In a baseband complex signal, I need to generate a BSPK modulated signal at a given frequency. If I generate a CW at this frequency, all is good. However, when I then BPSK modulate the CW, I get a signal at both the positive and negative frequencies.

Is there a way of implementing BPSK modulation without this negative frequency image, or must I simply implement a filter to remove the image?

  • $\begingroup$ What does BPSK modulating a chirp waveform (I assume that that's what you meant by "CW") mean? Normally you use BPSK modulation on digital data. $\endgroup$
    – Jim Clay
    Commented May 7, 2013 at 13:03
  • 1
    $\begingroup$ @JimClay CW usually means "continuous wave" among RF engineers, that is, the unmodulated sinusoid. When modulated by a digital signal, the continuity of the (mathematical representation of the) signal might be disrupted at the switching times between $0$ and $1$. Of course, all actual signals are continuous because actual voltages cannot change instantaneously. By careful choice of parameters, it can be arranged that the phase change occurs when the carrier signal has value $0$ so that (even the mathematical representation of) the signal is continuous (but its derivative is not). $\endgroup$ Commented May 7, 2013 at 13:48
  • $\begingroup$ @DilipSarwate I see, they're doing it at the carrier frequency instead of at baseband and then modulating it up to the carrier frequency. Thanks for the explanation. $\endgroup$
    – Jim Clay
    Commented May 7, 2013 at 14:14

2 Answers 2


This question apparently is not at all about negative frequencies which is the point I addressed earlier in a separate answer but about the signal at the image frequency that occurs as a result of the

It appears from the comments by the OP on an earlier answer of mine that his question is about an entirely different problem. The OP apparently creates a BPSK signal $x(t)\cos(2\pi f_0 t)$ at a low frequency $f_0$ Hz and then mixes it with a high-frequency signal at frequency $f_1$ Hz where $f_1 \gg f_0$. This creates two BPSK signals at carrier frequencies $f_1 \pm f_0$ respectively and the question is whether this duplication can be avoided somehow in the mixing process, or whether one must filter the output to leave only one BPSK signal.

The trigonometric identity $2\cos(A)\cos(B) = \cos(A+B) + \cos(A-B)$ shows that mixing (multiplying) the BPSK signal $x(t)\cos(2\pi f_0 t)$ with $2\cos(2\pi f_1 t)$ gives $$2\cos(2\pi f_1 t)x(t)\cos(2\pi f_0 t) = x(t)\cos(2\pi (f_1+f_0)t) + x(t)\cos(2\pi (f_1-f_0)t),$$ that is, two BPSK signals at carrier frequencies $f_1+f_0$ Hz and $f_1-f_0$ Hz. If one desires to have only the BPSK signal at $f_1+f_0 = f_c$ Hz, say, then the BPSK signal at the image frequency $f_1-f_0$ must be filtered out by passing the mixer output through a bandstop filter that removes the image signal.

Can the filtering to remove the image frequency be avoided by clever design of the mixer? Mathematically Yes, but the costs might be prohibitive and the physical circuitry requiring very careful design and constant retuning. The practical details are best discussed over on the sister site electronics.SE.

Suppose we create two BPSK signals $x(t)\cos(2\pi f_0 t)$ and $x(t)\sin(2\pi f_0 t)$ on phase-orthogonal carriers at frequency $f_0$ and mix them (separately) with phase-orthogonal carriers at frequency $f_1$. Then, their difference is $$x(t)\cos(2\pi f_0 t)\cos(2\pi f_1 t) - x(t)\sin(2\pi f_0 t)\sin(2\pi f_1 t) = x(t)\cos(2\pi (f_1+f_0)t)$$ which is a single BPSK signal at the desired frequency $f_1+f_0 = f_c$!! However, notice that we need two baseband modulators and two RF mixers instead of one of each plus a filter. Also, we need precise matching of hardware so that the two signals are generated with equal amplitudes, are amplified exactly equally, are modulated onto phase-orthogonal carrier signals of precisely equal amplitudes and precise phase difference that must be maintained through oscillator drift, ambient temperature changes, etc. Why is all this so important? Well, both mixer outputs $x(t)\cos(2\pi f_0 t)\cos(2\pi f_1 t)$ and $x(t)\sin(2\pi f_0 t)\sin(2\pi f_1 t)$ contain two BPSK signals, one at carrier frequency $f_1+f_0 = f_c$ and the other at carrier frequency $f_1-f_0$. When we take the difference, the BPSK signals at carrier frequency $f_1+f_0 = f_c$ add constructively and appear at the output of the subtractor, while the BPSK signals at at carrier frequency $f_1-f_0$ cancel out. This is easy to do mathematically or MATLABitacally or in software radio, but much more complicated to achieve with analog circuits where slight differences in gains, phase shifts, etc in the two allegedly identical circuits make it difficult to achieve the exact cancellation of the undesired BPSK signal. In short, getting a single BPSK signal via this method is by no means as simple in practice as the straightforward mathematical result suggests. As noted previously, questions about feasibility are best discussed over on electronics.SE.


An actual BPSK signal can be expressed as $$\begin{align} s(t) &= \operatorname{Re}\left[\sum_{n=-\infty}^\infty (-1)^{b_n}p(t-nT)e^{j(2\pi f_c t + \theta)}\right]\\ &= \left[\sum_{n=-\infty}^\infty (-1)^{b_n}p(t-nT)\right]\cos(2\pi f_c t + \theta) \end{align}$$ where

  • $b_n \in {0,1}$ is the $n$-th data bit (note that $(-1)^{b_n} = \pm 1$),

  • $p(t)$ is the baseband pulse shape. Typically $p(t)$ is nonzero only for $t \in \left[-\frac{T}{2},\frac{T}{2}\right)$, e.g. $\operatorname{rect}\left(\frac{t}{T}\right)$.

  • $f_c$ is the carrier frequency which is sometimes chosen to be a harmonic of the bit rate $T^{-1}$

  • $\theta$ is the carrier phase which might be arbitrary (depending on initial conditions in the local oscillator when it is turned on) or carefully controlled to have value $0$ or $\pm \frac{\pi}{2}$ etc. by MATLABi people.

Notice that the term in square brackets in the second displayed equation above is a series of nonoverlapping pulses $\pm p(t-nT)$ and that the value of $b_n$ affects the BPSK signal phase only during $\left[nT-\frac{T}{2},nT+\frac{T}{2}\right)$.

With that as a preliminary setting of the stage, note that the negative frequency image comes about from taking the real part in the first displayed equation above. We have $$\operatorname{Re}\left[e^{j(2\pi f_c t + \theta)}\right] = \frac{e^{j(2\pi f_c t + \theta)} + e^{-j(2\pi f_c t + \theta)}}{2} = \frac{e^{j(2\pi f_c t + \theta)} + e^{j(2\pi (-f_c) t - \theta)}}{2}.$$ Furthermore,

no filtering can remove the negative frequency image.

You might say "Well, let's not take the real part at all since that's the source of the trouble!" and say that $$\sum_{n=-\infty}^\infty (-1)^{b_n}p(t-nT)e^{j(2\pi f_c t + \theta)}$$ is a BPSK signal containing positive frequencies only. Unfortunately, that is a complex-valued signal that cannot be transmitted on a single wire or radiated from a single antenna. So you need to live with the negative frequency image: it is there to make your complex signal more real.

  • $\begingroup$ +1. The pulses often do overlap somewhat, but that doesn't affect the main point that a real signal is going to have negative frequencies. $\endgroup$
    – Jim Clay
    Commented May 7, 2013 at 14:16
  • $\begingroup$ Thanks for the expansive answer, Dilip. In this case, the complex signal is then being applied to a mixer to generate RF output. My intention is to create a BSPK modulated signal offset from the frequency of the mixer, but instead I'm ending up with the signal replicated at +/- my offset frequency. More reading required, I think. $\endgroup$
    – Darran
    Commented May 7, 2013 at 15:35
  • $\begingroup$ @Darran I am writing a different answer since it appears we have been talking at cross-purposes. $\endgroup$ Commented May 8, 2013 at 1:15

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