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.


1 Answer 1


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


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


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


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