Units, Symbols, and Terms, Oh My!

DonH50

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Jun 23, 2010
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I had intended this to be a short overview of basic terms. Didn't work out that way, sorry. At least it may help in defining some basic terms for people who have not seen them, or have seen them and never knew what they meant. More likely it will simply bore us all to tears. Oh, well! - Don


Before starting a discussion on sampling and such, I thought I should define some of the units, symbols, and terms most of us use but perhaps not everybody understands. I am paraphrasing many of these terms in an attempt to be clear and concise for the non-EE’s among us, not to be completely rigorous, so if you know better please bear with me…

I = current in Amperes (A). This tells how many electrons per second are flowing through a wire (let’s leave RF fields out of it for now).

R = resistance in ohms (). This is the resistance to current flow and is purely real, a physical thing. Often confused with impedance, which includes resistance and reactance (stay tuned).

V = voltage in volts (V), a measure of electromotive force. In the real world, current always meets resistance, and that generates a voltage. The symbol used is often E, for electromotive force, but I am going to stick with V for now.

Ohm’s Law: A current flowing into a resistance generates an equivalent voltage equal to the product of the current and the resistance, thus V = I * R = IR. This can be re-arranged to find whatever variable we want, e.g.

V = IR, I = V/R, R = V/I

P = power in Watts (W) and is the product of voltage and current, P = V*I. Since we have seen Ohm’s law, we can express power several ways:

P = VI = V^2 / R = I^2 * R, where “^2” means the square of the term.

It should be obvious that you can get lots of power with high voltage and very little current, or low voltage and lots of current. And everything in-between. More on power later.

Z = impedance and is like resistance in that it is the impedance to a current flow, but there are real and imaginary parts. (If you thought engineers had no imagination, now you know you were wrong. ;) ) Impedance is the sum of resistive (R) and reactive (X) components, with j = square root of -1 = an imaginary number: Z = R + jX.

The real part, resistance R, is what most of us are thinking when we calculate things, but in fact the reactive part often dominates. For an inductor, XL = jwL, and for a capacitor XC = 1/(jwC), where L and C are the values of the inductance and capacitance, and w = frequency in radians per second, or 2*pi*f with f in Hertz (Hz) – the latter is what most of us are used to seeing.

We usually work with the magnitude of Z, magZ = sqrt(R^2 + X^2) where X is wL or 1/wC or some combination, and treat it like it was just R in all the equations above. When a speaker says “8-ohm impedance”, what that really means is that for a goodly portion of the middle of the listening frequency band, the magnitude of Z is about 8 ohms. If you look at speaker impedance curves in a review, you may see numerous peaks and valleys, with our 8-ohm speaker dipping to 4 ohms or below at some points and into the 10’s of ohms at others. This is from the way the actual speaker drivers (usually a coil of wire around a magnet, so mostly L) and crossover (usually an RLC network) interact over frequency. It ain’t always pretty…

I realize this is clear as mud to many, but I want to do just a little more before moving on. I want to talk about how impedances combine in parallel and in series, since this answers the ever-popular “What happens if I connect two speakers to my amp?” question.

For resistors, or impedances, in parallel (plus to plus and minus to minus) the net resistance (impedance) is

Rptotal = 1 / (1/R1 + 1/R2 + 1/R3 + …), and you can replace R with Z if you wish.

For two resistors, this simplifies to Rptotal = R1 * R2 / (R1 + R2), which a little inspection will show is the same as Rptotal = R/2 if R1 and R2 are the same (and equal to just R). So, if you put two 8-ohm speakers in parallel across your amp’s terminals, it had better be able to drive a 4-ohm load.

Resistors in series simply add: Rstotal = (R1 + R2 + R3 + …) so clearly if you put those same two 8-ohm speakers in series (plus to minus to plus to minus) you now have a 16-ohm load.

What about power? Well, if the amplifier looks like a voltage source, as most do (or try), then remember P = V^2/R and let’s assume V doesn’t change. Then, if you halve the load by putting two speakers in parallel, the power doubles (but remember it is shared between the two speakers so each sees the same power as if it were alone). Of course, since I = V/R and we halved R, the current also doubled. Your amp had better be able to supply that current, and power, without overheating or current-limiting. In practice, most amps, including just about all AVRs, cannot put out that much power and so you typically see the 4-ohm rating (if there is one) is not twice the power of the 8-ohm rating. The output devices, power supply, and/or cooling system simply cannot handle that much power.

Eyes are glazing, but one last point: since XL = jwL you can plug inductance into the equations for R above and they will work just fine. That is, inductors in parallel divide, and inductors in series add. However, since XC = 1/ jwC (inverted), the equations are backwards for capacitors. Capacitors in series divide (lower C), and add in parallel (higher C).

The equations also show that inductors go up in impedance as frequency increases, and capacitors go down. Cables tend to exhibit series inductance and parallel capacitance, attenuating higher frequencies and passing the lower. However, these effects are very small at audio frequencies and for typical cable lengths and preamp and amplifier impedances. Naturally, “very small” is a matter of debate…

Frequency is measured in Hertz (Hz) and 1 Hz means the signal alternates back and forth (or positive and negative) once per second. 100 Hz means your woofer is moving back and forth 100 times a second, and 20 kHz means your tweeter is moving back and forth in one complete cycle 20,000 times a second.

We all learned about logarithms in school. Logarithms compress the dynamic range so we can work with smaller numbers. However, we can lose our sense of scale. Using decimals (units of ten), the exponent of a power of ten is the logarithm. That is,

1 = 10^0 so log(1) = log(10^0) = 0
100 = 10^2 and log(100) = log (10^2) = 2
10,000 = 10^4 and log(10^4) = 4

The numbers don’t have to be pretty. For example, 150 is between 100 (log is 2) and 1000 (log = 3) so the logarithm should be between 2 and 3. In fact, log(150) = 2.176. It should be clear that going from 2 to 3 in logarithms is a big step, i.e. from 100 to 1000. Unfortunately, it’s easy to forget this…

Logarithms are important because many things are expressed in units based upon logs. For example, the ubiquitous decibel, dB is 1/10 (“deci”) of a Bel, another unit seen frequently. Units of dB are used to express logarithmic ratios of numbers, like volts (dBV), milliwatts (dBm), femtowatts (dBf, seen in tuner specs), and watts (dBW).

dBW = 10 * log10(P) where P = power in Watts and log10 is a logarithm to base ten.
Alternatively, P = 10^(dBW/10)

dBV = 20 * log10(V) where V is the voltage in volts, and
V = 10^(dBV/20)

Why the difference? Remember power is related to voltage squared, so we can move the 2 out front when using voltages (or voltage ratios). Some ratios are unit-less, like the signal-to-noise ratio (SNR), since signal and noise are both expressed in the same units and disappear in the ratio:

SNR = 20 * log10 (signal voltage / noise voltage) = 10 * log10(signal power / noise power)

The key to remember is that logarithms compress the range of numbers. That is, the actual number is exponentially related to the logarithm: 10 dB => 10^(10/10) = 10 times the reference power! When you turn up the volume by 10 dB, you are requesting 10 times the power from your amp. 3 dB is not a large increase in volume, but requires twice the power (10^3/10 = 2). For music, an interesting note is that the peak-to-average ratio is 17 dB, which means you need about 50 times the average power to handle peaks without clipping. That’s why big amps are in vogue, and also why going from 100 Wpc to 120 Wpc is not really helping you any…

Bits in a digital system are also related to logarithms, but to base two :

Nbits = log2(quantization levels) where
Nbits = number of bits (resolution), and
quantization levels = the discrete values the signal can take on.

Since digital signals use binary math, i.e. 0 or 1, there are just two levels in each digit. We are used to counting in decimal digits, where each digit can take on ten values (0 through 9). Thus, 64 is 6 “tens” and 4 “ones”, or 6*10^1 + 4.*10^0. In binary, it takes more digits since each can only take on one value: 64 =2^6 or seven binary digits are needed. Thus 64 in binary is 1000000 = 1*2^6 + 0*2^5 + 0*2^4 + 0*2^3 + 0*2^2 + 0*2^1 + 0*2^0.

To find out how many bits are needed for a number, we can take the log of that number. For example, use the number 197. Note log10(197) = 2.294 as expected since it is between 100 and 1000. That is, 10^2.294 = 197 (ignoring rounding errors). Since 2.294 is greater than 2, we need three decimal digits to express 197. In binary, 197 = log2(197) = 7.622, or 2^7.622. We need 8 digits (bits) to express 197 as a binary number. (If your calculator does not do logs to base 2, then divide the base-10 log of the number by the log of 2, e.g. log10(197)/log10(2) gives the same answer. Ain’t math fun?) Digital numbers are often found in other formats, including octal (base 8) and hexadecimal (base 16), but binary digits are what most of us have in mind when we hear a digital number. More on digital stuff later…

Lastly (for now) let’s look into RMS values. You often see Volts or Watts RMS in a specification and perhaps have wondered what that meant. RMS stands for “root-mean-squared” and is just that: the square root of the mean (average) squared when computed (integrated) over time (usually one period). The RMS value causes the same heating in a resistor as a d.c. voltage (power) of the same value. That is, for a d.c. voltage, the measured d.c. and RMS values are the same. For single tone (frequency) sine wave, like a pure musical note (perhaps played on a flute), if we measure the voltage and get 1 Vrms then the peak value is 1.414 V (Vrms * sqrt(2)) or 2.828 V peak-to-peak (2*sqrt(2)).

The socket in your house (in the USA, anyway) puts out about 120 Vrms, or 170 Vpk, or 339 Vpp. When your amplifier is driving 100 Wrms into your 8-ohm speaker, that’s about 28.3 Vrms or 80 Vpp (remember P = V^2/R). No wonder the back says to not touch the speaker terminals when it’s playing!

I’m sure I left out something, and probably bobbled a few equations, but hopefully it’s a start. Next, on to sampling and why bits are not always just bits…

p.s. Is there any way to get a symbol font?
 
Last edited:

Johnny Vinyl

Member Sponsor & WBF Founding Member
May 16, 2010
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Nicely done Don

takes me back to high school physics

To this day I have a difficult time spelling FIZIKS or any of those other ones! :eek:

DON - Well done indeed and thank you!

John
 

DonH50

Member Sponsor & WBF Technical Expert
Jun 23, 2010
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More terms...

I decided to add a few terms thrown about recently just for grins.

Resistance and capacitance has already been mentioned, but we can put them together as an R in series with the signal and at the other end a capacitor to ground and define bandwidth:

BW = bandwidth = 1 / (2*pi * R *C) for a first-order (single R, single C) system driven by an ideal source and with infinite (no) load

The bandwidth of a system is where the value has dropped to one-half its initial ("d.c." or low-frequency) value, or -6 dB in voltage (or current) and -3 dB in power. At that frequency, the reactance of the capacitor (Xc = 1 / 2*pi*f*C ) is equal to the resistance. If you know your source (e.g. preamplifier output) and load (amplifier input) impedances and a little about the interconnect cables (e.g. typical interconnects are around 20 to 30 pF/ft) you can roughly calculate the bandwidth at the amplifier's input. This does not always work, as the inductance (neglected for simplicity in this example) may further decrease the bandwidth, and sometimes the cables are designed with specific RLC characteristics to provide a better match and then this simple equation does not hold.

If we know the bandwidth of a first-order system, calculating the rise time is pretty simple:

tr = rise time = 2.2 / (2*pi*BW)

This comes about because charging a capacitor through a resistor happens exponentially (to base e, e = 2718...), and to get to time we perform an inverse (sort of) operation using natural logarithms:

tr = R*C*ln(end_value/start_value) -- plugging in 10 and 90 for typical 10% - 90% rise time yields the 2.2 constant (ln(9) = 2.1972...)

A little fiddling with the numbers shows that BW = 0.35 / tr if we need to go the other way.

Slew rate tells how fast a signal changes in a given amount of time. The signal can be voltage, current, power, joules, or whatever, but in the audio world we mostly see the voltage slew rate, dV/dt. We have to know the type of signal, of course, because we have to know how fast the edges are to figure out the slew rate. For a sine wave, i.e. a single tone at frequency f, with amplitude A (say in V peak), we find:

SR - slew rate = 2*pi*f*A

A 1 kHz sine wave at 1 Vrms yields SR = 2*3.14159*1000 (Hz) * 1 (V) * sqrt(2) (to find the amplitude) = 8886 V/s or 8.9 kV/s (no, that is not very fast). An amp putting out 100 Wrms at 20 kHz into 8 ohm speakers (goodby, tweeters!) is slewing at 5.03 MV/s (5.03 million volts/second). If I did the math right (check it!) A bit more respectable. (In my day job I deal with slew rates well above 1 V/ns, or 1 GV/s, 1 billion volts per second. Now we're talking! Just not about audio...)

Propagation delay is something designers worry about (particularly that it is linear through the pass band) but not often discussed outside our Little world. Nevertheless, I have the formula on my little cheat sheet in front of me, so will pass it on. It derives the same way as rise time for a first-order system, but now we look at the center-crossing (ln(0.5)) and thus:

tpd = propagation delay = -RC(ln(0.5) = 0.694RC = 0.110 / f

Now you can find the delay at any frequency and decide if it matters. Remember in a wire signals travel about 0.7c, or about 0.7*1.86*10^5 = 130.2 miles/s = 687 million feet/second. Sound travels at around 1127 feet/second.

And lastly,

temperature in degrees F = (9/5) * (degrees C) + 32
degrees C = (5/9) * (degrees F - 32)

Only because I always forget those two little formulas... :)

HTH - Don
 

c1ferrari

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May 15, 2010
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Hello DonH50,

Thanks for the tutorial
!
 

flez007

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Aug 31, 2010
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I am glad I almost remembered them all!, but reading again after so many years had me understand them even better. It was usefull for me at school to understand electric current happening jusy as water flowing thru a hose... Thanks for restoring the primer Don.
 

Speedskater

Well-Known Member
Oct 1, 2010
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Don wrote:
Lastly (for now) let’s look into RMS values. You often see Volts or Watts RMS in a specification and perhaps have wondered what that meant. RMS stands for “root-mean-squared” and is just that: the square root of the mean (average) squared when computed (integrated) over time (usually one period). The RMS value causes the same heating in a resistor as a d.c. voltage (power) of the same value. That is, for a d.c. voltage, the measured d.c. and RMS values are the same. For single tone (frequency) sine wave, like a pure musical note (perhaps played on a flute), if we measure the voltage and get 1 Vrms then the peak value is 1.414 V (Vrms * sqrt(2)) or 2.828 V peak-to-peak (2*sqrt(2)).

The socket in your house (in the USA, anyway) puts out about 120 Vrms, or 170 Vpk, or 339 Vpp. When your amplifier is driving 100 Wrms into your 8-ohm speaker, that’s about 28.3 Vrms or 80 Vpp (remember P = V^2/R). No wonder the back says to not touch the speaker terminals when it’s playing!


While we use RMS Volts and Amps to determine power, Watts are not RMS Watts, they are just Watts.
 

DonH50

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Jun 23, 2010
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Sorry, I may have lapsed into techie-speak without being rigorous... We see RMS, average, peak, instantaneous, mean, real, imaginary, reactive, etc. Watts in communications all the time, but as I think I have said I am (and work with) hairy-knuckled engineers, not the real scientist types. My grammer ain't so great and definitions sometimes follow... :) While Wrms may not be rigorous, it tells us how the power was derived, just like Wavg, Wpk, etc.

Apologies - Don
 

DonH50

Member Sponsor & WBF Technical Expert
Jun 23, 2010
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I would say so, at least in "common English". Slang? And also judging by how many audio spec sheets I see that say "W RMS" or "RMS power"... In my experience, speaker companies seem to use it more than amplifier companies.
 

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