bandwidth of the system

[LM741]

hey all.

i read that the definition of the bandwidth of a system is the frequency range up until the signal's power (at DC) drops by -3dB.

This obviously only applies to a first order system , right?

Surely for a second order - it is defined as the range of frequency up until the power drops by -6dB?

thanks

[berkeman]

Your first statement applies to a low-pass function. If the transfer function is a bandpass, then the bandwidth is generally measured to the 3dB points on either side of the passband. It doesn't matter what order the system is, you usually use the -3dB points as the shoulders.

[berkeman]

Like in this figure:

http://en.wikipedia.org/wiki/Bandwidth

[LM741]

Thanks - but the reason why asked the above is because i am given the following system: H(s) = 1/ (s+4)^2, and asked to find the bandwidth of the system.
It can't be w=4 (If we wish to conform to the definition of bandwidth), because at this point we have a -6dB power drop.
On the other hand, if i was given the system as: H(s) = 1/(s+4), then the bandwidth would be equal to 4, i.e w=4

thanks again

[berkeman]

Is your H(s) a power or voltage transfer function? Remember that -3dB is not the 1/2 signal point, it's a 1/2 power point. The signal at -3dB is \frac{1}{\sqrt2}

[LM741]

power = signal drops by half DC value (or DC power??).
voltage or current = signal drops to 70 percent of DC value.
They are still both regarded as -3dB points by applying the corresponding equation:
for power : 10 log(P/2)
for voltage or current :20 log(V/srt(2)).

thanks

It can be seen as a voltage ...but i don't think in this case it will make a difference...

[LM741]

any sugestions guys...?

[Corneo]

I would solve for the value of s_0 such that H(s_0) = \frac {1}{\sqrt 2} H_{max}. Since the max is clearly 1, just solve for the denominator of H(s) (s+4)^2 = \sqrt {2}

[LM741]

thanks - also thought about doing it that way and sticking to the definition.