The Laplace transform of a cosine starting at $t=0$ is given by

$$F(s) = \frac{s}{s^2 + \omega_0^2}$$

If I sub in $s = j\omega$, I get the Fourier transform of a cosine starting at $t=0$:

$$F(j\omega) = \frac{j\omega}{\omega_0^2 - \omega^2}$$

As this is a purely imaginary result it shows that the input function is made up of only sine waves and must therefore be odd.

This doesn't make sense as a cosine wave starting at $t=0$ is neither odd nor even, so $F(j\omega)$ should have both real and imaginary parts.

I find a similar issue when using a sine wave starting at $t=0$ which returns a purely real function of frequency:

$$F_2(j\omega) = \frac{\omega_0}{\omega_0^2 - \omega^2}$$

What is it that I'm misunderstanding?


1 Answer 1


If I sub in $s=j\omega$, I get the Fourier transform of a cosine wave starting at $t=0$.

No, you don't. You can't just set $s=j\omega$ in an expression for the Laplace transform and expect the result to be the Fourier transform of the function. This only works if the imaginary axis is inside the region of convergence (ROC) of the Laplace transform. Since the given Laplace transform has poles on the imaginary axis, the imaginary axis is not part of the ROC (the ROC is $\textrm{Re}\{s\}>0$).

The Fourier transform of the function


exists in a distributional sense, i.e., if we allow Dirac impulses (and, possibly, its derivatives) in the expression of the Fourier transform.

Using the Fourier transform of the step function


and the modulation property of the Fourier transform


we can easily derive the Fourier transform of $(1)$ by noticing that $x(t)$ is the real part of the time-domain function on the left-hand side of $(3)$, so its Fourier transform must be the even part of $Y(\omega)$:



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