WorksheetsRevision Unit 3 A
Total questions: 10
Worksheet time: 5mins
The Fourier Transform of f(x) is F[f(x)]=
2π1∫−∞∞f(x) dx = F[s]
2π1∫−∞∞f(x)eisx dx = F[s]
2π1∫−∞∞f(x) ds = F[s]
2π1∫−∞∞f(x) cossx dx = F[s]
The inverse Fourier Transform of F[f(x)] is f(x) =
2π1∫−∞∞ F[s] dx = f(x)
2π1∫−∞∞F[s]eisx ds =f(x)
2π1∫−∞∞F[s] e−isxds =f(x)
2π1∫−∞∞F[s] cossx ds = f(x)
The Fourier sine Transform of f(x) is Fs[f(x)]=
2π1∫−∞∞f(x) dx = Fs[s]
2π1∫0∞f(x) sinsx dx = Fs[s]
π2∫0∞f(x) sinsx dx = Fs[s]
2π1∫−∞∞f(x) cossx dx = Fs[s]
The inverse Fourier sine Transform of F−1{Fs[f(x)]} = f(x)
2π1∫−∞∞f(x) ds= Fs[s]
2π1∫0∞ Fs[s]sinsx ds=f(x)
π2∫0∞Fs[s]sinsx ds = f(x)
2π1∫−∞∞ Fs[s] cossx dx =f(x)
The Fourier cosine Transform of f(x) is Fc[f(x)]=
2π1∫−∞∞f(x) cossx dx = Fc[s]
2π1∫0∞f(x) cossx dx = Fc[s]
π2∫0∞f(x) cossx dx = Fc[s]
2π1∫−∞∞f(x) cossx dx = Fs[s]
The Imaginary part of eisx is
sinsx
i sinsx
cossx
cosx
The Real part of eisx is
sinsx
i sinsx
cossx
cosx
The inverse Fourier cosine Transform of F−1{Fc[f(x)]} = f(x)
2π1∫−∞∞f(x) cossx ds= Fc[s]
π2∫0∞ Fc[s] cossx ds=f(x)
π2∫0∞Fs[s]sinsx ds = f(x)
2π1∫−∞∞ Fs[s] cossx dx =f(x)
The Fourier Sine Transform of
e−axFs[e−ax]=π2 a2+s2a
Fs[e−ax]=π2 a2+s2s
Fc[e−ax]=π2 a2+s2s
none of the above
The Fourier cosine Transform of
e−axFs[e−ax]=π2 a2+s2a
Fs[e−ax]=π2 a2+s2s
Fc[e−ax]=π2 a2+s2a
none of the above
