Fourier Transform Properties: Difference between revisions

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[[Max Woesner|<b><u>Max Woesner</u></b>]]<br><br>
[[Max Woesner|<b><u>Max Woesner</u></b>]]<br><br>
Find <math>\mathcal{F}[cos(w_0t)g(t)]\!</math><br>
'''Find <math>\mathcal{F}[cos(w_0t)g(t)]\!</math><br>'''
Recall <math> w_0 = 2\pi f_0\!</math>, so <math>\mathcal{F}[cos(w_0t)g(t)] = \mathcal{F}[cos(2\pi f_0t)g(t)] = \int_{-\infty}^{\infty}cos(2\pi f_0t)g(t)e^{-j2\pi ft}dt\!</math><br>
Recall <math> w_0 = 2\pi f_0\!</math>, so <math>\mathcal{F}[cos(w_0t)g(t)] = \mathcal{F}[cos(2\pi f_0t)g(t)] = \int_{-\infty}^{\infty}cos(2\pi f_0t)g(t)e^{-j2\pi ft}dt\!</math><br>
Also recall <math> cos(\theta) = \frac{1}{2}(e^{j\theta} + e^{-j\theta})\!</math>,so <math>\int_{-\infty}^{\infty}cos(2\pi f_0t)g(t)e^{-j2\pi ft}dt = \int_{-\infty}^{\infty} \frac{1}{2}[e^{j2\pi f_0t}+e^{-j2\pi f_0t}]g(t)e^{-j2\pi ft}dt\!</math><br>
Also recall <math> cos(\theta) = \frac{1}{2}(e^{j\theta} + e^{-j\theta})\!</math>,so <math>\int_{-\infty}^{\infty}cos(2\pi f_0t)g(t)e^{-j2\pi ft}dt = \int_{-\infty}^{\infty} \frac{1}{2}[e^{j2\pi f_0t}+e^{-j2\pi f_0t}]g(t)e^{-j2\pi ft}dt\!</math><br>

Revision as of 21:29, 17 October 2009

Max Woesner

Find ℱ[cos(w0t)g(t)]
Recall w0=2πf0, so ℱ[cos(w0t)g(t)]=ℱ[cos(2πf0t)g(t)]=∫−∞∞cos(2πf0t)g(t)e−j2πftdt
Also recall cos(θ)=12(ejθ+e−jθ),so ∫−∞∞cos(2πf0t)g(t)e−j2πftdt=∫−∞∞12[ej2πf0t+e−j2πf0t]g(t)e−j2πftdt
Now ∫−∞∞12[ej2πf0t+e−j2πf0t]g(t)e−j2πftdt=12∫−∞∞e−j2π(f−f0)tg(t)dt+12∫−∞∞e−j2π(f+f0)tg(t)dt=12G(f−f0)+12G(f+f0)
So ℱ[cos(w0t)g(t)]=12[G(f−f0)+G(f+f0)]


Nick Christman

Find ℱ[10tg(t)ej2πft0]

To begin, we know that

ℱ[10tg(t)ej2πft0]=∫−∞∞10tg(t)ej2πft0e−j2πftdt=∫−∞∞10tg(t)ej2πf(t0−t)dt

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


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

ℱ[10tg(t)ej2πft0]=∫−∞∞10tg(t)δ(t−t0)dt=10t0g(t0)

Therefore,

ℱ[10tg(t)ej2πft0]=10t0g(t0)