Tim Wescott <seemywebsite@myfooter.really> wrote: (big snip)> By your definition of orthogonality, p is orthogonal with itself, because > if you substitute p for q2 in your equation above, the result of the > integration is zero.>> I said p was orthogonal to q2.> Yes, and I checked your math by substituting p in place of q2, and found > it math wanting.>> As for "hitting the books", my "Engineer's Guide to DSP" book by Steven >> Smith mentions that a sine wave is orthogonal with a cosine wave. But >> he doesn't give an integral equation definition for orthogonality. >> There is no entry in the Index of my Opp & Schafer DSP book for the word >> orthogonal.Somewhere back in this thread, though maybe it got lost by now, I noted that to decode a quadrature modulated signal that you need a phase reference of some kind. There are a variety of ways to get that reference. I previously noted the color burst used for NTSC video. Stereo FM uses the 19kHz pilot to determine the phase of the 38kHz subcarrier. Stereo FM is another interesting example. A 38kHz carrier is AMSC (Amplitude Modulated with Suppressed Carrier) with the (R-L)/2 signal, This is then added to the mono (L+R)/2 signal, and the 19kHz pilot at 10% (of 100% modulation) is added. But another way to look at it is that the signal switches between L and R at 38kHz. Now, does it switch as a square wave or a sine? All the harmonics of a 38kHz square wave are multiples of 38kHz, with the higher ones presumably filtered out. They would not be audible in any case.>> I'm basing my definition of orthogonality on the definition given at:>> http://en.wikipedia.org/wiki/OrthogonalityBut the definition of orhogonality doesn't require a decoding reference. If you want to actually use it, it does. For digital signals, it is usual to modulate the change in phase, instead of absolute phase, which partly removes the need for a reference. (Harder to do for analog signals.) One can then generate a reference phase by squaring the original signal, which allows one to detect when the phase shifts occur.>> The same definition is also provided at:>> http://mathworld.wolfram.com/OrthogonalFunctions.html . >> Tim, I'll bet our disagreement is a problem of not agreeing upon the >> definition of orthogonality. >> Resolving such controversies is educational, at least they are for me. > > I think the difficulty comes about either with a definition of > orthogonality for complex signals, or with the validity of using the > concept of orthogonality for complex signals. I tried to resolve the > problem with suitable definition of "orthogonal".> The original question was, if you have two complex carriers, e^jwt and > e^j(wt + pi/2), can you transmit two independent signals?(snip) -- glen
What makes quadrature mixing possible?
Started by ●August 31, 2014
Reply by ●September 6, 20142014-09-06
Reply by ●September 6, 20142014-09-06
On Sat, 06 Sep 2014 17:15:48 -0500, Tim Wescott <seemywebsite@myfooter.really> wrote:>On Sat, 06 Sep 2014 13:20:25 -0700, Rick Lyons wrote: > >> On Sat, 06 Sep 2014 12:01:54 -0500, Tim Wescott >> <seemywebsite@myfooter.really> wrote: >> >>>On Sat, 06 Sep 2014 09:36:12 -0700, Rick Lyons wrote: >>> >>>> On Fri, 05 Sep 2014 14:35:45 -0500, tim <tim@seemywebsite.com> wrote: >>>> >>>>>On Fri, 05 Sep 2014 05:11:07 -0700, Rick Lyons wrote: >>>>> >>>>>> On Wed, 03 Sep 2014 17:54:38 -0500, Tim Wescott >>>>>> <seemywebsite@myfooter.really> wrote: >>>>>> >>>>>> [Snipped by Lyons] >>>>>> >>>>>>>I only just noticed this part of your question. >>>>>>> >>>>>>>e^jwt and e^(jwt + pi/2) are not orthogonal, because no signal is >>>>>>>orthogonal with itself, and e^(jwt + pi/2) is just e^jwt rotated by >>>>>>>90 degrees. >>>>>>> >>>>>>>In a sense e^jwt BY ITSELF already gives you a carrier for >>>>>>>transmitting two independent real signals -- asking for more is just >>>>>>>greedy. >>>>>> >>>>>> Hi Tim, >>>>>> miladsp described two complex signals: >>>>>> >>>>>> p = e^(j*w_c*t) and q1 = e^(j*w_c*t + pi/2). >>>>>> >>>>>> His q1 is equal to: >>>>>> >>>>>> q1 = e^(j*w_c*t)*e^(pi/2) >>>>>> >>>>>> where e^(pi/2) = 4.81. His q1 has a magnitude of 4.81. And his p >>>>>> and q1 = 4.81*p are orthogonal. >>>>>> >>>>>> >>>>>> What I think he MEANT to describe was: >>>>>> >>>>>> p = e^(j*w_c*t) and q2 = e^(j*(w_c*t + pi/2)). >>>>>> >>>>>> (Notice how q2 does not equal q1. q2 has a magnitude of one.) >>>>>> >>>>>> And as far as I can tell, p and q2 are orthogonal. >>>> >>>> Hi, >>>> >>>>>Over one cycle of p the integral of p and p* is a real, positive >>>>>number. >>>>>Hence, p is not orthogonal to itself. >>>> >>>> If by p* you mean the conjugate of p, then I agree. But we were not >>>> talkin' about the orthogonality of p and the conjugate of p. >>>> >>>>>Over one cycle of p (and q2), the integral of q2 * p* is a purely >>>>>imaginary number of absolute value distinctly greater than zero. >>>>>hence, >>>>>p is not orthogonal to q2. >>>> >>>> You have written p* again. The value p* was not part of our >>>> discussion. >>>> >>>> I claim that p = e^(j*w_c*t) and q2 = e^(j*(w_c*t + pi/2)) >>>> are orthogonal. That is, over one cycle the integral of the product >>>> of p times q2 is zero. >>>> >>>> Here's my thinking: >>>> Over one cycle we can evaluate the integral of p times q2 as: >>>> >>>> 2pi >>>> -- >>>> / >>>> / e^(jx)e^(jx + pi/2) dx >>>> / >>>> -- >>>> 0 >>>> >>>> 2pi >>>> -- >>>> / >>>> = / e^(jx)e^(jx)e^(pi/2) dx >>>> / >>>> -- >>>> 0 >>>> >>>> 2pi >>>> -- >>>> / >>>> = / je^(jx)e^(jx) dx >>>> / >>>> -- >>>> 0 >>>> >>>> 2pi - - 2pi >>>> -- | | >>>> / | e^(j2x) | >>>> = / je^(j2x) dx = j |----------| >>>> / | j2 | >>>> -- | | 0 - >>>> - 0 >>>> >>>> e^(j4pi) e^(0) >>>> = ---------- - ------- >>>> 2 2 >>>> >>>> 1 1 >>>> = --- - --- = 0. >>>> 2 2 >>>> >>>> >>>> If I've screwed up here, please let me know. >>>> >>>> [-Rick-] >>> >>>I think the problem is your definition of orthogonality. If you start >>>by saying that a signal with nonzero energy absolutely positively cannot >>>be orthogonal with itself, then your definition of orthogonality leads >>>to a contradiction, because the integral over N cycles of p * p is zero. >>> >>>So, I'm doing the usual engineer's stretch, and claiming that >>>orthogonality must be calculated using signal1 times the conjugate of >>>signal2. When used with p, this comes up with the sensible result that >>>the time average of p * p* is the power level of p. >>> >>>If you really insist that your definition of orthogonality is correct >>>then I think we both need to hit the books. >> >> Hi, >> Wow. I don't know why you and I are not >> communicating too well. >> >> I did not say a signal is not orthogonal with itself. > >By your definition of orthogonality, p is orthogonal with itself, because >if you substitute p for q2 in your equation above, the result of the >integration is zero. > >> I said p was orthogonal to q2. > >Yes, and I checked your math by substituting p in place of q2, and found >it math wanting. > >> As for "hitting the books", my "Engineer's Guide to DSP" book by Steven >> Smith mentions that a sine wave is orthogonal with a cosine wave. But >> he doesn't give an integral equation definition for orthogonality. >> There is no entry in the Index of my Opp & Schafer DSP book for the word >> orthogonal. >> >> I'm basing my definition of orthogonality on the definition given at: >> >> http://en.wikipedia.org/wiki/Orthogonality >> >> The same definition is also provided at: >> >> http://mathworld.wolfram.com/OrthogonalFunctions.html . >> Tim, I'll bet our disagreement is a problem of not agreeing upon the >> definition of orthogonality. >> Resolving such controversies is educational, at least they are for me. > >I think the difficulty comes about either with a definition of >orthogonality for complex signals, or with the validity of using the >concept of orthogonality for complex signals. I tried to resolve the >problem with suitable definition of "orthogonal".The definition I've always seen and used is that the inner product is zero. A zero-magnitude vector is magic, kind of like the number zero is magic, in that it is orthogonal to anything. That stretches the concept of "orthogonality", though, just like the goofy proofs discussed here lately where zero or division by zero is used to "prove" something that isn't true. Likewise using a zero-magnitude vector to test for orthogonality or the validity of a statement isn't very useful.>The original question was, if you have two complex carriers, e^jwt and >e^j(wt + pi/2), can you transmit two independent signals?Clearly you can. There are many, many ways to do it. There's time orthogonality, there's frequency orthogonality, and if you want to transmit them at the same time on the same frequency you can spread them with orthogonal codes or use MIMO in many cases (if the channels are faded independently).>The answer to THAT question depends, I guess, on whether you mean two >independent _complex_ signals, or two independent _real_ signals. Because >regardless of how you may define orthogonality, the answer to "can you use >these two carriers to transmit two independent complex signals over one >channel" is HELL NO, while the answer if the independent signals are real >is HELL YES.As I just mentioned, you can do it in either case.>I think you can see why -- the result of > >A * e^jwt + B * e^j(wt + pi/2) is (if I'm doing my math right) > >((Re(A) - Im(B)) + j(Im(A) + Re(B)) * e^jwt > >It should be obvious by inspection why an arbitrary A and B, with both >numbers complex, cannot be sent using the two cited carriers.Not always, but very often, you can. You can extend what you've written to an even more complex case: s1(t) = A(t)e^jwt and s2(t) = B(t)e^(jwt+pi/2). A(t) and B(t) could be two independent NRZ streams of symbols from the same bit clock, and then s1(t) and s2(t) can be interpreted as two BPSK signals, both at frequency = w. If the phase relationship between them created by the pi/2 offset in s2 is strictly maintained, then s1(t) + s2(t) = s3(t) just creates a QPSK signal, s3(t). This is done routinely. So, yes, as far as I can tell there's no reason for me to believe that your two example signals are not orthogonal and cannot be transmitted at the same time on the same channel and properly recovered. There are LOTS of things that could happen to make them not orthogonal and no longer separable, but I haven't seen any of those conditions specified. Maybe I missed it. Eric Jacobsen Anchor Hill Communications http://www.anchorhill.com
Reply by ●September 7, 20142014-09-07
On Sat, 06 Sep 2014 17:15:48 -0500, Tim Wescott <seemywebsite@myfooter.really> wrote:>On Sat, 06 Sep 2014 13:20:25 -0700, Rick Lyons wrote: > >> On Sat, 06 Sep 2014 12:01:54 -0500, Tim Wescott >> <seemywebsite@myfooter.really> wrote: >> >>>On Sat, 06 Sep 2014 09:36:12 -0700, Rick Lyons wrote: >>> >>>> On Fri, 05 Sep 2014 14:35:45 -0500, tim <tim@seemywebsite.com> wrote: >>>> >>>>>On Fri, 05 Sep 2014 05:11:07 -0700, Rick Lyons wrote: >>>>> >>>>>> On Wed, 03 Sep 2014 17:54:38 -0500, Tim Wescott >>>>>> <seemywebsite@myfooter.really> wrote: >>>>>> >>>>>> [Snipped by Lyons] >>>>>> >>>>>>>I only just noticed this part of your question. >>>>>>> >>>>>>>e^jwt and e^(jwt + pi/2) are not orthogonal, because no signal is >>>>>>>orthogonal with itself, and e^(jwt + pi/2) is just e^jwt rotated by >>>>>>>90 degrees. >>>>>>> >>>>>>>In a sense e^jwt BY ITSELF already gives you a carrier for >>>>>>>transmitting two independent real signals -- asking for more is just >>>>>>>greedy. >>>>>> >>>>>> Hi Tim, >>>>>> miladsp described two complex signals: >>>>>> >>>>>> p = e^(j*w_c*t) and q1 = e^(j*w_c*t + pi/2). >>>>>> >>>>>> His q1 is equal to: >>>>>> >>>>>> q1 = e^(j*w_c*t)*e^(pi/2) >>>>>> >>>>>> where e^(pi/2) = 4.81. His q1 has a magnitude of 4.81. And his p >>>>>> and q1 = 4.81*p are orthogonal. >>>>>> >>>>>> >>>>>> What I think he MEANT to describe was: >>>>>> >>>>>> p = e^(j*w_c*t) and q2 = e^(j*(w_c*t + pi/2)). >>>>>> >>>>>> (Notice how q2 does not equal q1. q2 has a magnitude of one.) >>>>>> >>>>>> And as far as I can tell, p and q2 are orthogonal. >>>> >>>> Hi, >>>> >>>>>Over one cycle of p the integral of p and p* is a real, positive >>>>>number. >>>>>Hence, p is not orthogonal to itself. >>>> >>>> If by p* you mean the conjugate of p, then I agree. But we were not >>>> talkin' about the orthogonality of p and the conjugate of p. >>>> >>>>>Over one cycle of p (and q2), the integral of q2 * p* is a purely >>>>>imaginary number of absolute value distinctly greater than zero. >>>>>hence, >>>>>p is not orthogonal to q2. >>>> >>>> You have written p* again. The value p* was not part of our >>>> discussion. >>>> >>>> I claim that p = e^(j*w_c*t) and q2 = e^(j*(w_c*t + pi/2)) >>>> are orthogonal. That is, over one cycle the integral of the product >>>> of p times q2 is zero. >>>> >>>> Here's my thinking: >>>> Over one cycle we can evaluate the integral of p times q2 as: >>>> >>>> 2pi >>>> -- >>>> / >>>> / e^(jx)e^(jx + pi/2) dx >>>> / >>>> -- >>>> 0 >>>> >>>> 2pi >>>> -- >>>> / >>>> = / e^(jx)e^(jx)e^(pi/2) dx >>>> / >>>> -- >>>> 0 >>>> >>>> 2pi >>>> -- >>>> / >>>> = / je^(jx)e^(jx) dx >>>> / >>>> -- >>>> 0 >>>> >>>> 2pi - - 2pi >>>> -- | | >>>> / | e^(j2x) | >>>> = / je^(j2x) dx = j |----------| >>>> / | j2 | >>>> -- | | 0 - >>>> - 0 >>>> >>>> e^(j4pi) e^(0) >>>> = ---------- - ------- >>>> 2 2 >>>> >>>> 1 1 >>>> = --- - --- = 0. >>>> 2 2 >>>> >>>> >>>> If I've screwed up here, please let me know. >>>> >>>> [-Rick-] >>> >>>I think the problem is your definition of orthogonality. If you start >>>by saying that a signal with nonzero energy absolutely positively cannot >>>be orthogonal with itself, then your definition of orthogonality leads >>>to a contradiction, because the integral over N cycles of p * p is zero. >>> >>>So, I'm doing the usual engineer's stretch, and claiming that >>>orthogonality must be calculated using signal1 times the conjugate of >>>signal2. When used with p, this comes up with the sensible result that >>>the time average of p * p* is the power level of p. >>> >>>If you really insist that your definition of orthogonality is correct >>>then I think we both need to hit the books. >> >> Hi, >> Wow. I don't know why you and I are not >> communicating too well. >> >> I did not say a signal is not orthogonal with itself.Hi Tim,>By your definition of orthogonality, p is orthogonal with itself, because >if you substitute p for q2 in your equation above, the result of the >integration is zero.Ah, yes. Now I see what you mean. I'm merely taking someone's word for the equation defining orthogonality and and plugging functions into that equation to see what result I obtain.> >> I said p was orthogonal to q2. > >Yes, and I checked your math by substituting p in place of q2, and found >it math wanting.I'm not exactly sure what you mean by "math wanting." Tim, did I make an error in my above mathematical integration? This thread has now changed from "mildly interesting" to "very interesting" for me. [-Rick-]
Reply by ●September 7, 20142014-09-07
On Sun, 07 Sep 2014 12:20:55 -0700, Rick Lyons wrote:> On Sat, 06 Sep 2014 17:15:48 -0500, Tim Wescott > <seemywebsite@myfooter.really> wrote: > >>On Sat, 06 Sep 2014 13:20:25 -0700, Rick Lyons wrote: >> >>> On Sat, 06 Sep 2014 12:01:54 -0500, Tim Wescott >>> <seemywebsite@myfooter.really> wrote: >>> >>>>On Sat, 06 Sep 2014 09:36:12 -0700, Rick Lyons wrote: >>>> >>>>> On Fri, 05 Sep 2014 14:35:45 -0500, tim <tim@seemywebsite.com> >>>>> wrote: >>>>> >>>>>>On Fri, 05 Sep 2014 05:11:07 -0700, Rick Lyons wrote: >>>>>> >>>>>>> On Wed, 03 Sep 2014 17:54:38 -0500, Tim Wescott >>>>>>> <seemywebsite@myfooter.really> wrote: >>>>>>> >>>>>>> [Snipped by Lyons] >>>>>>> >>>>>>>>I only just noticed this part of your question. >>>>>>>> >>>>>>>>e^jwt and e^(jwt + pi/2) are not orthogonal, because no signal is >>>>>>>>orthogonal with itself, and e^(jwt + pi/2) is just e^jwt rotated >>>>>>>>by 90 degrees. >>>>>>>> >>>>>>>>In a sense e^jwt BY ITSELF already gives you a carrier for >>>>>>>>transmitting two independent real signals -- asking for more is >>>>>>>>just greedy. >>>>>>> >>>>>>> Hi Tim, >>>>>>> miladsp described two complex signals: >>>>>>> >>>>>>> p = e^(j*w_c*t) and q1 = e^(j*w_c*t + pi/2). >>>>>>> >>>>>>> His q1 is equal to: >>>>>>> >>>>>>> q1 = e^(j*w_c*t)*e^(pi/2) >>>>>>> >>>>>>> where e^(pi/2) = 4.81. His q1 has a magnitude of 4.81. And his p >>>>>>> and q1 = 4.81*p are orthogonal. >>>>>>> >>>>>>> >>>>>>> What I think he MEANT to describe was: >>>>>>> >>>>>>> p = e^(j*w_c*t) and q2 = e^(j*(w_c*t + pi/2)). >>>>>>> >>>>>>> (Notice how q2 does not equal q1. q2 has a magnitude of one.) >>>>>>> >>>>>>> And as far as I can tell, p and q2 are orthogonal. >>>>> >>>>> Hi, >>>>> >>>>>>Over one cycle of p the integral of p and p* is a real, positive >>>>>>number. >>>>>>Hence, p is not orthogonal to itself. >>>>> >>>>> If by p* you mean the conjugate of p, then I agree. But we were not >>>>> talkin' about the orthogonality of p and the conjugate of p. >>>>> >>>>>>Over one cycle of p (and q2), the integral of q2 * p* is a purely >>>>>>imaginary number of absolute value distinctly greater than zero. >>>>>>hence, >>>>>>p is not orthogonal to q2. >>>>> >>>>> You have written p* again. The value p* was not part of our >>>>> discussion. >>>>> >>>>> I claim that p = e^(j*w_c*t) and q2 = e^(j*(w_c*t + pi/2)) >>>>> are orthogonal. That is, over one cycle the integral of the product >>>>> of p times q2 is zero. >>>>> >>>>> Here's my thinking: >>>>> Over one cycle we can evaluate the integral of p times q2 as: >>>>> >>>>> 2pi >>>>> -- >>>>> / >>>>> / e^(jx)e^(jx + pi/2) dx >>>>> / >>>>> -- >>>>> 0 >>>>> >>>>> 2pi >>>>> -- >>>>> / >>>>> = / e^(jx)e^(jx)e^(pi/2) dx >>>>> / >>>>> -- >>>>> 0 >>>>> >>>>> 2pi >>>>> -- >>>>> / >>>>> = / je^(jx)e^(jx) dx >>>>> / >>>>> -- >>>>> 0 >>>>> >>>>> 2pi - - 2pi >>>>> -- | | >>>>> / | e^(j2x) | >>>>> = / je^(j2x) dx = j |----------| >>>>> / | j2 | >>>>> -- | | 0 - >>>>> - 0 >>>>> >>>>> e^(j4pi) e^(0) >>>>> = ---------- - ------- >>>>> 2 2 >>>>> >>>>> 1 1 >>>>> = --- - --- = 0. >>>>> 2 2 >>>>> >>>>> >>>>> If I've screwed up here, please let me know. >>>>> >>>>> [-Rick-] >>>> >>>>I think the problem is your definition of orthogonality. If you start >>>>by saying that a signal with nonzero energy absolutely positively >>>>cannot be orthogonal with itself, then your definition of >>>>orthogonality leads to a contradiction, because the integral over N >>>>cycles of p * p is zero. >>>> >>>>So, I'm doing the usual engineer's stretch, and claiming that >>>>orthogonality must be calculated using signal1 times the conjugate of >>>>signal2. When used with p, this comes up with the sensible result >>>>that the time average of p * p* is the power level of p. >>>> >>>>If you really insist that your definition of orthogonality is correct >>>>then I think we both need to hit the books. >>> >>> Hi, >>> Wow. I don't know why you and I are not >>> communicating too well. >>> >>> I did not say a signal is not orthogonal with itself. > > Hi Tim, > >>By your definition of orthogonality, p is orthogonal with itself, >>because if you substitute p for q2 in your equation above, the result of >>the integration is zero. > > Ah, yes. Now I see what you mean. > > I'm merely taking someone's word for the equation defining orthogonality > and and plugging functions into that equation to see what result I > obtain. > > >>> I said p was orthogonal to q2. >> >>Yes, and I checked your math by substituting p in place of q2, and found >>it math wanting. > > I'm not exactly sure what you mean by "math wanting." > > Tim, did I make an error in my above mathematical integration? > > This thread has now changed from "mildly interesting" to "very > interesting" for me.Actually, I didn't go over what you wrote, because I did the same integration in my head and came up with the same answer. Unfortunately I think my sponge-o-insight has been wrung dry on this one. By "math wanting" I meant the use of the concept of orthogonality to the signal in question. I think, however, that the deficiency is not so much in the definition of orthogonality for complex-valued signals, but in the _use_ of that definition by the OP. -- www.wescottdesign.com
Reply by ●September 7, 20142014-09-07
On Sun, 07 Sep 2014 15:33:41 -0500, tim <tim@seemywebsite.com> wrote:>On Sun, 07 Sep 2014 12:20:55 -0700, Rick Lyons wrote: > >> On Sat, 06 Sep 2014 17:15:48 -0500, Tim Wescott >> <seemywebsite@myfooter.really> wrote: >> >>>On Sat, 06 Sep 2014 13:20:25 -0700, Rick Lyons wrote: >>> >>>> On Sat, 06 Sep 2014 12:01:54 -0500, Tim Wescott >>>> <seemywebsite@myfooter.really> wrote: >>>> >>>>>On Sat, 06 Sep 2014 09:36:12 -0700, Rick Lyons wrote: >>>>> >>>>>> On Fri, 05 Sep 2014 14:35:45 -0500, tim <tim@seemywebsite.com> >>>>>> wrote: >>>>>> >>>>>>>On Fri, 05 Sep 2014 05:11:07 -0700, Rick Lyons wrote: >>>>>>> >>>>>>>> On Wed, 03 Sep 2014 17:54:38 -0500, Tim Wescott >>>>>>>> <seemywebsite@myfooter.really> wrote: >>>>>>>> >>>>>>>> [Snipped by Lyons] >>>>>>>> >>>>>>>>>I only just noticed this part of your question. >>>>>>>>> >>>>>>>>>e^jwt and e^(jwt + pi/2) are not orthogonal, because no signal is >>>>>>>>>orthogonal with itself, and e^(jwt + pi/2) is just e^jwt rotated >>>>>>>>>by 90 degrees. >>>>>>>>> >>>>>>>>>In a sense e^jwt BY ITSELF already gives you a carrier for >>>>>>>>>transmitting two independent real signals -- asking for more is >>>>>>>>>just greedy. >>>>>>>> >>>>>>>> Hi Tim, >>>>>>>> miladsp described two complex signals: >>>>>>>> >>>>>>>> p = e^(j*w_c*t) and q1 = e^(j*w_c*t + pi/2). >>>>>>>> >>>>>>>> His q1 is equal to: >>>>>>>> >>>>>>>> q1 = e^(j*w_c*t)*e^(pi/2) >>>>>>>> >>>>>>>> where e^(pi/2) = 4.81. His q1 has a magnitude of 4.81. And his p >>>>>>>> and q1 = 4.81*p are orthogonal. >>>>>>>> >>>>>>>> >>>>>>>> What I think he MEANT to describe was: >>>>>>>> >>>>>>>> p = e^(j*w_c*t) and q2 = e^(j*(w_c*t + pi/2)). >>>>>>>> >>>>>>>> (Notice how q2 does not equal q1. q2 has a magnitude of one.) >>>>>>>> >>>>>>>> And as far as I can tell, p and q2 are orthogonal. >>>>>> >>>>>> Hi, >>>>>> >>>>>>>Over one cycle of p the integral of p and p* is a real, positive >>>>>>>number. >>>>>>>Hence, p is not orthogonal to itself. >>>>>> >>>>>> If by p* you mean the conjugate of p, then I agree. But we were not >>>>>> talkin' about the orthogonality of p and the conjugate of p. >>>>>> >>>>>>>Over one cycle of p (and q2), the integral of q2 * p* is a purely >>>>>>>imaginary number of absolute value distinctly greater than zero. >>>>>>>hence, >>>>>>>p is not orthogonal to q2. >>>>>> >>>>>> You have written p* again. The value p* was not part of our >>>>>> discussion. >>>>>> >>>>>> I claim that p = e^(j*w_c*t) and q2 = e^(j*(w_c*t + pi/2)) >>>>>> are orthogonal. That is, over one cycle the integral of the product >>>>>> of p times q2 is zero. >>>>>> >>>>>> Here's my thinking: >>>>>> Over one cycle we can evaluate the integral of p times q2 as: >>>>>> >>>>>> 2pi >>>>>> -- >>>>>> / >>>>>> / e^(jx)e^(jx + pi/2) dx >>>>>> / >>>>>> -- >>>>>> 0 >>>>>> >>>>>> 2pi >>>>>> -- >>>>>> / >>>>>> = / e^(jx)e^(jx)e^(pi/2) dx >>>>>> / >>>>>> -- >>>>>> 0 >>>>>> >>>>>> 2pi >>>>>> -- >>>>>> / >>>>>> = / je^(jx)e^(jx) dx >>>>>> / >>>>>> -- >>>>>> 0 >>>>>> >>>>>> 2pi - - 2pi >>>>>> -- | | >>>>>> / | e^(j2x) | >>>>>> = / je^(j2x) dx = j |----------| >>>>>> / | j2 | >>>>>> -- | | 0 - >>>>>> - 0 >>>>>> >>>>>> e^(j4pi) e^(0) >>>>>> = ---------- - ------- >>>>>> 2 2 >>>>>> >>>>>> 1 1 >>>>>> = --- - --- = 0. >>>>>> 2 2 >>>>>> >>>>>> >>>>>> If I've screwed up here, please let me know. >>>>>> >>>>>> [-Rick-] >>>>> >>>>>I think the problem is your definition of orthogonality. If you start >>>>>by saying that a signal with nonzero energy absolutely positively >>>>>cannot be orthogonal with itself, then your definition of >>>>>orthogonality leads to a contradiction, because the integral over N >>>>>cycles of p * p is zero. >>>>> >>>>>So, I'm doing the usual engineer's stretch, and claiming that >>>>>orthogonality must be calculated using signal1 times the conjugate of >>>>>signal2. When used with p, this comes up with the sensible result >>>>>that the time average of p * p* is the power level of p. >>>>> >>>>>If you really insist that your definition of orthogonality is correct >>>>>then I think we both need to hit the books. >>>> >>>> Hi, >>>> Wow. I don't know why you and I are not >>>> communicating too well. >>>> >>>> I did not say a signal is not orthogonal with itself. >> >> Hi Tim, >> >>>By your definition of orthogonality, p is orthogonal with itself, >>>because if you substitute p for q2 in your equation above, the result of >>>the integration is zero. >> >> Ah, yes. Now I see what you mean. >> >> I'm merely taking someone's word for the equation defining orthogonality >> and and plugging functions into that equation to see what result I >> obtain. >> >> >>>> I said p was orthogonal to q2. >>> >>>Yes, and I checked your math by substituting p in place of q2, and found >>>it math wanting. >> >> I'm not exactly sure what you mean by "math wanting." >> >> Tim, did I make an error in my above mathematical integration? >> >> This thread has now changed from "mildly interesting" to "very >> interesting" for me. > >Actually, I didn't go over what you wrote, because I did the same >integration in my head and came up with the same answer. > >Unfortunately I think my sponge-o-insight has been wrung dry on this one. > >By "math wanting" I meant the use of the concept of orthogonality to the >signal in question. > >I think, however, that the deficiency is not so much in the definition of >orthogonality for complex-valued signals, but in the _use_ of that >definition by the OP.I didn't see anything by the OP to suggest he was using the term improperly. He indicated that he wanted to modulate each of e^jwt and e^(jwt + pi/2) with two, independent, real signals. There's no problem with doing that, on the same channel at the same time, and separating them in a receiver. He can do that because the two real signals will remain orthogonal to each other as long as the pi/2 shift is maintained between the carriers. Eric Jacobsen Anchor Hill Communications http://www.anchorhill.com
Reply by ●September 7, 20142014-09-07
On Sun, 07 Sep 2014 15:33:41 -0500, tim <tim@seemywebsite.com> wrote:>On Sun, 07 Sep 2014 12:20:55 -0700, Rick Lyons wrote: >[Snipped by Lyons]>> I'm not exactly sure what you mean by "math wanting." >> >> Tim, did I make an error in my above mathematical integration? >> >> This thread has now changed from "mildly interesting" to "very >> interesting" for me. > >Actually, I didn't go over what you wrote, because I did the same >integration in my head and came up with the same answer. > >Unfortunately I think my sponge-o-insight has been wrung dry on this one. > >By "math wanting" I meant the use of the concept of orthogonality to the >signal in question. > >I think, however, that the deficiency is not so much in the definition of >orthogonality for complex-valued signals, but in the _use_ of that >definition by the OP.Hi, Well, I'm just about burned out on this topic. But I must say, today I spent a fair amount of time searching the web for useable definitions for the inner product equations defining orthogonality. I found two versions, one version that does not use conjugatation (the one I used) and one version that uses conjugation. [-Rick-]
Reply by ●September 7, 20142014-09-07
Eric Jacobsen <eric.jacobsen@ieee.org> wrote:> On Sun, 07 Sep 2014 15:33:41 -0500, tim <tim@seemywebsite.com> wrote:(snip)>>By "math wanting" I meant the use of the concept of orthogonality to the >>signal in question.>>I think, however, that the deficiency is not so much in the >> definition of orthogonality for complex-valued signals, but in >> the _use_ of that definition by the OP.> I didn't see anything by the OP to suggest he was using the term > improperly. He indicated that he wanted to modulate each of e^jwt > and e^(jwt + pi/2) with two, independent, real signals. There's no > problem with doing that, on the same channel at the same time, and > separating them in a receiver. He can do that because the two real > signals will remain orthogonal to each other as long as the pi/2 shift > is maintained between the carriers.Not being completely sure what you are calling e^(jwt), for one I will again note the need for a reference when decoding. The common form used to describe circularly polarized light is exp(jwt) and exp(-jwt). Conveniently in that case, there isn't a problem of a reference phase. -- glen
Reply by ●September 7, 20142014-09-07
On Sun, 07 Sep 2014 23:10:59 +0000, Eric Jacobsen wrote:> On Sun, 07 Sep 2014 15:33:41 -0500, tim <tim@seemywebsite.com> wrote: > >>On Sun, 07 Sep 2014 12:20:55 -0700, Rick Lyons wrote: >> >>> On Sat, 06 Sep 2014 17:15:48 -0500, Tim Wescott >>> <seemywebsite@myfooter.really> wrote: >>> >>>>On Sat, 06 Sep 2014 13:20:25 -0700, Rick Lyons wrote: >>>> >>>>> On Sat, 06 Sep 2014 12:01:54 -0500, Tim Wescott >>>>> <seemywebsite@myfooter.really> wrote: >>>>> >>>>>>On Sat, 06 Sep 2014 09:36:12 -0700, Rick Lyons wrote: >>>>>> >>>>>>> On Fri, 05 Sep 2014 14:35:45 -0500, tim <tim@seemywebsite.com> >>>>>>> wrote: >>>>>>> >>>>>>>>On Fri, 05 Sep 2014 05:11:07 -0700, Rick Lyons wrote: >>>>>>>> >>>>>>>>> On Wed, 03 Sep 2014 17:54:38 -0500, Tim Wescott >>>>>>>>> <seemywebsite@myfooter.really> wrote: >>>>>>>>> >>>>>>>>> [Snipped by Lyons] >>>>>>>>> >>>>>>>>>>I only just noticed this part of your question. >>>>>>>>>> >>>>>>>>>>e^jwt and e^(jwt + pi/2) are not orthogonal, because no signal >>>>>>>>>>is orthogonal with itself, and e^(jwt + pi/2) is just e^jwt >>>>>>>>>>rotated by 90 degrees. >>>>>>>>>> >>>>>>>>>>In a sense e^jwt BY ITSELF already gives you a carrier for >>>>>>>>>>transmitting two independent real signals -- asking for more is >>>>>>>>>>just greedy. >>>>>>>>> >>>>>>>>> Hi Tim, >>>>>>>>> miladsp described two complex signals: >>>>>>>>> >>>>>>>>> p = e^(j*w_c*t) and q1 = e^(j*w_c*t + pi/2). >>>>>>>>> >>>>>>>>> His q1 is equal to: >>>>>>>>> >>>>>>>>> q1 = e^(j*w_c*t)*e^(pi/2) >>>>>>>>> >>>>>>>>> where e^(pi/2) = 4.81. His q1 has a magnitude of 4.81. And his >>>>>>>>> p and q1 = 4.81*p are orthogonal. >>>>>>>>> >>>>>>>>> >>>>>>>>> What I think he MEANT to describe was: >>>>>>>>> >>>>>>>>> p = e^(j*w_c*t) and q2 = e^(j*(w_c*t + pi/2)). >>>>>>>>> >>>>>>>>> (Notice how q2 does not equal q1. q2 has a magnitude of one.) >>>>>>>>> >>>>>>>>> And as far as I can tell, p and q2 are orthogonal. >>>>>>> >>>>>>> Hi, >>>>>>> >>>>>>>>Over one cycle of p the integral of p and p* is a real, positive >>>>>>>>number. >>>>>>>>Hence, p is not orthogonal to itself. >>>>>>> >>>>>>> If by p* you mean the conjugate of p, then I agree. But we were >>>>>>> not talkin' about the orthogonality of p and the conjugate of p. >>>>>>> >>>>>>>>Over one cycle of p (and q2), the integral of q2 * p* is a purely >>>>>>>>imaginary number of absolute value distinctly greater than zero. >>>>>>>>hence, >>>>>>>>p is not orthogonal to q2. >>>>>>> >>>>>>> You have written p* again. The value p* was not part of our >>>>>>> discussion. >>>>>>> >>>>>>> I claim that p = e^(j*w_c*t) and q2 = e^(j*(w_c*t + pi/2)) >>>>>>> are orthogonal. That is, over one cycle the integral of the >>>>>>> product of p times q2 is zero. >>>>>>> >>>>>>> Here's my thinking: >>>>>>> Over one cycle we can evaluate the integral of p times q2 as: >>>>>>> >>>>>>> 2pi >>>>>>> -- >>>>>>> / >>>>>>> / e^(jx)e^(jx + pi/2) dx >>>>>>> / >>>>>>> -- >>>>>>> 0 >>>>>>> >>>>>>> 2pi >>>>>>> -- >>>>>>> / >>>>>>> = / e^(jx)e^(jx)e^(pi/2) dx >>>>>>> / >>>>>>> -- >>>>>>> 0 >>>>>>> >>>>>>> 2pi >>>>>>> -- >>>>>>> / >>>>>>> = / je^(jx)e^(jx) dx >>>>>>> / >>>>>>> -- >>>>>>> 0 >>>>>>> >>>>>>> 2pi - - 2pi >>>>>>> -- | | >>>>>>> / | e^(j2x) | >>>>>>> = / je^(j2x) dx = j |----------| >>>>>>> / | j2 | >>>>>>> -- | | 0 >>>>>>> - >>>>>>> - 0 >>>>>>> >>>>>>> e^(j4pi) e^(0) >>>>>>> = ---------- - ------- >>>>>>> 2 2 >>>>>>> >>>>>>> 1 1 >>>>>>> = --- - --- = 0. >>>>>>> 2 2 >>>>>>> >>>>>>> >>>>>>> If I've screwed up here, please let me know. >>>>>>> >>>>>>> [-Rick-] >>>>>> >>>>>>I think the problem is your definition of orthogonality. If you >>>>>>start by saying that a signal with nonzero energy absolutely >>>>>>positively cannot be orthogonal with itself, then your definition of >>>>>>orthogonality leads to a contradiction, because the integral over N >>>>>>cycles of p * p is zero. >>>>>> >>>>>>So, I'm doing the usual engineer's stretch, and claiming that >>>>>>orthogonality must be calculated using signal1 times the conjugate >>>>>>of signal2. When used with p, this comes up with the sensible >>>>>>result that the time average of p * p* is the power level of p. >>>>>> >>>>>>If you really insist that your definition of orthogonality is >>>>>>correct then I think we both need to hit the books. >>>>> >>>>> Hi, >>>>> Wow. I don't know why you and I are not >>>>> communicating too well. >>>>> >>>>> I did not say a signal is not orthogonal with itself. >>> >>> Hi Tim, >>> >>>>By your definition of orthogonality, p is orthogonal with itself, >>>>because if you substitute p for q2 in your equation above, the result >>>>of the integration is zero. >>> >>> Ah, yes. Now I see what you mean. >>> >>> I'm merely taking someone's word for the equation defining >>> orthogonality and and plugging functions into that equation to see >>> what result I obtain. >>> >>> >>>>> I said p was orthogonal to q2. >>>> >>>>Yes, and I checked your math by substituting p in place of q2, and >>>>found it math wanting. >>> >>> I'm not exactly sure what you mean by "math wanting." >>> >>> Tim, did I make an error in my above mathematical integration? >>> >>> This thread has now changed from "mildly interesting" to "very >>> interesting" for me. >> >>Actually, I didn't go over what you wrote, because I did the same >>integration in my head and came up with the same answer. >> >>Unfortunately I think my sponge-o-insight has been wrung dry on this >>one. >> >>By "math wanting" I meant the use of the concept of orthogonality to the >>signal in question. >> >>I think, however, that the deficiency is not so much in the definition >>of orthogonality for complex-valued signals, but in the _use_ of that >>definition by the OP. > > I didn't see anything by the OP to suggest he was using the term > improperly. He indicated that he wanted to modulate each of e^jwt and > e^(jwt + pi/2) with two, independent, real signals. There's no problem > with doing that, on the same channel at the same time, and separating > them in a receiver. He can do that because the two real signals will > remain orthogonal to each other as long as the pi/2 shift is maintained > between the carriers.The OP said "independent signals". Not "independent real signals". Yes, AS I POINTED OUT, those two carriers can be used to transmit two independent _real_ signals. -- www.wescottdesign.com
Reply by ●September 8, 20142014-09-08
On Sun, 07 Sep 2014 21:16:54 -0500, tim <tim@seemywebsite.com> wrote:>On Sun, 07 Sep 2014 23:10:59 +0000, Eric Jacobsen wrote: > >> On Sun, 07 Sep 2014 15:33:41 -0500, tim <tim@seemywebsite.com> wrote: >> >>>On Sun, 07 Sep 2014 12:20:55 -0700, Rick Lyons wrote: >>> >>>> On Sat, 06 Sep 2014 17:15:48 -0500, Tim Wescott >>>> <seemywebsite@myfooter.really> wrote: >>>> >>>>>On Sat, 06 Sep 2014 13:20:25 -0700, Rick Lyons wrote: >>>>> >>>>>> On Sat, 06 Sep 2014 12:01:54 -0500, Tim Wescott >>>>>> <seemywebsite@myfooter.really> wrote: >>>>>> >>>>>>>On Sat, 06 Sep 2014 09:36:12 -0700, Rick Lyons wrote: >>>>>>> >>>>>>>> On Fri, 05 Sep 2014 14:35:45 -0500, tim <tim@seemywebsite.com> >>>>>>>> wrote: >>>>>>>> >>>>>>>>>On Fri, 05 Sep 2014 05:11:07 -0700, Rick Lyons wrote: >>>>>>>>> >>>>>>>>>> On Wed, 03 Sep 2014 17:54:38 -0500, Tim Wescott >>>>>>>>>> <seemywebsite@myfooter.really> wrote: >>>>>>>>>> >>>>>>>>>> [Snipped by Lyons] >>>>>>>>>> >>>>>>>>>>>I only just noticed this part of your question. >>>>>>>>>>> >>>>>>>>>>>e^jwt and e^(jwt + pi/2) are not orthogonal, because no signal >>>>>>>>>>>is orthogonal with itself, and e^(jwt + pi/2) is just e^jwt >>>>>>>>>>>rotated by 90 degrees. >>>>>>>>>>> >>>>>>>>>>>In a sense e^jwt BY ITSELF already gives you a carrier for >>>>>>>>>>>transmitting two independent real signals -- asking for more is >>>>>>>>>>>just greedy. >>>>>>>>>> >>>>>>>>>> Hi Tim, >>>>>>>>>> miladsp described two complex signals: >>>>>>>>>> >>>>>>>>>> p = e^(j*w_c*t) and q1 = e^(j*w_c*t + pi/2). >>>>>>>>>> >>>>>>>>>> His q1 is equal to: >>>>>>>>>> >>>>>>>>>> q1 = e^(j*w_c*t)*e^(pi/2) >>>>>>>>>> >>>>>>>>>> where e^(pi/2) = 4.81. His q1 has a magnitude of 4.81. And his >>>>>>>>>> p and q1 = 4.81*p are orthogonal. >>>>>>>>>> >>>>>>>>>> >>>>>>>>>> What I think he MEANT to describe was: >>>>>>>>>> >>>>>>>>>> p = e^(j*w_c*t) and q2 = e^(j*(w_c*t + pi/2)). >>>>>>>>>> >>>>>>>>>> (Notice how q2 does not equal q1. q2 has a magnitude of one.) >>>>>>>>>> >>>>>>>>>> And as far as I can tell, p and q2 are orthogonal. >>>>>>>> >>>>>>>> Hi, >>>>>>>> >>>>>>>>>Over one cycle of p the integral of p and p* is a real, positive >>>>>>>>>number. >>>>>>>>>Hence, p is not orthogonal to itself. >>>>>>>> >>>>>>>> If by p* you mean the conjugate of p, then I agree. But we were >>>>>>>> not talkin' about the orthogonality of p and the conjugate of p. >>>>>>>> >>>>>>>>>Over one cycle of p (and q2), the integral of q2 * p* is a purely >>>>>>>>>imaginary number of absolute value distinctly greater than zero. >>>>>>>>>hence, >>>>>>>>>p is not orthogonal to q2. >>>>>>>> >>>>>>>> You have written p* again. The value p* was not part of our >>>>>>>> discussion. >>>>>>>> >>>>>>>> I claim that p = e^(j*w_c*t) and q2 = e^(j*(w_c*t + pi/2)) >>>>>>>> are orthogonal. That is, over one cycle the integral of the >>>>>>>> product of p times q2 is zero. >>>>>>>> >>>>>>>> Here's my thinking: >>>>>>>> Over one cycle we can evaluate the integral of p times q2 as: >>>>>>>> >>>>>>>> 2pi >>>>>>>> -- >>>>>>>> / >>>>>>>> / e^(jx)e^(jx + pi/2) dx >>>>>>>> / >>>>>>>> -- >>>>>>>> 0 >>>>>>>> >>>>>>>> 2pi >>>>>>>> -- >>>>>>>> / >>>>>>>> = / e^(jx)e^(jx)e^(pi/2) dx >>>>>>>> / >>>>>>>> -- >>>>>>>> 0 >>>>>>>> >>>>>>>> 2pi >>>>>>>> -- >>>>>>>> / >>>>>>>> = / je^(jx)e^(jx) dx >>>>>>>> / >>>>>>>> -- >>>>>>>> 0 >>>>>>>> >>>>>>>> 2pi - - 2pi >>>>>>>> -- | | >>>>>>>> / | e^(j2x) | >>>>>>>> = / je^(j2x) dx = j |----------| >>>>>>>> / | j2 | >>>>>>>> -- | | 0 >>>>>>>> - >>>>>>>> - 0 >>>>>>>> >>>>>>>> e^(j4pi) e^(0) >>>>>>>> = ---------- - ------- >>>>>>>> 2 2 >>>>>>>> >>>>>>>> 1 1 >>>>>>>> = --- - --- = 0. >>>>>>>> 2 2 >>>>>>>> >>>>>>>> >>>>>>>> If I've screwed up here, please let me know. >>>>>>>> >>>>>>>> [-Rick-] >>>>>>> >>>>>>>I think the problem is your definition of orthogonality. If you >>>>>>>start by saying that a signal with nonzero energy absolutely >>>>>>>positively cannot be orthogonal with itself, then your definition of >>>>>>>orthogonality leads to a contradiction, because the integral over N >>>>>>>cycles of p * p is zero. >>>>>>> >>>>>>>So, I'm doing the usual engineer's stretch, and claiming that >>>>>>>orthogonality must be calculated using signal1 times the conjugate >>>>>>>of signal2. When used with p, this comes up with the sensible >>>>>>>result that the time average of p * p* is the power level of p. >>>>>>> >>>>>>>If you really insist that your definition of orthogonality is >>>>>>>correct then I think we both need to hit the books. >>>>>> >>>>>> Hi, >>>>>> Wow. I don't know why you and I are not >>>>>> communicating too well. >>>>>> >>>>>> I did not say a signal is not orthogonal with itself. >>>> >>>> Hi Tim, >>>> >>>>>By your definition of orthogonality, p is orthogonal with itself, >>>>>because if you substitute p for q2 in your equation above, the result >>>>>of the integration is zero. >>>> >>>> Ah, yes. Now I see what you mean. >>>> >>>> I'm merely taking someone's word for the equation defining >>>> orthogonality and and plugging functions into that equation to see >>>> what result I obtain. >>>> >>>> >>>>>> I said p was orthogonal to q2. >>>>> >>>>>Yes, and I checked your math by substituting p in place of q2, and >>>>>found it math wanting. >>>> >>>> I'm not exactly sure what you mean by "math wanting." >>>> >>>> Tim, did I make an error in my above mathematical integration? >>>> >>>> This thread has now changed from "mildly interesting" to "very >>>> interesting" for me. >>> >>>Actually, I didn't go over what you wrote, because I did the same >>>integration in my head and came up with the same answer. >>> >>>Unfortunately I think my sponge-o-insight has been wrung dry on this >>>one. >>> >>>By "math wanting" I meant the use of the concept of orthogonality to the >>>signal in question. >>> >>>I think, however, that the deficiency is not so much in the definition >>>of orthogonality for complex-valued signals, but in the _use_ of that >>>definition by the OP. >> >> I didn't see anything by the OP to suggest he was using the term >> improperly. He indicated that he wanted to modulate each of e^jwt and >> e^(jwt + pi/2) with two, independent, real signals. There's no problem >> with doing that, on the same channel at the same time, and separating >> them in a receiver. He can do that because the two real signals will >> remain orthogonal to each other as long as the pi/2 shift is maintained >> between the carriers. > >The OP said "independent signals". Not "independent real signals".On Sep 2 the OP clarified: Hi Rick, Ok, let's say we're in Matlab world and I have two independent **real-value** baseband signals. If I multiply one by e^(j*w_c*t) and the other one by e^(j*w_c*t + pi/2) and add the results, can I still retrieve the two original messages from the this result? As Tim said, I'm trying to figure out if I can do all my quadrature modulation math without paying attention to the negative frequency part or not.>Yes, AS I POINTED OUT, those two carriers can be used to transmit two >independent _real_ signals.Not sure why all the diversion.>-- >www.wescottdesign.comEric Jacobsen Anchor Hill Communications http://www.anchorhill.com
Reply by ●September 8, 20142014-09-08
On Sun, 07 Sep 2014 16:43:36 -0700, Rick Lyons <R.Lyons@_BOGUS_ieee.org> wrote:>On Sun, 07 Sep 2014 15:33:41 -0500, tim <tim@seemywebsite.com> wrote: > >>On Sun, 07 Sep 2014 12:20:55 -0700, Rick Lyons wrote: >> > > [Snipped by Lyons] > >>> I'm not exactly sure what you mean by "math wanting." >>> >>> Tim, did I make an error in my above mathematical integration? >>> >>> This thread has now changed from "mildly interesting" to "very >>> interesting" for me. >> >>Actually, I didn't go over what you wrote, because I did the same >>integration in my head and came up with the same answer. >> >>Unfortunately I think my sponge-o-insight has been wrung dry on this one. >> >>By "math wanting" I meant the use of the concept of orthogonality to the >>signal in question. >> >>I think, however, that the deficiency is not so much in the definition of >>orthogonality for complex-valued signals, but in the _use_ of that >>definition by the OP. > >Hi, > Well, I'm just about burned out on this topic. > >But I must say, today I spent a fair amount of time >searching the web for useable definitions for the >inner product equations defining orthogonality. >I found two versions, one version that does not use >conjugatation (the one I used) and one version that >uses conjugation. > >[-Rick-]There is a fairly easy way to think about this that might help: There's a relationship between cross-correlation (which is a dot-product, aka inner-product), and orthogonality. If two vectors are orthogonal to each other, their cross-correlation is zero. If two vectors are identical, their cross-correlation is one. That's pretty much the basis for a lot of things, like using orthogonal vectors for spreading for CDMA and things like that. Remember that when doing cross-correlation of vectors of complex values, conjugation is required in one argument so that the phases will align and the output magnitude will be maximized if the vectors are the same. That can't be assured if one argument isn't conjugated. So I think for an orthogonality test conjugation of one argument would work. Eric Jacobsen Anchor Hill Communications http://www.anchorhill.com
