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Show that a series LR circuit is an LP filter if the output is taken across the resis

  1. Apr 11, 2007 #1
    1. The problem statement, all variables and given/known data

    Show that a series LR circuit is a Low Pass filter if the output is taken across the resistor. Calculate the corner frequency, [itex]f_c[/itex], if L = 2mH and R = 10k[itex]\Omega[/itex].



    2. Relevant equations

    [tex]H(\omega)\,=\,\frac{V_0(t)}{V_i(t)}[/tex]

    [tex]H\left(\omega_c\right)\,=\,\frac{1}{\sqrt{2}}[/tex]

    [tex]f\,=\,\frac{\omega}{2\pi}[/tex]



    3. The attempt at a solution

    [tex]H(\omega)\,=\,\frac{R}{R\,+\,j\omega L}[/tex]

    [tex]H(0)\,=\,1[/tex]

    [tex]H(\infty)\,=\,0[/tex]

    That does the "show" part. But now I don't know how to get the corner frequency.

    [tex]\frac{1}{\sqrt{2}}\,=\,\frac{R}{R\,+\,j\omega L}[/tex]
     
  2. jcsd
  3. Apr 11, 2007 #2
    Never mind! It's 796 kHz.
     
  4. Apr 11, 2007 #3
    Per your post on Bode plots, this transfer function would be written as:

    1/(1+jwL/R), you might want to plot this before tackling that much more complicated function, if you haven't done simple ones yet.
     
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