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Waveform notation

  1. Feb 20, 2009 #1
    This is my firs post to this forum.
    This may be a trivial question but got a bit confused in reading few texts.

    Usually periodic waveforms are denoted as
    A exp (jwt)

    However, as I understand the exponent consists of
    cos (jwt) + i sin (jwt) which is a complex number.

    So, shouldn't A exp (jwt) be Re (A exp (jwt)) ?
    Or is the Re() implicit here?
  2. jcsd
  3. Feb 20, 2009 #2


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    You can use either the real or the imaginary part; as long as you are consistent. What you use depends on how the waveform is written in the time-domain; i.e. using Sin or Cos.

    Hence, there "should" really be a Re or Im there, but since this is a well known convention you can leave it out. This is what is known as phasor notation and it is basically shorthand for what is really a Fourier transform.

    Btw, there shouldn't be any "j" in the trigonometric expressions, but I assume that is a typo.
    Last edited: Feb 20, 2009
  4. Feb 20, 2009 #3
    Hang on, you're confusing notation; i and j are the same thing, j represents i in electronics (Maxwell. Laplace, Fourier) because current is i in electromagnetics.

    A is amplitude., jwt is a complex f; start again with the trig expansion.
  5. Feb 20, 2009 #4
    Thanks for the advice.

    So I presume A exp (jwt) is in fact Re(A exp (jwt)). But is implicit. So, whenever i see A exp (jwt) in signals context I can safely assume it is actually A cos(wt) ?

    Also, exp (jwt) should be cos (wt) + j sin (wt). Had mistakenly include j inside cosines.
  6. Feb 20, 2009 #5
    Yep, that looks better; what you have is a general form for a sinusoidal wave. If A is a complex amplitude, and jwt a complex frequency; you now substitute A = A.exp (ja), in A.exp (jwt), and you have A.exp j(wt + a) (because exp is its own derivative, right?)

    Now the real and imaginary terms (cos + sin), can represent covariant 'variables' like voltage or current.
    You have a phasor representation, of the two covariants and how they diverge/converge over their rotational velocities (in time, of course).

    Woops, "whenever i see A exp (jwt) in signals context I can safely assume it is actually A cos(wt) ?" You can assume the A.cos(wt) is the Re{A.exp(jwt)} part of the complex frequency (with a complex amplitude).
    Last edited: Feb 20, 2009
  7. Feb 20, 2009 #6
    Thanks, but i couldn't get what you meant by,
    What is a complex amplitude and a complex frequency? As I understand a complex frequency is a result of a signal being represented in fourier exponent terms. ie. exp(-jwt) and exp(jwt) are frequencies of -w and w.
  8. Feb 21, 2009 #7
    It really just means its a complex number, it's 'on' the complex plane.

    Because current/voltage is "a complex number", or looks sorta exactly like one. Fourier for sure is the representation to use in/out of the time domain. Your -w,+w pair is a conjugate on the complex plane.
    The phasor rep extends to any rotating frame, not just in EM.
  9. Feb 22, 2009 #8


    Staff: Mentor

    In a signals context there is no implied Re or Im operator. Instead, if you have a purely real signal then it will be of the form A exp(jwt) + A* exp(-jwt) where the * denotes complex conjugation.

    There are many cases where the signal is complex rather than real. Google "quadrature detection" if you are interested. In such cases the negative frequency term will not be the complex conjugate of the positive frequency term. So one technique works for both cases and there is no need to implicitly insert a Re or Im.
  10. Feb 23, 2009 #9
    So, do you mean to say that -w is not in fact negative but negative with respect to a center frequency(wc). ( taking this as 0 will give -w otherwise would be wc - w ). But then again that is a result of some physical process eg. demodulation

    I was wondering why exp (jwt) is used to represent such a "real" signal ie. result of a physical phenomenon as,
    exp (jwt) = cos (wt) + j sin (wt) which has a complex sinuosoid. What would be the meaning of a complex sinusoid ?
    If there's no such thing as a complex sinusoid, I hoped see a Re(exp(jwt)) there.
  11. Feb 23, 2009 #10


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    The key here is to remember that a even "real world" signals needs be represented by TWO numbers in the general case: amplitude and phase.
    The reason why the "Re" is implicit in basic circuit theory is because the phase does not vary continuously as you move around the circuit at low frequencies (because the circuit is smaller than the wavelength) which simplifies things a great deal.
    However, in the more general case we need to keep track of BOTH the amplitude and phase.

    Now, in quadrature detection the signal really has two "components" that are in "quadrature" (90 degress out of phase): the in-phase (I) and quadrature phase (Q) component.
    The I-component is just Acos wt and the Q-component Asin wt.
    It should be fairly obvious why it is convenient to represent this using a complex number.

    Note that this is not just a "formal" definition, if you look at an IQ-demodulator (a special type of mixer) you will find that there really ARE I and Q outputs that give you the real and imaginary parts of the signal (referenced to a local oscillator).

    This is very useful since it means that you can e.g. calculate the amplitude of the down-mixed signal using sqrt(I^2+Q^2) and the phase using atan(I/Q).

    I am actually using exactly this technique in my lab at the moment for some microwave measurements, and the data I get from my experiments really is a set of complex numbers (usually as a function of frequency).
  12. Feb 23, 2009 #11


    Staff: Mentor

    No, -w is in fact a negative frequency, i.e. less than DC which is 0 frequency. There is nothing wrong with negative frequencies, and in fact artificially avoiding them can really mess you up. Remember that phase is frequency times time, so a negative frequency is simply a phase that decreases as time increases.

    Sure, but as I said above the sum of A exp(jwt) and A* exp(jwt) is purely real. So you can use the same exponentials to represent both real and complex signals.

    As f95toli already mentioned there are such things as complex sinusoids. He gave one example. Another example is in MRI, which is my specialty. In MRI you detect magnetic fields that oscillate in the RF range. To do that, the most efficient means of detecting the signal is to have two loops which are sensitive to magnetic fields that are perpendicular to each other. Such an antenna is called circularly polarized and has a real and an imaginary channel which are out of phase by 90º. The two channels actually do contain different information, so you cannot simply take the real or imaginary channel without losing some of the information that is present in the signal.

    Bottom line, there is no implied Re or Im. Real signals will have complex conjugate symmetry. And complex signals do in fact exist.
    Last edited: Feb 23, 2009
  13. Feb 24, 2009 #12
    Thanks for pointing me in the correct direction.

    I went through some books, namely
    The Scientist and Engineer's Guide to Digital Signal Processing
    By Steven W. Smith, Ph.D

    according to him

    to summarize what I understand from you and the texts,
    take a signal
    A cos (wt + p) = x cos (wt) + y sin (wt) ; after cos expansion
    = x + j y ; x is the real component, y is the imaginary component
    = B exp (p) ; B = A exp (jwt)

    Is this correct?
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