Please explain the term 3dB Bandwidth.

[Hamidzia]

3 dB Bandwidth

Assalam O Alekum !

Please explain the term '3dB Bandwidth'.

Thankx

 

[gam]

Re: 3 dB Bandwidth

We aleikom elsalam WRAWB,

Any amplifier has a certain frequency response. That is the gain has a certain variation with the frequency of the signal.

3dB means the signal is at half its power

3dB bandwidth is the range of frequency considered useful where the signal is above half of its maximum power.

see the attached frequency/power graph.

Best regards

gam

 

[Kevin Weddle]

Re: 3 dB Bandwidth

I like the graph. It answers the question nicely. The -3db is inherent to the equation db=20log Vout/Vin. Why use 70% or the sin45? This describes the unit circle which has a radius equal to one. The adjacent is equal to one, the opposite is equal to one, the hypotenuse is equal sqrt2. The sin45 is then 1/sqrt2 which is equal to 70.7% of the unit circle radius.

 

[Hamidzia]

Re: 3 dB Bandwidth

Assalam O Alekum !

Thank you very much to all
Specially to gam for a very comprehensive detail and graph.

Now i have a good idea of the term.

Tell me that can we call 3dB bandwidth as half power bandwidth ?

Is 3dB Bandwidth is nothing to do with the numeric value '3dB' ?

Is the numeric value just used for no reason ?

Thanks again

 

[Kevin Weddle]

Re: 3 dB Bandwidth

The half power comes from the fact that sin45 times V quantity squared is equal to .5 power while V squared over R is equal to power. So yes this is half the power, but what good is 1/2 as opposed to 1/3. I think the magic number deals with the fact that at 45 degrees, the radius and the heighth are equal to one, which is dimension of the unit circle. So you could say that %70.7 is just a number.

 

[maxwellequ]

Re: 3 dB Bandwidth

The ratio of powers (P1 and P2), expressed in dBs is defined as:

(P1/P2)dB=10 log10(P1/P2)

If P1=Pmax/2 and P2=Pmax you get:

((Pmax/2)/Pmax)dB=10 log10((Pmax/2)/Pmax)=10 log10(0.5)= -3.01 dB


Therefore a relation of powers of 0.5 corresponds, in log domain, to -3 dB.

Note that this is a ratio of two things with the same unit (watt). A ratio such as this does NOT have a unit (it is dimensionless). The "unit" dB is only used to stress the fact that you are giving the ratio of powers in log domain and not in "linear" domain.

Sometimes a ratio of voltages is also given in log domain (for example to refer to the gain of a voltage amplifier: A=Vo/Vi). Noting that the Power is proportional to the square of the voltage you get:

(V1/V2)dB=10 log10(V1^2/V2^2)=10 log10((V1/V2)^2)=
=20 log10((V1/V2))

Hope this helps

 

[Kevin Weddle]

Re: 3 dB Bandwidth

((Pmax/2)/Pmax)dB=10 log10((Pmax/2)/Pmax)=10 log10(0.5)= -3.01 dB


This equation is totally correct but it does not satisfy a key point. You would get a better feel if you used the equation

V squared/R = power. Then because you you use .707 reduction of the voltage it becomes (.707)V quantity squared/R = power. This becomes
.5 V squared / R = .5 power.

 

[maxwellequ]

Re: 3 dB Bandwidth

Kevin,

If you read my post carefully you also have there the explanation for the ratio of voltages (Power proportional to V^2...). Basically:

(V1/V2)dB=20 log10((V1/V2))
To have (V1/V2)dB=-3dB you must have (V1/V2)=0.707. FYI that’s were the 0.707 appears....

Now, reading again your previous post (the one that has "sin45 times V"), I still can not understand it. Can you explain it to us ?

 

[Kevin Weddle]

Re: 3 dB Bandwidth

The sin45V says that your discrete voltage level is .707 the original voltage. In a filter, you say that the -3db occurs at the point at which the voltage is reduced by 70.7%. Your equation just says that Pout/Pin = .5 which causes -3db. I reduce the voltage and set that equal to .5P in the equation sin45V quantity squared/R = .5 power. The sin45 squared is equal to .5.

 

[checkmate]

Re: 3 dB Bandwidth

The sin45V says that your discrete voltage level is .707 the original voltage. In a filter, you say that the -3db occurs at the point at which the voltage is reduced by 70.7%. Your equation just says that Pout/Pin = .5 which causes -3db. I reduce the voltage and set that equal to .5P in the equation sin45V quantity squared/R = .5 power. The sin45 squared is equal to .5.

The flaw in your analysis is that you fixed the circuit's transfer function to that of an ohmic one, whilst maxwellequ has given the general proof relevant to all transfer functions. Since filters are 99% non-ohmic, your derivation is correct but quite pointless. No offence here.

 

[maxwellequ]

Re: 3 dB Bandwidth

Kevin,

Please don't take what I am going to say (or write) personally, but you explanation is just completely wrong. The problem is not the only this post; I've seen others posts of yours that just do not make any sense.

As I wrote before the ratio of powers/voltages/currents are sometimes given in log domain. The dB is a (kind of) unit that is used to stress the fact that the ratio being considered is given in log domain - as you know such ratios do not really have a unit.

The correct definitions are the ones I've presented before. The fact that sin(45º)=0.707 is NOT RELATED AT ALL with the fact that 20*log10(0.707)=-3 dB. Then you use terms like "discrete voltage level" which are simply not applicable here ... do you even know what you are talking about ?

The problem is that there are beginners using this forum, people which really make an effort to learn a bit more about electronics. If everyone starts to talk about what they don't know, those guys will get more and more confused. To me you can write whatever you want, that I will certainly not suffer…

And yes, if you want to know I only write about what I know.

Regards

 

[Kevin Weddle]

Re: 3 dB Bandwidth

I am sure of what I have calculated. I stand by my statements. There is nothing wrong with your assertation that the 10Log.5=20log.707. This is indeed half power and it is equal to -3db. I would be hardpressed to explain the similarity between the -6db per octave and the -20db per decade. Could you explain this to me?

 

[checkmate]

Re: 3 dB Bandwidth

I've rethought the situation and I realised a need to revise my perception that your derivation is correct. The 3dB bandwidth is used to define an effective frequency range for a system defined by a transfer function, where an input is processed by the system to form a valid output.

Your derivation is largely based on P=sq(V)/R. One point I hate about equations is that they often lack detail. We can't just use any equation that is unit consistent and assume it has to work. You have to understand the implications behind them. That's why physical laws are usually expressed in words, and simplified in equations.
P here refers to the power consumed by an ohmic component.
V here refers to the voltage across that ohmic component.
R refers to the resistance of that ohmic component.
So there's no input or output in this case, hence no system. What is the R you defined in this case? If you take the special case of a purely resistive system, an ideal resistor is not supposed to change its output for an identical input while frequency is varied, so there's no such thing as a 3dB bandwidth at all! The very word "bandwidth" indicates that it's associated with frequency and distortion with frequency is the very last thing I'd associate a resistor with.

Anyway, these are just my views. You may have your own insights over this issue, please share them. I would like to know too if my electrical basics are flawed too.

Cheers!

 

[maxwellequ]

Re: 3 dB Bandwidth

Kevin,

In what concerns your questions about the relation between the –20dB/dec and –6 dB/oct, here goes:

Lets imagine a first order low pass filter, which has a pole at the frequency fp. For f<fp the gain of the filter is aprox constant and for f>fp it starts falling at a rate of 20 dB/dec (it decreases 20 dB for each increase of 10 times in frequency) or equivalently a rate of 6dB/octave (the gain decreases 6 dB for each increase of 2 times in frequency). Your question is the justification for this fact.

The gain of a low pass filter is given by
A(f)=A0/(1+jf/fp)
Where A0 is the low frequency gain, f is the frequency, fp is the pole frequency of the filter and j is the imaginary unit (j = sqrt(-1)).

From the equation shown above you can relate the magnitude and phase of the input and output sinewaves. To answer your question, we are interested on the magnitude response; the magnitude can be written has (remember your calculus classes ?):

|A(f)|=A0/sqrt(1+(f/fp)^2)

For f<< fp |A(f)| can be approximated by A0, which is why this is called the “low frequency gain”. For f>>fp |A(f)| can be approximated by

|A(f)|=aprox=A0/(f/fp)

Now you can use this simpler equation to see what happens to the magnitude response, when f>fp. When you compare the magnitude of the filter’s voltage gain for f1 and for f2=10 f1, you have:

(|A(f2)|/ |A(f1)|)dB =20 log10[(A0/(f2/fp))/(A0/(f1/fp))]=
=20 log10[(A0/(10f1/fp))/(A0/(f1/fp))]=
=20 log10[1/10]= -20 dB

Thus, when the frequency increases 10 times the gain of a first order low pass filter decreases 20 dB.

If you make the same calculations for f1 and f2=2 f1 (f2 is an octave above f1):

(|A(f2)|/ |A(f1)|)dB =20 log10[(A0/(f2/fp))/(A0/(f1/fp))]=
=20 log10[(A0/(2f1/fp))/(A0/(f1/fp))]=
=20 log10[1/2]=-6 dB

Therefore, when the frequency increases 2 times (an octave), the gain of the filter decreases 6 dB.

Finally, Kevin, I tell you again, sin45 has nothing to do with dBs …

 

[Kevin Weddle]

Re: 3 dB Bandwidth

I think what you did is took the -3db and multiplied it by 2. In filters, aren't you supposed to use voltage instead of power. Although your explaination exceeds my ability to understand. I had the idea to use the
-20db per decade and somehow arrive at the -6db per octave through the use of equalities. But it doesn't work that way. I feel that the best way to explain it is to use the formula Xc = 1/2piFC and to plug the filters values into this equation. I think that F being multiplied by 2 or 10 has something to with it.

 

[maxwellequ]

Re: 3 dB Bandwidth

Kevin,

The explanation I gave you is taught in any electrical engineering course. You asked how to relate the 20dB/dec and 6 dB/oct, and that’s what I explained to you.

It is incredible how you recognize that the "explaination exceeds my ability to understand" and then you suggest that what I wrote is wrong because it is not possible to relate the -20dB/dec and the dB/oct "through the use of equalities" – “it doesn't work that way” you say. If you don't understand, just ask people to explain better... Don't just start saying that they are wrong, especially when you lack the knowledge.

Grow up, study a little bit more and stop saying/writing stupid things.

 

[boot_strap]

Re: 3 dB Bandwidth

its the the frequncy which your highest frequncy response (A), reaches
A/(2^0.5)..!

 

[yeechyan]

3 dB Bandwidth

Kevin, why do you stressed on saying the value of sin45? Is there any relations between 45 degree and the output vs input voltage in this case?

 

[Kevin Weddle]

Re: 3 dB Bandwidth

The relationship is either half power if you like the sound of that or the voltage equation 10log vout/vin where vout/vin is equal to the sin45. Both equal -3db. I like the sound of sin45 as opposed to half power because it shows that you are using another mathematical option which states that the opposite and adjacent are equal to 1 which are the dimensions of the unit circle. The fact is you have an angle of 45 degrees when the dimensions are 1 by 1 according to triangle theory. The sin45 is then the opposite over the hypotenuse which is .707 which is a number represented by the sine wave at 45 degrees and it's a point on the unit circle.

In total, you have a real number that represents equal sides of a triangle and it is a point on the sine wave and it is a point on the unit circle. These three things combine to make the number relevant.

 

[Borber]

Re: 3 dB Bandwidth

First question that must be answered here is what is bandwidth?
For a communications or data signal, a measure of the amount of pectrum space the signal occupies. Usually, it is given as the difference between the frequencies at which the signal amplitude is nominally 3 dB down with
respect to the amplitude at the center frequency. These frequencies represent the half-power points of the amplitude-versus-frequency function
2. Also called NECESSARY BANDWIDTH. The minimum amount of spectrum space normally required for effective transmission and reception of a communications or data signal.
So bandwidth is usually determined by -3dB points in transfer function what is generally accepted. You may use other points like -6dB but you must mention it.