sinusoidal signals using complex conjugate exponentials

Started by June 30, 2013
```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
```
```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.com

Google 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

```
```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-]
```
```>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,

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
```
```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-]
```
```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!
```
```sentiments shared (to some extent) but i dunno if i wanna so *two*

> Sheeeesh!

--

r b-j                  rbj@audioimagination.com

"Imagination is more important than knowledge."

```
```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....
```
```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
```
```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,