Sirs, I would like to get a clarification on expressing cosine using complex exponentials ... It is given in texts that A * cos (pi) = A/2 * exp pow j * A/2 * pi + exp pow -(j * pi) When I plot the above two complex exponentials, I get 2 points on the complex plane but how these correspond to a single point of cosine signal, I am unable to get. And, how thesd eventually give a cosine signal? Can someone help me understand ... Thanks, manish _____________________________ Posted through www.DSPRelated.com
sinusoidal signals using complex conjugate exponentials
Started by ●June 30, 2013
Reply by ●June 30, 20132013-06-30
On 30.6.13 6:28 , manishp wrote:> Sirs, > > I would like to get a clarification on expressing cosine using complex > exponentials ... > > It is given in texts that A * cos (pi) = A/2 * exp pow j * A/2 * pi + exp > pow -(j * pi) > > When I plot the above two complex exponentials, I get 2 points on the > complex plane but how these correspond to a single point of cosine signal, > I am unable to get. And, how thesd eventually give a cosine signal? > > Can someone help me understand ... > > Thanks, manish > > _____________________________ > Posted through www.DSPRelated.comGoogle for 'Euler's formula', put it into your expression and clean up. If you cannot do ti, you're in a wrong place here. -- Tauno Voipio
Reply by ●June 30, 20132013-06-30
On Sat, 29 Jun 2013 22:28:31 -0500, "manishp" <58525@dsprelated> wrote:>Sirs, > >I would like to get a clarification on expressing cosine using complex >exponentials ... > >It is given in texts that A * cos (pi) = A/2 * exp pow j * A/2 * pi + exp >pow -(j * pi) > >When I plot the above two complex exponentials, I get 2 points on the >complex plane but how these correspond to a single point of cosine signal, >I am unable to get. And, how thesd eventually give a cosine signal? > >Can someone help me understand ... > >Thanks, manishHello manish, I suggest you visit: http://www.dsprelated.com/showarticle/192.php and focus on the section titled: "Representing Real Signals Using Complex Phasors" Good Luck, [-Rick-]
Reply by ●July 1, 20132013-07-01
>On Sat, 29 Jun 2013 22:28:31 -0500, "manishp" <58525@dsprelated> >wrote: > >>Sirs, >> >>I would like to get a clarification on expressing cosine using complex >>exponentials ... >> >>It is given in texts that A * cos (pi) = A/2 * exp pow j * A/2 * pi +exp>>pow -(j * pi) >> >>When I plot the above two complex exponentials, I get 2 points on the >>complex plane but how these correspond to a single point of cosinesignal,>>I am unable to get. And, how thesd eventually give a cosine signal? >> >>Can someone help me understand ... >> >>Thanks, manish > >Hello manish, > I suggest you visit: > >http://www.dsprelated.com/showarticle/192.php > >and focus on the section titled: "Representing Real >Signals Using Complex Phasors" > >Good Luck, >[-Rick-]Dear Rick, Thanks. My main confusion was imagining how addition of two phasors (e pow (j*pi*t) and e pow -(j*pi*t)) can result in a cosine signal. From an equation perspective, I think this makes perfect sense. That is, cos (pi*t) = (e pow (j*pi*t) + e pow -(j*pi*t) = cos (pi*t) + jsin (pi*t) + cos (pi*t) - jsin (pi*t) The figure 6 in your article clarifies this or at least I know the direction to take ... Thanks, manish _____________________________ Posted through www.DSPRelated.com
Reply by ●July 1, 20132013-07-01
On Mon, 01 Jul 2013 03:32:43 -0500, "manishp" <58525@dsprelated> wrote:>>On Sat, 29 Jun 2013 22:28:31 -0500, "manishp" <58525@dsprelated> >>wrote: >> >>>Sirs, >>> >>>I would like to get a clarification on expressing cosine using complex >>>exponentials ... >>> >>>It is given in texts that A * cos (pi) = A/2 * exp pow j * A/2 * pi + >exp >>>pow -(j * pi) >>> >>>When I plot the above two complex exponentials, I get 2 points on the >>>complex plane but how these correspond to a single point of cosine >signal, >>>I am unable to get. And, how thesd eventually give a cosine signal? >>> >>>Can someone help me understand ... >>> >>>Thanks, manish >> >>Hello manish, >> I suggest you visit: >> >>http://www.dsprelated.com/showarticle/192.php >> >>and focus on the section titled: "Representing Real >>Signals Using Complex Phasors" >> >>Good Luck, >>[-Rick-] > >Dear Rick,Hello Manish,> >Thanks. My main confusion was imagining how addition of two phasors (e pow >(j*pi*t) and e pow -(j*pi*t)) can result in a cosine signal. > >From an equation perspective, I think this makes perfect sense. >That is, cos (pi*t) = (e pow (j*pi*t) + e pow -(j*pi*t) > = cos (pi*t) + jsin (pi*t) + cos (pi*t) - jsin (pi*t)Oops! You forgot the factors of 1/2. The above equations should be: cos(pi*t) = [e pow(j*pi*t)]/2 + [e pow(-j*pi*t)]/2 = 0.5cos(pi*t) + j0.5sin(pi*t) + 0.5cos(pi*t) -0.5jsin (pi*t) = 0.5cos(pi*t) + 0.5cos(pi*t) = cos(pi*t). [-Rick-]
Reply by ●July 1, 20132013-07-01
On Saturday, June 29, 2013 10:28:31 PM UTC-5, manishp wrote:> Sirs, > > > > I would like to get a clarification on expressing cosine using complex > > exponentials ... > > > > It is given in texts that A * cos (pi) = A/2 * exp pow j * A/2 * pi + exp > > pow -(j * pi) > > > > When I plot the above two complex exponentials, I get 2 points on the > > complex plane but how these correspond to a single point of cosine signal, > > I am unable to get. And, how thesd eventually give a cosine signal? > > > > Can someone help me understand ... >Your questions get sillier each time you post a question on comp.dsp or on dsprelated.com. In **which text** did you read that A * cos (pi) = A/2 * exp pow j * A/2 * pi + exp pow -(j * pi)?? Do you even _bother_ to proof-read what you type? Do you know what the value of cos (pi) is, or do you know what buttons to press on your calculator to compute the value of cos (pi)? Do you ever bother to put brain in gear before putting fingers to keyboard? Sheeeesh!
Reply by ●July 1, 20132013-07-01
sentiments shared (to some extent) but i dunno if i wanna so *two* Vlad's posting here.> Sheeeesh!-- r b-j rbj@audioimagination.com "Imagination is more important than knowledge."
Reply by ●July 2, 20132013-07-02
On Monday, July 1, 2013 11:54:06 AM UTC-5, robert bristow-johnson quoted:> > Sheeeesh!Having been a faculty member for my entire professional life (and a son of a faculty member too), I have lived from childhood onwards with the firm conviction that there is no such thing as a stupid question. Now in my golden years, after encountering manishp's questions, I have begun to rethink the issue....
Reply by ●July 2, 20132013-07-02
dvsarwate <dvsarwate@yahoo.com> writes:> On Monday, July 1, 2013 11:54:06 AM UTC-5, robert bristow-johnson quoted: > >> > Sheeeesh! > > Having been a faculty member for my entire professional > life (and a son of a faculty member too), I have lived > from childhood onwards with the firm conviction that > there is no such thing as a stupid question. Now in my > golden years, after encountering manishp's questions, > I have begun to rethink the issue....Have you read Rony's? No, don't - you'll get a cardiac. -- Randy Yates Digital Signal Labs http://www.digitalsignallabs.com
Reply by ●July 2, 20132013-07-02
On Mon, 1 Jul 2013 20:33:01 -0700 (PDT), dvsarwate <dvsarwate@yahoo.com> wrote:>On Monday, July 1, 2013 11:54:06 AM UTC-5, robert bristow-johnson quoted: > >> > Sheeeesh! > >Having been a faculty member for my entire professional >life (and a son of a faculty member too), I have lived >from childhood onwards with the firm conviction that >there is no such thing as a stupid question. Now in my >golden years, after encountering manishp's questions, >I have begun to rethink the issue....Hi Dilip, Ha ha. Your post is interesting. Occasionally I teach a 3-day DSP class. And at the beginning of the class, in order to encourage questions from my students, I tell them something like, "There are no dumb questions regarding DSP, so please don't hesitate to ask questions." But your post, and some of the questions asked here, may well prove me wrong. Ha ha. [-Rick-]
