# Frequency Conversion Product Lower than 0 Hz (Negative Frequency?)

If I have a bandwidth of frequencies from 200 to 500 Hz made from cosine waves and I mix this with a oscillator at 400 Hz which is also a cosine wave, the mathematics of conversions show that I will get a signal at 600-900 Hz and -100 to 200 Hz. Does this negative frequency from the conversion math physically exist? I dont think it does - so where has some of the energy from -100 to 0 gone?

I understand the negative frequency fourier transform concept, but the above different and is tricking me.

Thanks for you help in advance.

• -100 to +200 I think you meant to say. Apr 30 '20 at 12:52
• Does this negative frequency from the conversion math physically exist? Oh yes, what if the phase is just different? On a "single" wire you cannot tell if a signal has a positive or a negative frequency. But look into quadrature signals, then you have an I and a Q signal, then we can tell if a signal has a positive or a negative frequency. Apr 30 '20 at 12:56
• If you draw a -100Hz waveform you'll find it looks a lot like a 100Hz waveform... Apr 30 '20 at 13:00
• so your saying that it just a duplication of frequencies at 0 to 100? Apr 30 '20 at 13:04
• Pretty much. There are differences but you have to look for them. For example: 300Hz and 500Hz both produce 100Hz lower sideband, but reducing by 1 Hz to 299Hz produces 101Hz while 499Hz produces 99Hz. The spectrum is kind of "folded" around 0 (and it is actually called spectral folding). May 1 '20 at 20:14

negative frequency is PHASE ROTATION of the other direction.

Had to assist in synthesizing GSM signals, some decades back. We soon understood we could produce whatever signals we wanted, using I+Q mixers followed by summation.

Frequency is typically the derivative of phase, and phase is the X in sin(X).

X typically is 2 * PI * Frequency * Time.

Since we don't reverse time, we do use ( -Frequency) and thus have a negative phase trajectory.

'Cosine waves' area real signal. If instead of a single real input signal, you break this down into two analytic signals, one at positive frequency, one at negative, and do the same for the local oscillator, then it's quite easy to see when mixing where the sums and differences come from, and where the resultant frequencies go to.

If you now discard the imaginary part of your resulting analytic output signal to get back to a real signal, you lose the information that allowed you to tell negative from positive frequency, and the 0 to -100 Hz part of the spectrum folds over onto the 0 to 100 Hz part. The energy is still there, just folded over.

Consider the following regards the phasor $$\e^{j\theta} = cos(\theta) +jsin(\theta)\$$: - It can be seen that sin and cos have 2 phasors rotating in opposite directions ($$\e^{j\theta}\$$ and $$\e^{-j\theta}\$$). We call these positive and negative frequencies.