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Complex signal

  1. Jul 31, 2009 #1
    Can a generic, not necessarily harmonic, signal of time be represented as a complex signal with a real and imaginary part?
    Usually the complex rappresentation is used for time harmonic signals and linear systems.
    The "real" time signal is transformed into a complex signal. At the end of the mathematical operations, it is possible to look at the real or imaginary part of the complex signal and get the correct results for the initial signal.
    But what if the starting real signal is not harmonic?

    thanks!
     
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  3. Jul 31, 2009 #2

    Born2bwire

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    We can almost always represent a signal in the frequency domain using Fourier transformations or series. In this case the signal would be a finite or infinite set of harmonic signals.
     
  4. Jul 31, 2009 #3

    jasonRF

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    Yes, this is done all the time in communications, radar, etc., for non-harmonic signals. Essentially, if you have a real signal, then the negative frequency Fourier components are the complex conjugate of the positive frequency components. Thus you only need half of the spectrum - setting the negative half to zero leaves you with a complex signal. In applications we always have band-limited signals and what is usually done is as follows:

    1. take the time-domain signal and multiply it by a phase ramp, this shifts the frequencies. Often we shift the frequency so that the middle of our band is centered on DC. This shifts the other half of the spectrum to higher frequency.

    2. Low-pass filter to get rid of the half of the spectrum that is at the higher frequency.

    Now you have a complex signal, which has less total bandwidth than the original signal. Since you know what operations you did on the signal, you can always reconstruct the original real signal.

    In the olden days this was done analog - take a signal, split it, mix one path with sin(wt) and the other with cos(wt), then use two A\D converters, one for each channel. One channel is "real" the other is "imaginary". What you save is that the total bandwidth is smaller so your required sample rate is lower (often giving you better dynamic range, less power consumption, etc.). These days we often do all this digitally. After the low-pass filter you can usually do some decimation to reduce the amount of computation required further up the chain.

    If you want more details, google "IQ sampling" and you will likely find more info. Communications books usually have nice discussions of this stuff, as do radar books.
     
  5. Aug 3, 2009 #4
    Hi JasonRF,

    thanks for the good response. Just one last clarification.
    A real signal has a Fourier transform which is a complex function of the real variable frequency. This function exists for the negative and positive frequency.
    By realizing that the Fourier transform is not only a complex function but also Hermitian, we can get rid of the negative frequencies (since their information is already contained in the positive frequencies). What we get is the complex analytic signal. What is the advantage of this operation? What do we gain? Are we saving bandwidth?
    There is something about the Hilbert transform here too...
    I guess any real signal can then be viewed as a complex function, meaning that it is equivalent to its Fourier transform, which is a complex function...
    thanks
    fisico30
     
  6. Aug 4, 2009 #5

    jasonRF

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    Exactly!

    Yes, you effectively save bandwidth. Once you only have 1/2 the spectrum, you can shift it in frequency (by multiplying by a phase ramp) to be centered at DC. You can decimate, and perform filtering, etc., on the complex signal. It also can make it "easy" to separate signals out by phase: applying phase shifts to complex signals is easy! So this is a practical thing to do, and is done all the time in communications equipment, radars, etc. It also makes the theoretical analysis of such systems easier, since we can just deal with the complex envelope, below ...

    Yes. If you have a real signal [tex]x(t)[/tex], then taking the Hilbert transform of the signal is equivalent to multiplying the Fourier transform of [tex]x(t)[/tex] by [tex]-i sgn(f)[/tex], where [tex]sgn(f)[/tex] is the "sign" function that is 1 for positive f and -1 for negative f. If the signal [tex]x(t)[/tex] is bandpass with center frequency [tex]f_c[/tex], then the complex envelope is usually written

    [tex]\tilde{x}(t)=e^{-i2\pi f_c t}\left[ x(t)+i\hat{x}(t)\right][/tex],

    where [tex]\hat{x}(t)[/tex] is the Hilbert transform of [tex]x(t)[/tex].
    The Fourier transform of the portion in brackets is zero for negative frequencies, and is twice the Fourier transform of [tex]x(t)[/tex] for positive frequencies. The phase ramp out in front simply shifts the remaining part of the spectrum to be centered at DC. You recover the original signal simply with

    [tex] x(t) = Re \left[\tilde{x}(t) e^{i 2 \pi f_c t} \right] [/tex].

    Just like for harmonic signals, this simplifies analysis. If we want apply a filter with an impulse response [tex]h(t)[/tex], make the complex envelope version (with an extra 2 to make the end result nicer) such that


    [tex] h(t) = Re \left[2 \tilde{h}(t) e^{i 2 \pi f_c t} \right] [/tex].

    Then, the convolution

    [tex]y(t) = h(t) \ast x(t)[/tex]

    can be written

    [tex] y(t) = Re \left[ \tilde{y}(t) e^{i 2 \pi f_c t} \right] [/tex]

    where

    [tex]\tilde{y}(t) = \tilde{h}(t) \ast \tilde{x}(t)[/tex].

    For analytical work this is much easier to deal with!
     
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