Hi, to all I've a question regarding negative frequency. Let: X(f) = delta(f+f0) + delta(f-f0) It's possible to see X(f) as 2 delta(f-f0) My trouble is started studying Super heterodyne receivers: http://www.microwaves101.com/encyclopedia/receivers_superhet.cfm Section: Image frequency This article tell that the minus sign is unimportant so -10Ghz signal is interfering with +10Ghz signal. I suppose a continuous line for frequency and negative frequency are distinct from positive frequency, like this: http://www.dsprelated.com/dspbooks/mdft/Sinusoid_Magnitude_Spectra.html Could you give me some help? Thank you Bye Ste
Negative frequency
Started by ●March 18, 2009
Reply by ●March 18, 20092009-03-18
Stephan wrote:> Hi, > > > > to all I've a question regarding negative frequency. > > > > Let: > > > > X(f) = delta(f+f0) + delta(f-f0) > > > > It's possible to see X(f) as 2 delta(f-f0) > > > > My trouble is started studying Super heterodyne receivers: > > http://www.microwaves101.com/encyclopedia/receivers_superhet.cfm > Section: Image frequency > > > > This article tell that the minus sign is unimportant so -10Ghz signal is > interfering with +10Ghz signal. > > > > I suppose a continuous line for frequency and negative frequency are > distinct from positive frequency, like this: > > http://www.dsprelated.com/dspbooks/mdft/Sinusoid_Magnitude_Spectra.html > > > > Could you give me some help?Negative frequencies are only semi-real. You cannot determine the sign of a single frequency with any instrument. Your -10GHz signal is labeled that way because of how it was formed. The +10GHz signal gets its label because of /its/ method of formation. Both are simply 10GHz signals, undistinguished by sign. Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������
Reply by ●March 18, 20092009-03-18
Thank you for your reply.> Both are simply 10GHz signals, undistinguished by sign.So can I say that: x(f) = cos(2*pi*fc*t) => X(f) = 1/2 [ delta(f-fc) + delta(f+fc) ] = 1/2 delta(f-fc) + 1/2 delta(f+fc) This spectrum is composed by 2 delta. One located in +fc and other in -fc. The sign + and - are labeled from math operation related to Fourier Transformation, but both represent fc. Now, if I analyze this spectrum using real instrument like spectrum analyzer or I modulate this signal for transmission, what I see is only one delta located in +fc having amplitude of 1 (1/2 + 1/2). Correct? So the negative frequency representation it's useful for math representation only, but in real word the are interfering with positive. Correct? Before ask I've tried to search in previous post, but I've not found one that can explain my doubt. Thank you very much for your help. Ste
Reply by ●March 18, 20092009-03-18
Stephan wrote:> Thank you for your reply. > > > >> Both are simply 10GHz signals, undistinguished by sign. > > > So can I say that: > > x(f) = cos(2*pi*fc*t) => X(f) = 1/2 [ delta(f-fc) + delta(f+fc) ] = 1/2 > delta(f-fc) + 1/2 delta(f+fc)Where does delta() enter into this? Spectra consist of sinusoids.> This spectrum is composed by 2 delta. One located in +fc and other in -fc. > > The sign + and - are labeled from math operation related to Fourier > Transformation, but both represent fc. > > > > Now, if I analyze this spectrum using real instrument like spectrum analyzer > or I modulate this signal for transmission, what I see is only one delta > located in +fc having amplitude of 1 (1/2 + 1/2). Correct?You see sinusoids, not deltas. You are muddling things together.> So the negative frequency representation it's useful for math representation > only, but in real word the are interfering with positive. Correct? > > > > Before ask I've tried to search in previous post, but I've not found one > that can explain my doubt. > > > > Thank you very much for your help.This needs more than I choose to type. Keep thinking. Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������
Reply by ●March 19, 20092009-03-19
On Mar 18, 5:04�pm, "Stephan" <n...@address.com> wrote:> Thank you for your reply. > > > Both are simply 10GHz signals, undistinguished by sign. > > So can I say that: > > x(f) = cos(2*pi*fc*t) => X(f) = 1/2 [ delta(f-fc) + delta(f+fc) ] = 1/2 > delta(f-fc) + 1/2 delta(f+fc) > > This spectrum is composed by 2 delta. One located in +fc and other in -fc. > > The sign + and - are labeled from math operation related to Fourier > Transformation, but both �represent fc. > > Now, if I analyze this spectrum using real instrument like spectrum analyzer > or I modulate this signal for transmission, what I see is only one delta > located in +fc having amplitude of 1 (1/2 + 1/2). Correct? > > So the negative frequency representation it's useful for math representation > only, but in real word the are interfering with positive. Correct? > > Before ask I've tried to search in previous post, but I've not found one > that can explain my doubt. > > Thank you very much for your help. > > SteStephan, It may help your intuition if you consider negative frequencies in the form of an I/Q phasor diagram. Think of the relationship of I and Q as the phasor rotates in each of the two directions. A counter-clockwise rotation is a positive frequency, the opposite rotation is a negative frequency. Negative frequencies have a "physical" existence insofar as they express the relationship between physically existing I and Q components. Darol Klawetter
Reply by ●March 19, 20092009-03-19
On Mar 19, 5:58�pm, Darol Klawetter <darol.klawet...@l-3com.com> wrote:> On Mar 18, 5:04�pm, "Stephan" <n...@address.com> wrote: > > > > > Thank you for your reply. > > > > Both are simply 10GHz signals, undistinguished by sign. > > > So can I say that: > > > x(f) = cos(2*pi*fc*t) => X(f) = 1/2 [ delta(f-fc) + delta(f+fc) ] = 1/2 > > delta(f-fc) + 1/2 delta(f+fc) > > > This spectrum is composed by 2 delta. One located in +fc and other in -fc. > > > The sign + and - are labeled from math operation related to Fourier > > Transformation, but both �represent fc. > > > Now, if I analyze this spectrum using real instrument like spectrum analyzer > > or I modulate this signal for transmission, what I see is only one delta > > located in +fc having amplitude of 1 (1/2 + 1/2). Correct? > > > So the negative frequency representation it's useful for math representation > > only, but in real word the are interfering with positive. Correct? > > > Before ask I've tried to search in previous post, but I've not found one > > that can explain my doubt. > > > Thank you very much for your help. > > > Ste > > Stephan, > > It may help your intuition if you consider negative frequencies in the > form of an I/Q phasor diagram. Think of the relationship of I and Q as > the phasor rotates in each of the two directions. A counter-clockwise > rotation is a positive frequency, the opposite rotation is a negative > frequency. �Negative frequencies have a "physical" existence insofar > as they express the relationship between physically existing I and Q > components. > > Darol KlawetterFuck - how does that help!
Reply by ●March 19, 20092009-03-19
On Mar 19, 11:04�am, "Stephan" <n...@address.com> wrote:> Thank you for your reply. > > > Both are simply 10GHz signals, undistinguished by sign. > > So can I say that: > > x(f) = cos(2*pi*fc*t) => X(f) = 1/2 [ delta(f-fc) + delta(f+fc) ] = 1/2 > delta(f-fc) + 1/2 delta(f+fc) > > This spectrum is composed by 2 delta. One located in +fc and other in -fc. > > The sign + and - are labeled from math operation related to Fourier > Transformation, but both �represent fc. > > Now, if I analyze this spectrum using real instrument like spectrum analyzer > or I modulate this signal for transmission, what I see is only one delta > located in +fc having amplitude of 1 (1/2 + 1/2). Correct? > > So the negative frequency representation it's useful for math representation > only, but in real word the are interfering with positive. Correct? > > Before ask I've tried to search in previous post, but I've not found one > that can explain my doubt. > > Thank you very much for your help. > > SteYou are confusing the time and freq domains. Please re-phrase the question. The deltas come in after you take the Fourier Transform of exp(jWt) terms. Hardy
Reply by ●March 19, 20092009-03-19
Darol Klawetter wrote:> On Mar 18, 5:04 pm, "Stephan" <n...@address.com> wrote: >> Thank you for your reply. >> >>> Both are simply 10GHz signals, undistinguished by sign. >> So can I say that: >> >> x(f) = cos(2*pi*fc*t) => X(f) = 1/2 [ delta(f-fc) + delta(f+fc) ] = 1/2 >> delta(f-fc) + 1/2 delta(f+fc) >> >> This spectrum is composed by 2 delta. One located in +fc and other in -fc. >> >> The sign + and - are labeled from math operation related to Fourier >> Transformation, but both represent fc. >> >> Now, if I analyze this spectrum using real instrument like spectrum analyzer >> or I modulate this signal for transmission, what I see is only one delta >> located in +fc having amplitude of 1 (1/2 + 1/2). Correct? >> >> So the negative frequency representation it's useful for math representation >> only, but in real word the are interfering with positive. Correct? >> >> Before ask I've tried to search in previous post, but I've not found one >> that can explain my doubt. >> >> Thank you very much for your help. >> >> Ste > > Stephan, > > It may help your intuition if you consider negative frequencies in the > form of an I/Q phasor diagram. Think of the relationship of I and Q as > the phasor rotates in each of the two directions. A counter-clockwise > rotation is a positive frequency, the opposite rotation is a negative > frequency. Negative frequencies have a "physical" existence insofar > as they express the relationship between physically existing I and Q > components."Physically existing I and Q components." Interesting. What kind of instrumentation is used with these? Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������
Reply by ●March 19, 20092009-03-19
On Mar 19, 8:58 am, Jerry Avins <j...@ieee.org> wrote:> Darol Klawetter wrote:. .> ... .> > Stephan, .>> > It may help your intuition if you consider negative frequencies in the.> > form of an I/Q phasor diagram. Think of the relationship of I and Q as .> > the phasor rotates in each of the two directions. A counter- clockwise .> > rotation is a positive frequency, the opposite rotation is a negative .> > frequency. Negative frequencies have a "physical" existence insofar .> > as they express the relationship between physically existing I and Q .> > components. fred harris used this approach to teach our in-house DSP classes at Spectral Dynamics/Scientific Atlanta. He would stand with his arms straight out at his sides and his wrists turned to point both index fingers vertically. Then he would turn both wrists either clockwise or counter-clockwise to rotate his index fingers around the axis of his extended arms. .> .> "Physically existing I and Q components." Interesting. What kind of .> instrumentation is used with these? .> .> Jerry ������ A dual channel oscilloscope for analog implementations and a logic analyzer or bus analyzer for digital implementations. Dale B. Dalrymple
Reply by ●March 19, 20092009-03-19
dbd wrote: I wrote::> .> "Physically existing I and Q components." Interesting. What kind of > .> instrumentation is used with these? > .> > .> Jerry > ������ > A dual channel oscilloscope for analog implementations and a logic > analyzer or bus analyzer for digital implementations.Do you need special I and Q probes? What happens to the displayed signal if you switch them? Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������
