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What is the logic behind the cutoff frequencies being defined at the 3dB points?
why do we use the 3dB points only. is there some thing to do with transient time?
these are some things we all study in our courses but what's the basis of these concepts. any one who can explain these basic concepts which we use everyday. ???

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In the Bode plot:

Cutoff frequency ( lower or upper ) are the frequency wherein the midrange gain is 3dB less.

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What is the logic behind the cutoff frequencies being defined at the 3dB points?
why do we use the 3dB points only. is there some thing to do with transient time?
these are some things we all study in our courses but what's the basis of these concepts. any one who can explain these basic concepts which we use everyday. ???

This is a method of calculating bandwidth. The bandwidth is the the difference between the upper and lower 3db points.

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Hi All,
I may not be 100% correct,but,a loss of 3db in gain does not hamper the overall response of the amplifier.i.e this loss is unnoticable in the o/p of amplifier.But we get higher bandwidth at peak- 3db gain,compared to bandwidth obtained considering only the flat portion of the gain-frequency plot.
I would like someone to give more explanation of it.

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Hi All,
I may not be 100% correct,but,a loss of 3db in gain does not hamper the overall response of the amplifier.i.e this loss is unnoticable in the o/p of amplifier.But we get higher bandwidth at peak- 3db gain,compared to bandwidth obtained considering only the flat portion of the gain-frequency plot.
I would like someone to give more explanation of it.

At -3dB the output power is half the midband power output. ( Yes it's noticable )

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Dear kebloks, Gazza
I think Sukhbinder already knows what you said.
His question (if I got it right) is: why do we use the -3dB figure for calculating our cut-off freq.s why not 2dB or any other value.
This has been on my mind for a long time.
My guess is there are some practical reasons behind it

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3dB is chosen as the cutoff frequency amplitude because it is half-power and is noticeable by most people as a small reduction in the level of sound or light.

6dB is half or double the amplitude.

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it depends on the type of filter -3db is just a number thrown out there, the actually cutoff depends on the application.

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Wikipedia agrees with the output at -3.01dB down (half-power) for a cutoff frequency.
It doesn't matter what type is the filter, Butterworth, Bessel or Chebyxxxx.

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*1 The cut-off frequency is not strictly defined as the frequency at which attenuation begins. In Butterworth type low-pass filter, the cut-off frequency is defined as the "–3 dB point of total characteristics". In a Chebyshev type low-pass filter, the cut-off frequency is defined as the "first point beyond the maximum ripple amplitude of a passband". Roughly, however, the cut-off frequency can be defined as the frequency at which attenuation begins.

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Thanks for the replies guys but i'm still in the dilemma that why the 3dB spec only.
Leaving aside filters, there is a cutoff freq for a device also (say a transistor). any body having any idea about transisent time concepts. i know these are the fundamentals but they are never thought in our courses as they should have been. we are more focussed on the applicationa than the theory behind it. ;)

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the cutoff in a transistor is determined by the base junction capacitance and the current gain bandwidth along with the beta gain and also the class in which it's biased or if purely switching.

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That value of 3dB is not "chosen".

And it's not exactly 3dB. Just like pi is not exactly 22/7. It's nature, and the maths we use to model it, that chose something close enough to 3dB that today we use that figure conventionally, as we use Pi.

Mathematically we say that a first order filter's response to a step input is related to e(-t/T). T is said to be the time constant (in seconds) and t is the time elasped after the step. Euler's number e is not chosen (unless you argue divine design), it's natural. The value of T, if anything, is the only arbitrary choice here. But for mathematical convenience we've decided that it's the time when the expression is equal to e-1, in the mathematical model of the system.

Now, it just so happens that this filter will attenuate high (or low) frequencies by 3dB at the frequency 1/(2.Pi.T). Nobody said 'let's use 3dB as the convention'. Rather nature says "in first order filters I will attenuate signals by 3dB at the frequency which is 1 divided by 2 divided by Pi divided by the time constant of the filter, and that time constant shall be the time it takes for blah blah blah'.

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Gazza,
Definition of the cut-off frequency has nothing to do with the sampling theory. In your diagram to allow a signal to have a bandwidth of f2-f1 we need to at least sample it at 2f2.

Also as quoted from 'All About Electronics', the amplitude at the 3dB frequency is around 70%(sqrt(2)/2) of the centre frequency; and the power is half.

Although not fully convinced, the most interesting is Cabwood's post.
Mathematical Convenience

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Although not fully convinced, the most interesting is Cabwood's post. Mathematical Convenience

LOL! Not entirely convinced! Interesting!

For a simple low-pass RC network, the cut-off frequency is 1/(2.Pi.R.C). The voltage gain for a simple sinusoidal input at that frequency is

Vout/Vin = 1 / Sqrt(2) = 0.707 = 71%

The power gain is the square of this (power is proportional to square of voltage):

Pout/Pin = 1 / 2 = one half. Very convenient.

Expressed as decibels:

10 log (1/2) = -3.01dB

Not exactly 3, as I said. But conventionally engineers have decided that it's close enough to 3 to be able to call the cut-off frequency the '3dB point'. It could be argued that '10 log(Pout / Pin)' stinks of arbitraryness, but Mr. Bell found that log-base-ten simplified the maths quite a lot where power (not amplitude) is concerned, and engineers subsequently noticed that multiplying the logarithm of gain by 10 made 10dB equal to a power gain of 10. All for convenience. If Mr. Bell had settled upon the natural logarithm as the way to go, then the numbers would be a nightmare.

I can imagine the meeting: 'Hey, 3 is nice round number, and half is a fantastically convenient gain, AND it happens to be roughly the frequency which is the inverse of the time constant of the system. Let's use it to define where a system stops or starts passing signals, and we can call it the 3dB point. Waddya reckon?" Surely it must have gone down to a vote.

It didn't escape these guys that if you created a second order filter by cascading two identical first order RC filters, at frequency 1/(2.Pi.R.C) the response would be -6dB (1/4), which wasn't so convenient. They still wanted to define the 'half power gain' point of their systems, and so Butterworth and Chebyshev and all the other second order filter folk, for convenience, worked out the frequency where their own designs had a power gain of a half (the 3dB point), and used those instead of the 6dB points to define the passbands of their designs. Just for consistency, you understand.

For 3rd order filters, which increased attentuation to 9dB at that f=1/(2.Pi.T) frequency, they did the same. "Let's not use the 9dB point to sell our designs", they said. "Stick with the conventional 3dBs" was their choice.

So, to eat my words, 3dB was chosen to describe higher-than-one order active designs, but only because nature had a convenient "nearly 3" response in it's own passive systems.

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Hi Cabwood,
Good summary of the "chosen" -3dB cutoff point. ;D

You forgot another "convenient" attribute of -3dB: It is a small drop in level that can be noticed by most people. Picture Bell cutting a speaker's power in half with a switch in his meeting, "Hear the small difference guys?".

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LOL! Not entirely convinced! Interesting!

For a simple low-pass RC network, the cut-off frequency is 1/(2.Pi.R.C). The voltage gain for a simple sinusoidal input at that frequency is

Vout/Vin = 1 / Sqrt(2) = 0.707 = 71%

The power gain is the square of this (power is proportional to square of voltage):

Pout/Pin = 1 / 2 = one half. Very convenient.

Expressed as decibels:

10 log (1/2) = -3.01dB

Not exactly 3, as I said. But conventionally engineers have decided that it's close enough to 3 to be able to call the cut-off frequency the '3dB point'. It could be argued that '10 log(Pout / Pin)' stinks of arbitraryness, but Mr. Bell found that log-base-ten simplified the maths quite a lot where power (not amplitude) is concerned, and engineers subsequently noticed that multiplying the logarithm of gain by 10 made 10dB equal to a power gain of 10. All for convenience. If Mr. Bell had settled upon the natural logarithm as the way to go, then the numbers would be a nightmare.

I can imagine the meeting: 'Hey, 3 is nice round number, and half is a fantastically convenient gain, AND it happens to be roughly the frequency which is the inverse of the time constant of the system. Let's use it to define where a system stops or starts passing signals, and we can call it the 3dB point. Waddya reckon?" Surely it must have gone down to a vote.

It didn't escape these guys that if you created a second order filter by cascading two identical first order RC filters, at frequency 1/(2.Pi.R.C) the response would be -6dB (1/4), which wasn't so convenient. They still wanted to define the 'half power gain' point of their systems, and so Butterworth and Chebyshev and all the other second order filter folk, for convenience, worked out the frequency where their own designs had a power gain of a half (the 3dB point), and used those instead of the 6dB points to define the passbands of their designs. Just for consistency, you understand.

For 3rd order filters, which increased attentuation to 9dB at that f=1/(2.Pi.T) frequency, they did the same. "Let's not use the 9dB point to sell our designs", they said. "Stick with the conventional 3dBs" was their choice.

So, to eat my words, 3dB was chosen to describe higher-than-one order active designs, but only because nature had a convenient "nearly 3" response in it's own passive systems.

That is an excellent answer. I am convinced.

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