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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

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

So we get

$$v_i(t)=v_o(t)+RC\frac{dv_o(t)}{dt}\tag{1}$$

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

$$RC\frac{dv_i(t)}{dt}=v_o(t)+RC\frac{dv_o(t)}{dt}\tag{2}$$

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

$$\frac{dv_i(t)}{dt}=\frac{dv_{oL}(t)}{dt}+\tau_L\frac{d^2v_{oL}(t)}{dt^2}\tag{3}$$

From (2) we have

$$\tau_H\frac{dv_{oL}(t)}{dt}=v_o(t)+\tau_H\frac{dv_o(t)}{dt}\tag{4}$$

and (by taking the derivative)

$$\tau_H\frac{d^2v_{oL}(t)}{dt^2}=\frac{dv_o(t)}{dt}+\tau_H\frac{d^2v_o(t)}{dt^2}\tag{5}$$

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

$$\tau_H\frac{dv_i(t)}{dt}=v_o(t)+(\tau_L+\tau_H)\frac{dv_o(t)}{dt}+\tau_L\tau_H\frac{d^2v_o(t)}{dt^2}$$

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closed as unclear what you're asking by Jason R, Matt L., Peter K. May 19 '14 at 13:31

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  • $\begingroup$ 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
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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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