On Jul 22, 7:00�am, "BCLIM" <boonchun_...@yahoo.com> wrote:> >On Jul 16, 9:45=A0pm, "BCLIM" <boonchun_...@yahoo.com> wrote: > >> >BCLIM wrote: > >> >> Hi All, > >> >> =A0 =A0 =A0 I have read out Richard Lyons's book about > approximation > >> envelope > >> >> detection. In the book got mentione after taking magniture of > analytic > >> >> signal, which is square root[square(I)+ square(Q)] and the result > need > >> to > >> >> be low pass filter to smoothen the signal. > > >> >> From the maths, the resultant of > >> >> the modulus of analytic signal will be only carrie(DC) plus message > >> >> signal(which is the envelope). > > >> >Then the math ignores the sampled nature of the signal > > >> >> In this case why we still need to low pass filter to smoothen out > the > >> signal. > > >> >To remove the images that result from the signal's sampled nature. > > >> >> I did a FFT plot of the signal before > >> >> low pass filtering, notice there are harmoincs of signal present. > Does > >> any > >> >> one know why the harmonic present? Thanks. > > >> >They are images. It is as if the baseband signal were modulated by > the > >> >sampling click. > > >> >Jerry > >> >-- > >> >Engineering is the art of making what you want from things you can > get. > > >> Hi Jerry, > >> Thanks for the feedback. > >> The below is the mathlab code I used to generate the modulated AM > signal. > >> Just added those sine wave. I did an FFT plot on this unprocess signal. > I > >> don't see any images occur. But after the modulus of the analytic > signal,= > > I > >> noticed the harmoinics occur. For example, 14kHz etc. > >> fs=3D312500; > >> T=3D41744; > >> n=3D(0:T-1); > >> f1=3D7000; > >> f2=3D7060; > >> f3=3D6940; > > xn_1=3D(1*sqrt(2))*sin(2*pi*f1*n/fs)+(0.1*sqrt(2))*sin(2*pi*f2*n/fs)+(0.1= > > > > >*sqr=ADt(2))*sin(2*pi*f3*n/fs); > > >> By maths, > >> =A0 =A0 =A0 =A0Analytic signal,Xc(t) =3D A(t)*e^(jwt) > >> A(t) is the envelope. > >> |xc(t)| =3D square root{(A(t)*cos(wt))^2+(A(t)*sin(wt))^2} > >> since cos^2(wt) + sin^2(wt) =3D 1 > >> =A0 =A0|xc(t)| =3D A(t), where A(t) =3D Ac + m*cos(wm*t) and > Ac=3Dcarrier > >> amplitude, > >> m=3Dmodulation index, wm=3Dw of message > >> Does the steps looks fine? If yes, then there shouldn't be any > hormonics > >> present. Weird.... > >> Regards, > >> BC- Hide quoted text - > > >> - Show quoted text - > > >BCLIM, > > >Please post the entire MATLAB code that is causing you the problem. > >You have the generation code above, I'd like to see it all. > > >Dirk Bell > >DSP Consultant > >Hi Dirk, > > � � � � The below is the whole code. > fs=3D312500; > T=41744; > n=(0:T-1); > f1=7000; > f2=7060; > f3=6940; > xn_1=3D(1*sqrt(2))*sin(2*pi*f1*n/fs)+(0.1*sqrt(2))*sin(2*pi*f2*n/fs)+(0.1 > sqrt(2))*sin(2*pi*f3*n/fs); > > fil=filter(HBT,1, xn_1); %Pass the Am signal thru hilbert transform filter > with 0dB gain in passband. > fil1=fil(129:41744); %remove group delay of filtered signal > xnnew=xn_1(129:41744); % syn the original signal by taking the same amount > of delay. > fil1sq=fil1.*fil1; %square Q > xn1sq=xnnew.*xnnew; �%square I > RSS=sqrt(fil1sq+xn1sq); %Square root the summing > %Then pass thru a LOW pass filter to remove the harmonics.- Hide quoted text - > > - Show quoted text -- Hide quoted text - > > - Show quoted text -BCLIM, Comment 1) I don't know what the 'D' is in fs and xn_1 (as in '3D...'), and some of the equations after your original post. It wasn't in your Jul 16 post, so I will assume someone's editor or other SW is acting up and the Jul 16 equations are correct. Tell us if the change actually means something. Comment 2) I think this is your problem. What you have as the AM signal is not the normal AM envelope that this method applies to. You have the information modulation signal xn_1, but you need to add a positive constant offset of greater value than the magnitude of the minimum xn_1 value to make x1_n correspond to an envelope modulation for this approach. So replace old xn_1 with (1+alpha*old xn_1) where alpha is such that (1+alpha*old xn_1)>0; maybe initially set alpha so 3/2> (1+alpha*old xn_1)>1/2. Then in your demodulated AM output you will have the DC offset of 1 that you can remove, and the old xn_1 scaled by alpha, which you can unscale for comparison to the old xn_1. Note, the added constant of '1' suggested could be some other positive number with corresponding changes to alpha, but it will do for now. Dirk Bell DSP Consultant
AM demodulation-envelope detection method
Started by ●July 16, 2009
Reply by ●July 22, 20092009-07-22
Reply by ●July 22, 20092009-07-22
On Jul 22, 10:30�am, Dirk Bell <bellda2...@cox.net> wrote:> On Jul 22, 7:00�am, "BCLIM" <boonchun_...@yahoo.com> wrote: > > > > > > > >On Jul 16, 9:45=A0pm, "BCLIM" <boonchun_...@yahoo.com> wrote: > > >> >BCLIM wrote: > > >> >> Hi All, > > >> >> =A0 =A0 =A0 I have read out Richard Lyons's book about > > approximation > > >> envelope > > >> >> detection. In the book got mentione after taking magniture of > > analytic > > >> >> signal, which is square root[square(I)+ square(Q)] and the result > > need > > >> to > > >> >> be low pass filter to smoothen the signal. > > > >> >> From the maths, the resultant of > > >> >> the modulus of analytic signal will be only carrie(DC) plus message > > >> >> signal(which is the envelope). > > > >> >Then the math ignores the sampled nature of the signal > > > >> >> In this case why we still need to low pass filter to smoothen out > > the > > >> signal. > > > >> >To remove the images that result from the signal's sampled nature. > > > >> >> I did a FFT plot of the signal before > > >> >> low pass filtering, notice there are harmoincs of signal present. > > Does > > >> any > > >> >> one know why the harmonic present? Thanks. > > > >> >They are images. It is as if the baseband signal were modulated by > > the > > >> >sampling click. > > > >> >Jerry > > >> >-- > > >> >Engineering is the art of making what you want from things you can > > get. > > > >> Hi Jerry, > > >> Thanks for the feedback. > > >> The below is the mathlab code I used to generate the modulated AM > > signal. > > >> Just added those sine wave. I did an FFT plot on this unprocess signal. > > I > > >> don't see any images occur. But after the modulus of the analytic > > signal,= > > > I > > >> noticed the harmoinics occur. For example, 14kHz etc. > > >> fs=3D312500; > > >> T=3D41744; > > >> n=3D(0:T-1); > > >> f1=3D7000; > > >> f2=3D7060; > > >> f3=3D6940; > > > xn_1=3D(1*sqrt(2))*sin(2*pi*f1*n/fs)+(0.1*sqrt(2))*sin(2*pi*f2*n/fs)+(0.1= > > > >*sqr=ADt(2))*sin(2*pi*f3*n/fs); > > > >> By maths, > > >> =A0 =A0 =A0 =A0Analytic signal,Xc(t) =3D A(t)*e^(jwt) > > >> A(t) is the envelope. > > >> |xc(t)| =3D square root{(A(t)*cos(wt))^2+(A(t)*sin(wt))^2} > > >> since cos^2(wt) + sin^2(wt) =3D 1 > > >> =A0 =A0|xc(t)| =3D A(t), where A(t) =3D Ac + m*cos(wm*t) and > > Ac=3Dcarrier > > >> amplitude, > > >> m=3Dmodulation index, wm=3Dw of message > > >> Does the steps looks fine? If yes, then there shouldn't be any > > hormonics > > >> present. Weird.... > > >> Regards, > > >> BC- Hide quoted text - > > > >> - Show quoted text - > > > >BCLIM, > > > >Please post the entire MATLAB code that is causing you the problem. > > >You have the generation code above, I'd like to see it all. > > > >Dirk Bell > > >DSP Consultant > > >Hi Dirk, > > > � � � � The below is the whole code. > > fs=3D312500; > > T=41744; > > n=(0:T-1); > > f1=7000; > > f2=7060; > > f3=6940; > > xn_1=3D(1*sqrt(2))*sin(2*pi*f1*n/fs)+(0.1*sqrt(2))*sin(2*pi*f2*n/fs)+(0.1 > > sqrt(2))*sin(2*pi*f3*n/fs); > > > fil=filter(HBT,1, xn_1); %Pass the Am signal thru hilbert transform filter > > with 0dB gain in passband. > > fil1=fil(129:41744); %remove group delay of filtered signal > > xnnew=xn_1(129:41744); % syn the original signal by taking the same amount > > of delay. > > fil1sq=fil1.*fil1; %square Q > > xn1sq=xnnew.*xnnew; �%square I > > RSS=sqrt(fil1sq+xn1sq); %Square root the summing > > %Then pass thru a LOW pass filter to remove the harmonics.- Hide quoted text - > > > - Show quoted text -- Hide quoted text - > > > - Show quoted text - > > BCLIM, > > Comment 1) > > I don't know what the 'D' is in fs and xn_1 (as in '3D...'), and some > of the equations after your original post. �It wasn't in your Jul 16 > post, so I will assume someone's editor or other SW is acting up and > the Jul 16 equations are correct. Tell us if the change actually means > something. > > Comment 2) > > I think this is your problem. > > What you have as the AM signal is not the normal AM envelope that this > method applies to. �You have the information modulation signal xn_1, > but you need to add a positive constant offset of greater value than > the magnitude of the minimum xn_1 value to make x1_n correspond to an > envelope modulation for this approach. So replace old xn_1 with > (1+alpha*old xn_1) where alpha is such that (1+alpha*old xn_1)>0; > maybe initially set alpha so 3/2> (1+alpha*old xn_1)>1/2. Then in your > demodulated AM output you will have the DC offset of 1 that you can > remove, and the old xn_1 scaled by alpha, which you can unscale for > comparison to the old xn_1. Note, the added constant of '1' suggested > could be some other positive number with corresponding changes to > alpha, but it will do for now. > > Dirk Bell > DSP Consultant- Hide quoted text - > > - Show quoted text -BTW, I assume the signal you are using falls within the passband of your Hilbert filter? Dirk
Reply by ●August 1, 20092009-08-01
On Thu, 16 Jul 2009 07:42:04 -0500, "BCLIM" <boonchun_lim@yahoo.com> wrote:>Hi All, > I have read out Richard Lyons's book about approximation envelope >detection. In the book got mentione after taking magniture of analytic >signal, which is square root[square(I)+ square(Q)] and the result need to >be low pass filter to smoothen the signal. From the maths, the resultant of >the modulus of analytic signal will be only carrie(DC) plus message >signal(which is the envelope). In this case why we still need to low pass >filter to smoothen out the signal. I did a FFT plot of the signal before >low pass filtering, notice there are harmoincs of signal present. Does any >one know why the harmonic present? Thanks.Hello BCLIM, Clay Turner's good reply answered your question. If you'd like to receive an errata for my book, please send me an E-mail. My E-mail address is in the back of the book. Regards, [-Rick-]
