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By Whyburn G.T.

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Unilateral exponential function, magnitude of the Fourier transform and phase (imaginary part/real part) Fig. 7. Real part and imaginary part of the Fourier transform of a unilateral exponential function 1/(λ2 + ω 2 ) is a Lorentzian again. This representation is often also called the 2 2 power representation: |F (ω)|2 = (real part) + (imaginary part) . e. when “in resonance”. Warning: The representation of the magnitude as well as of the squared magnitude does away with the linearity of the Fourier transformation!

Now, we calculate the Fourier transform of important functions. Let us start with the Gaussian. 2 (The normalised Gaussian). The prefactor is chosen in such a way that the area is 1. f (t) = 1 t2 1 √ e− 2 σ 2 . 16) −∞ +∞ 2 = √ σ 2π =e 1 t2 e− 2 σ2 cos ωt dt 0 − 12 σ 2 ω 2 . Again, the imaginary part is 0, as f (t) is even. The Fourier transform of a Gaussian results in another Gaussian. Note that the Fourier transform is not normalised to area 1. 177 × HWHM. f (t) has σ in the exponent’s denominator, F (ω) in the numerator: the slimmer f (t), the wider F (ω) and vice versa (cf.

Fig. 13). Please note the following: the interval, where f (t) ⊗ g(t) is unequal to 0, now is twice as big: 2T ! ), then also f (t) ⊗ g(t) would be symmetrical around 0. In this case we would have convoluted f (t) with itself. Now to a more useful example: let’s take a pulse that looks like a “unilateral” exponential function (Fig. 14 left): f (t) = e−t/τ for t ≥ 0 0 else . 3 Convolution, Cross Correlation, Autocorrelation, Parseval’s Theorem 49 h(t) T ✻ ✲ − T2 T 2 3T 2 t Fig. 13. Convolution h(t) = f (t) ⊗ g(t) Fig.

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