The Fourier and the Laplace transforms

Because feedback requires careful consideration of stability and causality, the transform tool of choice is almost always the Laplace transform. Yet, in many applications it is the Fourier transform that arises most naturally.

One such case is in the context of frequency response (Chapters 3 and 6). Without assuming causality or stability, let’s attempt to calculate the response of a time-invariant linear system to a complex exponential input

$$u(t) = e^{j \omega t}$$

Note that we assume a persistent input, not making the usual assumption that $u(t) = 0$, $t < 0$. The system response can be obtained from the convolution formula

y(t)=\int_{-\infty}^{\infty} g(\tau) \, u(t – \tau) \, d\tau
= \int_{-\infty}^{\infty} g(\tau) \, e^{j \omega (t – \tau)} \, d\tau

Since $t$ is not involved in the integration being carried over $\tau$

$$y(t)= e^{j \omega t} \int_{-\infty}^{\infty} g(\tau) \, e^{-j \omega \tau} \, d\tau$$


$$y(t)= G(j \omega) \, e^{j \omega t}, \quad G(j \omega) = \int_{-\infty}^{\infty} g(\tau) \, e^{-j \omega \tau} \, d\tau.$$

The quantity

$$G(j \omega) = \int_{-\infty}^{\infty} g(\tau) \, e^{-j \omega \tau} \, d\tau$$

is the Fourier transform of the impulse response signal $g$.

Either one of the following two conditions on the signal $g(t)$ are sufficient to ensure that $G(j \omega)$ is well defined:

  1. If $\int_{-\infty}^{\infty} |g(\tau)|^2 \, d\tau$ is bounded (bounded energy)
  2. If $\int_{-\infty}^{\infty} |g(\tau)| \, d\tau$ is bounded (absolutely integrable) and $g$ has a finite number of maxima and minima over any finite interval of time and a finite number of finite discontinuities over any finite interval of time (Dirichlet conditions)

So how do the above formula in terms of the Fourier transform relates to the similar concepts discussed in Chapter 3 and 6? One relevant case is that of a causal and stable system. In this case, $g(t) = 0$, $t < 0$ and $\int_{0}^{\infty} |g(\tau)|\, d\tau$ is bounded. Stability roughly means that the Fourier transform will converge. Causality reduces the above formula to

$$y(t)= e^{j \omega t} \int_{-\infty}^{\infty} g(\tau) \, e^{-j \omega \tau} \, d\tau= e^{j \omega t} \int_{0}^{\infty} g(\tau) \, e^{-j \omega \tau} \, d\tau $$

It is this form that ties together with the Laplace transform. Indeed, if

$$G(s) = \int_{0}^{\infty} g(\tau) \, e^{-s \tau} \, d\tau$$

is the Laplace transform of the causal and absolutely integrable signal $g$, then $G(j \omega)$ is indeed its Fourier transform, that is the coefficient of the response to an exponential input $u(t) = e^{j \omega t}$! See this post for more on that.

The above response, calculated for a persistent signal $u(t)=e^{j \omega t}$ defined over all time $t$, also coincides with the steady-state response from Section 3.8, calculated for a signal $u(t) = \cos(\omega t)$ defined for $t \geq 0$. A direct relationship with the frequency response formula from Section 3.8 can be derived by restricting our attention to the real part of the signals $u(t)$ and $y(t)$. Indeed, since

$$u(t) = \cos(\omega t) = \operatorname{Re}\left\{e^{j \omega t}\right\}$$


y_{\mathrm{ss}}(t) = \operatorname{Re}\left\{G(j \omega) \, e^{j \omega t}\right \}.

After writing $G(j \omega)$ in polar form, that is $G(j \omega) = |G(j \omega)|e^{j \angle G(j \omega)}$ one obtains

y_{\mathrm{ss}}(t) = |G(j \omega)| \operatorname{Re}\left\{e^{j \angle G(j \omega)} e^{j \omega t}\right\}
= |G(j \omega)| \cos(\omega t + \angle G(j \omega))

which is precisely the frequency response formula from Section 3.8.

The links between the Fourier and Laplace transforms go well beyond the frequency response. We shall delve into some of those relationships in future posts.

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