by dervast
 P: 133 Hi i was reading somewhere and i have found the following: loss: 6db/octave 20db/decade I know what db is but i dont know what octave is...
 P: 2,056 Isn't an octave every eight notes on the piano, with each one having half or double the frequency as the one before it?
 Emeritus Sci Advisor PF Gold P: 6,236 I'll move this to electrical engineering. An octave is normally doubling signal intensity, while a decade is multiplying by 10 signal intensity.
 P: 133 Octave vs decade I am not sure i really understand what 6db/octave means.. The classic db i have learned it is used for decades.. right?
 Mentor P: 40,664 An octave is a doubling of frequency. A decade is 10x frequency. If you have a single-pole network, the gain through it ratios directly with frequency. So at double or half of some frequency, you get double or half the gain through the network (assuming you are on the rolloff or rollup side of the pole). To convert from amplitude to dB, use: A[dB] = 20 log (A/A0) where A0 is some amplitude reference like uV. To convert from power to dB, use: P[dB] = 10 log (P/P0) where P0 is some power reference like mW. So using the first formula for amplitude in dB, you can see that a 10X increase over A0 will give you 20dB, and hence for a single pole rolloff, you will get 20dB per decade of frequency change. For a doubling, you will get 6dB per octave of frequency change.
HW Helper
P: 1,326
 Quote by berkeman An octave is a doubling of frequency
 Quote by Mk Isn't an octave every eight notes on the piano, with each one having half or double the frequency as the one before it?
Note what berkeman has described is consistent with what Mk referred to in music. The 8th note of a major scale is the octave which is 2X the frequency of the starting note. It has the same meaning. The 880Hz A is an octave above 440Hz A and 220Hz A is an octave below.

 Quote by berkeman To convert from power to dB, use: P[dB] = 10 log (P/P0) where P0 is some power reference like mW.
An illustrative example comparing dB to power; if a transmitting antenna gives you 3dB of gain it translates to 2X (double) the effective radiated power (ERP). If your transmitter outputs 100W, with 3dB gain of the antenna, you effectively transmit 200W. With 9dB antenna, the ERP is (2X)^3 or 800W.
P: n/a
 Quote by Ouabache An illustrative example comparing dB to power; if a transmitting antenna gives you 3dB of gain it translates to 2X (double) the effective radiated power (ERP). If your transmitter outputs 100W, with 3dB gain of the antenna, you effectively transmit 200W. With 9dB antenna, the ERP is (2X)^3 or 800W.
Not really! The gain of a directive antenna is its power output in the direction it radiates relative to the power output of an omnidirectional antenna.
If you have a 100 W transmitter plugged to an omnidirectional antenna, it will radiate 100 W in all directions (a sphere). If you plug it to a 3 dB gain antenna, it will still radiate 100 W, but in an hemisphere. A 9 dB antenna will radiate 100 W in a solid angle of $$\frac{2\pi}{8}strd$$.
HW Helper
P: 1,326
 Quote by SGT Not really! The gain of a directive antenna is its power output in the direction it radiates relative to the power output of an omnidirectional antenna.
You're correct, directional antennas (e.g. Yagi, Quad, beam etc), have more gain (and ERP) in one direction. It would be clearer to footnote my discussion compared to a unity gain standard antenna such as an omnidirectional. It is also a good approximation for a dipole antenna.
See the table at the bottom of this page for an illustration of gain versus effective radiated power. It is a good rule-of-thumb.

Just try it in the power formula, a variation of the one berkeman gave us.
He had typo, it should be
$$dB_{power}=10log[\frac{P_{out}}{P_{in}}]$$
applied to erp becomes:
$$dB_{power} = 10log[\frac{P_{erp}}{P_{xmtr}}]$$

Solving for $P_{erp}$ becomes
$$P_{erp}=P_{xmtr}{log^{-1}[0.1(dB_{power})]}$$
Try it, you will see you get a doubling in effective radiated power for each 3dB gain.
P: 2,251
 Quote by vanesch I'll move this to electrical engineering.
dunno why, but even though i'm an EE, i almost never come here.

 An octave is normally doubling signal intensity, while a decade is multiplying by 10 signal intensity.
don't you mean?:

An octave is normally doubling signal frequency, while a decade is multiplying by 10 signal frequency.
Mentor
P: 6,228
 Quote by rbj don't you mean?: An octave is normally doubling signal frequency, while a decade is multiplying by 10 signal frequency.
Yes, but see post #5 by berkeman.

Regards,
George
 P: 133 Thanks lot reading all the above posts now i know that each octave is 6db larger thatn the previous octave... Still i cant understand that 20db/decade!!! What thats mean?
Mentor
P: 40,664
 Quote by dervast Thanks lot reading all the above posts now i know that each octave is 6db larger thatn the previous octave...
No. That is not what we said. The octave itself is not changing. The value of some quantity is changing by some amount with respect to frequency, and that amount can be measured with respect to octave or decade changes in frequency.

 Still i cant understand that 20db/decade!!! What thats mean?
Okay, let's try it this way. Do you have some log-log graph paper handy? If not, download some from here:

http://www.csun.edu/~vceed002/ref/me...aph_paper.html

Now on a piece of log-log graph paper, label the horizontal axis frequency, and the vertical axis impedance. Now I would like you to draw the plot of the |Z(f)| impedance of a capacitor. Let's use some semi-round numbers like a 1uF capacitor and frequencies ranging from 1kHz to 10MHz or so. What are the impedances of the 1uF cap at those two frequency extremes? Using those impedances, figure out how to label the vertical axis of the plot, and draw the plot of the |Z(f)|. Hint -- it will be a straight line on the log-log graph, with a slope of -1.

Now look at that plot and think about what it means. Remember that the impedance of a capacitor goes as 1/2PI*f*C, so it is inversely proportional to frequency, right? That means that a 10x increase in frequency gives you a 10x lower impedance. Recall from the dB equations earlier in the thread that for a non-power measurement you express the change in value in dB by using this:

change [dB] = 20 log (new value / reference value)

A decade is 10x or 1/10x in frequency, so you can say that the impedance of a capacitor changes at +/-20dB per decade. Do you see that now?

Now look again at the graph, and see how many dB the line changes for each doubling or halving of the frequency (that's the octave change in frequency that we talked about). You should get about 6dB (6.021... actually).

Hope that makes better sense now. You will use this concept a lot, so please spend enough time with the log-log plots to get an intuitive feel for how quantities can be expressed this way.

Quiz question -- if you have a 2-pole lowpass filter, at what rate does the transfer function Vo/Vi roll off once you get a decade past the break frequency? How many dB per octave? How many per decade?
 P: 4 Hi folks: Be careful here. If the original post refers to audio/sound (20Hz to 20kHz), then dB refers to voltage, not power. When referring to sound applications 6dB is half voltage (3dB is half power). dB(voltage) = 20 log (out V/in V). Power depends on load/impedance, which depends on the gear you use (and matching), so sound guys use voltage, not power as their dB reference. All audio equipment (mixers, sound boards, and most gear except amplifiers) will refer to voltage. The faders on every sound board are calibrated in Voltage (-6dB for half, not 3) i.e. 6dB atten per octave - 1 octave could be from 2kHz to 4kHz - Voltage down from .7 V to .35 Volts in that octave 20 dB atten per decade- in Voltage/sound, 20dB is an attenuation of 10 (not 100), so a voltage drop from .7 to .07 (not .007). Hope this helps.
 Sci Advisor PF Gold P: 3,515 since power is proportional to square of voltage and a decibel is logarithm; and, when you square something you double its logarithm so 20 db of voltage is same as 10 db of power since the base unit 'bel' is a ratio it begs the question "ratio to what?" often electric sound signals are referred to a power of one milliwatt into 600 ohms which works out to 0.775 volts RMS. that's because the unit grew out of the telephone industry, hence the name Bel after Alexander Graham. On better analog multimeters you'll see a db scale. it's logarithmic with with zero db at that AC voltage.
 P: 311 20dB/Decade means that the signal is increasing 20dB every time the frequency gains an order of magnitude. For instance if you plot frequency response and the signal gains 20dB from 1Hz to 10Hz then the slope is 20dB/decade.
 P: 4 Hi Jim: The 0.775V reference actually came from CBS during the early days of television. Yes, 1mW over a 600 ohm load. It is the heart of the old VU (volume units) meter. It is now referred to as the -10dBV relative to 1V as 0dBV (do the math, you get .775V) reference. This is also known as the 0dBu reference level. Most digital and pro audio systems now use a +4dBu reference which is 1.23V. -10dBV=0dBu, so do the math on +4dB gain with .775V as the 0dB reference (you get 1.23V at output). Your home stereo equipment is all referenced to the -10dBV system (or 0dBu, same thing) level as "0 dB", where as the sound gear, including the mixing board in the studio, or at a concert uses 0dB gain as referenced to +4dBu (or 1.23V). Play pro-audio on your home stereo, and you'll notice how loud it is or blow the speakers. Play a CD player into a pro audio mixer/amp combination and find out how quiet it is - turn up the pre-amp!. It is amazing, since the units do correlate, and yes power is relative to i squared or V squared. However, 20dB per decade if your talking power, is 100x, but only 10x if your referring to voltage. Don't draw voltage response on a power logarithmic graph however, or you'll be wrong. That 1/2 "x" (x=power OR voltage) point will be at 6, when it should be at 3. Whenever I hear 6dB per octave, I'm quite sure you're talking voltage. If I hear 3dB per octave, I'm quite sure your talking power. Typical roll off of common op-amps.
 P: 4 ooops, got that wrong. -6dB is half voltage and quarter power at the same time. They are all equivalent points on the same log curve. Amazing how hard it is to break a bad habit. Glad I got that mistake out of my head, finally.
 Sci Advisor PF Gold P: 3,515 Jack Thanks for that post! i am interested in audio but have no familiarity with pro-stuff. does pro video use a higher signal too?

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