# Filtering out negative frequencies

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?

• 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. – robert bristow-johnson Oct 25 '16 at 21:40
• But why is the often said that e.g. DMT systems filter out negative frequencies? – Cali Oct 25 '16 at 21:48
• what'sa DMT? (used to mean "dimethyltryptamine".) – robert bristow-johnson Oct 25 '16 at 21:55
• Sorry my mistake.... DMT=Discrete Multitone Transmission – Cali Oct 25 '16 at 21:59
• 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. – 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 $\omega > 0$.

• "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. – msm Oct 26 '16 at 7:56
• 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. – Matt L. Oct 26 '16 at 9:56
• 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. – Matt L. Oct 26 '16 at 9:58
• @MattL. How about constructing a band-limited continuous function first (with a continuous equivalent of $k$) and sampling that to get the sequence? – Olli Niemitalo Oct 26 '16 at 10:04
• 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)? – 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.

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