# Inverse Fourier of Two-Pole Transfer Function

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

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:

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

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.

• 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. – Matt L. May 7 '19 at 14:27
• Indeed! I had misplaced it. Thanks. – MrUser May 7 '19 at 14:47
• The subtraction part is due to partial fraction expansion which converts multiplication terms into addition of fractions. – Ch.Siva Ram Kishore May 7 '19 at 14:49

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

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

$$\tilde{H}(\omega) = \Big(\frac{1}{\tau_2 - \tau_1}\Big) \Big[ \frac{\tau_2}{1 + i \omega \tau_2} -\frac{\tau_1}{1 + i \omega \tau_1}\Big]$$
$$\tilde{H}(\omega)$$ 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:
$$\mathcal{F}^{-1}\Big\{\Big(\frac{\tau}{1 + i \omega \tau}\Big) e^{-i\omega t_d}\Big\} = \Theta(t) ~ e^-\frac{t - t_d}{\tau}}$$