[7.2-1 Connection Between the Fourier and Laplace Transforms](#page-12-0)

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7.2-1 Connection Between the Fourier and Laplace Transforms

The general (bilateral) Laplace transform of a signal x(t), according to Eq. (4.1), is

$$X(s) = \int_{-\infty}^{\infty} x(t)e^{-st} dt$$

(7.24)

Setting s = jω in this equation yields

$$X(j\omega) = \int_{-\infty}^{\infty} x(t)e^{-j\omega t} dt$$

where X(jω) = X(s)|^s=jω. But, the right-hand-side integral defines X(ω), the Fourier transform of x(t). Does this mean that the Fourier transform can be obtained from the corresponding Laplace transform by setting s = jω? In other words, is it true that X(jω) = X(ω)? Yes and no. Yes, it is true in most cases. For example, when x(t) = e−atu(t), its Laplace transform is 1/(s + a), and X(jω) = 1/(jω +a), which is equal to X(ω) (assuming a < 0). However, for the unit step function u(t), the Laplace transform is

$$u(t) \Longleftrightarrow \frac{1}{s} \qquad \text{Re}\, s > 0$$