I would appreciate if someone could walk me through this derivation.

I have a transfer function in the frequency domain, which has two poles

$$\tilde{H}(\omega) = \Big(\frac{1}{1 + i \omega \tau_1}\Big)\Big(\frac{1}{1 + i \omega \tau_2}\Big)$$

For a single pole, the inverse Fourier transform is: $$\mathcal{F}^{-1}\Big\{\Big(\frac{\tau}{1 + i \omega \tau}\Big)\Big\} = \Theta(t) ~ e^{\displaystyle -\frac{t}{\tau}} $$

where $\Theta(t)$ is the Heaviside function since this is a causal filter. (See this answer.) I also plugged the above $\tilde{H}$ into Wolfram Alpha and got the expected result:

$$\mathcal{F}^{-1}\Big\{\tilde{H}(\omega)\Big\} = \frac{1}{\tau_2 - \tau_1} \Bigg( ~ e^{\displaystyle -\frac{t - t_d}{\tau_2}} - e^{\displaystyle -\frac{t - t_d}{\tau_1}} \Bigg)\Theta(t - t_d)$$

where $t_d$ is the time after which the causal filter is effective. I'm not sure if it's a reversal of integration limits, integration by parts, or simply some of the terms cancelling after integration and plugging in the limits that might be responsible for the subtraction.

And I'm also not 100% sure how to work in the $t_d$ term.

  • 1
    $\begingroup$ There should be a term $e^{-j\omega t_d}$ in the expression of your frequency response, otherwise there can't be any $t_d$ in the inverse transform. $\endgroup$
    – Matt L.
    May 7, 2019 at 14:27
  • $\begingroup$ Indeed! I had misplaced it. Thanks. $\endgroup$
    – MrUser
    May 7, 2019 at 14:47
  • 1
    $\begingroup$ The subtraction part is due to partial fraction expansion which converts multiplication terms into addition of fractions. $\endgroup$ May 7, 2019 at 14:49

1 Answer 1


The comments to the question already hint at the solution, but I'll write it out.

$\tilde{H}(\omega)$ can converted from a product into a sum using Partial Fraction Expansion:

$$\tilde{H}(\omega) = \frac{1}{1 + i \omega \tau_1} \frac{1}{1 + i \omega \tau_2} = \frac{A}{1 + i \omega \tau_1} +\frac{B}{1 + i \omega \tau_2}$$

$$ 1 \omega ^0 + 0 \omega ^1 = A(1 + i \omega \tau_2) + B(1 + i \omega \tau_1)$$

Which gives $$\implies A + B = 1$$ and \begin{equation} \begin{split} A i \omega \tau_2 + B + i \omega \tau_1 &= 0\\ (1 - B) \tau_2 + B \tau_1 &= 0\\ B(\tau_1 - \tau_2) &= -\tau_2\\ B = \frac{\tau_2}{\tau_2 - \tau_1} \end{split} \end{equation}

$$A = 1 - B = \frac{- \tau_1}{\tau_2 - \tau_1}$$

Leading to:

$$\tilde{H}(\omega) = \Big(\frac{1}{\tau_2 - \tau_1}\Big) \Big[ \frac{\tau_2}{1 + i \omega \tau_2} -\frac{\tau_1}{1 + i \omega \tau_1}\Big]$$

$\tilde{H}(\omega)$ also contains a term $e^{-i\omega t_d}$, so now for each of the terms in the sum above the inverse Fourier transform is:

$$\mathcal{F}^{-1}\Big\{\Big(\frac{\tau}{1 + i \omega \tau}\Big) e^{-i\omega t_d}\Big\} = \Theta(t) ~ e^{\displaystyle -\frac{t - t_d}{\tau}} $$

and their sum has the desired form in the question.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.