Fourier series - by Ray Betz - Revision history

Smitry at 21:51, 8 October 2006

2006-10-08T21:51:27Z

← Older revision		Revision as of 14:51, 8 October 2006
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			*[[Signals and systems\|Signals and Systems]]
	==Fourier Series==		==Fourier Series==
	If		If

Andrew: /* Fourier Series */

2006-10-04T05:30:41Z

Fourier Series

← Older revision		Revision as of 22:30, 3 October 2006
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	If		If
	# <math> x(t) = x(t + T)</math>		# <math> x(t) = x(t + T)</math>
	# [~~http://en.wikipedia.org/wiki/Dirichlet_boundary_condition~~ Dirichlet ~~conditions~~] are satisfied		*[[Dirichlet Conditions]] are satisfied
	then we can write		then we can write
	<center>		<center>

SDiver: /* FIR Filters */

2005-12-04T21:06:04Z

FIR Filters

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	Example: Design a FIR low-pass filter to pass between <math> -\frac{1}{4T} < f < \frac{1}{4T} </math> and reject the rest.		Example: Design a FIR low-pass filter to pass between <math> -\frac{1}{4T} < f < \frac{1}{4T} </math> and reject the rest.

	Our desired response is: <math> H_{hat} = 1 </math>, if \|f\| is less then or equal to <math> \frac{1}{4T} <\math> or <math> H_{hat} = 0 </math> otherwise.		Our desired response is: <math> H_{hat} = 1 </math>, if \|f\| is less then or equal to <math> \frac{1}{4T} </math> or <math> H_{hat} = 0 </math> otherwise.

	So, <math> h(mT) = T \int_{ . . . </math>		So, <math> h(mT) = T \int_{} . . . </math>

	Note: From the Circular Convolution we get: <math> y(n) = \sum_{m=0}^{N-1}h(m)x(n-m)</math>		Note: From the Circular Convolution we get: <math> y(n) = \sum_{m=0}^{N-1}h(m)x(n-m)</math>

SDiver: /* FIR Filters */

2005-12-04T21:04:17Z

FIR Filters

← Older revision		Revision as of 14:04, 4 December 2005
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	A finite impulse response filter (FIR filter) is a digital filter that is applied to data before sending it out a D/A converter. This type of filter allows for compensation of the signal before is it destorted so that it will look as it was originally recorded. Using an FIR filter also allows us to put a cheap low-pass filter on after the D/A converter because the signal has been compensated so it doesn't take an expensive low-pass filter, as it would without the FIR filter.		A finite impulse response filter (FIR filter) is a digital filter that is applied to data before sending it out a D/A converter. This type of filter allows for compensation of the signal before is it destorted so that it will look as it was originally recorded. Using an FIR filter also allows us to put a cheap low-pass filter on after the D/A converter because the signal has been compensated so it doesn't take an expensive low-pass filter, as it would without the FIR filter.

			The coefficients that are sent out to the D/A converter are:

			<math>
			h_m = { T } \int_{T} H(f)e^{j2 \pi m f T}\,df
			</math>

			where <math> H(f)=\sum_{m=-M}^{M}h(mT)e^{-j 2 \pi f m T} </math>

			Example: Design a FIR low-pass filter to pass between <math> -\frac{1}{4T} < f < \frac{1}{4T} </math> and reject the rest.

			Our desired response is: <math> H_{hat} = 1 </math>, if \|f\| is less then or equal to <math> \frac{1}{4T} <\math> or <math> H_{hat} = 0 </math> otherwise.

			So, <math> h(mT) = T \int_{ . . . </math>

	Note: From the Circular Convolution we get: <math> y(n) = \sum_{m=0}^{N-1}h(m)x(n-m)</math>		Note: From the Circular Convolution we get: <math> y(n) = \sum_{m=0}^{N-1}h(m)x(n-m)</math>

SDiver: /* FIR Filters */

2005-12-04T19:47:24Z

FIR Filters

← Older revision		Revision as of 12:47, 4 December 2005
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	==FIR Filters==		==FIR Filters==

	A finite impulse response filter (FIR filter) is a digital filter that is applied to data before sending to out a D/A ~~filter~~. This type of filter allows for compensation of the signal before is it destorted so that it will look as it was originally recorded. Using an FIR filter also allows us to put a cheap low-pass filter on after the D/A converter because the signal has been compensated so it doesn't take ~~as good a~~ low-pass filter as it would without the FIR filter.		A finite impulse response filter (FIR filter) is a digital filter that is applied to data before sending it out a D/A converter. This type of filter allows for compensation of the signal before is it destorted so that it will look as it was originally recorded. Using an FIR filter also allows us to put a cheap low-pass filter on after the D/A converter because the signal has been compensated so it doesn't take an expensive low-pass filter, as it would without the FIR filter.

	Note: From the Circular Convolution we get: <math> y(n) = \sum_{m=0}^{N-1}h(m)x(n-m)</math>		Note: From the Circular Convolution we get: <math> y(n) = \sum_{m=0}^{N-1}h(m)x(n-m)</math>

SDiver: /* CD Player */

2005-12-04T19:27:28Z

CD Player

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	'''In Time Domain:'''		'''In Time Domain:'''

	Let's start with a signal <math> h(t) </math>, as shown in the below picture. In this signal there is an infinite amount of information. Obviously, we can't hold it all in a computer, but we could take samples every <math> T </math>. Lets do that by multiplying <math> h(t) </math> by <math> \sum_{n=-\infty}^\infty \delta (t-nT) </math>. Since the magnitude of our delta function is one, we get a series of delta functions that record the value of <math> h(t) </math> at intervals of <math> T </math>. This gives us a result that looks like: <math> h(t)\sum_{n=-\infty}^\infty \delta (t-nT) = \sum_{n=-\infty}^\infty x(nt) \delta (t-nT)</math>		Let's start with a signal <math> h(t) </math>, as shown in the below picture. In this signal there is an infinite amount of information. Obviously, we can't hold it all in a computer, but we could take samples every <math> T </math> seconds. Lets do that by multiplying <math> h(t) </math> by <math> \sum_{n=-\infty}^\infty \delta (t-nT) </math>. Since the magnitude of our delta function is one, we get a series of delta functions that record the value of <math> h(t) </math> at intervals of <math> T </math>. This gives us a result that looks like: <math> h(t)\sum_{n=-\infty}^\infty \delta (t-nT) = \sum_{n=-\infty}^\infty h(t) \delta (t-nT)</math>

	'''In Frequency Domain:'''		'''In Frequency Domain:'''

SDiver: /* Fourier Transform */

2005-12-04T19:17:14Z

Fourier Transform

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	<math> = \frac{1}{2} \int_{-\infty}^\infty x(t) e^{-j 2 \pi (f-f_0) t} dt + \frac{1}{2} \int_{-\infty}^\infty x(t) e^{j 2 \pi (f+f_0) t} dt = \frac{1}{2} X(f-f_0) + \frac{1}{2} X(f+f_0)</math>		<math> = \frac{1}{2} \int_{-\infty}^\infty x(t) e^{-j 2 \pi (f-f_0) t} dt + \frac{1}{2} \int_{-\infty}^\infty x(t) e^{j 2 \pi (f+f_0) t} dt = \frac{1}{2} X(f-f_0) + \frac{1}{2} X(f+f_0)</math>

	What would happen if we multiplied our time by a constant in <math> x(t) </math>? We will substitute <math> u=at </math> and <math> du = adt </math>. If <math> a \ne 0 </math>:		What would happen if we multiplied our time (time scaling) by a constant in <math> x(t) </math>? We will substitute <math> u=at </math> and <math> du = adt </math>. If <math> a \ne 0 </math>:

	<math> \mathcal{F} [x(a t)] = \int_{-\infty}^\infty x(at) e^{-j 2 \pi f t} dt = \int_{-\infty}^\infty x(u) e^\frac{-j 2 \pi f u}{a} \frac{du}{\|a\|} = \frac{1}{\|a\|} X(\frac{f}{a})</math>		<math> \mathcal{F} [x(a t)] = \int_{-\infty}^\infty x(at) e^{-j 2 \pi f t} dt = \int_{-\infty}^\infty x(u) e^\frac{-j 2 \pi f u}{a} \frac{du}{\|a\|} = \frac{1}{\|a\|} X(\frac{f}{a})</math>

SDiver: /* Fourier Transform */

2005-12-04T19:16:01Z

Fourier Transform

← Older revision		Revision as of 12:16, 4 December 2005
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	==Fourier Transform==		==Fourier Transform==

	Fourier transforms emerge because we want to be able to make Fourier ~~expressions~~ of non-periodic functions. We can ~~take~~ the limit of ~~those non-periodic functions to get a fourier expression for the function~~.		Fourier transforms emerge because we want to be able to make Fourier expransions of non-periodic functions. We can accomplish this by taking the limit of x(t).

	Remember that:		Remember that:
	<math>x(t)=x(t+T)= \sum_{k=-\infty}^\infty \alpha_k e^ \frac {j 2 \pi k t}{T} = \sum_{k=-\infty}^\infty 1/T \int_{-\frac{T}{2}}^\frac{T}{2} x(u)e^ \frac {-j 2 \pi k u }{T} du e^ \frac {j 2 \pi k t}{T} </math>		<math>x(t)=x(t+T)= \sum_{k=-\infty}^\infty \alpha_k e^ \frac {j 2 \pi k t}{T} = \sum_{k=-\infty}^\infty 1/T \int_{-\frac{T}{2}}^\frac{T}{2} x(u)e^ \frac {-j 2 \pi k u }{T} du e^ \frac {j 2 \pi k t}{T} </math>

			Let's substitute in x(t) for k/T substitute f, for 1/T substitute df, and for the summation substitute the integral.

	So,		So,
	<math> \lim_{x \to \infty}x(t)= \int_{-\infty}^\infty (\int_{-\infty}^\infty x(u) e^{-j 2 \pi f u} du) e^{j 2 \pi f t} df</math>		<math> \lim_{T \to \infty}x(t)= \int_{-\infty}^\infty (\int_{-\infty}^\infty x(u) e^{-j 2 \pi f u} du) e^{j 2 \pi f t} df</math>

	From the above limit we define <math> x(t)</math> and <math> X(f) </math>.		From the above limit we define <math> x(t)</math> and <math> X(f) </math>.

SDiver: /* Fourier Series (indepth) */

2005-12-04T18:59:27Z

Fourier Series (indepth)

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	<math> x(t) = \alpha_0 +\sum_{n=1}^\infty 2Re(\alpha_n e^ \frac {j 2 \pi n t}{T}) </math>		<math> x(t) = \alpha_0 +\sum_{n=1}^\infty 2Re(\alpha_n e^ \frac {j 2 \pi n t}{T}) </math>

			In terms of cosine <math> x(t) = \alpha_0 +\sum_{n=1}^\infty 2 \|\alpha_n\| cos(\frac{2 \pi n t}{T} + \omega_n) </math> where <math> \omega_n </math> is an angle.

	==Fourier Transform==		==Fourier Transform==

SDiver: /* Linear Systems */

2005-12-04T18:40:43Z

Linear Systems

← Older revision		Revision as of 11:40, 4 December 2005
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	What if we now multiplied our impulse by a coefficient? Since our system is linear, the proportionality property applies. If we put <math> x(u)\delta(t-u)</math> into our system then we should get out <math>x(u)h(t-u)</math>.		What if we now multiplied our impulse by a coefficient? Since our system is linear, the proportionality property applies. If we put <math> x(u)\delta(t-u)</math> into our system then we should get out <math>x(u)h(t-u)</math>.

	By the superposition property(because we have a linear system) we may put into the system the integral of <math> x(u)\delta(t-u)</math> with respect to u and we would get out <math> \int_{-\infty}^\infty x(u)h(t-u) du</math>. What would we get if we put <math> e^{j 2 \pi f t} </math> into our system? We could find out by plugging <math> e^{j 2 \pi f t} </math> in for <math> x(u) </math> in the integral that we just found the output for above. If we do a change of variables (<math> v = t-u </math>, and <math> dv = -du </math>) we get <math> \int_{-\infty}^\infty x(u)h(t-u) du = \int_{-\infty}^\infty e^{j 2 \pi f t} h(t-u) du = -\int_{\infty}^{-\infty} e^{j 2 \pi f (t-v)} h(v) dv = e^{j 2 \pi f t} \int_{-\infty}^\infty h(v)e^{-j 2 \pi f v} dv</math>. By pulling <math> e^{j 2 \pi f t} </math> out of the integral and calling the remaining integral <math> ~~B_k~~ </math> we get <math> e^{j 2 \pi f t} ~~B_k~~</math>.		By the superposition property(because we have a linear system) we may put into the system the integral of <math> x(u)\delta(t-u)</math> with respect to u and we would get out <math> \int_{-\infty}^\infty x(u)h(t-u) du</math>. This is because What would we get if we put <math> e^{j 2 \pi f t} </math> into our system? We could find out by plugging <math> e^{j 2 \pi f t} </math> in for <math> x(u) </math> in the integral that we just found the output for above. If we do a change of variables (<math> v = t-u </math>, and <math> dv = -du </math>) we get <math> \int_{-\infty}^\infty x(u)h(t-u) du = \int_{-\infty}^\infty e^{j 2 \pi f t} h(t-u) du = -\int_{\infty}^{-\infty} e^{j 2 \pi f (t-v)} h(v) dv = e^{j 2 \pi f t} \int_{-\infty}^\infty h(v)e^{-j 2 \pi f v} dv</math>. By pulling <math> e^{j 2 \pi f t} </math> out of the integral and calling the remaining integral <math> H_f </math> we get <math> e^{j 2 \pi f t} H_f</math>.


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	\|-		\|-
	\|<math> e^{j 2 \pi f t} </math>		\|<math> e^{j 2 \pi f t} </math>
	\|<math> e^{j 2 \pi f t} ~~B_k~~</math>		\|<math> e^{j 2 \pi f t} H_f</math>
	\|Superposition (from above)		\|Superposition (from above)
	\|}		\|}