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I'm struggling with a signal analysis electrical engineering college assignment. We need to a derive differential equation for a low-pass filter, high-pass filter and a band pass filter (made by connecting the output of a low pass filter to the input of a high pass filter). Here are diagrams for reference:

Low pass filter Low Pass Filter Image

High pass filter High Pass Filter Image

Band Pass Filter$$$$ Band pass filter (buffer can be removed)

The equations I derived are:

  • low pass filter: $$V_o=V_i-RC \frac{dV_i}{dt}$$
  • high pass filter: $$V_o=RC \frac{dV_i}{dt}$$
  • band pass filter: $$V_o = R_2C_2 ( (1-C_1-R_1)- R_1C_1 \frac{d^2V_i}{dt^2} )$$

The equation for the band pass filter I found by making the input of the high pass filter the output of the low pass filter. So I derived the equation of the low pass filter with respect to time and got: $$ \frac{dV_o}{dt} = \frac{dV_i}{dt} - R_1\frac{dVi}{dt} - C_1\frac{dV_i}{dt} - R_1C_1\frac{d^2V_i}{dt^2} $$

I substituted in this rate of change of voltage into the equation of the high pass filter to get the equation I derived for the band pass filter.

Could anyone please tell me if what I have done is valid and if not, please explain to me what the correct strategy is. Thank you kindly.

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I'm afraid your results are not correct. Let's have a look at the low-pass filter. The current $i(t)$ through $R$ and $C$ must be

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

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

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

By the way, if you were allowed to use the Laplace or Fourier transform, everything would be a lot easier.

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