# Function in the Fourier domain of a low-pass filter

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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"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? –  Dilip Sarwate Mar 31 '12 at 13:41
@DilipSarwate I use $Hz$ as unit of measure for frequency. –  Mazzy Mar 31 '12 at 14:24
I repeat: Does the prefix "k" to "Hz" have any meaning? –  Dilip Sarwate Mar 31 '12 at 14:56
yes it means ${3}\ kHz = 3 * 10^3 Hz$ –  Mazzy Mar 31 '12 at 15:11
-2 seems a little harsh. Why the minus votes? –  Jim Clay Apr 2 '12 at 1:24