ASN4 -Fourier Transform property: Difference between revisions
Jump to navigation
Jump to search
Jodi.Hodge (talk | contribs) No edit summary |
Jodi.Hodge (talk | contribs) No edit summary |
||
Line 13: | Line 13: | ||
<math> = \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} [\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} \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} \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} \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> |
||
Line 20: | Line 18: | ||
<math> =\int_{-\infty}^{\infty}\frac{1}{2}e^{-j2\pi (f-f_0)t}g(t)dt \ + \int_{-\infty}^{\infty}\frac{1}{2}e^{-j2\pi (f+f_0)t}g(t)dt \!</math> |
<math> =\int_{-\infty}^{\infty}\frac{1}{2}e^{-j2\pi (f-f_0)t}g(t)dt \ + \int_{-\infty}^{\infty}\frac{1}{2}e^{-j2\pi (f+f_0)t}g(t)dt \!</math> |
||
Note the Inverse Fourier Transform expressions in the above equation. With substitution the result 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 10:18, 19 December 2009
Find the Fourier transform of
Applying the forward Fourier transform
Applying Euler's cosine identity
Note the Inverse Fourier Transform expressions in the above equation. With substitution the result is