Assignment: Difference between revisions
(New page: Summery of the class notes from Oct. 5: What if a periodic signal had an infinite period? We would no longer be able to tell the difference between it and a non periodic signal. We can ...) |
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Begining with a Fourier Series: |
Begining with a Fourier Series: |
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<math>x(t) = x(t+T) = \sum_{n=- \infty}^{\infty} \alpha _n e^{\frac{j2 \pi nt}{T}}</math> |
(1) <math>x(t) = x(t+T) = \sum_{n=- \infty}^{\infty} \alpha _n e^{\frac{j2 \pi nt}{T}}</math> |
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where |
where |
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We then take the limit of a Fourier series as its period T approaches infinity: |
We then take the limit of a Fourier series as its period T approaches infinity: |
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<math>\lim_{T \to \infty}\sum_{n=-\infty}^\infty (\frac{1}{T}\int_{-\frac{T}{2}}^{\frac{T}{2}}x(t^')e^{\frac{-j2\pi nt^'}{T}}dt^')e^{\frac{j2\pi nt}{T}}\!</math> |
(2)<math>\lim_{T \to \infty}\sum_{n=-\infty}^\infty (\frac{1}{T}\int_{-\frac{T}{2}}^{\frac{T}{2}}x(t^')e^{\frac{-j2\pi nt^'}{T}}dt^')e^{\frac{j2\pi nt}{T}}\!</math> |
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In order to evaluate this limit we need the following relationships: |
In order to evaluate this limit we need the following relationships: |
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<table> |
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<tr> |
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<td width=100> |
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<math>\frac{1}{T}</math> |
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</td> |
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<td width=50> |
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<math>\rightarrow</math> |
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</td> |
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<td widt=100> |
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<math>\,df</math> |
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</td> |
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</tr> |
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<tr> |
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<td> |
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<math>\frac{n}{T}</math> |
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</td> |
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<td> |
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<math>\rightarrow</math> |
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</td> |
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<td> |
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<math>\,f</math> |
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</td> |
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</tr> |
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<tr> |
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<td> |
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<math>\sum_{n=- \infty}^{\infty} \frac{1}{T}</math> |
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</td> |
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<td> |
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<math>\rightarrow</math> |
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</td> |
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<td> |
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<math>\int_{-\infty}^{\infty}(\mbox{ })\,df</math> |
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</td> |
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</tr> |
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</table> |
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We can now write out the following: |
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(3)<math>x(t) = \lim_{T \to \infty} \left[ \sum_{n=- \infty}^{\infty} \left(\frac{1}{T} \int_{-\frac{T}{2}}^{\frac{T}{2}} x(t')e^{- \frac{j2 \pi nt'}{T}} \,dt' \right) e^{\frac{j2 \pi nt}{T}} \right] </math> |
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which can also be written as: |
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(4)<math>x(t) =\int_{- \infty}^{\infty} \left( \int_{-\infty}^{\infty} \,x(t') e^{-j2 \pi ft'} \,dt' \right) e^{j2 \pi tf} \,df</math> |
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using, |
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<math>\alpha_n\rightarrow X(f) \!</math> |
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we now have |
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(5)<math>\,X(f) = \int_{-\infty}^{\infty} \,x(t') e^{-j2 \pi ft'} \,dt' |
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</math> |
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We can now relate a signal in the time domain to a signal in the frequency domain. Using vector notation we can show this relationship as such: |
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<table> |
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<tr> |
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<td width=100><math>\,X(f) = \int_{-\infty}^{\infty} \,x(t) e^{-j2 \pi ft} \,dt</math></td> |
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<td width=100 align="center"><math>\equiv</math></td> |
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<td width=100><math>\langle x(t) | e^{j2 \pi tf} \rangle </math></td> |
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<td width=200><math>\,x(t) \mbox{ projected onto } e^{j2 \pi tf}</math></td> |
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</tr> |
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<tr> |
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<td><math>\,x(t) = \int_{-\infty}^{\infty} \,X(f) e^{-j2 \pi ft} \,df</math></td> |
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<td align="center"><math>\equiv</math></td> |
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<td><math>\langle X(f) | e^{j2 \pi tf} \rangle </math></td> |
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<td><math>\,x(f) \mbox{ projected onto } e^{j2 \pi tf}</math></td> |
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</tr> |
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</table> |
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Where, |
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<math><x(t)|e^{j2\pi ft}>\!</math> Is the <b>Fourier transform</b>, or <math>\mathcal{F}[x(t)]\!</math><br> |
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<math><X(f)|e^{j2\pi ft}>\!</math> Is the <b>inverse Fourier transform</b>, or <math>\mathcal{F}^{-1}[X(f)]\!</math><br> |
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Now we can use our new tool, the Fourier transform on equation (2) to give us the following: |
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<table> |
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<tr> |
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<td width=100> |
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<math>\int_{- \infty}^{\infty} \left( \int_{-\infty}^{\infty} \,x(t') e^{-j2 \pi ft'} \,dt' \right) e^{j2 \pi tf} \,df</math> |
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</td> |
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<td width=100 align="center"> |
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<math>\equiv</math> |
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</td> |
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<td width=100> |
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<math>\int_{- \infty}^{\infty} \,x(t') \left( \int_{-\infty}^{\infty} e^{j2 \pi f(t-t')} \,df \right) \,dt'</math> |
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</td> |
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</tr> |
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</table> |
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Notice |
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<math>e^{j2 \pi f(t-t')} \equiv \delta (t-t')</math> |
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Similarly, |
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<table> |
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<tr> |
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<td width=100> |
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<math>\int_{- \infty}^{\infty} \left( \int_{-\infty}^{\infty} \,X(f') e^{j2 \pi f't} \,dt' \right) e^{-j2 \pi tf} \,df</math> |
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</td> |
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<td width=100 align="center"> |
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<math>\equiv</math> |
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</td> |
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<td width=100> |
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<math>\int_{- \infty}^{\infty} \,X(f') \left( \int_{-\infty}^{\infty} e^{j2 \pi t(f'-f)} \,df \right) \,dt'</math> |
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</td> |
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</tr> |
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</table> |
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Again, notice |
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<math>e^{j2 \pi f(f'-f)} \equiv \delta (f'-f) = \delta (f-f')</math> |
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Both the time-domain and frequency domain have non-zero integrals when |
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<math> \,t = \,t' \mbox{ and } \,f = \,f'</math> respectively. |
Latest revision as of 14:58, 15 October 2009
Summery of the class notes from Oct. 5:
What if a periodic signal had an infinite period? We would no longer be able to tell the difference between it and a non periodic signal. We can use this property to look at signals that do not have a period (an observable one at least).
Begining with a Fourier Series:
(1)
where
We then take the limit of a Fourier series as its period T approaches infinity:
(2)
In order to evaluate this limit we need the following relationships:
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We can now write out the following:
(3)
which can also be written as:
(4)
using,
we now have
(5)
We can now relate a signal in the time domain to a signal in the frequency domain. Using vector notation we can show this relationship as such:
Where,
Is the Fourier transform, or
Is the inverse Fourier transform, or
Now we can use our new tool, the Fourier transform on equation (2) to give us the following:
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Notice
Similarly,
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Again, notice
Both the time-domain and frequency domain have non-zero integrals when respectively.