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Spectrum of BPSK Signal

  1. Jul 11, 2008 #1
    I did some simulations where I created a BPSK signal, and then found the spectrum of the signal. The spectrum had a peak at the carrier frequency. It also had peaks at regular intervals frequencies higher than the carrier frequency. There were no peaks at frequencies lower than the carrier frequency.

    People around me, people who have PhDs in the field of digital communications give me conflicting information. They tell me that there are spectral peaks on both sides of the carrier frequency. In other words, they say that the spectrum is symmetrical about the carrier frequency. This outright contradicts my simulation, and I don't think that it can be proven any other way than by doing a simulation, and I'm generally better and simulation and spectral analysis than they are.

    Can anyone confirm or deny that the spectrum of a BPSK wave is indeed asymmetrical about the carrier frequency?
  2. jcsd
  3. Jul 11, 2008 #2


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    Don't know much about BPSK, but

    this -> http://www.elec.mq.edu.au/~cl/files_pdf/elec321/lect_mpsk.pdf
    Edit: so it would not be asymmetrical.
  4. Jul 11, 2008 #3


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    Here are some spectra of a BPSK signal. Note the 1st, 2nd and last plot, the spectra all have symmetry about the central transmitting frequency.

    I have transmitted/received BPSK over the airwaves, though I never took a close look at the frequency spectra before. I was more interested in encoding/decoding the content of the transmission. In your simulation, did you modulate the BPSK signal to RF or did you do it at base band?
  5. Jul 11, 2008 #4


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    If you mean power spectrum, it is symmetric as seen by calculating it. The signal is

    [tex]s(t)=e^{i\pi x_n}cos(\omega t)[/tex]

    where the binary modulation sequence [tex]x_n[/tex] consists of 1's and 0's during time intervals [tex]nT \leq x < (n+1)T[/tex].

    The exponential term simply switches between -1 and 1, so we can rewrite write the signal [tex]s(t)=y_n cos(\omega t)[/tex] in terms of a related modulation sequence [tex]y_n=2x_n-1[/tex].

    It follows that the transform [tex]S(\omega)=\mathcal{F}\{s(t)\}[/tex] is a convolution of transforms of the parts, [tex]S(\omega)=\mathcal{F}\{y_n\}\star\mathcal{F}\{cos(\omega t)\}[/tex]. So the RF spectrum is the spectrum of the modulation translated to the carrier frequency.

    Now for the key part: y is purely real so its Fourier transform is conjugate anti-symmetric, that is, [tex]\mathcal{F}\{y_n\}=Y(\omega)=Y^{*}(-\omega)[/tex] (see Bracewell or any other book on Fourier transforms). If you calculate the complex transform, then, the real part is symmetric and the imaginary part antisymmetric. It is unusual, however, to look at the complex spectrum. One usually considers the power spectrum, and that indeed is the only thing measured in the lab. The power spectrum for the present case
    is strictly symmetric.

    So the power spectrum, which is what is commonly meant by "spectrum", must be symmetric about the carrier.

    As for NoTime's quote, the carrier is suppressed only in the special case of a square wave modulation. In general the carrier is present.
  6. Jul 17, 2008 #5
    My signal is just a BPSK (or QPSK) signal. The bits are randomly encoded.

    What I don't yet see, is how the abrupt phase changes would cause the signal to have frequency components that are lower than the carrier frequency.
  7. Jul 17, 2008 #6


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    In the complex signal domain, both positive and negative frequencies are allowed. The spectrum of your discontinuous binary information has high frequencies of both signs. When you modulate the carrier with the signal, its spectrum surrounds the carrier both above and below.
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