Revision as of 21:39, 22 April 2018

Table of CT Fourier Series Coefficients and Properties

Fourier series Coefficients

Function

Fourier Series

Coefficients

sin(\omega_0t)

\frac{1}{2j}e^{j\omega_0 t} - \frac{1}{2j}e^{-j\omgea_0 t}

a_1 = \frac{1}{2j} a_{-1} = -\frac{1}{2j} and a_k = 0 for k not 1,-1

cos(\omega_0t)

\frac{1}{2}e^{j\omega_0 t} + \frac{1}{2}e^{-j\omgea_0 t}

a_1 = \frac{1}{2} a_{-1} = \frac{1}{2} and a_k = 0 for k not 1,-1

sinc(

}

Properties of CT Fourier systems

}

Property Name	Property	Proof
Linearity	$\mathfrak{F}(c_1g(t) + c_2h(t) = c_1G(f) + c_2H(f)$	$\mathfrak{F}(c_1g(t) + c_2h(t) = \int_{-\infty}^\infty c_1g(t) dt + \int_{-\infty}^\infty c_2h(t) dt$ $=c_1\int_{-\infty}^\infty g(t)e^{i2\pi ft} dt + c_2 \int_{-\infty}^\infty g(t)e^{i2\pi ft} dt$ $$ =c_1G(f) + c_2H(f) $$
Time Shifting	$\mathfrak{F}(g(t - a)) = e^{-i2\pi fa}*G(f)$	$\mathfrak{F}(g(t - a)) = \int_{-\infty}^\infty g(t-a)e^{-2\pi ft}dt$ $=\int_{-\infty}^\infty g(u)e^{-i2\pi f(u+a)}du$ $=e^{-i2\pi fa}\int_{-\infty}^\infty g(u)e^{-i2\pi fu}du$ $=e^{-i2\pi fa} G(f)$
Time Scaling	$\mathfrak{F}(g(ct)) = \frac{G(\frac{f}{c})}{\|c\|}$	$\mathfrak{F}(g(ct)) = \int_{-\infty}^\infty g(ct)e^{-i2\pi ft}dt$ subtitute : u = ct, du = cdt $\mathfrak{F}(g(ct)) = \int_{-c\infty}^{c\infty} \frac{g(u)}{c}e^{-i2\pi f\frac{u}{c}}du$ if c is greater than 0: then no signs change. if c is less than 0: the integration must be flipped as well as the negative from the c so you still get the same equation. therefore the absolute value of c is obtained.
Frequency Shifting	$\mathfrak{F}^{-1}[X(j\omega + omega_0)] = x(t)e^{-j\omega_0t}$	$\mathfrak{F}^{-1}[X(j\omega + omega_0)] = \frac{1}{2\pi} \int_{-\infty}^{\infty}X(j(\omega +\omega_0))e^{j\omega t} d\omega$ $=\frac{1}{2\pi} \int_{-\infty}^{\infty} X(j\omega ')e^{j\omega (\omega ' + \omega_0)} d\omega$ $=e^{j\omega_0 t}\frac{1}{2\pi} \int_{-\infty}^{\infty} X(j\omega ')e^{j\omega '} d\omega'$ $=x(t)e^{j\omega_0 t}$
Time Reversal	$\mathfrak{F}[g(-t)] = G(-\omega)$	$\mathfrak{F}[g(-t)] = \int_{-\infty}^{\infty}g(-t)e^{-j\omgea t} dt$ replace t with -t $\mathfrak{F}[g(-t)] = -\int_{\infty}^{-\infty} g(t')e^{-j\omega t' } dt'$ $=\int_{-\infty}^{\infty}g(t') e^{-j\omega t'} dt'$ $=G(-\omega)$
Complex Conjugate	$\mathfrak{F}(g(t) = G(-j\omega)$	$g(t) = [\frac{1}{2\pi} \int_{-\infty}^{\infty} G(-\omega)e^{j\omega t} d\omega]^$ $=[\frac{1}{2\pi} \int_{-\infty}^{\infty} G(-\omega)e^{j\omega' t} d\omega]^$ $x(t) = \frac{1}{2\pi} \int_{-\infty}^{\infty} G(-\omega')e^{j\omega' t} d\omega$ $= \mathfrak{F}[G*(-\omega)]$

@@ Line 12: / Line 12: @@
 ! Coefficients
 |-
-|
-|
+|<math>sin(\omega_0t) </math>
-|
+|<math>\frac{1}{2j}e^{j\omega_0 t} - \frac{1}{2j}e^{-j\omgea_0 t} </math>
-|}
+|<math>a_1 = \frac{1}{2j} a_{-1} = -\frac{1}{2j} and a_k = 0 for k not 1,-1 </math>
+|-
+|<math>cos(\omega_0t) </math>
+|<math>\frac{1}{2}e^{j\omega_0 t} + \frac{1}{2}e^{-j\omgea_0 t} </math>
+|<math>a_1 = \frac{1}{2} a_{-1} = \frac{1}{2} and a_k = 0 for k not 1,-1 </math>
+|-
+|sinc(
+}

Difference between revisions of "CT Fourier Proofs Pranith Lomada Spring18 Boutin" - Rhea

Revision as of 21:39, 22 April 2018

Table of CT Fourier Series Coefficients and Properties

Fourier series Coefficients

Properties of CT Fourier systems

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