HT method still requires low pass filtering to > remove the 4x carrier component but that the HT method is considered > better because the 4x carrier component is easier to filter (remove) > than a 2x or 1x component? > > thanks > Mark >

Reply by ●April 11, 20052005-04-11

Ulrich Thanks... yes I see now how the 90 deg shifted signals add to produce a flat envelope with no ripple. Wonderful !!!! I like it when there is a truly elegant solution to a problem. thanks Mark

Reply by ●April 13, 20052005-04-13

On 11 Apr 2005 06:48:44 -0700, "Mark" <makolber@yahoo.com> wrote:>Ulrich > >Thanks... yes I see now how the 90 deg shifted signals add to produce a >flat envelope with no ripple. > >Wonderful !!!! > >I like it when there is a truly elegant solution to a problem. > >thanks > >MarkHi Mark, you've received some terrific replies here. The darned problem is that to compute the Hilbert transform requires a fair amount of processing. The wider your signal bandwidth, the more computations you must perform. Once you have the I & Q samples, you square them, add the products, and then ya' have to perform a dog-gone square root. Now, ... there are several algorithms for estimating the magnitude of a complex sample, in a fairly efficient way, if you can tolerate a few percent error in your magnitude result. Ya' might take a look at the March 2005 "DSP Tips & Tricks" column of the IEEE Signal Processing magazine. It's a good article on this exact topic. I just ran across an AM demod algorithm that operates on real-only signals (no HT required), but the signal must be narrowband and centered exactly at 1/4th the sample rate, and it also requires a square root operation. Good Luck, [-Rick-]

Reply by ●April 18, 20052005-04-18

R.Lyons@_BOGUS_ieee.org (Rick Lyons) wrote in news:425d221a.45744156 @news.sf.sbcglobal.net:> On 11 Apr 2005 06:48:44 -0700, "Mark" <makolber@yahoo.com> wrote: > >>Ulrich >> >>Thanks... yes I see now how the 90 deg shifted signals add to produce a >>flat envelope with no ripple. >> >>Wonderful !!!! >> >>I like it when there is a truly elegant solution to a problem. >> >>thanks >> >>Mark > > Hi Mark, > > you've received some terrific replies here. > > The darned problem is that to compute the > Hilbert transform requires a fair amount of processing. > The wider your signal bandwidth, the more > computations you must perform. Once you have the > I & Q samples, you square them, add the > products, and then ya' have to perform a > dog-gone square root. > > Now, ... there are several algorithms for > estimating the magnitude of a complex sample, > in a fairly efficient way, if you can > tolerate a few percent error in your > magnitude result. Ya' might take a look > at the March 2005 "DSP Tips & Tricks" column > of the IEEE Signal Processing magazine. > It's a good article on this exact topic. > > I just ran across an AM demod algorithm > that operates on real-only signals > (no HT required), but the signal must > be narrowband and centered exactly at 1/4th > the sample rate, and it also requires > a square root operation. > > Good Luck, > [-Rick-] > > > > > >Something to consider.... A lot of times you can use an envelope detector to determine if a signal is present or above a certain level. In this case, you can skip the square root function. Furthermore, if your signal is far enough from the band edge (DC for example), the hilbert filter can be fairly short. This makes it fairly inexpensive to implement from a computation point of view. The catch to the hilbert filter is that its magnitude response will have ripple (in the passband). This means that even though the phase is separated by 90 degrees, the magnitude squared calculation (I^2 + R^2) is not exact. You can improve the magnitude response with a larger filter. No doubt, everything I just said is in Rick's excellent book. Buy the second edition! (I have a copy of both) -- Al Clark Danville Signal Processing, Inc. -------------------------------------------------------------------- Purveyors of Fine DSP Hardware and other Cool Stuff Available at http://www.danvillesignal.com

Reply by ●April 19, 20052005-04-19

On Mon, 18 Apr 2005 21:57:38 GMT, Al Clark <dsp@danvillesignal.com> wrote: (snipped)> >Something to consider.... > >A lot of times you can use an envelope detector to determine if a signal >is present or above a certain level. In this case, you can skip the >square root function. Furthermore, if your signal is far enough from the >band edge (DC for example), the hilbert filter can be fairly short. This >makes it fairly inexpensive to implement from a computation point of >view. > >The catch to the hilbert filter is that its magnitude response will have >ripple (in the passband). This means that even though the phase is >separated by 90 degrees, the magnitude squared calculation (I^2 + R^2) is >not exact. You can improve the magnitude response with a larger filter. > >No doubt, everything I just said is in Rick's excellent book. Buy the >second edition! (I have a copy of both) > > >-- >Al ClarkHi Al, dog gone it. I sure wish you'd write an article for the "DSP Tips & Tricks" column of the IEEE Signal Processing magazine. See Ya', [-Rick-]

Reply by ●April 19, 20052005-04-19

On Tue, 19 Apr 2005 05:04:07 +0000, Rick Lyons wrote:> On Mon, 18 Apr 2005 21:57:38 GMT, Al Clark <dsp@danvillesignal.com> > wrote: > > (snipped) >> >>Something to consider.... >> >>A lot of times you can use an envelope detector to determine if a signal >>is present or above a certain level. In this case, you can skip the >>square root function. Furthermore, if your signal is far enough from the >>band edge (DC for example), the hilbert filter can be fairly short. This >>makes it fairly inexpensive to implement from a computation point of >>view. >> >>The catch to the hilbert filter is that its magnitude response will have >>ripple (in the passband). This means that even though the phase is >>separated by 90 degrees, the magnitude squared calculation (I^2 + R^2) is >>not exact. You can improve the magnitude response with a larger filter. >> >>No doubt, everything I just said is in Rick's excellent book. Buy the >>second edition! (I have a copy of both) >> >> >>-- >>Al Clark > > Hi Al, > > dog gone it. I sure wish you'd write > an article for the "DSP Tips & Tricks" column of the > IEEE Signal Processing magazine.There's a "Tips and Tricks" article [letter] on this subject (sort of) in IEEE TAS around '94 (by yours truly et al) and a summary of same in the comp.dsp faq. Related to Al's point: You don't *have* to have a ripply pass-band. It's just a filter design problem: you can opt for a maximally flat response if you like, by using the appropriate design rules. The Hilbert filter (particularly one that doesn't attempt to stretch all the way to the frequency limits) is just a modulated low-pass filter, and you can design any low-pass filter that you like. Cheers, -- Andrew

Reply by ●April 19, 20052005-04-19

Andrew Reilly <andrew-newspost@areilly.bpc-users.org> wrote in news:pan.2005.04.19.05.38.47.404944@areilly.bpc-users.org:> On Tue, 19 Apr 2005 05:04:07 +0000, Rick Lyons wrote: > >> On Mon, 18 Apr 2005 21:57:38 GMT, Al Clark <dsp@danvillesignal.com> >> wrote: >> >> (snipped) >>> >>>Something to consider.... >>> >>>A lot of times you can use an envelope detector to determine if a >>>signal is present or above a certain level. In this case, you can >>>skip the square root function. Furthermore, if your signal is far >>>enough from the band edge (DC for example), the hilbert filter can be >>>fairly short. This makes it fairly inexpensive to implement from a >>>computation point of view. >>> >>>The catch to the hilbert filter is that its magnitude response will >>>have ripple (in the passband). This means that even though the phase >>>is separated by 90 degrees, the magnitude squared calculation (I^2 + >>>R^2) is not exact. You can improve the magnitude response with a >>>larger filter. >>> >>>No doubt, everything I just said is in Rick's excellent book. Buy the >>>second edition! (I have a copy of both) >>> >>> >>>-- >>>Al Clark >> >> Hi Al, >> >> dog gone it. I sure wish you'd write >> an article for the "DSP Tips & Tricks" column of the >> IEEE Signal Processing magazine. > > There's a "Tips and Tricks" article [letter] on this subject (sort of) > in IEEE TAS around '94 (by yours truly et al) and a summary of same in > the comp.dsp faq. > > Related to Al's point: > > You don't *have* to have a ripply pass-band. It's just a filter > design problem: you can opt for a maximally flat response if you like, > by using the appropriate design rules. The Hilbert filter > (particularly one that doesn't attempt to stretch all the way to the > frequency limits) is just a modulated low-pass filter, and you can > design any low-pass filter that you like. > > Cheers, >I agree that you can have very small ripple in the passband. The implied point that I was trying to make is that even though the phase is perfect, the magnitude is never quite perfect. Obviously, outside the band edges the magnitude calculation falls apart. The filter needs to be long if you want to extend to low frequencies. This is fairly obvious if you think about how much delay you would need to shift "DC" by 90 degrees. -- Al Clark Danville Signal Processing, Inc. -------------------------------------------------------------------- Purveyors of Fine DSP Hardware and other Cool Stuff Available at http://www.danvillesignal.com