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Working out Fourier Transforms Pairs Fourier Transform Pairs The Fourier transform transforms a function of time, f (t), into a function of frequency, F(s): ∞ π F { f (t)}(s)= F(s)= f (t)e− j2 stdt. Z−∞ The inverse Fourier transform transforms a func- tion of frequency, F(s), into a function of time, f (t): ∞ π F −1{F(s)}(t)= f (t)= F(s)e j2 stds. Z−∞ The inverse Fourier transform of the Fourier trans- form is the identity transform: ∞ ∞ π τ π f (t)= f (τ)e− j2 s dτ e j2 stds. Z−∞ Z−∞ Fourier Transform Pairs (contd). Because the Fourier transform and the inverse Fourier transform differ only in the sign of the exponential’s argument, the following recipro- cal relation holds between f (t) and F(s): F f (t) −→ F(s) is equivalent to F F(t) −→ f (−s). This relationship is often written more econom- ically as follows: F f (t) ←→ F(s) where f (t) and F(s) are said to be a Fourier transform pair. Fourier Transform of Gaussian Let f (t) be a Gaussian: π 2 f (t)= e− t . By the definition of Fourier transform we see that: ∞ π 2 π F(s) = e− t e− j2 stdt Z ∞ −∞ π 2 = e− (t + j2st)dt. Z−∞ Now we can multiply the right hand side by 2 2 e−πs eπs = 1: ∞ π 2 π 2 π 2 F(s) = e− s e− (t + j2st)+ s dt Z ∞ −∞ π 2 π 2 2 = e− s e− (t + j2st−s )dt Z ∞ −∞ π 2 π = e− s e− (t+ js)(t+ js)dt Z ∞ −∞ π 2 π 2 = e− s e− (t+ js) dt Z−∞ Fourier Transform of Gaussian (contd.) ∞ π 2 π 2 F(s) = e− s e− (t+ js) dt Z−∞ After substituting u for t + js and du for dt we see that: ∞ π 2 π 2 F(s)= e− s e− u du. Z−∞ 1 It follows that the Gaussian| {z is its} own Fourier transform: π 2 F π 2 e− t ←→ e− s . Fourier Transform of Dirac Delta Function To compute the Fourier transform of an impulse we apply the definition of Fourier transform: ∞ − j2πst F {δ(t −t0)}(s)= F(s)= δ(t −t0)e dt Z−∞ which, by the sifting property of the impulse, is just: π e− j2 st0. It follows that: F − j2π st0 δ(t −t0) −→ e . Fourier Transform of Harmonic Signal What is the inverse Fourier transform of an im- pulse located at s0? Applying the definition of inverse Fourier transform yields: ∞ −1 j2πst F {δ(s−s0)}(t)= f (t)= δ(s−s0)e ds Z−∞ which, by the sifting property of the impulse, is just: π e j2 s0 t. It follows that: F j2π s0 t e −→ δ(s − s0). Fourier Transform of Sine and Cosine We can compute the Fourier transforms of the sine and cosine by exploiting the sifting prop- erty of the impulse: ∞ f (x)δ(x − x0)dx = f (x0). Z−∞ • Question What is the inverse Fourier trans- form of a pair of impulses spaced symmetri- cally about the origin? −1 F {δ(s + s0)+ δ(s − s0)} • Answer By definition of inverse Fourier trans- form: ∞ j2πst f (t)= [δ(s + s0)+ δ(s − s0)]e ds. Z−∞ Fourier Transform of Sine and Cosine (contd.) Expanding the above yields the following ex- pression for f (t): ∞ ∞ j2πst j2πst δ(s + s0)e ds + δ(s − s0)e ds Z−∞ Z−∞ Which by the sifting property is just: π π f (t) = e j2 s0 t + e− j2 s0 t = 2cos(2π s0 t). Fourier Transform of Sine and Cosine (contd.) It follows that F 1 cos(2π s t) ←→ [δ(s + s )+ δ(s − s )]. 0 2 0 0 A similar argument can be used to show: F j sin(2π s t) ←→ [δ(s + s ) − δ(s − s )]. 0 2 0 0 Fourier Transform of the Pulse To compute the Fourier transform of a pulse we apply the definition of Fourier transform: ∞ π F(s) = Π(t)e− j2 stdt Z−∞ 1 2 π = e− j2 stdt Z 1 −2 1 1 π 2 = e− j2 st − j2πs 1 −2 1 π π = e− j s − e j s − j2πs π π 1 e j s − e− j s = πs 2 j (e jx−e− jx) Using the fact that sin(x) = 2 j we see that: sin(πs) F(s) = . πs Fourier Transform of the Shah Function Recall the Fourier series for the Shah function: ∞ 1 t 1 jω t. πIII π = π ∑ e 2 2 2 ω=−∞ By the sifting property, t ∞ ∞ III = ∑ δ(s − ω)e jstds. π Z ∞ 2 ω=−∞ − Changing the order of the summation and the integral yields t ∞ ∞ III = ∑ δ(s − ω)e jstds. π Z ∞ 2 − ω=−∞ Factoring out e jst from the summation t ∞ ∞ III = e jst ∑ δ(s − ω)ds π Z ∞ 2 − ω=−∞ ∞ = e jstIII(s)ds. Z−∞ Fourier Transform of the Shah Function Substituting 2πτ for t yields ∞ π τ III(τ) = III(s)e j2 s ds Z−∞ = F −1{III(s)}(τ). Consequently we see that F {III} = III..
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