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LPF Magnitude slope

  1. Apr 12, 2012 #1
    I have read that LPF (or HPF) have magnitudes which fall at the rate of 20dB/decade from the corner frequency onwards. Why is it assumed this way although the slope is changing with frequency?

  2. jcsd
  3. Apr 12, 2012 #2


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    Staff: Mentor

    The -20dB/decade is the rate of fall for each zero in the denominator of the transfer function. It's the slope of the straight line tangent that the actual response always approaches for frequencies well beyond the corner frequency, as seen on a gain (dB) vs. frequency plot.

    A single-pole LPF transfer function takes the form: A₀/(1 + j ω/ω₀)

    A two-pole LPF transfer function takes the form: A₀/❲(1 + j ω/ω₀) (1 + j ω/ω₁)❳
    With two poles in its transfer function, the two-pole LPF falls off at -40dB/decade
  4. Apr 12, 2012 #3
    Hi Salil, stop reading at any point, but I'll give you a lot of information you may or may not know already.

    First, bode plots have a corresponding transfer function. A transfer function can have poles and zeroes (when the function = 0 or is undefined). The bode magnitude plot you're referring to is the magnitude of the transfer function in dB. An example of a transfer function with 1 pole and 1 zero is below at frequencies [itex]\tau_{1}[/itex] and [itex]\tau_{2}[/itex] and [itex]s = j\omega[/itex]. Note a transfer function can be written in different looking but equivalent algebraic forms depending on what you want to see (time constant, poles/zeros, frequency, DC gain, etc), but this can also be confusing. I wrote it out in the form I did to show that when [itex]s = -\frac{1}{\tau1}[/itex], the function is undefined at that pole.

    [itex]tf = \frac{s+\frac{1}{\tau_{1}}}{s+\frac{1}{\tau_{2}}}[/itex]

    The frequency of a pole or zero marks where the affect of its slope begins. For any pole, the slope will decrease at a rate of 20dB/decade, and when multiple poles are in effect, you multiply this slope by how many poles have accumulated . Likewise, each zero will add 20dB/decade. e.g.: 2 poles = -40db, 3 poles = -60db, 2 zeroes = 40db, 1 pole and 1 zero = 0db.

    About your question of why this is assumed; it is assumed this way because it is a quickhand method for plotting the approximate frequency response asymptotically, and so they usually only talk about the rules without the reasoning behind it.

    But its not that much math to see how it works. The value of 20dB/decade slope is a consequence of using logarithm and using voltage rather than power, which results in using 20log instead of 10log since voltage squared is proportional to power. If you consider that the log of anything has a slope of 1 per a decade, then its easy to see how this number comes about, e.g. log(1) = 0, log(10) = 1, log(100) = 2, etc.

    The slope is demonstrated by picking any decade interval:

    [itex]\frac{log(100) - log(10)} {1\:decade} = \frac{2 -1}{1\:decade} = \frac{1}{1\:decade}[/itex]

    The 20 comes into play as I mentioned previously, because we take 20log instead of just log. So for example, if we have a transfer function with 1 pole, the division of the pole turns into a logarithmic subtraction. If you plug in two frequencies that are a decade apart and calculate the change in magnitude over frequency, you get the -20dB/decade.

    If the above explanation isnt enough, keep reading and consider an example given 1 pole transfer function:

    [itex]tf = \frac{1}{s +\frac{1}{\tau}}[/itex]
    subsituting the laplace s variable, its magnitude is found as:

    [itex]20log\left(\left|tf\right|\right) = 20log\left(\left|\frac{1}{jw + (\frac{1}{\tau})^{2}}\right|\right) = 20log\left(\frac{1}{\sqrt{\omega^{2}+(\frac{1}{\tau})^{2}}}\right)[/itex]

    if you use the property that multiplication/division translates to addition/subtraction in log, and find the slope of the expression at a decade interval between [itex]\omega[/itex] and [itex]10\omega[/itex], you get:


    [itex]= \frac{20log\left(\sqrt{\omega^{2}+(\frac{1}{\tau})^{2}}\right) - 20log\left(\sqrt{(10\omega)^{2}+(\frac{1}{\tau})^{2}}\right)}{1\:decade}[/itex]

    Now the tricky part comes in trying to make the slope asymptotic, rather than smooth, which is purely so you can draw these down fast without doing the math. When you short hand draw a bode, you approximate and plot it asymptotically, however if you look at the real bode plot, you will notice its smooth. The assumption made is that the terms [itex]\omega^{2}[/itex] and [itex](10\omega)^{2}[/itex] are relatively much larger than the [itex](\frac{1}{\tau})^{2}[/itex]. Around the frequencies where the pole is, this assumption is not as accurate, and that's where you see the deviation of the real bode plot from the asymptotic plot. So, if we go the asymptotic route, and assume [itex](\frac{1}{\tau})^{2}[/itex] is negligible then the slope becomes:

    [itex]= \frac{20log\left(\sqrt{\omega^{2}}\right) - 20log\left(\sqrt{(10\omega)^{2}}\right)}{1\:decade}[/itex]

    again, using logs to seperate the multiplication into addition,

    [itex]= \frac{20log\left(\omega\right) - 20log\left(10\omega\right)}{1\:decade}= \frac{20log\left(\omega\right) - 20log\left(\omega\right) - 20log\left(10\right)}{1\:decade} = \frac{- 20log\left(10\right)}{1\:decade}[/itex]

    [itex]= \frac{- 20dB}{1\:decade}[/itex]
    Last edited: Apr 12, 2012
  5. Apr 14, 2012 #4
    Thanks a lot DragonPetter for your wonderful explanation. I really couldn't stop reading. Actually I tried a lot using derivatives and stuff. Must have messed it. Thanks again.
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