Homework Four

Fourier Transform Properties

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Nick Christman

1. Find ${\mathcal {F}}\left[\int _{-\infty }^{\infty }10^{t}g(t)e^{j2\pi ft_{0}}\,dt\right]$

To begin, we know that

${\mathcal {F}}\left[\int _{-\infty }^{\infty }10^{t}g(t)e^{j2\pi ft_{0}}\,dt\right]=\int _{-\infty }^{\infty }\left(\int _{-\infty }^{\infty }10^{t}g(t)e^{j2\pi ft_{0}}\,dt\right)e^{-j2\pi ft}\,dt$

After some factoring and combinting of like terms we get:

${\mathcal {F}}\left[\int _{-\infty }^{\infty }10^{t}g(t)e^{j2\pi ft_{0}}\,dt\right]=\int _{-\infty }^{\infty }\left(\int _{-\infty }^{\infty }10^{t}g(t)\,dt\right)e^{j2\pi f(t_{0}-t)}\,dt$

But recall that $\int _{-\infty }^{\infty }e^{j2\pi f(t_{0}-t)}\,dt\equiv \delta (t_{0}-t){\mbox{ or }}\delta (t-t_{0})$

Because of this definition (and some "math magic") our problem has been simplified significantly:

${\mathcal {F}}\left[\int _{-\infty }^{\infty }10^{t}g(t)e^{j2\pi ft_{0}}\,dt\right]=\int _{-\infty }^{\infty }10^{t}g(t)\delta (t-t_{0})\,dt=10^{t_{0}}g(t_{0})$

Therefore,

${\mathcal {F}}\left[\int _{-\infty }^{\infty }10^{t}g(t)e^{j2\pi ft_{0}}\,dt\right]=10^{t_{0}}g(t_{0})$