# How do i set the band filter such that it has a pass band of 100Hz to 300Mhz? [closed]

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

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

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

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

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

• I don't see a question here. – Jason R May 19 '14 at 11:25
• Why do you copy my answer to an older question into your question without actually asking a question??? dsp.stackexchange.com/questions/15646/… – 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 $$R_2 \cdot C_2 = \frac{1}{2\cdot \pi \cdot 100MHz}$$
• Shouldn't the second term have 100Hz rather than 100MHz ? – Peter K. May 19 '14 at 18:52