If you feel that this post is more suitable for sci.math, allow me some latitude: I am rasing this issue in the context of computing the Fourier transform. At least two DSP books I've looked at have stated that the integral of an odd function is zero. The example under question is the FT of x(t)=1/t, where the integrand of the real part is cos(wt)/t. But the point glossed over is that the integral is an improper one (of the third kind). It may well be that the integral of cos(t)/t from -infinity to +infinity is zero, but not because of the superficial reasoning that it is an odd function. An example where the argument fails is when the integrand is, say, x(1+x^2). The integral diverges and one would be wrong to conclude that the result is zero. What _is_ zero is the Cauchy Principal Value (for a very nice description see http://mathforum.org/library/drmath/view/61246.html). CPV is a weaker condition. ObDSP quiz: We know that if two systems are in cascade that the overall transfer function is H1(w)H2(w). Consider the simple voltage divider with two equal resistances. Vout/Vin is 1/2. Cascade two such systems. You would expect the overall transfer function to be 1/4, but, applying KVL/KCL you can show it is 1/5. What gives?!
integral of odd function
Started by ●September 12, 2003
Reply by ●September 12, 20032003-09-12
Vanamali wrote:>...> > ObDSP quiz: We know that if two systems are in cascade that the > overall transfer function is H1(w)H2(w). Consider the simple voltage > divider with two equal resistances. Vout/Vin is 1/2. Cascade two such > systems. You would expect the overall transfer function to be 1/4, > but, applying KVL/KCL you can show it is 1/5. What gives?!Bring me up to speed please. What are KVL and KCL? Although a voltage divider may divide by two when looking into an open circuit, when two are cascaded, the first one is loaded. Is that your problem? Many device transfer functions depend on the load. It's a rare device than can develop much voltage across a short circuit. Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������
Reply by ●September 12, 20032003-09-12
"Vanamali" <vanamali@netzero.net> wrote in message news:8d4ba7e3.0309120606.4979709e@posting.google.com...> If you feel that this post is more suitable for sci.math, allow me > some latitude: I am rasing this issue in the context of computing the > Fourier transform. At least two DSP books I've looked at have stated > that the integral of an odd function is zero. The example under > question is the FT of x(t)=1/t, where the integrand of the real part > is cos(wt)/t. But the point glossed over is that the integral is an > improper one (of the third kind). It may well be that the integral of > cos(t)/t from -infinity to +infinity is zero, but not because of the > superficial reasoning that it is an odd function.In comp.dsp you would be discussing discrete transforms, and sums instead of integrals. The more popular DCT's don't evaluate at zero, so would not even notice the singularity. Sometimes you know what the right answer is, and don't worry about what the math might say. -- glen
Reply by ●September 12, 20032003-09-12
KVL = Kirchhoff's Voltage Law, i.e.: Around any loop in a circuit, the voltages algebraically sum to zero. KCL = Kirchhoff's Current Law, i.e.: At any node of a circuit, the currents algebraically sum to zero. Just your basic stuff from 1st year circuits class. "Jerry Avins" <jya@ieee.org> wrote in message news:3F61F4D5.7F269123@ieee.org...> Vanamali wrote: > > > ... > > > > ObDSP quiz: We know that if two systems are in cascade that the > > overall transfer function is H1(w)H2(w). Consider the simple voltage > > divider with two equal resistances. Vout/Vin is 1/2. Cascade two such > > systems. You would expect the overall transfer function to be 1/4, > > but, applying KVL/KCL you can show it is 1/5. What gives?! > > Bring me up to speed please. What are KVL and KCL? Although a voltage > divider may divide by two when looking into an open circuit, when two > are cascaded, the first one is loaded. Is that your problem? Many device > transfer functions depend on the load. It's a rare device than can > develop much voltage across a short circuit. > > Jerry > -- > Engineering is the art of making what you want from things you can get. > �����������������������������������������������������������������������
Reply by ●September 12, 20032003-09-12
In that case, it's as I surmised. The open-circuit transfer function was = mistaken used instead of the loaded transfer function. Actually, the=20 transfer function of the cascaded pair can be made almost any value,=20 including with a peak or a notch, depending on the impedance levels of=20 the two dividers. The question is loaded with unwarranted assumptions. Jerry --=20 Engineering is the art of making what you want from things you can get. =AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF= =AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF= =AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF Jon Harris wrote:> KVL =3D Kirchhoff's Voltage Law, i.e.: Around any loop in a circuit, th=e> voltages algebraically sum to zero. > KCL =3D Kirchhoff's Current Law, i.e.: At any node of a circuit, the cu=rrents> algebraically sum to zero. >=20 > Just your basic stuff from 1st year circuits class. >=20 > "Jerry Avins" <jya@ieee.org> wrote in message > news:3F61F4D5.7F269123@ieee.org... >=20 >>Vanamali wrote: >> >> ... >> >>>ObDSP quiz: We know that if two systems are in cascade that the >>>overall transfer function is H1(w)H2(w). Consider the simple voltage >>>divider with two equal resistances. Vout/Vin is 1/2. Cascade two such=>>>systems. You would expect the overall transfer function to be 1/4, >>>but, applying KVL/KCL you can show it is 1/5. What gives?! >> >>Bring me up to speed please. What are KVL and KCL? Although a voltage >>divider may divide by two when looking into an open circuit, when two >>are cascaded, the first one is loaded. Is that your problem? Many devic=e>>transfer functions depend on the load. It's a rare device than can >>develop much voltage across a short circuit. >> >>Jerry >>-- >>Engineering is the art of making what you want from things you can get.=>>=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF==AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF= =AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF
Reply by ●September 12, 20032003-09-12
"Jerry Avins" <jya@ieee.org> wrote in message news:bjta97$552$1@bob.news.rcn.net... In that case, it's as I surmised. The open-circuit transfer function was mistaken used instead of the loaded transfer function. Actually, the transfer function of the cascaded pair can be made almost any value, including with a peak or a notch, depending on the impedance levels of the two dividers. The question is loaded with unwarranted assumptions. Jerry -- Engineering is the art of making what you want from things you can get. ����������������������������������������������������������������������� Jon Harris wrote:> KVL = Kirchhoff's Voltage Law, i.e.: Around any loop in a circuit, the > voltages algebraically sum to zero. > KCL = Kirchhoff's Current Law, i.e.: At any node of a circuit, thecurrents> algebraically sum to zero. > > Just your basic stuff from 1st year circuits class. > > "Jerry Avins" <jya@ieee.org> wrote in message > news:3F61F4D5.7F269123@ieee.org... > >>Vanamali wrote: >> >> ... >> >>>ObDSP quiz: We know that if two systems are in cascade that the >>>overall transfer function is H1(w)H2(w). Consider the simple voltage >>>divider with two equal resistances. Vout/Vin is 1/2. Cascade two such >>>systems. You would expect the overall transfer function to be 1/4, >>>but, applying KVL/KCL you can show it is 1/5. What gives?!Why do you say ObDSP? ObElectronics maybe....>> >>Bring me up to speed please. What are KVL and KCL? Although a voltage >>divider may divide by two when looking into an open circuit, when two >>are cascaded, the first one is loaded. Is that your problem? Many device >>transfer functions depend on the load. It's a rare device than can >>develop much voltage across a short circuit. >> >>Jerry >>-- >>Engineering is the art of making what you want from things you can get. >>�����������������������������������������������������������������������
Reply by ●September 12, 20032003-09-12
Jerry Avins <jya@ieee.org> wrote in message news:<3F61F4D5.7F269123@ieee.org>...> Bring me up to speed please. What are KVL and KCL? Although a voltage > divider may divide by two when looking into an open circuit, when two > are cascaded, the first one is loaded. Is that your problem? Many device > transfer functions depend on the load. It's a rare device than can > develop much voltage across a short circuit.KVL and KCL are the Kirchoff's voltage and current laws. BTW, is that guy's name pronounced "keer--koff" (keer similar to dear)? Jerry is right on the money in his answer. Yes, it is loading that is the reason. If we connect the two voltage dividers by using a voltage follower in between, then the overall transfer function will be 1/4. Many DSP books talk about interconnection of systems but don't mention this important point. Systems, after all, are practical circuits and students should be alerted to issues like this.
Reply by ●September 13, 20032003-09-13
vanamali@netzero.net (Vanamali) wrote in message news:<8d4ba7e3.0309121735.1cc7c31e@posting.google.com>...> Jerry Avins <jya@ieee.org> wrote in message news:<3F61F4D5.7F269123@ieee.org>... > > > Bring me up to speed please. What are KVL and KCL? Although a voltage > > divider may divide by two when looking into an open circuit, when two > > are cascaded, the first one is loaded. Is that your problem? Many device > > transfer functions depend on the load. It's a rare device than can > > develop much voltage across a short circuit. > > KVL and KCL are the Kirchoff's voltage and current laws. BTW, is that > guy's name pronounced "keer--koff" (keer similar to dear)? Jerry is > right on the money in his answer. Yes, it is loading that is the > reason. If we connect the two voltage dividers by using a voltage > follower in between, then the overall transfer function will be 1/4. > Many DSP books talk about interconnection of systems but don't mention > this important point. Systems, after all, are practical circuits and > students should be alerted to issues like this.Eh... how do you "load" a digital filter? To the best of my knowledge, digital signal processing (as opposed to analog signal processing) is about manipulating numbers. What you use any number for, once computed, should not affect the outcome of the computation itself. Economists may, of course, have different opinions. Rune
Reply by ●September 13, 20032003-09-13
Vanamali wrote:> > If you feel that this post is more suitable for sci.math, allow me > some latitude: I am rasing this issue in the context of computing the > Fourier transform. At least two DSP books I've looked at have stated > that the integral of an odd function is zero.If f(t) = -f(-t) [f(t) is odd], then \int_{-\infty}^{+\infty} f(t) dt = \int_{-\infty}^{0} f(t) dt + \int_{0}^{+\infty} f(t) dt = \int_{0}^{+\infty} f(-t) dt + \int_{0}^{+\infty} f(t) dt = \int_{0}^{+\infty} -f(t) dt + \int_{0}^{+\infty} f(t) dt = -\int_{0}^{+\infty} f(t) dt + \int_{0}^{+\infty} f(t) dt [1] = 0 (if the integrals exist). If the integrals don't exist in this form, you can try a limit: \int_{-\infty}^{+\infty} f(t) dt = lim_{T-->\infty} -\int_{0}^{T} f(t) dt + lim_{T-->\infty} \int_{0}^{T} f(t) dt [2] = 0 (if the limits exist)> The example under > question is the FT of x(t)=1/t, where the integrand of the real part > is cos(wt)/t. But the point glossed over is that the integral is an > improper one (of the third kind).Yes, so you can't use the fundamental theorem of calculus (FTC) directly. Probably a simple limit towards the point t=0 will do.> It may well be that the integral of > cos(t)/t from -infinity to +infinity is zero, but not because of the > superficial reasoning that it is an odd function. An example where > the argument fails is when the integrand is, say, x(1+x^2). The > integral diverges and one would be wrong to conclude that the result > is zero.Yes, that is a good observation. If you used the limit form in [2], then you won't get to conclude the result is 0 since the individual limits don't exist.> What _is_ zero is the Cauchy Principal Value (for a very > nice description see > http://mathforum.org/library/drmath/view/61246.html). CPV is a weaker > condition.I didn't care much for it.> ObDSP quiz: We know that if two systems are in cascade that the > overall transfer function is H1(w)H2(w). Consider the simple voltage > divider with two equal resistances. Vout/Vin is 1/2. Cascade two such > systems. You would expect the overall transfer function to be 1/4, > but, applying KVL/KCL you can show it is 1/5. What gives?!Ob? Obstetrics? (Isn't this month national gynecological month???) Duh how about inter-stage coupling considerations? Last I checked the simple combination you provided was in context of a digital system, not analog. -- % Randy Yates % "...the answer lies within your soul %% Fuquay-Varina, NC % 'cause no one knows which side %%% 919-577-9882 % the coin will fall." %%%% <yates@ieee.org> % 'Big Wheels', *Out of the Blue*, ELO http://home.earthlink.net/~yatescr
Reply by ●September 13, 20032003-09-13
Rune Allnor wrote:> > vanamali@netzero.net (Vanamali) wrote in message news:<8d4ba7e3.0309121735.1cc7c31e@posting.google.com>... > > Jerry Avins <jya@ieee.org> wrote in message news:<3F61F4D5.7F269123@ieee.org>... > > > > > Bring me up to speed please. What are KVL and KCL? Although a voltage > > > divider may divide by two when looking into an open circuit, when two > > > are cascaded, the first one is loaded. Is that your problem? Many device > > > transfer functions depend on the load. It's a rare device than can > > > develop much voltage across a short circuit. > > > > KVL and KCL are the Kirchoff's voltage and current laws. BTW, is that > > guy's name pronounced "keer--koff" (keer similar to dear)? Jerry is > > right on the money in his answer. Yes, it is loading that is the > > reason. If we connect the two voltage dividers by using a voltage > > follower in between, then the overall transfer function will be 1/4. > > Many DSP books talk about interconnection of systems but don't mention > > this important point. Systems, after all, are practical circuits and > > students should be alerted to issues like this. > > Eh... how do you "load" a digital filter? To the best of my knowledge, > digital signal processing (as opposed to analog signal processing) is > about manipulating numbers. What you use any number for, once computed, > should not affect the outcome of the computation itself. > > Economists may, of course, have different opinions. > > RuneWhat digital filter? The quizzer had in mind two R/R analog dividers, making 4 equal Rs in all. That's made evident by the overall response he computed. If the dividers has been 1K-1K followed by 1-1 or 1-1 followed by 1K-1K, the numbers would have been different. If they had been 1F-1F follower by 1H-1H, the overall transfer function wouldn' even have had a flat spectrum. Here's one for him: It is well known that to get the maximum power out of a source with a resistive internal impedance, the load should match that impedance. For a given resistive load, what should the generator's resistance be for maximum power transfer? Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������
