From this derivation to get the differential of a band pass filter: Id like to find out how we set the upper frequency cut-off and the lower frequency cut-off.

Band Pass Filter


So we get


For the high-pass filter we have

$$i(t)=\frac{v_o(t)}{R}=C\frac{d(v_i(t)-v_o(t))}{dt}$$ which gives


Let's call the time constants of the low-pass and high-pass filters $\tau_L=R_LC_L$ and $\tau_H=R_HC_H$, respectively. For the band-pass filter we need a relation between $v_i(t)$ of the low-pass filter and $v_o(t)$ of the high-pass filter. If we use $v_{oL}(t)$ to denote the output of the low-pass filter, which equals the input of the high-pass filter, we get from (1) by taking the derivative


From (2) we have


and (by taking the derivative)


Plugging (4) and (5) into (3) we finally get for the band-pass filter


  • $\begingroup$ I don't see a question here. $\endgroup$ – Jason R May 19 '14 at 11:25
  • $\begingroup$ Why do you copy my answer to an older question into your question without actually asking a question??? dsp.stackexchange.com/questions/15646/… $\endgroup$ – Matt L. May 19 '14 at 12:18

That's much easier in the frequency domain. You simply get $$V_o(\omega) = V_i(\omega)\cdot \frac{1}{1+j\cdot \omega\cdot R_1 \cdot C_1}\cdot \frac{j\cdot \omega\cdot R_2 \cdot C_2}{1+j\cdot \omega\cdot R_2 \cdot C_2}$$

The first term is the lowpass and the second term is the highpass. To set the frequencies you need $$R_1 \cdot C_1 = \frac{1}{2\cdot \pi \cdot 300MHz}$$ and $$R_2 \cdot C_2 = \frac{1}{2\cdot \pi \cdot 100MHz}$$

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  • $\begingroup$ Shouldn't the second term have 100Hz rather than 100MHz ? $\endgroup$ – Peter K. May 19 '14 at 18:52
  • $\begingroup$ second term is 100Hz. Thats correct. Thanks! $\endgroup$ – Sach May 20 '14 at 9:07

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