Is it actually possible to filter out the negative frequencies for the baseband signal is we take into consideration that the filter's impulse response is symmetric around the zero frequency?

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    $\begingroup$ not physically. there are no physical quantities having complex (or imaginary) values and a non-zero signal having only positive frequency components is a complex signal. physically, one can represent a complex signal with two real signals and one can interpret that pair as a single complex signal. in that case the "imaginary part" must be the Hilbert transform of the "real part" for the negative-frequency components to have zero amplitude. $\endgroup$ – robert bristow-johnson Oct 25 '16 at 21:40
  • $\begingroup$ But why is the often said that e.g. DMT systems filter out negative frequencies? $\endgroup$ – Cali Oct 25 '16 at 21:48
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    $\begingroup$ what'sa DMT? (used to mean "dimethyltryptamine".) $\endgroup$ – robert bristow-johnson Oct 25 '16 at 21:55
  • $\begingroup$ Sorry my mistake.... DMT=Discrete Multitone Transmission $\endgroup$ – Cali Oct 25 '16 at 21:59
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    $\begingroup$ The baseband representation of a DMT signal is a sum of complex exponentials; each complex exponential has only positive (or negative) frequencies. This is similar to a baseband OFDM symbol. But this is not really related to your question as stated AFAICT. $\endgroup$ – MBaz Oct 25 '16 at 22:17

I'm not sure if you ask about symmetry of the impulse response or symmetry of the frequency response. All filters with a real impulse response have conjugate symmetric frequency responses and symmetrical magnitude frequency responses. A symmetrical magnitude frequency response would affect the magnitudes of both frequencies $\omega$ and $-\omega$ identically so it could not possibly remove or attenuate a negative frequency while keeping the positive frequency intact. So the impulse response has to be complex.

The discrete impulse response $b_k$ of an ideal do-nothing filter is:

$$b_k = \frac{\int_{-\pi}^{\pi}e^{i\omega k}d\omega}{2\pi} = \left\{\begin{array}{ll}1 &\text{ if }k = 0\\\frac{\sin(\pi k)}{\pi k}&\text{ if } k \ne 0\end{array}\right.\\ = [\dots, 0, 0, 0, 1, 0, 0, 0, \dots]$$

The middle value $1$ is at $b_0$. Changing where the integral starts gives the ideal negative frequency removal filter:

$$b_k = \frac{\int_{0}^{\pi}e^{i\omega k}d\omega}{2\pi} = \left\{\begin{array}{ll}1/2&\text{ if } k = 0\\\frac{\sin(\pi k)}{2 \pi k} + i\left(\frac{1 - \cos(\pi k)}{2\pi k}\right)&\text{ if }k \ne 0\end{array}\right.= [\dots, 0, -\frac{i}{5\pi}, 0, -\frac{i}{3\pi}, 0, -\frac{i}{\pi}, \frac{1}{2}, \frac{i}{\pi}, 0, \frac{i}{3\pi}, 0, \frac{i}{5\pi}, 0, \dots]$$

The middle value $1/2$ is at $b_0$. The real part of the impulse response is zero at other values of $k$. The imaginary part has odd symmetry with respect to $\operatorname{Im}(b_0) = 0$. It is plain to see from the integral that the frequency response is zero for frequency $\omega < 0$ and constant ($1$) for $\omega > 0$.

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    $\begingroup$ "All filters with a real impulse response have symmetrical frequency responses." "symmetric" alone is not accurate here. For real filters, the frequency response is conjugate symmetric. It means $H(-\omega)=H^*(\omega)$. Of course, this implies even symmetry for the real part and the magnitude, and odd symmetry for imaginary part and the phase. $\endgroup$ – msm Oct 26 '16 at 7:56
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    $\begingroup$ It's maybe worth pointing out that the second filter in your answer is actually what is sometimes called a "phase splitter", with (complex) impulse response $b[n]=\frac12(\delta[n]+jh[n])$, where $h[n]$ is the impulse response of an ideal Hilbert transformer. $\endgroup$ – Matt L. Oct 26 '16 at 9:56
  • $\begingroup$ One thing that bugs me in your otherwise very good answer is the story about taking the limit $k\rightarrow 0$. I think this is meaningless, because by definition, $k$ is an integer, and must be treated as such. In my opinion, the only correct way to deal with the problem is to distinguish cases: Case 1: $k=0$, which in your first example will simply give you $$b_0=\frac{\int_{-\pi}^{\pi}d\omega}{2\pi}=1$$ Case 2: $k\neq 0$, etc. We can't just divide by $k$ if $k=0$. Annoyingly, taking the limit $k\rightarrow 0$ (as if $k$ were a continuous variable) actually gives the correct result. $\endgroup$ – Matt L. Oct 26 '16 at 9:58
  • $\begingroup$ @MattL. How about constructing a band-limited continuous function first (with a continuous equivalent of $k$) and sampling that to get the sequence? $\endgroup$ – Olli Niemitalo Oct 26 '16 at 10:04
  • $\begingroup$ Well, yes, that is what's happening. But nobody tells the non-initiated people that secretly we change $k$ to a continuous variable, compute the function in continuous-time, and then go back to discrete-time by sampling the continuous function. So I think that's pedagogically (and notation-wise) questionable, and furthermore, why complicate things in such a way if everything can be done very easily leaving $k$ an integer (as most people believe it has been throughout)? $\endgroup$ – Matt L. Oct 26 '16 at 10:10

A complex filter's response might not be symmetric around f=0, and thus could filter out just the negative frequency components of a complex IQ baseband signal (such as what some USB SDR dongles provide).


It's perfectly possible to filter out negative frequencies using a complex filter. In fact, Reilly, Fraser, and Boashash suggest that this is a really good way to generate a good quality analytic signal.

The effective transfer function they try to generate is as per their Figure 2, included below.

There is a pre-publication version of the paper here.

enter image description here

  • $\begingroup$ Peter, can you be more clear what the signals are like going into and outa the "complex filter" referred to? $\endgroup$ – robert bristow-johnson Oct 27 '16 at 1:50

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