Fourier Transform Properties: Difference between revisions

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<math> \mathcal{F}[10^{t}g(t)e^{j2 \pi ft_0}] = \int_{-\infty}^{\infty}10^{t}g(t)e^{j2 \pi ft_0}e^{-j2 \pi ft}\,dt = \int_{-\infty}^{\infty}10^{t}g(t)e^{j2 \pi f(t_{0}-t)}\,dt</math>
<math> \mathcal{F}[10^{t}g(t)e^{j2 \pi ft_0}] = \int_{-\infty}^{\infty}10^{t}g(t)e^{j2 \pi ft_0}e^{-j2 \pi ft}\,dt = \int_{-\infty}^{\infty}10^{t}g(t)e^{j2 \pi f(t_{0}-t)}\,dt</math>


But recall that, <math>e^{j2 \pi f(t_{0}-t)} \equiv \delta (t_{0}-t) \mbox{ or } \delta (t-t_{0})</math>
But recall that <math>e^{j2 \pi f(t_{0}-t)} \equiv \delta (t_{0}-t) \mbox{ or } \delta (t-t_{0})</math>





Revision as of 15:13, 16 October 2009

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(t0t)δ(t0t) or δ(tt0)


Because of this definition, our problem has now been simplified significantly:

[10tg(t)ej2πft0]=10tg(t)δ(tt0)dt=10t0g(t0)

Therefore,

[10tg(t)ej2πft0]=10t0g(t0)