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- fD = fNYQ
- fD = 0,5 fNYQ
- fD = 2 fNYQ
D. Lineárna interpolácia vzoriek z úlohy C.
Riešenie
1a)
1b)
Maximálna frekvencia prítomná v signály: fM = 3 Hz
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Vzorkovanie signálu: f(t) →f(nTD ) = cos(2πf1.nTD )+1/4 cos(2πf2.nTD)
1c)
fD = fNYQ
n | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |
nTd | 0 | 0.167 | 0.333 | 0.5 | 0.667 | 0.833 | 1 | 1.167 | 1.333 | 1.5 | 1.667 | 1.833 | 1.167 |
f(nTd) | 1.25 | 0.25 | -0.25 | -1.25 | -0.25 | 0.25 | 1.25 | 0.25 | -0.25 | -1.25 | -0.25 | -1.25 | -0.25 |
fD = 0,5 fNYQ
n | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |
nTd | 0 | 0.333 | 0.667 | 1 | 1.333 | 1.667 | 2 | 2.333 | 2.667 | 3 | 3.333 | 3.667 | 4 |
f(nTd) | 1.25 | -0.25 | -0.25 | 1.25 | -0.25 | -0.25 | 1.25 | -0.25 | -0.25 | 1.25 | -0.25 | -0.25 | 1.25 |
fD = 2 fNYQ
n | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |
nTd | 0 | 0.083 | 0.167 | 0.25 | 0.333 | 0.417 | 0.5 | 0.583 | 0.667 | 0.75 | 0.833 | 0,917 | 1 |
f(nTd) | 1.25 | -0.25 | -0.25 | 1.25 | -0.25 | -0.25 | 1.25 | -0.25 | -0.25 | 1.25 | -0.25 | -0.25 | 1.25 |
1d)
fD = fNYQ
fD = 0.5fNYQ
fD = 0.5fNYQ
Úloha 2
Vykonajte rekonštrukciu vzoriek pre signál f(t) z úlohy 1c) pomocou Shannon-Kotelníkového rádu.
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