# D dt f t jωf ω 16 jtf t d dω f ω 17 f t t e jωt

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d dt f ( t ) jωF ( ω ) (1.6) jtf ( t ) d F ( ω ) (1.7) f ( t t ) e jωt F ( ω ) (1.8) e jωt f ( t ) F ( ω ω ) (1.9) f ( at ) 1 | a | F ( ω/a ) (1.10) f ( t ) F ( ω ) (1.11) Convolution theorem One of the most important properties of the Fourier transform is the way that the convolution operator transforms. The convolution of two functions f and g is defined by f g = integraldisplay t 0 f ( t τ ) g ( τ ) Convolution appears in a number of signal processing and radar contexts, particularly where filters are concerned. While the integral can sometimes be carried out simply, it is just as often difficult to calculate analytically and compu- tationally. The most expedient means of calculation usually involves Fourier analysis. Consider the product of the frequency-domain representations of f and g : F ( ω ) G ( ω ) = integraldisplay −∞ f ( τ ) e jωτ integraldisplay −∞ g ( τ ) e jωτ = integraldisplay −∞ integraldisplay −∞ e ( τ + τ ) f ( τ ) g ( τ ) With the change of variables t = τ + τ , dt = , this becomes F ( ω ) G ( ω ) = integraldisplay −∞ integraldisplay −∞ e jωt f ( t τ ) g ( τ ) dtdτ = integraldisplay −∞ e jωt bracketleftbiggintegraldisplay −∞ f ( t τ ) g ( τ ) bracketrightbigg dt The quantity in the square brackets is clearly the convolution of f ( t ) and g ( t ) . Evidently then, the Fourier transform of the convolution of two functions is the product of the Fourier transforms of the functions. This suggests an easy and efficient means of evaluating a convolution integral. It also suggests another way of interpreting the convolution operation, which translates into a product in the frequency domain. 16

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Parseval’s theorem The convolution theorem proven above can be used to reformulate Parseval’s theorem for continuous functions. First, it is necessary to establish how the complex conjugate operation behaves under Fourier transformation. Let f ( t ) = 1 2 π integraldisplay −∞ F ( ω ) e jωt Taking the complex conjugate of both sides gives f ( t ) = 1 2 π integraldisplay −∞ F ( ω ) e jωt Next, replace the variable t with t : f ( t ) = 1 2 π integraldisplay −∞ F ( ω ) e jωt which is then the inverse Fourier transform of F ( ω ) . Let us use this information to express the product F ( ω ) G ( ω ) , borrowing from results obtained earlier. F ( ω ) G ( ω ) = integraldisplay −∞ e jωt bracketleftbiggintegraldisplay −∞ f [ ( t τ )] g ( τ ) bracketrightbigg dt Taking the inverse Fourier transform gives: 1 2 π integraldisplay −∞ F ( ω ) G ( ω ) e jωt = integraldisplay −∞ f ( τ t ) g ( τ ) This expression represents a general form of Parseval’s theorem. A more narrow form is given considering t = 0 and taking f = g , F = G : 1 2 π integraldisplay −∞ | F ( ω ) | 2 = integraldisplay −∞ | f ( t ) | 2 dt (1.12) where t has replaced τ as the dummy variable on the right side of the last equation.
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