complex exponential signals

Thread Starter

bhuvanesh

Joined Aug 10, 2013
268
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
 
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?
 

Thread Starter

bhuvanesh

Joined Aug 10, 2013
268
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?
 

WBahn

Joined Mar 31, 2012
30,234
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?
'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).
 

Papabravo

Joined Feb 24, 2006
21,256
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:
 1 * j =  j
 j * j = -1
-1 * j = -j
-j * j =  1
 
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.
 

WBahn

Joined Mar 31, 2012
30,234
What makes them special is that they make the math a LOT easier.
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.
 

Thread Starter

bhuvanesh

Joined Aug 10, 2013
268
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.
\(
y \; = \; 10\cos \( \omega t + 20^{\circ} \) + \( -20 \sin \( \omega t + 70^{\circ} \)/w \)
\)
 
Last edited:

WBahn

Joined Mar 31, 2012
30,234
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)
 

Thread Starter

bhuvanesh

Joined Aug 10, 2013
268
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
 

Thread Starter

bhuvanesh

Joined Aug 10, 2013
268
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)

sorry i corrected the post differentiation of cosx is - sinx

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

WBahn

Joined Mar 31, 2012
30,234
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
I'm trying to walk you through an example which might show you that there is a strong reason to agree with that.
 

Thread Starter

bhuvanesh

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

my last reply about that is

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

WBahn

Joined Mar 31, 2012
30,234
sorry i corrected the post differentiation of cosx is - sinx

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

Thread Starter

bhuvanesh

Joined Aug 10, 2013
268
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 \( \omega t + \phi \)
\)

\(
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?
 

WBahn

Joined Mar 31, 2012
30,234
\(
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?
No. ω = 5 r/s. Replace ω with 5 r/s. There is no π there. And what happened to the variable t?
 

Thread Starter

bhuvanesh

Joined Aug 10, 2013
268
No. ω = 5 r/s. Replace ω with 5 r/s. There is no π there. And what happened to the variable t?
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 \( \(5r/s) t + 70^{\circ} \) \)
\)
 

WBahn

Joined Mar 31, 2012
30,234
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 \( \(5r/s) t + 70^{\circ} \) \)
\)
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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