Why in frequency domain analysis of control systems, bandwidth is defined as 3 dB down from the zero frequency value of magnitude and not from maximum value of magnitude?
Dear Ankur
3 dB because the half power point is approximately 3 dB because \log_{10} 2 = .3010... \approx .3, so 10 \log_{10} 2 = 3.010... \approx 3 - (latex editor format)
regards
Gerro
Dear Prinsloo
I want to know if this 3 dB down value is from max value of the curve or 3 dB down from the zero frequency value of the curve?
i have read everywhere (like in filters) that it is taken from the max value. But in freq domain analysis of control systems, everywhere it is taken from zero frequency value.
@Ankur Goyal, yes - you are right!
There is no fixed definition of the bandwidth for a lowpass. Of course, in many cases (first order filters, second-order Butterworth filters) the corner frequency (BW) is defined at a frequency where the amplitude is 3dB down with respect to the value at DC.
But even this "definition" is an arbritrary one and it is up to you for a specific application to define the BW at the -1 dB or -2 dB point.
However, there are other lowpass approximations (Chebyshev, for example) where it makes no sense to define the band edge as described above.
Example: Let a second-order Chebyshev response with an allowed ripple of 2dB start at 0 dB. At a certain frequency the magnitude reaches +2dB. If the pass region would be defined at the -3dB point (3dB below 0 dB) we would have an amplitude variation of 2+3=5 dB within the defined bandwidth. This does not sound reasonable.
For this reason, the bandwidth of Chebyshev filters is defined NOT at the point where the magnitude is 3dB down if compared with DC. Instead, it is the allowed "ripple" which defines the bandwidth.
That means: The bandwidth is defined at that frequency where the magnitude starts to fall below the specified ripple. In our example (+2dB ripple) this is the frequency where the magnitude again crossses the 0 dB line.
Note that for a second-order filter the "ripple" reduces to one single peak only.
Indeed, in control theory the definition of bandwidth differs from signal processing or audio. There are a main reason to that : in most case you need at least one integrator in the system and thus the maximum gain is infinite (at 0 frequency) and thus the usual bandwidth is 0 !
In control the bandwidth is defined à 0 dB (absolute gain) of OPEN-LOOP. But if you consider the corresponding CLOSED-LOOP, you find again the usual definition, i.e. -3dB.
To be quite precise, lets denote the open loop transfer function by G(f) (where f is the frequency). The corresponding closed-loop have transfer function H(f) = G(f) / ( 1 + G(f) ).
If the low frequency gain of the open loop is sufficiently high and G is a decreasing function of the frequency (this hypothesis are always satisfied for control systems), you have the equivalence between the 2 following frequency intervals :
(1) { f >= 0 | 20 log G(f) >= 0dB }
(2) { f >= 0 | 20 log H(f) > = -3dB }
Note that (1) is the definition of bandwidth of the control system in the meaning of "control"
Note that (2) is the usual definition of the bandwidth of the closed-loop system in the "usual" meaning.
Of course, (1) is the definition of the closed-loop systems
Quote: "in most case you need at least one integrator in the system and thus the maximum gain is infinite (at 0 frequency) and thus the usual bandwidth is 0 ! "
The "maximum gain" of the system with feedback? No - the gain of the (idealized !) integrator only.
Quote: "H(f) = G(f) / ( 1 + G(f) ) ".
I suppose, you have suppressed the "j" (complex frequency) not intentionally? More than that, it should be noted that this equation assumes 100% feedback.
Your first statement gives BW for low pass systems. your question statement gives BW for band pass systems.
It is quite right for the natural reasons. The ideal system should not make attenuation, the ideal level of 0. Unfortunately, the real system are always makes distortions and bandwidth is defined at the level of -3 dB.
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