Homework Four: Difference between revisions

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After some factoring and combinting of like terms we get:  
After some factoring and combinting <math> \Longleftarrow </math> "combining-Kevin" of like terms we get:  


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<math> \mathcal{F} \left[ \int_{- \infty}^{\infty}10^{t}g(t)e^{j2 \pi ft_0} \,dt \right] = 10^{t_0}g(t_0) </math>
<math> \mathcal{F} \left[ \int_{- \infty}^{\infty}10^{t}g(t)e^{j2 \pi ft_0} \,dt \right] = 10^{t_0}g(t_0) </math>
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Other than the one typo looks good - Kevin

Revision as of 20:41, 2 November 2009

Fourier Transform Properties


Nick Christman

1. Find ℱ[∫−∞∞10tg(t)ej2πft0dt]

To begin, we know that

ℱ[∫−∞∞10tg(t)ej2πft0dt]=∫−∞∞(∫−∞∞10tg(t)ej2πft0dt)e−j2πftdt

After some factoring and combinting ⟸ "combining-Kevin" of like terms we get:


ℱ[∫−∞∞10tg(t)ej2πft0dt]=∫−∞∞(∫−∞∞10tg(t)dt)ej2πf(t0−t)dt

But recall that ∫−∞∞ej2πf(t0−t)dt≡δ(t0−t) or δ(t−t0)


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

ℱ[∫−∞∞10tg(t)ej2πft0dt]=∫−∞∞10tg(t)δ(t−t0)dt=10t0g(t0)

Therefore,

ℱ[∫−∞∞10tg(t)ej2πft0dt]=10t0g(t0)
Other than the one typo looks good - Kevin