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Bandwidth of the system

  1. Dec 11, 2006 #1
    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
     
  2. jcsd
  3. Dec 11, 2006 #2

    berkeman

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    Staff: Mentor

    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.
     
  4. Dec 11, 2006 #3

    berkeman

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    Staff: Mentor

  5. Dec 11, 2006 #4
    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
     
  6. Dec 11, 2006 #5

    berkeman

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    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 [tex]\frac{1}{\sqrt2}[/tex]
     
  7. Dec 11, 2006 #6
    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...
     
  8. Dec 12, 2006 #7
    any sugestions guys...?
     
  9. Dec 13, 2006 #8
    I would solve for the value of [itex]s_0[/itex] such that [itex]H(s_0) = \frac {1}{\sqrt 2} H_{max}[/itex]. Since the max is clearly 1, just solve for the denominator of H(s) [tex](s+4)^2 = \sqrt {2}[/tex]
     
  10. Dec 15, 2006 #9
    thanks - also thought about doing it that way and sticking to the definition.
     
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