ASN4 -Fourier Transform property: Difference between revisions
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Using Euler's cosine identity | 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}+\frac{1}{2}e^{-j2\pi f_0t}g(t)e^{-j2\pi ft}dt\!</math> | <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}+\frac{1}{2}e^{-j2\pi f_0t}g(t)e^{-j2\pi ft}dt\!</math> | ||
<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 ft} dt + \int_{-\infty}^{\infty}\frac{1}{2}e^{-j2\pi f_0t}g(t)e^{-j2\pi ft}dt\!</math> | <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 ft} dt + \int_{-\infty}^{\infty}\frac{1}{2}e^{-j2\pi f_0t}g(t)e^{-j2\pi ft}dt\!</math> | ||
<math> \ | <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-f_0)t}g(t)dt \ + \ \frac{1}{2}\int_{-\infty}^{\infty}\frac{1}{2}e^{-j2\pi (f+f_0)t}g(t)dt \!</math> | ||
Identifying that the above equation contains Fourier Transforms the solution is | Identifying that the above equation contains Fourier Transforms the solution is | ||
<math>\mathcal{F}[cos(2\pi f_0t)g(t)] = \frac{1}{2}G(f-f_0)+ \frac{1}{2}[G(f+f_0)\!</math> | <math>\mathcal{F}[cos(2\pi f_0t)g(t)] = \frac{1}{2}G(f-f_0)+ \frac{1}{2}[G(f+f_0)\!</math> |
Revision as of 00:02, 19 December 2009
Using Euler's cosine identity
Identifying that the above equation contains Fourier Transforms the solution is