Fourier Transform Properties

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Max Woesner

Find [cos(w0t)g(t)]
Recall w0=2πf0, so [cos(w0t)g(t)]=[cos(2πf0t)g(t)]=cos(2πf0t)g(t)ej2πftdt
Also recall cos(θ)=12(ejθ+ejθ),so cos(2πf0t)g(t)ej2πftdt=12[ej2πf0t+ej2πf0t]g(t)ej2πftdt
Now 12[ej2πf0t+ej2πf0t]g(t)ej2πftdt=12ej2π(ff0)tg(t)dt+12ej2π(f+f0)tg(t)dt=12G(ff0)+12G(f+f0)
So [cos(w0t)g(t)]=12[G(ff0)+G(f+f0)]


Nick Christman

Find [10tg(t)ej2πft0]

To begin, we know that

[10tg(t)ej2πft0]=10tg(t)ej2πft0ej2πftdt=10tg(t)ej2πf(t0t)dt


But recall that, ej2πf(t0tδ(t0t) or δ(tt0)
Also recall cos(θ)=12(ejθ+ejθ),so cos(2πf0t)g(t)ej2πftdt=12[ej2πf0t+ej2πf0t]g(t)ej2πftdt
Now 12[ej2πf0t+ej2πf0t]g(t)ej2πftdt=12ej2π(ff0)tg(t)dt+12ej2π(f+f0)tg(t)dt=12G(ff0)+12G(f+f0)
So [cos(w0t)g(t)]=12[G(ff0)+G(f+f0)]