Relation between laplace transform and fourier transform pairs

A 2M - 1 point triangle in the time domain can be formed by convolving an M point rectangular pulse with itself. Since convolution in the time domain results in multiplication in the frequency domain, convolving a waveform with itself will square the frequency spectrum. Is there a waveform that is its own Fourier Transform? The answer is yes, and there is only one: the Gaussian.

Figure e shows a Gaussian curve, and f shows the corresponding frequency spectrum, also a Gaussian curve. This relationship is only true if you ignore aliasing. While only one side of a Gaussian is shown in f , the negative frequencies in the spectrum complete the full curve, with the center of symmetry at zero frequency. Figure g shows what can be called a Gaussian burst.

It is formed by multiplying a sine wave by a Gaussian. For example, g is a sine wave multiplied by the same Gaussian shown in e. The corresponding frequency domain is a Gaussian centered somewhere other than zero frequency. As before, this transform pair is not as important as the reason it is true. The Laplace transform is widely used for solving differential equations since the Laplace transform exists even for the signals for which the Fourier transform does not exist.

The Fourier transform is rarely used for solving the differential equations since the Fourier transform does not exists for many signals. The Laplace transform has a convergence factor and hence it is more general. The Fourier transform does not have any convergence factor. The Fourier transform is equivalent to the Laplace transform evaluated along the imaginary axis of the s-plane.

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