Ee610-ep-[farfan]-[marco].docx

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1) Grafique de manera aproxiamada la respuesta en magnitud del filtro no-recursivo cuyo diagrama de bloques se muestra. 1 1 π‘₯[𝑛] + π‘₯[π‘›βˆ’1] = 𝑦[𝑛] 2 2 Hallamos la funciΓ³n de transferencia:

1 1 𝑋[Ω] ( + 𝑒 βˆ’π‘—β„¦ ) = π‘Œ[Ω] 2 2 𝐻(Ω) =

1 + 𝑒 βˆ’π‘—β„¦ 1 = (√2 + 2 cos(Ω)) 2 2

1 1 1 π‘₯[𝑛] + π‘₯[π‘›βˆ’1] + π‘₯[π‘›βˆ’2] = 𝑦[𝑛] 2 2 2 Hallamos la funciΓ³n de transferencia:

1 1 1 𝑋[Ω] ( + 𝑒 βˆ’π‘—β„¦ + 𝑒 βˆ’π‘—2Ω ) = π‘Œ[Ω] 2 2 2 𝐻(Ω) =

1 + 𝑒 βˆ’π‘—β„¦ + 𝑒 βˆ’π‘—2Ω 2

2) Indique la expresiΓ³n de la respuesta en frecuencia del sistema cuya ecuaciΓ³n de diferencias se muestra

𝑦[𝑛] βˆ’ 1.1314𝑦[π‘›βˆ’1] + 0.64𝑦[π‘›βˆ’2] = π‘₯[π‘›βˆ’2] + π‘₯[𝑛] 𝑋[Ω] (1 + 𝑒 βˆ’π‘—2Ω ) = π‘Œ[Ω] (1 βˆ’ 1.1314𝑒 βˆ’π‘—β„¦ + 0.64𝑒 βˆ’π‘—2Ω ) 𝐻(Ω) =

1 + 𝑒 βˆ’π‘—2Ω 1 βˆ’ 1.1314𝑒 βˆ’π‘—β„¦ + 0.64𝑒 βˆ’π‘—2Ω

𝑦[𝑛] βˆ’ 1.1314𝑦[π‘›βˆ’1] + 0.64𝑦[π‘›βˆ’2] = 𝐴π‘₯[π‘›βˆ’2] + π‘₯[𝑛] 𝑋[Ω] (𝐴 + 𝑒 βˆ’π‘—2Ω ) = π‘Œ[Ω] (1 βˆ’ 1.1314𝑒 βˆ’π‘—β„¦ + 0.64𝑒 βˆ’π‘—2Ω ) 𝐻(Ω) =

𝐴 + 𝑒 βˆ’π‘—2Ω 1 βˆ’ 1.1314𝑒 βˆ’π‘—β„¦ + 0.64𝑒 βˆ’π‘—2Ω

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