> >> Hi Rick, >> >> Ok, let's say we're in Matlab world and I have two independentreal-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 partor>> >> not. >> >> I'll relate them to figures in your book once I get home. >> >Maybe this will help you: >You have to real signals (with equal bandwidth, BW), x1(t) and x2(t).Combine them to a complex signal, x12(t) = x1(t) + j*x2(t)>Translate this signal, using a mixer: x12t(t) = x12(t) * exp(j*2*pi*f*t) >This signal has no negative frequencies if f > BW, and you can take thereal part of this signal like:>real(x12t(t)) = x1(t) * cos(2*pi*f*t) - x2(t) * sin(2*pi*f*t) > >To avoid aliasing the sampling rate, Fs, must be: Fs > 2 * (BW+f) >--- >ww >Ooops. I was making a silly mistake. Everything is suddenly making sense. Thanks everyone _____________________________ Posted through www.DSPRelated.com
What makes quadrature mixing possible?
Started by ●August 31, 2014
Reply by ●September 3, 20142014-09-03
Reply by ●September 3, 20142014-09-03
On Tue, 02 Sep 2014 15:02:59 -0500, miladsp wrote:>>On Sun, 31 Aug 2014 17:33:39 -0500, "miladsp" <99479@dsprelated> >>wrote: >> >>>By the way, almost all articles on the subject have references to > "Complex >>>Signals" series by "N. Boutin" in "RF Design" but I can't find any >>>links > to >>>them. Does any one know where they can be found? >> >>Hello miladsp, >> I couldn't find an electronic copy of >>Noel Boutin's "Complex Signals" article on the web. (Actually, it was a >>series of four articles.) That's too bad. >> >>It appears that Prof. Boutin is still with the University of Sherbrooke >>in Quebec. >> >>You might send him an e-mail and very respectfully ask if he is willing >>to send you an electronic copy of his four-part article. (If you are >>successful miladsp, PLEASE LET US KNOW, OK?) >> >>See: >> >>http://www.usherbrooke.ca/ssf/documentation/grandes-entrevues/noel-boutin/>> >> >>Good Luck, >>[-Rick-] >> >> > Hi Rick! I'm waiting for a reply from him. I'll let you guys know if I > get a copy. > > Now some further questions about quadrature modulation. First let's > assume our physical channel consists of two coax cables (just like the > example in your book) which means we can physically transmit complex > signals. Now, am I right in thinking that two complex signals (e^jwt) > and (e^jwt+pi/2) are orthogonal?? If they are, can we use them as > carriers for transmitting two independent signals??<< snip >> 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. -- Tim Wescott Wescott Design Services http://www.wescottdesign.com
Reply by ●September 3, 20142014-09-03
On Wed, 03 Sep 2014 17:54:38 -0500, Tim Wescott <seemywebsite@myfooter.really> wrote:>On Tue, 02 Sep 2014 15:02:59 -0500, miladsp wrote: > >>>On Sun, 31 Aug 2014 17:33:39 -0500, "miladsp" <99479@dsprelated> >>>wrote: >>> >>>>By the way, almost all articles on the subject have references to >> "Complex >>>>Signals" series by "N. Boutin" in "RF Design" but I can't find any >>>>links >> to >>>>them. Does any one know where they can be found? >>> >>>Hello miladsp, >>> I couldn't find an electronic copy of >>>Noel Boutin's "Complex Signals" article on the web. (Actually, it was a >>>series of four articles.) That's too bad. >>> >>>It appears that Prof. Boutin is still with the University of Sherbrooke >>>in Quebec. >>> >>>You might send him an e-mail and very respectfully ask if he is willing >>>to send you an electronic copy of his four-part article. (If you are >>>successful miladsp, PLEASE LET US KNOW, OK?) >>> >>>See: >>> >>>http://www.usherbrooke.ca/ssf/documentation/grandes-entrevues/noel- >boutin/ >>> >>> >>>Good Luck, >>>[-Rick-] >>> >>> >> Hi Rick! I'm waiting for a reply from him. I'll let you guys know if I >> get a copy. >> >> Now some further questions about quadrature modulation. First let's >> assume our physical channel consists of two coax cables (just like the >> example in your book) which means we can physically transmit complex >> signals. Now, am I right in thinking that two complex signals (e^jwt) >> and (e^jwt+pi/2) are orthogonal?? If they are, can we use them as >> carriers for transmitting two independent signals?? > ><< snip >> > >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.Aren't sin and cos orthogonal? They're just rotated by 90 degrees from each other.>In a sense e^jwt BY ITSELF already gives you a carrier for transmitting >two independent real signals -- asking for more is just greedy. > >-- > >Tim Wescott >Wescott Design Services >http://www.wescottdesign.comEric Jacobsen Anchor Hill Communications http://www.anchorhill.com
Reply by ●September 4, 20142014-09-04
On Wed, 03 Sep 2014 23:50:53 +0000, Eric Jacobsen wrote:> On Wed, 03 Sep 2014 17:54:38 -0500, Tim Wescott > <seemywebsite@myfooter.really> wrote: > >>On Tue, 02 Sep 2014 15:02:59 -0500, miladsp wrote: >> >>>>On Sun, 31 Aug 2014 17:33:39 -0500, "miladsp" <99479@dsprelated> >>>>wrote: >>>> >>>>>By the way, almost all articles on the subject have references to >>> "Complex >>>>>Signals" series by "N. Boutin" in "RF Design" but I can't find any >>>>>links >>> to >>>>>them. Does any one know where they can be found? >>>> >>>>Hello miladsp, >>>> I couldn't find an electronic copy of >>>>Noel Boutin's "Complex Signals" article on the web. (Actually, it was >>>>a series of four articles.) That's too bad. >>>> >>>>It appears that Prof. Boutin is still with the University of >>>>Sherbrooke in Quebec. >>>> >>>>You might send him an e-mail and very respectfully ask if he is >>>>willing to send you an electronic copy of his four-part article. (If >>>>you are successful miladsp, PLEASE LET US KNOW, OK?) >>>> >>>>See: >>>> >>>>http://www.usherbrooke.ca/ssf/documentation/grandes-entrevues/noel- >>boutin/ >>>> >>>> >>>>Good Luck, >>>>[-Rick-] >>>> >>>> >>> Hi Rick! I'm waiting for a reply from him. I'll let you guys know if I >>> get a copy. >>> >>> Now some further questions about quadrature modulation. First let's >>> assume our physical channel consists of two coax cables (just like the >>> example in your book) which means we can physically transmit complex >>> signals. Now, am I right in thinking that two complex signals (e^jwt) >>> and (e^jwt+pi/2) are orthogonal?? If they are, can we use them as >>> carriers for transmitting two independent signals?? >> >><< snip >> >> >>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. > > Aren't sin and cos orthogonal? They're just rotated by 90 degrees from > each other.In the sense that I was speaking, sin rotated by 90 degrees is j * sin. -- www.wescottdesign.com
Reply by ●September 5, 20142014-09-05
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. [-Rick-]
Reply by ●September 5, 20142014-09-05
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.Over one cycle of p the integral of p and p* is a real, positive number. Hence, p is not orthogonal to itself. 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. -- www.wescottdesign.com
Reply by ●September 6, 20142014-09-06
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-]
Reply by ●September 6, 20142014-09-06
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. -- Tim Wescott Wescott Design Services http://www.wescottdesign.com
Reply by ●September 6, 20142014-09-06
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. I said p was orthogonal to q2. 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. [-Rick-]
Reply by ●September 6, 20142014-09-06
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 original question was, if you have two complex carriers, e^jwt and e^j(wt + pi/2), can you transmit two independent signals? 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. 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. -- Tim Wescott Wescott Design Services http://www.wescottdesign.com
