ASN4 -Fourier Transform property: Difference between revisions

Revision as of 11:07, 19 December 2009

Find the Fourier transform of $c o s (2 π f_{0} t) g (t) =$

\mathcal{F}[cos(2\pi f_0t)g(t)]=

Using Euler's cosine identity

$ℱ [c o s (2 π f_{0} t) g (t)] = \int_{- \infty}^{\infty} c o s (2 π f_{0} t) g (t) e^{- j 2 π f t} d t$

$= \int_{- \infty}^{\infty} [\frac{1}{2} e^{j 2 π f_{0} t} + \frac{1}{2} e^{- j 2 π f_{0} t}] g (t) e^{- j 2 π f t} d t$

$= \int_{- \infty}^{\infty} \frac{1}{2} e^{j 2 π f_{0} t} + \frac{1}{2} e^{- j 2 π f_{0} t} g (t) e^{- j 2 π f t} d t$

$= \int_{- \infty}^{\infty} \frac{1}{2} e^{j 2 π f_{0} t} e^{- j 2 π f t} d t + \int_{- \infty}^{\infty} \frac{1}{2} e^{- j 2 π f_{0} t} g (t) e^{- j 2 π f t} d t$

$= \int_{- \infty}^{\infty} \frac{1}{2} e^{- j 2 π (f - f_{0}) t} g (t) d t + \int_{- \infty}^{\infty} \frac{1}{2} e^{- j 2 π (f + f_{0}) t} g (t) d t$

Identifying that the above equation contains Fourier Transforms the solution is

$ℱ [c o s (2 π f_{0} t) g (t)] = \frac{1}{2} G (f - f_{0}) + \frac{1}{2} [G (f + f_{0})$

@@ 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>
+\mathcal{F}[cos(2\pi f_0t)g(t)]=
 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)]=\int_{-\infty}^{\infty}cos(2\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}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} \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} \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}\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>
 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>

ASN4 -Fourier Transform property: Difference between revisions

Revision as of 11:07, 19 December 2009

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