Neste caso, n = 10 e ω = 2π rd/s. Assim, fmax = 1 Hz e a condição
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{\displaystyle t_{a}<{\frac {1}{2f_{max}}}}
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{\displaystyle Q(k)\;=\;\{p_{k}\}\;|\;p_{k}\;=\;cos(2\pi k\cdot t_{a}),\;k\;=\;\{0,1,2...9\}}
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{\displaystyle Q(k)\;=\;\{p_{k}\}\;|\;p_{k}\;=\;cos(0.4\cdot \pi k),\;k\;=\;\{0,1,2...9\}}
editar
Vamos obter a DHT primeiro através da expressão da convolução. Para isso, calcula-se a matriz H-1 (k-j) e depois o produto matricial H-1 (k-j) · Q(j)
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{\displaystyle H^{-1}(k-j)\;=\;[p_{k,j}]\;|\;p_{k,j}\;=\;-\;{\frac {2}{n}}\cdot sin^{2}\left({\frac {\pi }{2}}\;(k\;\;j)\right)\cdot cot\left({\frac {\pi }{n}}\;(k\;-\;j)\right),\;k,j\;=\;\{0,1,2...9\}}
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[
0.000
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0.145
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{\displaystyle H^{-1}(k-j)\;=\;{\begin{bmatrix}0.000&0.616&0.000&0.145&0.000&0.000&0.000&-0.145&0.000&-0.616\\-0.616&0.000&0.616&0.000&0.145&0.000&0.000&0.000&-0.145&0.000\\0.000&-0.616&0.000&0.616&0.000&0.145&0.000&0.000&0.000&-0.145\\-0.145&0.000&-0.616&0.000&0.616&0.000&0.145&0.000&0.000&0.000\\0.000&-0.145&0.000&-0.616&0.000&0.616&0.000&0.145&0.000&0.000\\0.000&0.000&-0.145&0.000&-0.616&0.000&0.616&0.000&0.145&0.000\\0.000&0.000&0.000&-0.145&0.000&-0.616&0.000&0.616&0.000&0.145\\0.145&0.000&0.000&0.000&-0.145&0.000&-0.616&0.000&0.616&0.000\\0.000&0.145&0.000&0.000&0.000&-0.145&0.000&-0.616&0.000&0.616\\0.616&0.000&0.145&0.000&0.000&0.000&-0.145&0.000&-0.616&0.000\\\end{bmatrix}}}
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{\displaystyle H^{-1}(k-j)\cdot Q(j)\;=\;{\begin{bmatrix}0.000&0.190&0.000&-0.118&0.000&0.000&0.000&0.118&0.000&-0.190\\-0.616&0.000&-0.498&0.000&0.045&0.000&0.000&0.000&0.118&0,000\\0.000&-0.190&0.000&-0.498&0.000&0.145&0.000&0.000&0.000&-0.045\\-0.145&0.000&0.498&0.000&0.190&0.000&0.045&0.000&0.000&0.000\\0.000&-0.045&0.000&0.498&0.000&0.616&0.000&-0.118&0.000&0.000\\0.000&0.000&0.118&0.000&-0.190&0.000&0.190&0.000&-0.118&0.000\\0.000&0.000&0.000&0.118&0.000&-0.616&0.000&-0.498&0.000&0.045\\0.145&0.000&0.000&0.000&-0.045&0.000&-0.190&0.000&-0.498&0.000\\0.000&0.045&0.000&0.000&0.000&-0.145&0.000&0.498&0.000&0.190\\0.616&0.000&-0.118&0.000&0.000&0.000&-0.045&0.000&0.498&0.000\\\end{bmatrix}}}
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{\displaystyle DHT(k)\;=\;\{p_{k}\}\;|\;p_{k}\;=\;\sum _{j\;=\;0}^{9}\left[H^{-1}(k\;-\;j)\cdot Q(j)\right]\;k,j\;=\;\{0,1,2...9\}}
Vamos obter a DHT agora através da transformada discreta de Fourier. H(k) será dada por
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{\displaystyle H(k)\;=\;\{p_{k}\}\;|\;p_{k}\;=\;i\cdot sgn\left({\frac {n}{2}}\;-\;k\right)\cdot sgn(k),\;k\;=\;\{0,1,2...9\}}
-1 > são dados, por definição (ver Transformada de Fourier , por
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{\displaystyle DFT\{F(k)\}\;=\;\sum _{j\;=\;0}^{n\;-\;1}\left[F[j]\cdot e^{-i\;{\frac {2\pi }{n}}\;jk}\right]}
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{\displaystyle DFT^{-1}\{F(k)\}\;=\;{\frac {1}{n}}\cdot \sum _{j\;=\;0}^{n\;-\;1}\left[F[j]\cdot e^{i\;{\frac {2\pi }{n}}\;jk}\right]}
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{\displaystyle DFT\{Q(k)\}\;=\;\sum _{j\;=\;0}^{n\;-\;1}\left[Q[j]\cdot e^{-i\;{\frac {2\pi }{n}}\;jk}\right]\;=\;\sum _{j\;=\;0}^{9}\left[cos(0.4\cdot \pi j)\cdot e^{-\;0.2\cdot i\pi jk}\right]}
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{\displaystyle DFT\{Q(k)\}\;=\;\sum _{j\;=\;0}^{9}\left[cos(0.4\cdot \pi j)\cdot \left(cos(-0.2\cdot \pi jk)\;+\;i\;sin(-0.2\cdot \pi jk)\right)\right]}
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{\displaystyle DHT(k)\;=\;DFT^{-1}\{H(k)\cdot DFT(k)\}\;=\;DFT^{-1}\{F(k)\}\;=\;{\frac {1}{n}}\cdot \sum _{j\;=\;0}^{n\;-\;1}\left[F[j]\cdot e^{i\;{\frac {2\pi }{n}}\;jk}\right]}
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{\displaystyle DHT(k)\;=\;{\frac {1}{n}}\cdot \sum _{j\;=\;0}^{n\;-\;1}\left[F[j]\left(cos(-0.2\cdot \pi jk)\;+\;i\;sin(-0.2\cdot \pi jk)\right)\right]}
Uma vez que a transformada de Hilbert û(t) em forma fechada é conhecida, pode-se amostrá-la a intervalos ta de maneira a obter os coeficientes exatos de DHT.
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{\displaystyle DHT(k)\;=\;\{p_{k}\}\;|\;p_{k}\;=\;{\hat {u}}(k\cdot t_{a}),\;k\;=\;\{0,1,2...9\}}
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{\displaystyle DHT(k)\;=\;\{p_{k}\}\;|\;p_{k}\;=\;-\;sin(2\pi k\cdot t_{a}),\;k\;=\;\{0,1,2...9\}}
![{\displaystyle H^{-1}(k-j)\;=\;[p_{k,j}]\;|\;p_{k,j}\;=\;-\;{\frac {2}{n}}\cdot sin^{2}\left({\frac {\pi }{2}}\;(k\;\;j)\right)\cdot cot\left({\frac {\pi }{n}}\;(k\;-\;j)\right),\;k,j\;=\;\{0,1,2...9\}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ac7fd015728327925334dca730361a75daa9aa3b)
![{\displaystyle DHT(k)\;=\;\{p_{k}\}\;|\;p_{k}\;=\;\sum _{j\;=\;0}^{9}\left[H^{-1}(k\;-\;j)\cdot Q(j)\right]\;k,j\;=\;\{0,1,2...9\}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6ee12baf7a0015dd86e3a6bf8639cdf556d1d7c4)
![{\displaystyle DFT\{F(k)\}\;=\;\sum _{j\;=\;0}^{n\;-\;1}\left[F[j]\cdot e^{-i\;{\frac {2\pi }{n}}\;jk}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5153af9b46c31ad191a3ace8cbc37496da0b2607)
![{\displaystyle DFT^{-1}\{F(k)\}\;=\;{\frac {1}{n}}\cdot \sum _{j\;=\;0}^{n\;-\;1}\left[F[j]\cdot e^{i\;{\frac {2\pi }{n}}\;jk}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f45e48daa34188516b0369289f8e784a923b42fd)
![{\displaystyle DFT\{Q(k)\}\;=\;\sum _{j\;=\;0}^{n\;-\;1}\left[Q[j]\cdot e^{-i\;{\frac {2\pi }{n}}\;jk}\right]\;=\;\sum _{j\;=\;0}^{9}\left[cos(0.4\cdot \pi j)\cdot e^{-\;0.2\cdot i\pi jk}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bd60e334d0aa6f88a237c49fb3ed0e70af043989)
![{\displaystyle DFT\{Q(k)\}\;=\;\sum _{j\;=\;0}^{9}\left[cos(0.4\cdot \pi j)\cdot \left(cos(-0.2\cdot \pi jk)\;+\;i\;sin(-0.2\cdot \pi jk)\right)\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/06cb36385a155865c16779cda4738b6c34282f52)
![{\displaystyle DHT(k)\;=\;DFT^{-1}\{H(k)\cdot DFT(k)\}\;=\;DFT^{-1}\{F(k)\}\;=\;{\frac {1}{n}}\cdot \sum _{j\;=\;0}^{n\;-\;1}\left[F[j]\cdot e^{i\;{\frac {2\pi }{n}}\;jk}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a883f8048fbff44a13aa5fb2dd991eaa5946fd59)
![{\displaystyle DHT(k)\;=\;{\frac {1}{n}}\cdot \sum _{j\;=\;0}^{n\;-\;1}\left[F[j]\left(cos(-0.2\cdot \pi jk)\;+\;i\;sin(-0.2\cdot \pi jk)\right)\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/492c6bbd31c0c3460a9b7b896f7cdcbd19dacf8b)
