complex exponential signals

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

  1. bhuvanesh

    Thread Starter Member

    Aug 10, 2013
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    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
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    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.

    [​IMG]

    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
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    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
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    Too funny!

    <br />
j=sqrt{-1}<br />

    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
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    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
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    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
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    Consider the following problem.

    What is

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

    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
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    <br />
y \; = \; 10\cos \( \omega t + 20^{\circ} \) +  \( -20 \sin \( \omega t + 70^{\circ} \)/w \)<br />
     
    Last edited: Mar 4, 2015
  11. WBahn

    Moderator

    Mar 31, 2012
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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
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    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
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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
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    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
     
  16. WBahn

    Moderator

    Mar 31, 2012
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    Forget complex exponentials for now. Using trig only, write

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

    with

    <br />
\omega \; = \; 5 \frac{r}{s}<br />

    in the form

    <br />
y \; = \; A\cos \( \omega t + \phi \)<br />
     
  17. bhuvanesh

    Thread Starter Member

    Aug 10, 2013
    268
    2

    <br />
y \; = \; 10\cos \( \omega t + 20^{\circ} \) +  \(- 20 \sin \( \omega t + 70^{\circ} \) \)<br />
    w=5r/s=5*2pi/s
    so

    <br />
y \; = \; 10\cos \( \10\pi + 20^{\circ} \) +  \(- 20 \sin \( \10\pi  + 70^{\circ} \) \)<br />
<br />

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

    am i going right?
     
  18. WBahn

    Moderator

    Mar 31, 2012
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    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
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    not getting here r represent radian and radian mean 2pI and seconds and t are cancel

    okay by your way replacing we get

    <br />
y \; = \; 10\cos \( \(5r/s)t + 20^{\circ} \) + \(- 20 \sin \( \(5r/s) t + 70^{\circ} \) \)<br />
     
  20. WBahn

    Moderator

    Mar 31, 2012
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    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

    <br />
d \; = \; v \cdot t<br />
\;<br />
d \; = \; \(40 \frac{m}{s}\) t<br />

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

    <br />
d \; = \; \(40 \frac{m}{s}\) \(10s\)<br />
\;<br />
d \; = \; 400m<br />

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