Pridma's Article on Fourier Transforms - Revision history

Pridma: /* Useful Applications */

2007-10-12T03:29:12Z

Useful Applications

← Older revision		Revision as of 20:29, 11 October 2007
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	To build this into a circuit, we need to find the transfer function in the time domain. We can inverse Fourier transform <math>H(f) \!</math>.		To build this into a circuit, we need to find the transfer function in the time domain. We can inverse Fourier transform <math>H(f) \!</math>.

	:<math>h(t) = \mathcal{F}^{-1}\left[H(f)\right] = \int_{-f_{0}}^{f_{0}} 1 e^{j2\pi ft} df = \left[\frac{e^{j2\pi ft}}{j2\pi ft}\right]_{f=-f_{0}}^{f_{0}}		:<math>h(t) = \mathcal{F}^{-1}\left[H(f)\right] = \int_{-f_{0}}^{f_{0}} 1 e^{j2\pi ft} df = \left[\frac{e^{j2\pi ft}}{j2\pi t}\right]_{f=-f_{0}}^{f_{0}}

	= \left(\frac{e^{j2\pi ft} - e^{-j2\pi f_{0}t}}{2j}\right)\frac{1}{\pi t} = \frac{sin(\pi t)}{\pi t}		= \left(\frac{e^{j2\pi ft} - e^{-j2\pi f_{0}t}}{2j}\right)\frac{1}{\pi t} = \frac{sin(\pi t)}{\pi t}
	</math>		</math>

	If we take a closer look at the transfer function, we'd notice that if we input an impulse function, the output would have to be non-causal: the output would be a sine wave extending from <math>-\infty</math> to <math>\infty</math>! So, it is impossible to build an exact "brick-wall" low pass filter, unless we know the input an infinite amount of time before it even happened!		If we take a closer look at the transfer function, we'd notice that if we input an impulse function, the output would have to be non-causal: the output would be a sine wave extending from <math>-\infty</math> to <math>\infty</math>! So, it is impossible to build an exact "brick-wall" low pass filter, unless we know the input an infinite amount of time before it even happened!

Pridma at 03:18, 12 October 2007

2007-10-12T03:18:50Z

← Older revision		Revision as of 20:18, 11 October 2007
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	:<math>h(t) = \mathcal{F}^{-1}\left[H(f)\right] = \int_{-f_{0}}^{f_{0}} 1 e^{j2\pi ft} df = \left[\frac{e^{j2\pi ft}}{j2\pi ft}\right]_{f=-f_{0}}^{f_{0}}		:<math>h(t) = \mathcal{F}^{-1}\left[H(f)\right] = \int_{-f_{0}}^{f_{0}} 1 e^{j2\pi ft} df = \left[\frac{e^{j2\pi ft}}{j2\pi ft}\right]_{f=-f_{0}}^{f_{0}}

	= \left(\frac{e^{j2\pi ft} - e^{-j2\pi f_{0}t}}{2j}\right)\frac{1}{\pi t} = \frac{sin(\pi t)}{\pi t}		= \left(\frac{e^{j2\pi ft} - e^{-j2\pi f_{0}t}}{2j}\right)\frac{1}{\pi t} = \frac{sin(\pi t)}{\pi t}
	</math>		</math>

	If we take a closer look at the transfer function, we'd notice that if we input an impulse function, the output would have to be non-causal: the output would be a sine wave extending from <math>-\infty</math> to <math>\infty</math>! So, it is impossible to build an exact "brick-wall" low pass filter, unless we know the input an infinite amount of time before it even happened!		If we take a closer look at the transfer function, we'd notice that if we input an impulse function, the output would have to be non-causal: the output would be a sine wave extending from <math>-\infty</math> to <math>\infty</math>! So, it is impossible to build an exact "brick-wall" low pass filter, unless we know the input an infinite amount of time before it even happened!

Pridma: /* Useful Applications */

2007-10-12T03:16:59Z

Useful Applications

← Older revision		Revision as of 20:16, 11 October 2007
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	An ideal low pass filter would keep all the frequencies below a certain frequency, <math> f_{0} \!</math> and would discard all higher frequencies.		An ideal low pass filter would keep all the frequencies below a certain frequency, <math> f_{0} \!</math> and would discard all higher frequencies.

	The transfer function of said system would have to be <math>H(f) = \{_{0, \|f\| > f_{0}~~}^{~~1, \|f\| < f_{0}}</math>.		The transfer function of said system would have to be <math>H(f) = \begin{Bmatrix} 0, \|f\| > f_{0} \\ 1, \|f\| < f_{0} \end{Bmatrix}</math>.

	To build this into a circuit, we need to find the transfer function in the time domain. We can inverse Fourier transform <math>H(f) \!</math>.		To build this into a circuit, we need to find the transfer function in the time domain. We can inverse Fourier transform <math>H(f) \!</math>.

Pridma: /* Useful Applications */

2007-10-12T03:02:25Z

Useful Applications

← Older revision		Revision as of 20:02, 11 October 2007
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	To build this into a circuit, we need to find the transfer function in the time domain. We can inverse Fourier transform <math>H(f) \!</math>.		To build this into a circuit, we need to find the transfer function in the time domain. We can inverse Fourier transform <math>H(f) \!</math>.

	:<math>h(t) = \mathcal{F}^{-1}\left[H(f)\right] = \int_{-f_{0}}^{f_{0}} 1 e^{j2\pi ft} df = \left[\frac{e^{j2\pi ft}}{j2\pi ft}\right]_{f=f_{0}}^{f_{0}}		:<math>h(t) = \mathcal{F}^{-1}\left[H(f)\right] = \int_{-f_{0}}^{f_{0}} 1 e^{j2\pi ft} df = \left[\frac{e^{j2\pi ft}}{j2\pi ft}\right]_{f=-f_{0}}^{f_{0}}

	= \left(\frac{e^{j2\pi ft} - e^{-j2\pi f_{0}t}}{2j}\right)\frac{1}{\pi t} = \frac{sin(\pi t)}{\pi t}		= \left(\frac{e^{j2\pi ft} - e^{-j2\pi f_{0}t}}{2j}\right)\frac{1}{\pi t} = \frac{sin(\pi t)}{\pi t}
	</math>		</math>

	If we take a closer look at the transfer function, we'd notice that if we input an impulse function, the output would have to be non-causal: the output would be a sine wave extending from <math>-\infty</math> to <math>\infty</math>! So, it is impossible to build an exact "brick-wall" low pass filter, unless we know the input an infinite amount of time before it even happened!		If we take a closer look at the transfer function, we'd notice that if we input an impulse function, the output would have to be non-causal: the output would be a sine wave extending from <math>-\infty</math> to <math>\infty</math>! So, it is impossible to build an exact "brick-wall" low pass filter, unless we know the input an infinite amount of time before it even happened!

Pridma: /* Even and Odd Parts */

2007-10-12T02:32:50Z

Even and Odd Parts

← Older revision		Revision as of 19:32, 11 October 2007
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	Once you have the function split up into it's even and odd parts, you can use the following Fourier Transform formula:		Once you have the function split up into it's even and odd parts, you can use the following Fourier Transform formula:

	:<math>\mathcal{F}[x(t)] = \int_{-\infty}^{\infty} x(t) e^{-j2\pi ft} dt = \int_{-\infty}^{\infty} [x_{e}(t) + x_{o}(t)](cos(2\pi ft) - j sin(2\pi ft)) dt</math>		:<math>\mathcal{F}[x(t)] = \int_{-\infty}^{\infty} x(t) e^{-j2\pi ft} dt = \int_{-\infty}^{\infty} [x_{e}(t) + x_{o}(t)]\left(cos(2\pi ft) - j sin(2\pi ft)\right) dt</math>

	::::<math>= \int_{-\infty}^{\infty} x_{e}(t) cos(2\pi ft) dt - j \int_{-\infty}^{\infty} x_{o}(t) sin(2\pi ft) dt</math>		::::<math>= \int_{-\infty}^{\infty} x_{e}(t) cos(2\pi ft) dt - j \int_{-\infty}^{\infty} x_{o}(t) sin(2\pi ft) dt</math>

Pridma: /* Time Shift */

2007-10-12T02:30:55Z

Time Shift

← Older revision		Revision as of 19:30, 11 October 2007
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	The time shift property of the Fourier Transform is fairly straight-forward as well.		The time shift property of the Fourier Transform is fairly straight-forward as well.

	:<math>x(t-t_{0}) = \int_{-\infty}^{\infty} X(f) e^{j2\pi ft} df = \int_{-\infty}^{\infty} e^{-j2\pi ft_{0}} X(f) e^{j2\pi ft} df = \mathcal{F}^{-1}[e^{-j2\pi ft_{0}} X(f)]</math>		:<math>x(t-t_{0}) = \int_{-\infty}^{\infty} X(f) e^{j2\pi ft} df = \int_{-\infty}^{\infty} e^{-j2\pi ft_{0}} X(f) e^{j2\pi ft} df = \mathcal{F}^{-1}\left[e^{-j2\pi ft_{0}} X(f)\right]</math>

	This formula is actually fairly obvious... a time delay is really just a linear phase delay, which can be seen from the <math>e^{-j2\pi ft_{0}}</math> part of the inverse Forier Transform shown above.		This formula is actually fairly obvious... a time delay is really just a linear phase delay, which can be seen from the <math>e^{-j2\pi ft_{0}}</math> part of the inverse Forier Transform shown above.


	===Even and Odd Parts===		===Even and Odd Parts===

Pridma: /* The Derivative of x(t) */

2007-10-12T02:25:28Z

The Derivative of x(t)

← Older revision		Revision as of 19:25, 11 October 2007
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	If you take the derivative of <math>x(t) \!</math> and if you know it's transform <math>X(f) \!</math>, you can easily find the transform of the derivative:		If you take the derivative of <math>x(t) \!</math> and if you know it's transform <math>X(f) \!</math>, you can easily find the transform of the derivative:

	:<math>\frac{dx(t)}{dt} = \frac{d}{dt}\int_{-\infty}^{\infty} X(f) e^{j2\pi ft} df = \int_{-\infty}^{\infty} X(f) \frac{d}{dt} e^{j2\pi ft} df = \int_{-\infty}^{\infty} j2\pi fX(f) e^{j2\pi ft}df = \mathcal{F}^{-1}[j2\pi fX(f)]</math>		:<math>\frac{dx(t)}{dt} = \frac{d}{dt}\int_{-\infty}^{\infty} X(f) e^{j2\pi ft} df = \int_{-\infty}^{\infty} X(f) \frac{d}{dt} \left[e^{j2\pi ft}\right] df = \int_{-\infty}^{\infty} j2\pi fX(f) e^{j2\pi ft}df = \mathcal{F}^{-1}\left[j2\pi fX(f)\right]</math>


	===Time Shift===		===Time Shift===

Pridma at 02:21, 12 October 2007

2007-10-12T02:21:01Z

← Older revision		Revision as of 19:21, 11 October 2007
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	:<math>h(t) = \mathcal{F}^{-1}\left[H(f)\right] = \int_{-f_{0}}^{f_{0}} 1 e^{j2\pi ft} df = \left[\frac{e^{j2\pi ft}}{j2\pi ft}\right]_{f=f_{0}}^{f_{0}}		:<math>h(t) = \mathcal{F}^{-1}\left[H(f)\right] = \int_{-f_{0}}^{f_{0}} 1 e^{j2\pi ft} df = \left[\frac{e^{j2\pi ft}}{j2\pi ft}\right]_{f=f_{0}}^{f_{0}}
			= \left(\frac{e^{j2\pi ft} - e^{-j2\pi f_{0}t}}{2j}\right)\frac{1}{\pi t} = \frac{sin(\pi t)}{\pi t}
	= (\frac{e^{j2\pi ft} - e^{-j2\pi f_{0}t}}{2j})\frac{1}{\pi t} = \frac{sin(\pi t)}{\pi t}
	</math>		</math>

	If we take a closer look at the transfer function, we'd notice that if we input an impulse function, the output would have to be non-causal: the output would be a sine wave extending from <math>-\infty</math> to <math>\infty</math>! So, it is impossible to build an exact "brick-wall" low pass filter, unless we know the input an infinite amount of time before it even happened!		If we take a closer look at the transfer function, we'd notice that if we input an impulse function, the output would have to be non-causal: the output would be a sine wave extending from <math>-\infty</math> to <math>\infty</math>! So, it is impossible to build an exact "brick-wall" low pass filter, unless we know the input an infinite amount of time before it even happened!

Pridma: /* Useful Applications */

2007-10-12T02:18:37Z

Useful Applications

← Older revision		Revision as of 19:18, 11 October 2007
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	To build this into a circuit, we need to find the transfer function in the time domain. We can inverse Fourier transform <math>H(f) \!</math>.		To build this into a circuit, we need to find the transfer function in the time domain. We can inverse Fourier transform <math>H(f) \!</math>.

	:<math>h(t) = \mathcal{F}^{-1}[H(f)] = \int_{-f_{0}}^{f_{0}} 1 e^{j2\pi ft} df = \frac{e^{j2\pi ft}}{j2\pi ft}\|_{f=f_{0}}^{f_{0}}		:<math>h(t) = \mathcal{F}^{-1}\left[H(f)\right] = \int_{-f_{0}}^{f_{0}} 1 e^{j2\pi ft} df = \left[\frac{e^{j2\pi ft}}{j2\pi ft}\right]_{f=f_{0}}^{f_{0}}

	= (\frac{e^{j2\pi ft} - e^{-j2\pi f_{0}t}}{2j})\frac{1}{\pi t} = \frac{sin(\pi t)}{\pi t}		= (\frac{e^{j2\pi ft} - e^{-j2\pi f_{0}t}}{2j})\frac{1}{\pi t} = \frac{sin(\pi t)}{\pi t}
	</math>		</math>

	If we take a closer look at the transfer function, we'd notice that if we input an impulse function, the output would have to be non-causal: the output would be a sine wave extending from <math>-\infty</math> to <math>\infty</math>! So, it is impossible to build an exact "brick-wall" low pass filter, unless we know the input an infinite amount of time before it even happened!		If we take a closer look at the transfer function, we'd notice that if we input an impulse function, the output would have to be non-causal: the output would be a sine wave extending from <math>-\infty</math> to <math>\infty</math>! So, it is impossible to build an exact "brick-wall" low pass filter, unless we know the input an infinite amount of time before it even happened!

Pridma: /* Examples */

2007-10-12T02:15:03Z

Examples

← Older revision		Revision as of 19:15, 11 October 2007
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	Here are some examples of Fourier Transforms:		Here are some examples of Fourier Transforms:

	:<math> \mathcal{F}[cos(2\pi f_{0}t)] = \int_{-\infty}^{\infty} (\frac{e^{j2\pi f_{0}t} + e^{-j2\pi f_{0}t}}{2}) e^{-j2\pi ft} dt = \frac{1}{2}(\int_{-\infty}^{\infty} e^{-j2\pi(f-f_{0})t} dt + \int_{-\infty}^{\infty} e^{j2\pi(f+f_{0})t} dt)</math>		:<math> \mathcal{F}\left[cos(2\pi f_{0}t)\right] = \int_{-\infty}^{\infty} \left(\frac{e^{j2\pi f_{0}t} + e^{-j2\pi f_{0}t}}{2}\right) e^{-j2\pi ft} dt = \frac{1}{2}\left(\int_{-\infty}^{\infty} e^{-j2\pi(f-f_{0})t} dt + \int_{-\infty}^{\infty} e^{j2\pi(f+f_{0})t} dt\right)</math>

	::::::<math> = \frac{1}{2} \delta(f-f_{0}) + \frac{1}{2} \delta(f+f_{0})</math>		::::::<math> = \frac{1}{2} \delta\left(f-f_{0}\right) + \frac{1}{2} \delta\left(f+f_{0}\right)</math>

	==Useful Applications==		==Useful Applications==