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I have to represent a low-pass filter in the Fourier domain having a cut-off frequency of $f_c={3}\ kHz$.

Could all low-pass filters could be represented with this function: $H(f)=\frac{1}{1+j\left(\frac{f}{f_c}\right)}$?

If so, my filter could be represented with this function: $H(f)=\frac{1}{1+j\left(\frac{f}{3*10^3}\right)}$. Is this correct?

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  • $\begingroup$ "right or not?" Not! What would the answer be if $f_c = 3$ Hz? What would the answer be if $f_c = 3$ rad/s ? Does the prefix "k" to "Hz" have any meaning? $\endgroup$
    – Dilip Sarwate
    Mar 31, 2012 at 13:41
  • $\begingroup$ @DilipSarwate I use $Hz$ as unit of measure for frequency. $\endgroup$
    – Mazzy
    Mar 31, 2012 at 14:24
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    $\begingroup$ I repeat: Does the prefix "k" to "Hz" have any meaning? $\endgroup$
    – Dilip Sarwate
    Mar 31, 2012 at 14:56
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    $\begingroup$ yes it means ${3}\ kHz = 3 * 10^3 Hz$ $\endgroup$
    – Mazzy
    Mar 31, 2012 at 15:11
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    $\begingroup$ -2 seems a little harsh. Why the minus votes? $\endgroup$
    – Jim Clay
    Apr 2, 2012 at 1:24

1 Answer 1

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Not right. Your basic assumption that all lowpass filters can be represent through your formula is wrong. What you have is a first order IIR low pass but there are many more options for a low pass filter, such as higher order IIR filter of different types (butterworth, bessel, chebycheff 1+2, elliptic) and plus various FIR filters as well.

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