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I am a software developer (using high level languages like .NET,C,C++ etc) trying to understand how computers work at a lower level.

I understand that amplitude is always positive because it is calculated by (top-bottom)/2. However, I don't understand what a negative amplitude actually is i.e. what does it mean if the wave falls below the equilibrium (0).

The negative values that appear confusing are given in decibels (dB).

This is probably more of a physics question but I am trying to understand analogue circuits.

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Do you mean the amplitude of something like a sine wave in dB? – Oli Glaser Jan 13 '13 at 11:27
@Oli Glaser,yes. – w0051977 Jan 13 '13 at 11:29

A decibel (\$dB\$) is a way to express a ratio. Most practical uses of decibels are measuring some thing in relation to some other thing. A negative number of decibels indicates that the thing being measured is less than the reference thing.

Let's consider as an example \$dBm\$, a unit that measures a power \$p\$ relative to \$1mW\$. Thus:

\$ P_{dB} = 10 \log_{10}\left(\dfrac{p}{1mW}\right) \$

So 1mW is:

\$ 10 \log_{10}\left(\dfrac{1mW}{1mW}\right) = 10 \log_{10}(1) = 0 dBm \$

What about \$100mW\$?

\$ 10 \log_{10}\left(\dfrac{100mW}{1mW}\right) = 10 \log_{10}(100) = 20 dBm \$

What about \$2\mu W\$?

\$ 10 \log_{10}\left(\dfrac{2\mu W}{1mW}\right) = 10 \log_{10}(0.002) \approx -26.99 dBm \$

When we are considering something like voltage, it's customary to consider the ratio of the squares of the values, because power is proportional to the square of amplitude. For example, \$1V\$ on a \$1\Omega\$ load is \$(1V)^2 / 1\Omega = 1W\$, but if the voltage is 2V then \$(2V)^2 / 1\Omega = 4W\$. I think this is a screwy convention, and if you want your measurements expressed in decibels to be like power, then you should measure power. But, it's the convention, and you can probably blame the engineers who developed the telephone network.

Anyway, let's consider \$dBV\$, which uses 1V as the reference. Here's an example with \$1V\$:

\$ 10 \log_{10}\left(\dfrac{(1V)^2}{(1V)^2}\right) = 20 \log_{10}\left(\dfrac{1V}{1V}\right) = 20 \log_{10}(1) = 0 dBV \$

Notice that rather than squaring both voltages in the fraction, we can multiply the logarithm by 2. The two are mathematically equivalent, but multiplying by 2 is easier than squaring.

\$ 20 \log_{10}\left(\dfrac{120V}{1V}\right) = 20 \log_{10}(120) \approx 41.58 dBV \$

\$ 20 \log_{10}\left(\dfrac{3mV}{1V}\right) = 20 \log_{10}(0.003) \approx -50.47 dBV \$

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If we are talking about an amplitude level such as voltage, then the formula is 20 * log10(Vsig/Vref) rather than 10 * log10(Vsig/Vref). So for example 100V is 40dBV, not 20dBV. The 10 * log10 is for power quantities, so 100mW in dBm would be 20dBm. – Oli Glaser Jan 13 '13 at 17:04
@OliGlaser right you are. Edited. – Phil Frost Jan 13 '13 at 18:43
There's a typo in the 2µW formula (s/V/W/) - I'd fix it but third party edits need to be at least 6 characters. – Paul R Jan 13 '13 at 21:06

The level for something like a sine wave is generally given as the RMS (Root Mean Square) value, which (for a sine wave) is 0.707 of the peak value.

For example, 240VAC mains voltage is actually (1/0.707) * 240V = 340V peak to peak - the RMS is used as this is the equivalent of the DC value power wise (i.e. 240VDC would provide the same power as 340VAC pk-pk) Since the RMS value is usually assumed, if you mean peak tp peak you should write e.g. 240VAC pk-pk if the highest pont is +/- 240V

Negative amplitude means the signal is attenuated relative to a reference point, so if you see e.g. -20dB, it means the signal is 1/10th of the reference value. dB on it's own is unitless, so you will see things like dBm (relative to 1mW → 0dB = 1mW), or dBV (relative to 1V → 0dB = 1V)

So if you see -3dBV, this means the level is 0.707 * 1V = 0.707V and -20dBV would be 0.1V.

Similarly 20dBV would mean 10V.

(In the below calculations log10 refers to the base 10 logarithm, as opposed to the natural logarithm or e.g. log2 for base 2 logarithm) The calculation for dB is 20 * log10(signal/ref), so for the above:

20 * log10(10/1) = 20dBV

For the 0.707 case:

20 * log10(0.707) = -3dBV

1mV in dBV would be:

20 * log10(0.001/1) = -60dBV

For measurements of power, the calculation is:

10 * log10(power_level/ref_power_level) so for example, 100W in dBW would be:

10 * log10(100/1) = 20dBW

So a negative amplitude means a reduction in amplitude relative to a reference point.

See the Wikipedia page on Decibels.

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Technically that's negative log amplitude, the amplitude itself is still positive. But in common usage, you're right. – Brian Drummond Jan 13 '13 at 12:04
"Amplitude in something like a sine wave is generally given as the RMS (Root Mean Square) value" - couldn't this be misleading? Consider a sine wave, zero offset, 2Vpkpk. The amplitude of this signal is 1V, the RMS value is 0.7V. It's clear where you're heading when you continue on to dB, but amplitude and RMS are not necessarily the same thing. Not saying you're wrong, but it really depends a lot on the context, and someone could confuse amplitude and RMS. Maybe level would be better than amplitude? – zebonaut Jan 13 '13 at 12:19
@zebonaut - fair point, I will alter the wording to "level". – Oli Glaser Jan 13 '13 at 12:34
@Brian - yes, the amplitude is always positive, so the question confused me initially (which is why I asked about dB) whether it was about the negative swing or the amplitude in reference to something. After the comment/edit to question, it appears the confusion is over negative dB values. – Oli Glaser Jan 13 '13 at 12:39

The question is a little unclear to me : but if you mean, how is amplitude measured or defined while the signal is below 0V, then remember the difference between speed and velocity : amplitude (like speed) is a magnitude, and is either zero or positive.

The signal (like velocity) is a vector : velocity is defined by speed and direction; signal (restricting the discussion to cosines for the moment) is defined by amplitude and phase. Thus the negative peak -V of the signal is defined as amplitude V and phase Pi (or 180 degrees).

More complex signals can be represented as a sum of different cosine waves with different frequencies, amplitude and phases, the Fourier transform is a technique for translating an arbitrary waveform into such a representation (and back again)

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Decibels describe the ratio of signal strengths, according to how many factors of ten a new signal (such as some circuit's output) is compared to the original or some standard reference signal.

When the output is smaller than the input, you'll have to divide by some factors of ten -- the same as multiplying by 1/10, which is (10)^(-1). Thus negative decibels.

In the illustration, the big signal is an input to some gadget, and I made up the value 15.0V for its peak (from zero) amplitude. For a sine, the RMS voltage is 1/sqrt(2) of the peak amplitude. The peak-to-peak is double. The second sine wave has a smaller amplitude. If we imagine applying these sine waves to a simple load (the resistor), currents will flow in proportion to the voltages.

Power is voltage times current, so the smaller signal's power (heating the resistor) is (0.4)^2 of the original's power. This power ratio is what engineers usually care about.

Engineers, being fond of slide rules and easy math, use base ten logarithms for a lot of things. A chain of amplifiers and lossy filters can be dealt with more easily by adding logarithms of gains and losses, instead of multiplying the gains and loss factors. A factor of 10 is one "Bel" but since we're often dealing with fractional quantities like 0.3 Bel (a doubling of power), for ages we've been using decibels to shift that decimal point over.

Note that dB always (usually) refer to power and not voltages. Note also that it doesn't matter if we use peak amplitude, peak-to-peak, or RMS, as long as we're consistent measuring the input and output the same way.

Illustration showing relation between amplitudes, power, decibels.

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Zero decibels means unity gain, or no change in signal level, because \$10^0 = 1\$.

Decibels are usually some relative measure, like output related to input.Positive decibel values are increases in signal level (amplification), and negative decibel values are decreases (attenuation).

I recently created a panel where some knobs are labeled as going from \$-\infty\$ to \$0\$ decibels, with a gradation of negatively valued ticks in between. This reflects the fact that the knob is a linear potentiometer which attenuates the input signal. \$-\infty\$ means that the signal is completely trimmed to zero, and \$0\$ means that the full signal is passed through. The midpoint is marked \$-6\$ because the voltage is cut in half. Voltage cut in half means power is reduced to a quarter, which is about six decibels down: \$20\times\log_{10}(0.5)\$.

There exist scales of measure in which decibels are associated with some absolute level. In those scales, zero decibels will refer to a specific absolute voltage or wattage or other quantity. For example, in the dBm scale, 0 dB is one milliwatt. In the dBu scale, zero decibels is 0.775 VRMS.

Regarding \$-\infty\$ dB: that's a bit of an abuse of notation that appears on instrumentation, which everyone understands. Logarithms are not defined for zero, but grow large as their argument approaches zero from above. Of course, infinity is not a number, and a zero signal has no defined decibel value.

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I wouldn't consider the −∞ as an abuse of notation any worse than the ∞ mark on camera focus dial. Basically, it marks a limiting case. On a lens with a 100mm focal length, if the subject is at distance d, from the lens, the film should be distance 1/(1/(100mm)-1/d). If one wanted to mark the lens for 10m, 100m, 1km, 10km, and 100km, the marks for the larger distances would be so close that resolving a focal distance to even within an order of magnitude would be tough. Simply using a ∞ mark is clearer. – supercat Jan 18 '13 at 2:40

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