For the differential equation $$\frac{d^2y(t)}{dt^2}+\frac{3dy(t)}{dt} + 2y = x(t),$$

I was able to find the frequency response as $$H(j\omega)= \frac{1}{-\omega^2+3j\omega+2}.$$ However, I am not sure how to find the impulse response, which is the next part. Please help, and thank you in advance.

  • 6
    $\begingroup$ Hint: the impulse response is the inverse Fourier transform of the transfer function. $\endgroup$ Mar 26, 2013 at 3:32

1 Answer 1


As Dilip pointed out in the comment above, you can get the impulse response using the inverse Fourier transform. However, a slightly easier method might be to use the Laplace domain instead; it's more amenable to easy inverse transforming via transform tables. First, recall that the frequency response is really just the $s$-plane transfer function evaluated along the imaginary axis. That is:

$$ H(j\omega) = H(s)|_{s=j\omega} $$

Go through your frequency response expression and substitute $s$ back in for $j\omega$:

$$ \begin{align} H(j\omega) &= \frac{1}{-\omega^2+3j\omega+2} \\ &= \frac{1}{(j\omega)^2+3j\omega+2} \\ \end{align} $$

$$ H(s) = \frac{1}{s^2+3s+2} $$

So now we have the transfer function in the $s$-plane. To get from here to an easily-inverse-transformable expression, we use a partial fraction decomposition on the transfer function (an online calculator is available here:

$$ H(s) = \frac{1}{s+1} + \frac{1}{s+2} $$

Given this form, we now refer to a table of common Laplace transforms to find:

$$ e^{-\alpha t} u(t) \Leftrightarrow \frac{1}{s+\alpha} $$

where $u(t)$ is the Heaviside step function. The impulse response of the system is the inverse Laplace transform of its transfer function. Using the linearity of the Laplace transform and the above relation, we can therefore conclude that:

$$ h(t) = u(t) \left(e^{-t} + e^{-2t} \right) $$


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