I'm trying to do the following conversion:
Re{2*exp(j*2*pi*520*t) + 3*exp(j*2*pi*504*t)}
= e(t)*cos(f*t + phase(t))
where e(t) is an envelope function, phase(t) denotes time-varying phase.
please help.
Simple Beat Waveform Problem
Started by ●February 13, 2005
Reply by ●February 14, 20052005-02-14
I don't think this can be done easily: The source signal is a sum of two sine waves of different frequencies (520, 504) and different amplitudes. If the two components had the SAME AMPLITUDE, then they could be represented as multiplication of frequencies (f1+f2)/2 and (f1-f2)/2, i.e. 512 and 8. Then your envelope e(t)=sin(2*pi*8*t+phi). You could try experimenting numerically with phase(t)=Phase0*sin(2*pi*8*t+alpha). That is, try having phase oscilate with the same frequency as the envelope.
Reply by ●February 14, 20052005-02-14
Navraj Singh wrote:> I'm trying to do the following conversion: > > Re{2*exp(j*2*pi*520*t) + 3*exp(j*2*pi*504*t)} > > = e(t)*cos(f*t + phase(t)) > > where e(t) is an envelope function, phase(t) denotes time-varying phase. > > please help.Why do you think it's simple? Stripped of your textbook's gobbledygook, You want to express A*cos(at) + B*cos(bt) as a product. Forget the rest for now. Can you at least describe the envelope? What is its maximum value? Minimum? Can it ever be zero? Negative? What would a negative envelope mean? An envelope is usually the result of a non-linear process, but your process is linear. It there really an envelope here? Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������
Reply by ●February 14, 20052005-02-14
Jerry Avins wrote:> Navraj Singh wrote: > > >>I'm trying to do the following conversion: >> >>Re{2*exp(j*2*pi*520*t) + 3*exp(j*2*pi*504*t)} >> >>= e(t)*cos(f*t + phase(t)) >> >>where e(t) is an envelope function, phase(t) denotes time-varying phase. >> >>please help. > > > Why do you think it's simple? Stripped of your textbook's gobbledygook, > You want to express A*cos(at) + B*cos(bt) as a product. Forget the rest > for now. Can you at least describe the envelope? What is its maximum > value? Minimum? Can it ever be zero? Negative? What would a negative > envelope mean? > > An envelope is usually the result of a non-linear process, but your > process is linear. It there really an envelope here? > > JerryWe may be fighting definition problems. Changing value of frequencies to make things clearer/cleaner. Experimental setup: 1. tuning fork "a" resonates at 1000 Hz. 2. tuning fork "b" resonates at 1001 Hz. 3. detector "1" consists of a. microphone connected to b. oscilloscope with 1 horizontal scan per 5 seconds 4. detector "2" consists of a. human ear Wouldn't the both detectors identify an envelope varying at 1 Hz.? Would not both detectors be linear?
Reply by ●February 14, 20052005-02-14
By 'simple' I did not mean that it is easy (if it was, I would have solved it)...only that its a pretty standard problem and that it is clear what is wanted. The simplicity is in the face that a sum has to be written as a product (pretty common beat waveform or AM modulation problem). I'm pretty sure e(t) is a sinusoid - so the envelope is also sinusoidal. It will have a much lesser frequency than the carrier, which apparently has a time-varying phase. I think the following description fits the envelope: max value: A+B min value: -(A+B) (this is probably the min of the negative envelope) can the envelope be zero? probably not - I think there's a DC term in the envelope. Not sure what a negative envelope would mean. I know that the complete envelope consists of the positive and negative versions of e(t)...but not sure why you multiple the negative.> An envelope is usually the result of a non-linear process, but your > process is linear. It there really an envelope here?A sum of two sinusoids that don't differ very much in frequency (as in this problem) can be expressed as a product of a center frequency and deviation frequency (512 and 8 Hz in this case). So the linear prob is pretty much turned into a non-linear prob - and hence gives rise to the envelope. Now I have a question: Does a time varying change the frequency of a sinusoid? I'm not sure, but I think it does. The frequency is given by the derivative of a sinusoid's argument. If you have a signal: x(t) = cos(2*pi*f*t + p*t) where p*t is a linearly varying phase the instantaneous frequency is given by: (derivative of 2*pi*f*t + p*t)/(2*pi) - right?. if so, then frequency = (2*pi*f + p)/(2*pi) = f + p/(2*pi) If the phase was fixed, the frequency would just be f. If the above is true, maybe its possible to breakdown one of the frequencies in some way and integrate to find the phase function? ----- Original Message ----- From: "Jerry Avins" <jya@ieee.org> Newsgroups: comp.dsp Sent: Monday, February 14, 2005 2:51 PM Subject: Re: Simple Beat Waveform Problem> Navraj Singh wrote: > > > I'm trying to do the following conversion: > > > > Re{2*exp(j*2*pi*520*t) + 3*exp(j*2*pi*504*t)} > > > > = e(t)*cos(f*t + phase(t)) > > > > where e(t) is an envelope function, phase(t) denotes time-varying phase. > > > > please help. > > Why do you think it's simple? Stripped of your textbook's gobbledygook, > You want to express A*cos(at) + B*cos(bt) as a product. Forget the rest > for now. Can you at least describe the envelope? What is its maximum > value? Minimum? Can it ever be zero? Negative? What would a negative > envelope mean? > > An envelope is usually the result of a non-linear process, but your > process is linear. It there really an envelope here? > > Jerry > -- > Engineering is the art of making what you want from things you can get. > �����������������������������������������������������������������������
Reply by ●February 14, 20052005-02-14
Here's the question again: Does a time varying phase change the frequency of a sinusoid? forgot 'phase' in the quesiton i posed below. sorry for the confusion "Navraj Singh" <gtg256q@mail.gatech.edu> wrote in message news:curmpj$1f7$1@news-int.gatech.edu...> By 'simple' I did not mean that it is easy (if it was, I would have solved > it)...only that its a pretty standard problem and that it is clear what is > wanted. The simplicity is in the face that a sum has to be written as a > product (pretty common beat waveform or AM modulation problem). > > I'm pretty sure e(t) is a sinusoid - so the envelope is also sinusoidal.It> will have a much lesser frequency than the carrier, which apparently has a > time-varying phase. I think the following description fits the envelope: > > max value: A+B > min value: -(A+B) (this is probably the min of the negative envelope) > can the envelope be zero? probably not - I think there's a DC term in the > envelope. > Not sure what a negative envelope would mean. I know that the complete > envelope consists of the positive and negative > versions of e(t)...but not sure why you multiple the negative. > > > An envelope is usually the result of a non-linear process, but your > > process is linear. It there really an envelope here? > A sum of two sinusoids that don't differ very much in frequency (as inthis> problem) can be expressed as a product of a center frequency and deviation > frequency (512 and 8 Hz in this case). So the linear prob is pretty much > turned into a non-linear prob - and hence gives rise to the envelope. > > Now I have a question: > > Does a time varying change the frequency of a sinusoid? I'm not sure, butI> think it does. The frequency is given by the derivative of a sinusoid's > argument. If you have a signal: > > x(t) = cos(2*pi*f*t + p*t) where p*t is a linearly varying phase > > the instantaneous frequency is given by: (derivative of 2*pi*f*t + > p*t)/(2*pi) - right?. > > if so, then frequency = (2*pi*f + p)/(2*pi) = f + p/(2*pi) > > If the phase was fixed, the frequency would just be f. > > If the above is true, maybe its possible to breakdown one of thefrequencies> in some way and integrate to find the phase function? > > ----- Original Message ----- > From: "Jerry Avins" <jya@ieee.org> > Newsgroups: comp.dsp > Sent: Monday, February 14, 2005 2:51 PM > Subject: Re: Simple Beat Waveform Problem > > > > Navraj Singh wrote: > > > > > I'm trying to do the following conversion: > > > > > > Re{2*exp(j*2*pi*520*t) + 3*exp(j*2*pi*504*t)} > > > > > > = e(t)*cos(f*t + phase(t)) > > > > > > where e(t) is an envelope function, phase(t) denotes time-varyingphase.> > > > > > please help. > > > > Why do you think it's simple? Stripped of your textbook's gobbledygook, > > You want to express A*cos(at) + B*cos(bt) as a product. Forget the rest > > for now. Can you at least describe the envelope? What is its maximum > > value? Minimum? Can it ever be zero? Negative? What would a negative > > envelope mean? > > > > An envelope is usually the result of a non-linear process, but your > > process is linear. It there really an envelope here? > > > > Jerry > > -- > > Engineering is the art of making what you want from things you can get. > > ����������������������������������������������������������������������� > >
Reply by ●February 14, 20052005-02-14
Richard Owlett wrote:> Jerry Avins wrote: > >> Navraj Singh wrote: >> >> >>> I'm trying to do the following conversion: >>> >>> Re{2*exp(j*2*pi*520*t) + 3*exp(j*2*pi*504*t)} >>> >>> = e(t)*cos(f*t + phase(t)) >>> >>> where e(t) is an envelope function, phase(t) denotes time-varying phase. >>> >>> please help. >> >> >> >> Why do you think it's simple? Stripped of your textbook's gobbledygook, >> You want to express A*cos(at) + B*cos(bt) as a product. Forget the rest >> for now. Can you at least describe the envelope? What is its maximum >> value? Minimum? Can it ever be zero? Negative? What would a negative >> envelope mean? >> >> An envelope is usually the result of a non-linear process, but your >> process is linear. It there really an envelope here? >> >> Jerry > > > We may be fighting definition problems. > > Changing value of frequencies to make things clearer/cleaner. > > Experimental setup: > 1. tuning fork "a" resonates at 1000 Hz. > 2. tuning fork "b" resonates at 1001 Hz. > 3. detector "1" consists of > a. microphone connected to > b. oscilloscope with 1 horizontal scan per 5 seconds > 4. detector "2" consists of > a. human ear > > Wouldn't the both detectors identify an envelope varying at 1 Hz.? > Would not both detectors be linear?I'll talk to you privately. Either by email of behind the woodshed. :-) Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������
Reply by ●February 14, 20052005-02-14
Navraj Singh wrote:> By 'simple' I did not mean that it is easy (if it was, I would have solved > it)...only that its a pretty standard problem and that it is clear what is > wanted. The simplicity is in the face that a sum has to be written as a > product (pretty common beat waveform or AM modulation problem). > > I'm pretty sure e(t) is a sinusoid - so the envelope is also sinusoidal. It > will have a much lesser frequency than the carrier, which apparently has a > time-varying phase. I think the following description fits the envelope: > > max value: A+B > min value: -(A+B) (this is probably the min of the negative envelope) > can the envelope be zero? probably not - I think there's a DC term in the > envelope.An envelope normally consists of two curves, one joining the positive peaks of the waveform, the other joining the negative peaks. What do you mean by a DC term? Under what conditions might it disappear?> Not sure what a negative envelope would mean. I know that the complete > envelope consists of the positive and negative > versions of e(t)...but not sure why you multiple the negative.I wasn't clear. Near any instant, the waveform goes between a positive and negative value at approximately the higher frequency. At one time it varies between +/-(a+b) at another, it varies between +/-(a-b). When b increases to equal a. the envelope at times becomes zero. When b exceeds a, the positive and negative envelopes cross over; it can be said that "the" envelope is negative. It will be noticed by a careful observer that the phase of the high frequency is inverted in that interval.>>An envelope is usually the result of a non-linear process, but your >>process is linear. It there really an envelope here? > > A sum of two sinusoids that don't differ very much in frequency (as in this > problem) can be expressed as a product of a center frequency and deviation > frequency (512 and 8 Hz in this case). So the linear prob is pretty much > turned into a non-linear prob - and hence gives rise to the envelope.That's an arm-waving leap! What turns a linear system non-linear? Can any linear system do that? Whenever? The product of a center frequency and deviation frequency (512 and 8 Hz in this case) produces three terms, and that will be amplitude modulation. You have the sum, which has only two terms. You haven't grasped the nature of your problem.> Now I have a question: > > Does a time varying change the frequency of a sinusoid?Something is missing from that sentence that I can't fill in with confidence. Please try again.> I'm not sure, but I > think it does. The frequency is given by the derivative of a sinusoid's > argument. If you have a signal: > > x(t) = cos(2*pi*f*t + p*t) where p*t is a linearly varying phase > > the instantaneous frequency is given by: (derivative of 2*pi*f*t + > p*t)/(2*pi) - right?. > > if so, then frequency = (2*pi*f + p)/(2*pi) = f + p/(2*pi) > > If the phase was fixed, the frequency would just be f.And if the phase varies linearly -- with time, I must assume -- it merely offsets f to another fixed value. So what?> If the above is true, maybe its possible to breakdown one of the frequencies > in some way and integrate to find the phase function?I'm not sure enough of what you have in mind to hazard a guess. Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������
Reply by ●February 14, 20052005-02-14
Navraj Singh wrote:> Here's the question again: > > Does a time varying phase change the frequency of a sinusoid? > > forgot 'phase' in the quesiton i posed below. > > sorry for the confusionNo sweat. The simple answer is yes. If the variation is cyclic, that is phase (or frequency) modulation, in which the average frequency is not affected, but the instantaneous frequency is. Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������
Reply by ●February 15, 20052005-02-15
"Jerry Avins" <jya@ieee.org> wrote in message news:37d968F4m7m6gU1@individual.net...> Navraj Singh wrote: > >> By 'simple' I did not mean that it is easy (if it was, I would have >> solved >> it)...only that its a pretty standard problem and that it is clear what >> is >> wanted. The simplicity is in the face that a sum has to be written as a >> product (pretty common beat waveform or AM modulation problem). >> >> I'm pretty sure e(t) is a sinusoid - so the envelope is also sinusoidal. >> It >> will have a much lesser frequency than the carrier, which apparently has >> a >> time-varying phase. I think the following description fits the envelope: >> >> max value: A+B >> min value: -(A+B) (this is probably the min of the negative envelope) >> can the envelope be zero? probably not - I think there's a DC term in the >> envelope. > > An envelope normally consists of two curves, one joining the positive > peaks of the waveform, the other joining the negative peaks. What do you > mean by a DC term? Under what conditions might it disappear? > >> Not sure what a negative envelope would mean. I know that the complete >> envelope consists of the positive and negative >> versions of e(t)...but not sure why you multiple the negative. > > I wasn't clear. Near any instant, the waveform goes between a positive > and negative value at approximately the higher frequency. At one time it > varies between +/-(a+b) at another, it varies between +/-(a-b). When b > increases to equal a. the envelope at times becomes zero. When b exceeds > a, the positive and negative envelopes cross over; it can be said that > "the" envelope is negative. It will be noticed by a careful observer > that the phase of the high frequency is inverted in that interval. > >>>An envelope is usually the result of a non-linear process, but your >>>process is linear. It there really an envelope here? >> >> A sum of two sinusoids that don't differ very much in frequency (as in >> this >> problem) can be expressed as a product of a center frequency and >> deviation >> frequency (512 and 8 Hz in this case). So the linear prob is pretty much >> turned into a non-linear prob - and hence gives rise to the envelope. > > That's an arm-waving leap! What turns a linear system non-linear? Can > any linear system do that? Whenever? > > The product of a center frequency and deviation frequency (512 and 8 Hz > in this case) produces three terms, and that will be amplitude > modulation. You have the sum, which has only two terms. You haven't > grasped the nature of your problem. > >> Now I have a question: >> >> Does a time varying change the frequency of a sinusoid? > > Something is missing from that sentence that I can't fill in with > confidence. Please try again. > >> I'm not sure, but I >> think it does. The frequency is given by the derivative of a sinusoid's >> argument. If you have a signal: >> >> x(t) = cos(2*pi*f*t + p*t) where p*t is a linearly varying phase >> >> the instantaneous frequency is given by: (derivative of 2*pi*f*t + >> p*t)/(2*pi) - right?. >> >> if so, then frequency = (2*pi*f + p)/(2*pi) = f + p/(2*pi) >> >> If the phase was fixed, the frequency would just be f. > > And if the phase varies linearly -- with time, I must assume -- it > merely offsets f to another fixed value. So what? > >> If the above is true, maybe its possible to breakdown one of the >> frequencies >> in some way and integrate to find the phase function? > > I'm not sure enough of what you have in mind to hazard a guess. > > Jerry > -- > Engineering is the art of making what you want from things you can get. >Dear Navraj, cos x - cos y = -2 sin( (x - y)/2 ) sin( (x + y)/2 ) or you could find some other table of trig identities. Best of Luck - Mike
