ASN4 -Fourier Transform property: Difference between revisions

Latest revision as of 11:37, 19 December 2009

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

$ℱ [c o s (2 π f_{0} t) g (t)]$

Applying the forward Fourier transform

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

Applying Euler's cosine identity

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

Distribting to both terms in side the brackets

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

Combining exponential terms

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

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

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

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+[[Jodi Hodge| Back to my home page]]
+Find the Fourier transform of <math> cos(2\pi f_0t)g(t) \!</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 \!</math>
+Applying Euler's cosine identity
-[[Jodi Hodge| Back to my home page]]
+<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>
-'''Find '''<math>\mathcal{F}[cos(w_0t)g(t)]\!</math>
+Distribting to both terms in side the brackets
-Recall <math> w_0 = 2\pi f_0\!</math>, so
+<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>\mathcal{F}[cos(w_0t)g(t)] = \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>
+Combining exponential terms
-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>
+<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>
-Now <math>\int_{-\infty}^{\infty} \frac{1}{2}[e^{j2\pi f_0t}+e^{-j2\pi f_0t}]g(t)e^{-j2\pi ft}dt = \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 = \frac{1}{2}G(f-f_0) \ + \ \frac{1}{2}G(f+f_0)\!</math><br><br>
+Note that there are forward Fourier Transform expressions in the above equation. With substitution the result is
-So <math>\mathcal{F}[cos(w_0t)g(t)] = \frac{1}{2}[G(f-f_0)+ G(f+f_0)]\!</math><br><br>
+<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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