ASN4 -Fourier Transform property: Difference between revisions

Latest revision as of 11:37, 19 December 2009

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Find the Fourier transform of $cos(2\pi f_{0}t)g(t)\!$

${\mathcal {F}}[cos(2\pi f_{0}t)g(t)]\!$

Applying the forward Fourier transform

$=\int _{-\infty }^{\infty }cos(2\pi f_{0}t)g(t)e^{-j2\pi ft}dt\!$

Applying Euler's cosine identity

$=\int _{-\infty }^{\infty }[{\frac {1}{2}}e^{j2\pi f_{0}t}+{\frac {1}{2}}e^{-j2\pi f_{0}t}]g(t)e^{-j2\pi ft}dt\!$

Distribting to both terms in side the brackets

$=\int _{-\infty }^{\infty }{\frac {1}{2}}e^{j2\pi f_{0}t}e^{-j2\pi ft}dt+\int _{-\infty }^{\infty }{\frac {1}{2}}e^{-j2\pi f_{0}t}g(t)e^{-j2\pi ft}dt\!$

Combining exponential terms

$=\int _{-\infty }^{\infty }{\frac {1}{2}}e^{-j2\pi (f-f_{0})t}g(t)dt\ +\int _{-\infty }^{\infty }{\frac {1}{2}}e^{-j2\pi (f+f_{0})t}g(t)dt\!$

Note that there are forward Fourier Transform expressions in the above equation. With substitution the result is

${\mathcal {F}}[cos(2\pi f_{0}t)g(t)]={\frac {1}{2}}G(f-f_{0})+{\frac {1}{2}}[G(f+f_{0})\!$

@@ Line 1: / Line 1: @@
 [[Jodi Hodge| Back to my home page]]
-<math>\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>
+Find the Fourier transform of <math> cos(2\pi f_0t)g(t) \!</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}+\frac{1}{2}e^{-j2\pi f_0t}g(t)e^{-j2\pi ft}dt\!</math>
+<math> \mathcal{F}[cos(2\pi f_0t)g(t)] \!</math>
+Applying the forward Fourier transform
-<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-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>
+<math> =\int_{-\infty}^{\infty}cos(2\pi f_0t)g(t)e^{-j2\pi ft}dt \!</math>
+Applying Euler's cosine identity
-Identifying that the above equation contains Fourier Transforms the solution is
+<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>
+Distribting to both terms in side the brackets
+<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>
+Combining exponential terms
+<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 that there are forward 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>

ASN4 -Fourier Transform property: Difference between revisions

Latest revision as of 11:37, 19 December 2009

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