What is the reason for filtering out the negative frequencies of a signal?

Question

I am reading this tutorial. Quoting the lines from the topic "Analytic signals and Hilbert transform filters":

the corresponding analytic signal $ z(t)=x(t) + j {\cal H}_t\{x\}$ has the property that all ``negative frequencies'' of $ x(t)$ have been "filtered out''

I have a few questions about this:

1. what is the purpose of filtering out negative frequencies?
2. What is the interpretation of filtering out negative frequency?
3. What are negative frequencies in physical world? I read that negative frequencies doesn't exist and are introduced to simplify the mathematics behind theory.
4. If negative frequencies don't exist, then why do we all plot two sided spectrum?
5. If negative frequencies exist, why do we only consider the first half of an N-point FFT?

Answer 1

I'll answer your questions out of order:

3. Negative frequencies do exist. There is nothing controversial about them.

• For a cosine signal: since $\cos(2\pi(-f)t)=\cos(2\pi ft)$, a cosine of negative frequency $-f$ is equal to a cosine of frequency $f$.

• Since $\sin(2\pi(-f)t)=-\sin(2\pi ft)$, a sine of frequency $-f$ is $\pi$ radians out of phase with respect to a sine of frequency $f$.

• More importantly, a complex exponential $e^{2\pi ft}$ can be represented by a point in the complex plane, which rotates counter-clockwise if $f$ is positive and clockwise if $f$ is negative.

5. It turns out that, for all physical signals (whose imaginary part is zero), the magnitude spectrum is even (the negative frequencies are a mirror image of the positive frequencies). There is no need to display or calculate them. That's the reason a spectrum analyzer will only display positive frequencies. Complex signals do not have an even magnitude spectrum, and you need to calculate it for negative as well as positive frequencies.

2. Filtering out the negative frequencies is just that: remove the negative frequencies of a signal. The result is a complex signal, because the resulting spectrum is not even. The Hilbert transform you mention at the start of your question is an easy way to implement such a filter. Without it, you'd need a complex filter, which are not trivial to implement and use.

1. There can be many reasons to filter out the negative frequencies of a signal. In digital communications, it is used as one step in obtaining the complex envelope of a modulated signal. In general, a modulated signal looks like $s(t)=\Re[a(t)e^{2\pi f_ct}]$, where $a(t)$ is a quadrature signal, and is complex, and where $f_c$ is very large (up to several gigahertz). However, the information is contained in $a(t)$, so we would like to recover it from $s(t)$. This is accomplished by:

• Filtering out the negative frequencies of $s(t)$

• Downconverting the result to baseband