I know that when the frequency is 0, the voltage will be pure DC. But in DSP and Digital Communication, I have seen mentioning of negative frequencies which I dont quite understand. For example, like \$ -f_{0}\$ to \$ f_{0}\$ frequency range. How could frequency become negative?
Electronic – Negative frequencies: what is that
frequency
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Frequency is the inverse of time for repeating events. If a single cycle of your mains is 1/50 of a second in duration (0.02 seconds), then there will be 50 cycles in a second (1 / 0.02). We say the frequency is 50 Hz.
The unit for frequency is the Hertz (Hz). 1 Hz is equal to 1 cycle per second, an older name for it (cps). It's a convenient unit, even for very short cycles we use, with a prefix: MHz, GHz. For longer cycles (near or longer than 1 Hz) we sometimes use the minute as unit: a heart rate of 70 beats per minute (BPM), a metronome setting of 100 BPM.
Still longer cycles are often expressed as a period of time (1 / frequency).
Every system has its typical (range of) frequencies. A heartbeat will be around 1-2 Hz, and the resonance frequency of a trampoline is also in the order of 1 Hz. Radio waves have a broad range of frequencies: 30 kHz (for a corresponding wavelength of 10 km) is VLF (very low frequency), while the magnetron of a microwave oven "transmits" at 2.45 GHz. And while 30 kHz is low for radio, it's already well beyond what we acoustically can perceive.
A higher frequency (top track in the picture) of a signal will be shown on an oscilloscope as a faster repetition than a lower frequency (bottom track).
"Direct current" (DC) as a constant voltage has a frequency of 0 Hz.
Bandwidth indicates a range of frequencies, going from the lower limit to the higher. If your scope can handle signals from DC (0 Hz) up to 100 MHz, its bandwidth is 100 MHz - 0 Hz = 100 MHz.
FFTs work by treating signals as 2-dimensional -- with real and imaginary parts. Remember the unit circle? Positive frequencies are when the phasor spins counter-clockwise, and negative frequencies are when the phasor spins clockwise.
If you throw away the imaginary part of the signal, the distinction between positive and negative frequencies will be lost.
For example (source):
If you were to plot the imaginary part of the signal, you would get another sinusoid, phase shifted with regards to the real part. Notice how if the phasor were spinning the other way, the top signal would be exactly the same but the phase relationship of the imaginary part to the real part would be different. By throwing away the imaginary part of the signal you have no way of knowing if a frequency is positive or negative.
Best Answer
The derivation of
\$cos(\omega t) = \frac{1}{2} \left(e^{j\omega t} + e^{-j\omega t}\right)\$
is all very nice and such (thanks, Mark), but it's not very intuitive.
A sine can be presented in the complex plane as a rotating vector:
You can see how the vector consists of a real and an imaginary part. But what you see when you watch the signal on your scope is a real signal, so how can you get rid of the imaginary part, such that the vector stays on the x-axis, increasing and decreasing? The solution is to add a mirror image of the rotating vector, rotating clockwise instead of counterclockwise.
The imaginary parts have the same magnitude, but opposite signs, so when you add both vectors the imaginary parts cancel each other, leaving a purely real signal.
If counterclockwise rotation stands for positive frequency, clockwise rotation has to stand for negative frequency.