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?
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9\$\begingroup\$ Frequency is kind of a modular concept. When talking about negative frequencies it's not really referring to the rate of change anymore (that can be thought of as the absolute value) but often a direction is implied as a result of the sign. So for example, a wheel spinning backwards might have a negative number of revolutions per second, but the wheel is spinning at the same "frequency" as if it were going forward. Not sure if that analogy would hold for everything since I'm hardly a DSP expert, but I think it's a good way to think about it. \$\endgroup\$– NickHaldenCommented Jun 17, 2011 at 5:28
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1\$\begingroup\$ That might be important in practice when you have more than one phase, for example for motors. \$\endgroup\$– starblueCommented Jun 17, 2011 at 7:52
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\$\begingroup\$ dsp.stackexchange.com/questions/431/… \$\endgroup\$– endolithCommented Oct 18, 2011 at 0:57
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3 Answers
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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.
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4\$\begingroup\$ I was never a fan of the graphical phasor approach but to each his own. You've got your clockwise / counter-clockwise backwards though, counter-clockwise is 'positive frequency'. \$\endgroup\$– MarkCommented Jun 17, 2011 at 6:43
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1\$\begingroup\$ @JGord, product-to-sum:
cos(x) * cos(y) = 0.5 * cos(x - y) + 0.5 * cos(x + y). I plotted0.5 * cos(99*t) + 0.5 * cos(101*t). WRT to signal processing, the spectrum of a 1 Hz cosine is two delta functions at +/- 1 Hz with weight 0.5. Multiplication in time is convolution in frequency, and convolving with a delta is a shift. When modulated by a 100 Hz carrier, the deltas at +/- 1 Hz shift to 99, 101 Hz and -99, -101 Hz, each with magnitude 0.25. That's 4 complex exponentials, or 2 cosines. \$\endgroup\$– Eryk SunCommented Jun 21, 2011 at 20:10 -
1\$\begingroup\$ @JGord your correct, it is just two waves multiplied together which can be explained completely in the time(real) domain. Where negative frequency comes in is that if you model that multiplication using a complex domain representation of those signals, you can think of it as just shifting the complex representation of the 1Hz wave up in frequency maintaining its positive and negative frequency components. Once your comfortable thinking about it in the complex domain this is a much simpler calculation than doing it in the time domain as the math the @eryksun provided shows. \$\endgroup\$– MarkCommented Jun 21, 2011 at 21:59
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2\$\begingroup\$ @JGord - while superimposed and multiplied (AM modulated) look similar you can easily tell them apart when looking at the positive and negative envelope. When superimposed the envelopes are in phase, when multiplied the negative envelope is a mirror image of the positive. \$\endgroup\$– stevenvhCommented Jun 22, 2011 at 5:48
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1\$\begingroup\$ @JGord - Sorry, I forgot the factor of
2*pi. I plotted0.5 * cos(2*pi*99*t) + 0.5 * cos(2*pi*101*t). The 1 Hz envelope emerges from the sum of the shifted positive and negative frequency components (-1 + 100 and 1 + 100). \$\endgroup\$– Eryk SunCommented Jun 22, 2011 at 9:03
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It can't in reality.
A full answer would take an entire text book but the basic answer is:
In signal processing signals are often discussed as a sum of complex sinusoids (\$e^{j\omega t}\$) because its mathematically convenient.
This leads to Euler's formula:
\$e^{j\omega t} = cos(wt) + j \cdot sin(\omega t)\$
Which leads to its inverse:
\$cos(\omega t) = \frac{1}{2} \cdot (e^{j\omega t} + e^{-j\omega t})\$
Which implies that both positive and negative frequency is present which is where it pops up in signal processing discussion.
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1\$\begingroup\$ But it should be clearly stated that "negative frequencies" do not exist in reality. However, its introduction simplifies many mathematic manipulations. \$\endgroup\$– LvWCommented Jul 22, 2015 at 15:01
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\$\begingroup\$ I rolled back the last edit. My point here is that negative frequency doesn't existing in 'reality', as in the 'real physical world', nothing to do with 'real valued' sinusoids. \$\endgroup\$– MarkCommented May 9, 2016 at 13:51
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The way I see it:
This is a complex sinusoid (\$e^{i\omega t}\$):
It can also be drawn less intuitively like this (left side), and has a one-sided spectrum like this (right side):
Negative frequency just means the helix is rotating the opposite way, and the spectrum is a delta function on the negative side of the frequency axis instead.
If you add a complex sinusoid of positive frequency with one of the same but negative frequency, the counter-rotating imaginary parts cancel out and it produces a real sine wave.
In this case, it's meaningless to talk about a sine wave with negative frequency, since a sine wave contains both positive and negative frequencies.
(I'd really like to make better illustrations of this, instead of copying these old poor-quality ones, but I've tried and it isn't easy. I think the 3D diagram of the spectra above is actually wrong. The delta functions should be parallel to the real/imaginary plane, and perpendicular to the frequency axis.)
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\$\begingroup\$ Hm. That third dimension didn't help for me. \$\endgroup\$– stevenvhCommented Jul 15, 2011 at 17:46
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\$\begingroup\$ @stevenvh: I rephrased it on DSP.se: dsp.stackexchange.com/a/449/29 \$\endgroup\$– endolithCommented May 22, 2012 at 21:12