ASN4 -Fourier Transform property: Difference between revisions
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Jodi.Hodge (talk | contribs) New page: Back to my home page '''Find '''<math>\mathcal{F}[cos(w_0t)g(t)]\!</math> Recall <math> w_0 = 2\pi f_0\!</math>, so <math>\mathcal{F}[cos(w_0t)g(t)] = \mathcal{F}[cos(... |
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[[Jodi Hodge| Back to my home page]] | [[Jodi Hodge| Back to my home page]] | ||
'''Find '''<math>\mathcal{F}[cos( | '''Find '''<math>\mathcal{F}[cos(2\pi f_0t)g(t)]\!</math> | ||
Recall <math> w_0 = 2\pi f_0\!</math>, so | Recall <math> w_0 = 2\pi f_0\!</math>, so | ||
<math>\ | Using Euler's cosine identity <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> | ||
Now <math> \mathcal{F}[cos(2\pi f_0t)g(t)]= \frac{1}{2}\int_{-\infty}^{\infty}e^{-j2\pi (f-f_0)t}g(t)dt \ + \ \frac{1}{2}\int_{-\infty}^{\infty}e^{-j2\pi (f+f_0)t}g(t)dt \!</math> | |||
So <math>\mathcal{F}[cos(2\pi f_0t)g(t)] = \frac{1}{2}[G(f-f_0)+ G(f+f_0)]\!</math> | |||
So <math>\mathcal{F}[cos( |