If I take the Fourier Transform of a cosine wave at freq w0 I get two impulses in the Freq domain. If I then take an inverse FT I get back to the cosine wave. What happens if I remove one of the impulses in the Freq domain ie say the neg freq one and then inverse transform? We should get a complex term in the time-domain exp(-jw0t). What exactly does this look like? In the S domain this is a pure time delay but in this case it is a signal not a system and we are in the time-domain of course. It is obviously complex frequency but what does it look like? What happens if I repeat the above but remove the positive impulse in the freq domain before inverse transforming? I would get exp(+jwot). Thanks Rimmer
Simple Fourier TF Question
Started by ●June 11, 2005
Reply by ●June 11, 20052005-06-11
Rimmer wrote:> If I take the Fourier Transform of a cosine wave at freq w0 I get two > impulses in the Freq domain. > If I then take an inverse FT I get back to the cosine wave. > > What happens if I remove one of the impulses in the Freq domain ie say the > neg freq one and then inverse transform? > > We should get a complex term in the time-domain exp(-jw0t). What exactly > does this look like? In the S domain this is a pure time delay but in this > case it is a signal not a system and we are in the time-domain of course. > > It is obviously complex frequency but what does it look like? What happens > if I repeat the above but remove the positive impulse in the freq domain > before inverse transforming? I would get exp(+jwot). > > Thanks > > RimmerTransformations only change the viewing platform. The two "impulses" in frequency are(e^jwt + e^-jwt)/2 = cosh(jwt). Can you take it from there? I'm in the middle of cooking supper. Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������
Reply by ●June 12, 20052005-06-12
"Jerry Avins" <jya@ieee.org> wrote in message news:JuudnRg8F946yDbfRVn-2A@rcn.net...> Rimmer wrote: > > If I take the Fourier Transform of a cosine wave at freq w0 I get two > > impulses in the Freq domain. > > If I then take an inverse FT I get back to the cosine wave. > > > > What happens if I remove one of the impulses in the Freq domain ie saythe> > neg freq one and then inverse transform? > > > > We should get a complex term in the time-domain exp(-jw0t). What exactly > > does this look like? In the S domain this is a pure time delay but inthis> > case it is a signal not a system and we are in the time-domain ofcourse.> > > > It is obviously complex frequency but what does it look like? Whathappens> > if I repeat the above but remove the positive impulse in the freq domain > > before inverse transforming? I would get exp(+jwot). > > > > Thanks > > > > Rimmer > > Transformations only change the viewing platform. The two "impulses" in > frequency are(e^jwt + e^-jwt)/2 = cosh(jwt). Can you take it from there? > I'm in the middle of cooking supper.Still not with you.... Rimmer
Reply by ●June 12, 20052005-06-12
Rimmer wrote:> If I take the Fourier Transform of a cosine wave at freq w0 I get two > impulses in the Freq domain. > If I then take an inverse FT I get back to the cosine wave. > > What happens if I remove one of the impulses in the Freq domain ie say the > neg freq one and then inverse transform? > > We should get a complex term in the time-domain exp(-jw0t). What exactly > does this look like? In the S domain this is a pure time delay but in this > case it is a signal not a system and we are in the time-domain of course. > > It is obviously complex frequency but what does it look like? What happens > if I repeat the above but remove the positive impulse in the freq domain > before inverse transforming? I would get exp(+jwot).Rimmer, you already answered your own question: you get a complex phasor. It looks like the sum of a real and an imaginary harmonic wave. That's probably not what you want to hear, but I can't extract any other question from your post to answer. Regards, Andor
Reply by ●June 13, 20052005-06-13
"Rimmer" <rimmerx@reddwarf2.com> wrote in message news:RMHqe.6659$U4.961559@news.xtra.co.nz...> If I take the Fourier Transform of a cosine wave at freq w0 I get two > impulses in the Freq domain. > If I then take an inverse FT I get back to the cosine wave. > > What happens if I remove one of the impulses in the Freq domain ie say the > neg freq one and then inverse transform? > > We should get a complex term in the time-domain exp(-jw0t). What exactly > does this look like? In the S domain this is a pure time delay but in this > case it is a signal not a system and we are in the time-domain of course. > > It is obviously complex frequency but what does it look like? What happens > if I repeat the above but remove the positive impulse in the freq domain > before inverse transforming? I would get exp(+jwot). > > Thanks > > RimmerI like to get back to basics first.... The original is even and real in frequency. It's inverse transform is even and real in time. When you remove one part in frequency you destroy the evenness in frequency. Accordingly, the inverse transform isn't any longer real and there are all sorts of details, etc. that you might ponder or examine. What does it look like? Well a cosine is the sum of two exponentials / rotating vectors in a complex plane. They rotate in opposite directions around the orgin. The sum that forms the cosine happens to cancel the imaginary parts so that the result is real. Similarly with a sinusoid, the sum that forms the sine happens to cancel the imaginary parts so that the result is real but is pi/2 changed in time relative to the cosine. Fred
Reply by ●June 13, 20052005-06-13
"Rimmer" <rimmerx@reddwarf2.com> ha scritto nel messaggio news:RMHqe.6659$U4.961559@news.xtra.co.nz...> If I take the Fourier Transform of a cosine wave at freq w0 I get two > impulses in the Freq domain. > If I then take an inverse FT I get back to the cosine wave. > > What happens if I remove one of the impulses in the Freq domain ie say the > neg freq one and then inverse transform? > > We should get a complex term in the time-domain exp(-jw0t). What exactly > does this look like? In the S domain this is a pure time delay but in this > case it is a signal not a system and we are in the time-domain of course. > > It is obviously complex frequency but what does it look like? What happens > if I repeat the above but remove the positive impulse in the freq domain > before inverse transforming? I would get exp(+jwot). > > Thanks > > Rimmer > >In the real world all signals are real. Imaginary signals, and even negative frequencies are all math tricks for doing calculations in an easier and often more elegant way. So, if you want to visualize (e.g. in an oscilloscope) a complex signal you need to represent two real signals or functions. In our case (cos(w0t), sin(w0t)) and you get a Lissajous figure, here a circumference. Angelo Note:Antitransforming the spectrum of cos(w0t), if you suppress the neg freq you get 1/2*exp(jw0t), the positive freq 1/2*exp(-jw0t).
Reply by ●June 13, 20052005-06-13
"Angelo Ricotta" <a.ricotta@isac.cnr.it> wrote in message news:k3are.147207$IN.2544574@twister2.libero.it...> "Rimmer" <rimmerx@reddwarf2.com> ha scritto nel messaggio > news:RMHqe.6659$U4.961559@news.xtra.co.nz... > > If I take the Fourier Transform of a cosine wave at freq w0 I get two > > impulses in the Freq domain. > > If I then take an inverse FT I get back to the cosine wave. > > > > What happens if I remove one of the impulses in the Freq domain ie saythe> > neg freq one and then inverse transform? > > > > We should get a complex term in the time-domain exp(-jw0t). What exactly > > does this look like? In the S domain this is a pure time delay but inthis> > case it is a signal not a system and we are in the time-domain ofcourse.> > > > It is obviously complex frequency but what does it look like? Whathappens> > if I repeat the above but remove the positive impulse in the freq domain > > before inverse transforming? I would get exp(+jwot). > > > > Thanks > > > > Rimmer > > > > > In the real world all signals are real. Imaginary signals, and evennegative> frequencies are all math tricks for doing calculations in an easier and > often more elegant way. So, if you want to visualize (e.g. in an > oscilloscope) a complex signal you need to represent two real signals or > functions. In our case (cos(w0t), sin(w0t)) and you get a Lissajousfigure,> here a circumference. > Angelo > Note:Antitransforming the spectrum of cos(w0t), if you suppress the negfreq> you get 1/2*exp(jw0t), the positive freq 1/2*exp(-jw0t). > >I am still no wiser I am afraid. I can see if I had teh sum of a cosine and sine eg acos(x)+bsin(x) = ccos(x-phi) but I have the j in between and I have been told that all signals are real (previous poster). So nobody has been able to tell me what I would actually see on the scope.Would it just be a phase-shifted sine-wave? Rimmer
Reply by ●June 13, 20052005-06-13
Rimmer wrote:> "Angelo Ricotta" <a.ricotta@isac.cnr.it> wrote in message > news:k3are.147207$IN.2544574@twister2.libero.it... > >>"Rimmer" <rimmerx@reddwarf2.com> ha scritto nel messaggio >>news:RMHqe.6659$U4.961559@news.xtra.co.nz... >> >>>If I take the Fourier Transform of a cosine wave at freq w0 I get two >>>impulses in the Freq domain. >>>If I then take an inverse FT I get back to the cosine wave. >>> >>>What happens if I remove one of the impulses in the Freq domain ie say > > the > >>>neg freq one and then inverse transform? >>> >>>We should get a complex term in the time-domain exp(-jw0t). What exactly >>>does this look like? In the S domain this is a pure time delay but in > > this > >>>case it is a signal not a system and we are in the time-domain of > > course. > >>>It is obviously complex frequency but what does it look like? What > > happens > >>>if I repeat the above but remove the positive impulse in the freq domain >>>before inverse transforming? I would get exp(+jwot). >>> >>>Thanks >>> >>>Rimmer >>> >>> >> >>In the real world all signals are real. Imaginary signals, and even > > negative > >>frequencies are all math tricks for doing calculations in an easier and >>often more elegant way. So, if you want to visualize (e.g. in an >>oscilloscope) a complex signal you need to represent two real signals or >>functions. In our case (cos(w0t), sin(w0t)) and you get a Lissajous > > figure, > >>here a circumference. >>Angelo >>Note:Antitransforming the spectrum of cos(w0t), if you suppress the neg > > freq > >>you get 1/2*exp(jw0t), the positive freq 1/2*exp(-jw0t). >> >> > > I am still no wiser I am afraid. I can see if I had teh sum of a cosine and > sine eg > > acos(x)+bsin(x) = ccos(x-phi) > > but I have the j in between and I have been told that all signals are real > (previous poster). So nobody has been able to tell me what I would actually > see on the scope.Would it just be a phase-shifted sine-wave? > > Rimmer >Rimmer, The "j" indicates that this 'imaginary' signal-component must be carried by a separate wire from the other, 'real' signal-component. The two wires together are needed to carry one complex signal. You could connect the two wires to a dual beam oscilloscope, and you would then see a cos wave and a sin wave, or you could connect the imaginary ("j")wire to the X input of the 'scope and leave the real wire connected to a Y input and you would see a circular Lissajou pattern, as Angelo says. Why do you suddenly need two wires? (After all, all you did originally was to eliminate the negative frequency.) Both the negative-frequency part of your original signal and the positive-frequency part have real and imaginary components, but in this case the imaginary components are equal and opposite, so cancel out. The resulting 'real' components are present together on one wire. Any 'imaginary' wire, if provided, has zero signal. If you disturb this arrangement by eliminating the positive or the negative part of the original signal, you are left with a complex signal, which needs two wires. Regards, John .
Reply by ●June 13, 20052005-06-13
"Rimmer" <rimmerx@reddwarf2.com> wrote in message news:S7are.6929$U4.991166@news.xtra.co.nz...> > "Angelo Ricotta" <a.ricotta@isac.cnr.it> wrote in message > news:k3are.147207$IN.2544574@twister2.libero.it... >> "Rimmer" <rimmerx@reddwarf2.com> ha scritto nel messaggio >> news:RMHqe.6659$U4.961559@news.xtra.co.nz... >> > If I take the Fourier Transform of a cosine wave at freq w0 I get two >> > impulses in the Freq domain. >> > If I then take an inverse FT I get back to the cosine wave. >> > >> > What happens if I remove one of the impulses in the Freq domain ie say > the >> > neg freq one and then inverse transform? >> > >> > We should get a complex term in the time-domain exp(-jw0t). What >> > exactly >> > does this look like? In the S domain this is a pure time delay but in > this >> > case it is a signal not a system and we are in the time-domain of > course. >> > >> > It is obviously complex frequency but what does it look like? What > happens >> > if I repeat the above but remove the positive impulse in the freq >> > domain >> > before inverse transforming? I would get exp(+jwot). >> > >> > Thanks >> > >> > Rimmer >> > >> > >> In the real world all signals are real. Imaginary signals, and even > negative >> frequencies are all math tricks for doing calculations in an easier and >> often more elegant way. So, if you want to visualize (e.g. in an >> oscilloscope) a complex signal you need to represent two real signals or >> functions. In our case (cos(w0t), sin(w0t)) and you get a Lissajous > figure, >> here a circumference. >> Angelo >> Note:Antitransforming the spectrum of cos(w0t), if you suppress the neg > freq >> you get 1/2*exp(jw0t), the positive freq 1/2*exp(-jw0t). >> >> > I am still no wiser I am afraid. I can see if I had teh sum of a cosine > and > sine eg > > acos(x)+bsin(x) = ccos(x-phi) > > but I have the j in between and I have been told that all signals are real > (previous poster). So nobody has been able to tell me what I would > actually > see on the scope.Would it just be a phase-shifted sine-wave? > > RimmerI think I already gave you the necessary information. I'll repeat it more pedantically in the hope that it will make sense: Perhaps if we get past a notation, graphical representation first: There are two ways to depict signals: One is a graph of amplitude as a function of time as you would see on a common oscilloscope trace. Another is a a 2-D graph in a complex plane where we look at rotating vectors to represent sinusoids. This is very commonly used in analysing AC power because of various phases that are in use. I note that it's mostly used for single-frequency situations. There are a couple of things to say about this: - The most common form of this is on a sheet of paper and the vectors are shown relative to one another in phase. "Time" is only imagined if we imagine that the vectors are rotating. However, one can make a 3-D plot with the third dimension being time. Then, the vectors still rotate in Re and jIm but they also progress in a corkscrew fashion in time. Showing the time axis doesn't ususally help very much so we avoid the complication of a 3-D plot. Also, note that using Re and jIm is just a convention for showing 2 dimensions. We could just as easily use x and y as long as scalars have x and y components to them and we use the proper rules for combining them - as we have rules for Re and jIm. So much for the graphical views. Now on to the mathematical representations: A rotating vector is of the form e^j(w)t or e^j(-w)t where I show w and -w explicitly to represent vector rotation clockwise or counterclockwise. A useful identity is: e^ja=cos(a) + jsin(a) So a rotating vector of the form e^jwt has a real part and an imaginary part. How do we deal with the imaginary part if we are interested in only real signals? Let's use another useful identity: cos(wt)=[e^jwt + e^-jwt]/2 Here we have two rotating vectors that coincide when t=0 and coincide when wt=2*pi. They oppose when wt=k*pi where k is an odd integer. The imaginary parts always oppose and sum to zero; so there is zero imaginary part always. It is handy visually to think of this expression as being represented by a *single* rotating vector and to *assume* that we are only interested in its projection on the real axis in the 2_D complex plane. The projection of the rotating vector on the real axis - *if plotted as a function of time* - is a cosine. In the 3-D plots with time as an added axis, that would be the projection of the rotating vector on the x-t plane. Similarly, the sine is represented by: sin(wt)=[e^jwt - e^-jwt]/j2 = 0.5*j*e^-jwt - 0.5*j*e^jwt One can envision these rotating vectors that are rotated by pi/2. The counterclockwise vector starts at t=0 at -j0.5 and the clockwise vector starts at j0.5. They sum to zero at t=0 as we would expect from a sine. At wt=pi/2 they coincide on the x axis .... and so forth. Once more, the imaginary parts of the vector sum is zero but now the peaks occur shifted pi/2 relative to the cosine. It is *convenient to envision* the sine as a rotating vector starting at pi/2 and rotating clockwise and only considering its projection on the real or x axis. In the 3-D plots with time as an added axis, that would be the projection of the rotating vector on the x-t plane. If you remove one of the unit sample terms in the FFT of a cosine then you are left with only one of the two rotating vectors mentioned above. In that case the imaginary part does not cancel out (as in the two cases above) and all you have is a single rotating vector with complex values. Note, a single rotating vector of this form is not a real signal. It is a mathematical abstraction. All of the sums above end up purely real..... Further, what if we have an FFT of a cosine with phase of something like pi/4? Then what? Then there is: cos(wt+pi/4) = 0.5*e^j(wt-pi/4) + 0.5*e^-j(wt-pi/4) We see two rotating vectors that coincide on the real axis when t=pi/4w. The imaginary parts still cancel as in the examples above. Thus this remains a real function. We also note that It is *convenient to envision* this as a single counterclockwise rotating vector starting at -pi/4 and rotating clockwise and only considering its projection on the real or x axis. We also note that in the 3-D plots with time as an added axis, that would be the projection of the rotating vector on the x-t plane that peaks at t=pi/4. Whew! I sure hope *this* helps. Fred
Reply by ●June 13, 20052005-06-13
Angelo Ricotta wrote:> [snip1] > In the real world all signals are real. Imaginary signals, and even negative > frequencies are all math tricks for doing calculations in an easier and > often more elegant way. So, if you want to visualize (e.g. in an > oscilloscope) a complex signal you need to represent two real signals or > functions. In our case (cos(w0t), sin(w0t)) and you get a Lissajous figure, > here a circumference. > Angelo > [snip comment related to contents of snip1] >Relating phasors etc to Lissajous figures strikes me as an interesting idea. One of the reasons I suspect I never completed a BSEE is I just don't(can't?) think in formal math terms. My 1st semester calculus instructor was a *PURE* math grad student burdened with a bunch of engineers. I did not begin to understand what he was trying to communicate until late in 2nd semester physics. I can "mechanically" use FFT etc. As in I "can drive" a car with no comprehension of Carnot cycle. [sic or possibly sick ;] Could Lissajous figures be used to illustrate/explain why the terms "real and even", "real and odd", "complex and even", and "complex and odd" frequently appear in discussions? Might Lissajous figures illustrate other "things" to we math challenged? [ aside to those tending to chide me for not explicitly stating "what I'm looking for". I *EXPLICITLY STATE*, I've *NOT* the foggiest ;]
