# complex exponential signals

Discussion in 'Homework Help' started by bhuvanesh, Mar 2, 2015.

1. ### bhuvanesh Thread Starter Member

Aug 10, 2013
268
2
what makes complex exponential special.its just an addition of cosine and sine with imaginary amplitude ,isn't it?.In my book the waveform for real exponential is given and for complex exponential signal the graph given is just sinusoidal signal .so i cannot differentiate complex exponential and sinusoidal signal.Thank you in advance

2. ### WBahn Moderator

Mar 31, 2012
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What makes them special is that they make the math a LOT easier.

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3. ### StayatHomeElectronics Well-Known Member

Sep 25, 2008
864
40
The equation provides a relationship between exponential and sinusoidal signals. The real and complex portions are just sinusoidal signals; however, both portions are occurring at the same time.

Try plotting for different x on the real and imaginary plane. Does the plot still look sinusoidal?

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4. ### bhuvanesh Thread Starter Member

Aug 10, 2013
268
2
actually i don't know the purpose of j(i).Does cos(x)+ sin(x) and cos(x)+jsin(x) same?what does j does to sin(x).is j just an amplitude of sin?

5. ### shteii01 AAC Fanatic!

Feb 19, 2010
3,507
512
Too funny!

$
j=sqrt{-1}
$

Most people don't use square root of negative one because it is just TOO HARD to use.

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6. ### WBahn Moderator

Mar 31, 2012
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'j' is used in electrical engineering and 'i' is used in math (because 'i' in EE is too tightly bound to electrical current).

Both are simply the number that, when squared, yields a value of -1.

And, yes, j·sin(x) just means that the amplitude is the sqrt(-1).

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7. ### Papabravo Expert

Feb 24, 2006
10,340
1,850
If you have trouble wrapping your head around j as the square root of -1, it might(?) help to think of it as a 90° counter-clockwise rotation operator in the complex plane.

Code (Text):
1.
2.  1 * j =  j
3.  j * j = -1
4. -1 * j = -j
5. -j * j =  1
6.

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8. ### StayatHomeElectronics Well-Known Member

Sep 25, 2008
864
40
Another way of thinking about it is to consider a plot with axes x and y. For the complex exponential equation, the x axis is going to be considered the REAL axis and the y axis is the IMAGINARY axis, indicated by the j or i. We will evaluate the two terms REAL and IMAGINARY at the same time keeping them separate. So, for the following equation:

e^(j*angle) = cos(angle) + j*sin(angle)

When we evaluate the cos(angle) portion, we get a value to plot along the x (or REAL) axis. When we evaluate the sin(angle) portion, we get a value to plot along the y (or IMAGINARY) axis.

e^(j*angle) = x + j*y = REAL + j*IMAGINARY

If you evaluate cos(angle) + sin (angle), both of these would add on the REAL,or x, axis of the plot.

9. ### WBahn Moderator

Mar 31, 2012
18,087
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Consider the following problem.

What is

$
y \; = \; 10\cos $$\omega t + 20^{\circ}$$ + \frac{d}{dt} $$20 \cos \( \omega t + 70^{\circ}$$ \)
$

Do that using trig and show your work.

Then we'll see what is involved in doing it using complex exponentials.

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10. ### bhuvanesh Thread Starter Member

Aug 10, 2013
268
2
$
y \; = \; 10\cos $$\omega t + 20^{\circ}$$ + $$-20 \sin \( \omega t + 70^{\circ}$$/w \)
$

Last edited: Mar 4, 2015
11. ### WBahn Moderator

Mar 31, 2012
18,087
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Is the derivative of cos(x) equal to sin(x)?

Let's pick a specific value of ω, namely ω=5 r/s.

I'm looking for an answer of the form

y = A·cos(ωt+φ)

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12. ### bhuvanesh Thread Starter Member

Aug 10, 2013
268
2
my professor stressed the line"complex exponential play big role in signals and system and it makes the things easier".i need strong reason to agree that.Thank you in advance

13. ### bhuvanesh Thread Starter Member

Aug 10, 2013
268
2

sorry i corrected the post differentiation of cosx is - sinx

and we can represent A·cos(ωt+φ) as e^jwφ-e^-jwφ

14. ### WBahn Moderator

Mar 31, 2012
18,087
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I'm trying to walk you through an example which might show you that there is a strong reason to agree with that.

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15. ### bhuvanesh Thread Starter Member

Aug 10, 2013
268
2
sure i am waiting to learn that by myself.Thank you

we can represent A·cos(ωt+φ) as e^jwφ-e^-jwφ

16. ### WBahn Moderator

Mar 31, 2012
18,087
4,917
Forget complex exponentials for now. Using trig only, write

$
y \; = \; 10\cos $$\omega t + 20^{\circ}$$ + \frac{d}{dt} $$20 \cos \( \omega t + 70^{\circ}$$ \)
$

with

$
\omega \; = \; 5 \frac{r}{s}
$

in the form

$
y \; = \; A\cos $$\omega t + \phi$$
$

17. ### bhuvanesh Thread Starter Member

Aug 10, 2013
268
2

$
y \; = \; 10\cos $$\omega t + 20^{\circ}$$ + $$- 20 \sin \( \omega t + 70^{\circ}$$ \)
$

w=5r/s=5*2pi/s
so

$
y \; = \; 10\cos $$\10\pi + 20^{\circ}$$ + $$- 20 \sin \( \10\pi + 70^{\circ}$$ \)

$

cos(2pi+x)=cos x
so
$
y \; = \; 10\cos $$20^{\circ}$$ + $$- 20 \sin \( 70^{\circ}$$ \)

$

am i going right?

18. ### WBahn Moderator

Mar 31, 2012
18,087
4,917
No. ω = 5 r/s. Replace ω with 5 r/s. There is no π there. And what happened to the variable t?

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19. ### bhuvanesh Thread Starter Member

Aug 10, 2013
268
2
not getting here r represent radian and radian mean 2pI and seconds and t are cancel

okay by your way replacing we get

$
y \; = \; 10\cos $$\(5r/s)t + 20^{\circ}$$ + $$- 20 \sin \( \(5r/s) t + 70^{\circ}$$ \)
$

20. ### WBahn Moderator

Mar 31, 2012
18,087
4,917
radian is a unit of angular measure, just like degrees are.

seconds are a unit of time -- it is not the variable t.

Just like in the distance equals velocity times time formula, d = v·t. If the velocity is 40 meters/second, you would (should) write

$
d \; = \; v \cdot t
\;
d \; = \; $$40 \frac{m}{s}$$ t
$

The 't' does NOT cancel out. If you now wanted to know how far the object traveled in 10 seconds, you would have

$
d \; = \; $$40 \frac{m}{s}$$ $$10s$$
\;
d \; = \; 400m
$

Now the units of seconds cancel out, leaving you with just units of distance, which is good.

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