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C.Hodnoty funkcie pre vzorkovacie frekvencie pre n ∈⟨0,12⟩:
- fD = fNYQ2fMAX
- fD = 0,5 fNYQMAX
- fD = 2 fNYQ
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- 4fMAX
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 = fNYQ2fMAX
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 fNYQMAX
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 fNYQ4fMAX
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
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Úloha 2
Vykonajte rekonštrukciu vzoriek pre signál f(t) z úlohy 1c) pomocou Shannon-Kotelníkového rádu.
fD = fNYQ2fMAX
DOMÁCA ÚLOHA
fD = 0,5fNYQfMAX
fD =2fNYQ4fMAX
