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.
AM demodulation-envelope detection method
Started by ●July 16, 2009
Reply by ●July 16, 20092009-07-16
BCLIM 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).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. �����������������������������������������������������������������������
Reply by ●July 16, 20092009-07-16
Jerry Avins wrote:> They are images. It is as if the baseband signal were modulated by the > sampling click.sampling *clock*. Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������
Reply by ●July 16, 20092009-07-16
On Jul 16, 8:42�am, "BCLIM" <boonchun_...@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.Aliasing? When you square the I and Q signals their bandwidths double. Do you have enough headroom between your signal content and the sample rate to avoid aliasing? Clay
Reply by ●July 16, 20092009-07-16
>BCLIM wrote: >> Hi All, >> I have read out Richard Lyons's book about approximationenvelope>> detection. In the book got mentione after taking magniture of analytic >> signal, which is square root[square(I)+ square(Q)] and the result needto>> 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 thesignal.> >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. Doesany>> 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=312500; T=41744; n=(0:T-1); f1=7000; f2=7060; f3=6940; xn_1=(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); By maths, Analytic signal,Xc(t) = A(t)*e^(jwt) A(t) is the envelope. |xc(t)| = square root{(A(t)*cos(wt))^2+(A(t)*sin(wt))^2} since cos^2(wt) + sin^2(wt) = 1 |xc(t)| = A(t), where A(t) = Ac + m*cos(wm*t) and Ac=carrier amplitude, m=modulation index, wm=w of message Does the steps looks fine? If yes, then there shouldn't be any hormonics present. Weird.... Regards, BC
Reply by ●July 19, 20092009-07-19
>On Jul 16, 8:42=A0am, "BCLIM" <boonchun_...@yahoo.com> wrote: >> Hi All, >> =A0 =A0 =A0 I have read out Richard Lyons's book about approximationenve=>lope >> detection. In the book got mentione after taking magniture of analytic >> signal, which is square root[square(I)+ square(Q)] and the result needto>> be low pass filter to smoothen the signal. From the maths, theresultant =>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 lowpass>> filter to smoothen out the signal. I did a FFT plot of the signalbefore>> low pass filtering, notice there are harmoincs of signal present. Doesan=>y >> one know why the harmonic present? Thanks. > >Aliasing? When you square the I and Q signals their bandwidths double. >Do you have enough headroom between your signal content and the sample >rate to avoid aliasing? > >Clay > >Hi Clay,The I and Q signal are digitized signal, the square of these signals are done in the software. Therefore there shouldn't be any aliasing problem. But definetely sqaure of each the signal will have second harmonis, in this case I thought the mathematical shown at the end i will only have the DC term(carrier) plus the message signal so I have no idea why the harmonics still present. Thanks. Regards, BC
Reply by ●July 20, 20092009-07-20
On Jul 19, 10:06�pm, "BCLIM" <boonchun_...@yahoo.com> wrote:> >On Jul 16, 8:42=A0am, "BCLIM" <boonchun_...@yahoo.com> wrote: > >> Hi All, > >> =A0 =A0 =A0 I have read out Richard Lyons's book about approximation > enve= > >lope > >> 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 > an= > >y > >> one know why the harmonic present? Thanks. > > >Aliasing? When you square the I and Q signals their bandwidths double. > >Do you have enough headroom between your signal content and the sample > >rate to avoid aliasing? > > >Clay > > >Hi Clay, > > � � � � The I and Q signal are digitized signal, the square of these > signals are done in the software. Therefore there shouldn't be any aliasing > problem. But definetely sqaure of each the signal will have second > harmonis, in this case I thought the mathematical shown at the end i will > only have the DC term(carrier) plus the message signal so I have no idea > why the harmonics still present. Thanks. > > Regards, > BC- Hide quoted text - > > - Show quoted text -Hello BC, I think you missed the point. For example let's say you have a 1000 Hz sinewave sampled at 2500 Hz. Here there is no aliasing. Now square the 1000 Hz sinewave. A simple trig theorem tells us that we now have a mix of 0Hz and 2000 Hz. Thus the 2000 Hz component sampled at 2500Hz becomes a 500 Hz component. The aliasing has folded the signal's spectrum back on itself. It doesn't matter that this is done in software after the original sampling process. Now back to the main question. Is your sampling rate more than four times the maximum frequency? (before any squaring) Clay
Reply by ●July 20, 20092009-07-20
On Jul 20, 10:44�am, Clay <c...@claysturner.com> wrote:> On Jul 19, 10:06�pm, "BCLIM" <boonchun_...@yahoo.com> wrote: > > > > > > > >On Jul 16, 8:42=A0am, "BCLIM" <boonchun_...@yahoo.com> wrote: > > >> Hi All, > > >> =A0 =A0 =A0 I have read out Richard Lyons's book about approximation > > enve= > > >lope > > >> 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 > > an= > > >y > > >> one know why the harmonic present? Thanks. > > > >Aliasing? When you square the I and Q signals their bandwidths double. > > >Do you have enough headroom between your signal content and the sample > > >rate to avoid aliasing? > > > >Clay > > > >Hi Clay, > > > � � � � The I and Q signal are digitized signal, the square of these > > signals are done in the software. Therefore there shouldn't be any aliasing > > problem. But definetely sqaure of each the signal will have second > > harmonis, in this case I thought the mathematical shown at the end i will > > only have the DC term(carrier) plus the message signal so I have no idea > > why the harmonics still present. Thanks. > > > Regards, > > BC- Hide quoted text - > > > - Show quoted text - > > Hello BC, > > I think you missed the point. For example let's say you have a 1000 Hz > sinewave sampled at 2500 Hz. Here there is no aliasing. Now square the > 1000 Hz sinewave. A simple trig theorem tells us that we now have a > mix of 0Hz and 2000 Hz. Thus the 2000 Hz component sampled at 2500Hz > becomes a 500 Hz component. The aliasing has folded the signal's > spectrum back on itself. It doesn't matter that this is done in > software after the original sampling process. > > Now back to the main question. Is your sampling rate more than four > times the maximum frequency? (before any squaring) > > Clay- Hide quoted text - > > - Show quoted text -And the impact on the AM and FM signals obtained by from the IQ method will be? Dirk Bell DSP Consultant
Reply by ●July 20, 20092009-07-20
On Jul 20, 10:44�am, Clay <c...@claysturner.com> wrote:> On Jul 19, 10:06�pm, "BCLIM" <boonchun_...@yahoo.com> wrote: > > > > > > > >On Jul 16, 8:42=A0am, "BCLIM" <boonchun_...@yahoo.com> wrote: > > >> Hi All, > > >> =A0 =A0 =A0 I have read out Richard Lyons's book about approximation > > enve= > > >lope > > >> 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 > > an= > > >y > > >> one know why the harmonic present? Thanks. > > > >Aliasing? When you square the I and Q signals their bandwidths double. > > >Do you have enough headroom between your signal content and the sample > > >rate to avoid aliasing? > > > >Clay > > > >Hi Clay, > > > � � � � The I and Q signal are digitized signal, the square of these > > signals are done in the software. Therefore there shouldn't be any aliasing > > problem. But definetely sqaure of each the signal will have second > > harmonis, in this case I thought the mathematical shown at the end i will > > only have the DC term(carrier) plus the message signal so I have no idea > > why the harmonics still present. Thanks. > > > Regards, > > BC- Hide quoted text - > > > - Show quoted text - > > Hello BC, > > I think you missed the point. For example let's say you have a 1000 Hz > sinewave sampled at 2500 Hz. Here there is no aliasing. Now square the > 1000 Hz sinewave. A simple trig theorem tells us that we now have a > mix of 0Hz and 2000 Hz. Thus the 2000 Hz component sampled at 2500Hz > becomes a 500 Hz component. The aliasing has folded the signal's > spectrum back on itself. It doesn't matter that this is done in > software after the original sampling process. > > Now back to the main question. Is your sampling rate more than four > times the maximum frequency? (before any squaring) > > Clay- Hide quoted text - > > - Show quoted text -And the impact on the signals obtained from the IQ demod method will be? Dirk Bell DSP Consultant
Reply by ●July 20, 20092009-07-20
Clay wrote:> On Jul 19, 10:06 pm, "BCLIM" <boonchun_...@yahoo.com> wrote: >>> On Jul 16, 8:42=A0am, "BCLIM" <boonchun_...@yahoo.com> wrote: >>>> Hi All, >>>> =A0 =A0 =A0 I have read out Richard Lyons's book about approximation >> enve= >>> lope >>>> 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 >> an= >>> y >>>> one know why the harmonic present? Thanks. >>> Aliasing? When you square the I and Q signals their bandwidths double. >>> Do you have enough headroom between your signal content and the sample >>> rate to avoid aliasing? >>> Clay >>> Hi Clay, >> The I and Q signal are digitized signal, the square of these >> signals are done in the software. Therefore there shouldn't be any aliasing >> problem. But definetely sqaure of each the signal will have second >> harmonis, in this case I thought the mathematical shown at the end i will >> only have the DC term(carrier) plus the message signal so I have no idea >> why the harmonics still present. Thanks. >> >> Regards, >> BC- Hide quoted text - >> >> - Show quoted text - > > Hello BC, > > I think you missed the point. For example let's say you have a 1000 Hz > sinewave sampled at 2500 Hz. Here there is no aliasing. Now square the > 1000 Hz sinewave. A simple trig theorem tells us that we now have a > mix of 0Hz and 2000 Hz. Thus the 2000 Hz component sampled at 2500Hz > becomes a 500 Hz component. The aliasing has folded the signal's > spectrum back on itself. It doesn't matter that this is done in > software after the original sampling process. > > Now back to the main question. Is your sampling rate more than four > times the maximum frequency? (before any squaring)Clay, I'm confused. We aren't squaring the signal, we're squaring individual samples as part of calculating sqrt(i[n]^2 +q[n]^2), or instantaneous magnitude. How do aliases arise? Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������
