Eserciziario di Elaborazione dei Segnali Inf Musicale Pedersini

Esercizio 4.2

y(n) = 0.6y(n-1) - 0.08y(n-2) + x(n) - x(n-1)

Y(z) = 0.6z^-1Y(z) - 0.08z^-2Y(z) + X(z) - z^-1X(z)

Y(z)(1-0.6z^-1 - 0.08z^-2) = X(z)(1-z^-1)

H(z) = ^1-z-1/_{1-0.6z^-1 + 0.08z^-2}

POLI = z²0.6z + 0.08 = 0

Z₁ = 0.4, Z₂ = 0.2

^{0.6 ± √(0.36 - 0.08)}/₂, ^{0.6 + 0.2}/₂

|H(i)| = ^√2/_{√0.86i + 2i} = ^√2/_√2.3664 = 1.287

∠H(i) = -π/4 → ω(rad) (11.882^o)

1 - z^-1 = ^{1 - z-1}/_{(1-0.6z^-1)(1-0.2z^-1)} = ^{A + B}/_{(1-0.6z^-1)(1-0.2z^-1)}

• A - 0.2z^-1A + B - 0.2z^-1B = -z^-1(0.2A + 0.2B) + A + B = 1 - z^-1
• A + B - 1{_{0.2A + 0.2B}}
• A = 1 - B
• 0.2 - 0.2B + 0.4B = 1
• B = 0.8, A = 0.2
• A = 3

^{4 * 1}/_{1 - 0.2z^-1} - ^{3 * 1}/_{1 - 0.6z^-1} = ( ^{(0.2) n (0.4)}/_n)

S(t) = 5t , -1/2 ≤ t < 1/2

can T = 2s

0 Hz ⇒ m/2 = 0 ⇒ m = 0

0.5 Hz ⇒ m/2 = 1/2 ⇒ m = 1

1 Hz ⇒ m/2 = 1 ⇒ m = 2

S(t) = Σ B_m e^-i2πmt

C_m = 1/2 ∫ _-1 ¹ 5t e^-i2πmt dt

C₀ = 1/2 ∫ _-1 ¹ 5t e^{-i2π 0 t} dt = 1/2 ∫ _-1 ¹ 5t dt = [5/2 (1/2 t²)] _-1 ¹ = 5/2 (1/2 - 1/2) = 0

C_n = 1/2 ∫ _-1 ¹ 5t e^-i2πnt dt = 1/2 ∫ _-1 ¹ 5t e^-iπt dt = 5/2 ∫ _-1 ¹ t e^-iπt dt

= 5/2 { [t / iπ e^-iπt] _-1 ¹ - ∫ _-1 ¹ e^-iπt dt } = 5/2 { 1/-iπ e^-iπ 1/iπ e^-iπ - [1/iπ e^-iπt ] _-1 ¹ }

= 5/2 { 1/iπ [ e^-iπ + e^iπ ] + 1/iπ (e^-iπ - e^iπ) } = 5/2 { -2/iπ } = = - 5/iπ

C₂ = 1/2 ∫ _-1 ¹ 5t e^{-i2π (2/2) t} dt = 5/2 ∫ _-1 ¹ t e^-i2πt dt

= 5/2 { [t / i2π e^-i2πt ] _-1 ¹ - ∫ _-1 ¹ e^-i2πt dt } = 5/2 { -1/i2π e^-i2π 1/i2π e^i2π + 1/i2π [ e^-i2πt ] _-1 ¹ }

= 5/2 { -2/i2π + 1/i2π [ e^-i2π i2π ¹ - -1 ^-1 ] } = - 5/i2π = 5/2π i

es.2.2

s(t) = 12 sin(8πt)

f_s = 48 Hz

Δf = 0.25 Hz

Δf = f_s / N → N = f_s / Δf = 48 / 0.25 = 192

DURATA = 192 / 48 = 4 sec.

CORRISPONDE A FREQUENZA 24 * 0.25 = 6 Hz

NORMALIZZATA 6 / 48 = 1 / 8 = f / f_s

X(k) = ∑ x(m) e^{-j2π km / N}

1 / 48 → 58º 1 / 6 = 12

Y(m)=0.8y(m-1)-0.81y(m-2)+x(m)

DETERMINARE H(z), POLI, ZERI, ROC, RISPN FREQP ERF: ₁/₂,1/₂

1. Y(z)=0.8z^-1Y(z)-0.81z^-2Y(z)+X(z)

   Y(z)(1-0.8z^-1+0.81z^-2)=X(z)

   H(z)=¹/_{1-0.8z^-1+0.81z^-2}

2. 2/₂

   POLI=z²-0.8z+0.81=0

   z_1/2=^{0.8±sqrt(0.81-4∙0.81)}/₂

   =^{0.8±sqrt(-1.3∙0.81)}/₂

   =0.3±i∙0.81^1/23/₂

   |z_1/2|=0.45∙sqrt(1+3)=0.8

   ∠z₁=atan(^0.45√3/_0.45)=atan(√3)∙60

   ∠z₂=atan(^-0.45√3/_0.45)=atan(-√3)∙-60

   =0.45±i∙0.45√3

   =0,45(1±i√3)

3. ROC: CERCHIO ESTERNO A RAGGIO 0.8

4. H(e^i2π)=H(1)=¹/_1+0.8-0.81=¹/_2.7

   • |H(1)|=¹/_2.7
   • ∠H(1)=0

5. H(e^iπ)=H(i)=¹/_1+0.8i-0.81=¹/_0.9+0.8i

   • |H(i)|=¹/_0.9²+0.8²=¹/_0.5858≈1.08
   • ∠H(i)=0-atan(^0.8/_0.9)≈-78°

1.a Determinare spettro di x(t) = cos⁵(2πt)

x(t) = cos(2πt)cos(2πt)cos(2πt)

X(f) = [₁/₂ δ(f−1) + ₁/₂ δ(f+1)] ⊗ [₁/₂ δ(f−1) + ₁/₂ δ(f+1)]

= ₃/₈ δ(f−1) + ₃/₈ δ(f+1) + ₁/₈ δ(f+3) + ₁/₈ δ(f−3)

x(t) = ₁/₂ cos(6πt) + ₃/₅ cos(2πt)

τ = ₁/₃

x_τ(t) = Σ_m=-∞^∞ C_m e^i2π3t

C_m = ₃/₄ ∫_-1/2^1/₂ cos(6πt) e^-i2πm3t dt

C₁ = ₃/₄ ∫[e^i6πt + e^-i6πt] / 2 e^-i2π3t dt

= ₃/₈ ∫₀^τ e^-i12πt dt

= ₃/₈ [_e^-i12πt/_-i12π]

= ₃/₈ + ₃/₈ [_e^-i12πt/_-i12π − e⁰]

= ₁/₈

C₂ = C_m = ⌀

γ(t) =

∫_-1⁰ (t+τ)(4+τ-t) dτ + ∫₀^t (4-τ)(1-τ+t) dτ + ∫_t¹ (1-τ)(1-τ+t) dτ

= 1/6 (-t³ + 3t + 2) - 1/6 (t² 6t + 6) + 1/6 (-t³ + 3t + 2)

= 1/3 (-t³ + 3t + 2) - 1/6 (t² 6t + 6)

0 ≤ t < 1

γ(t) =

∫_t-1^t (4-τ-t)(4-τ-t) dτ + ∫₀^t (4-τ)(1-τ+t) dτ + ∫_t¹ (1-τ)(1-τ+t) dτ

= 1/3 (-t³ 3t + 2) + 1/6 (t² 6t + 6)

1 ≤ t < 2

γ(t) =

∫_t-1¹ (t-τ)(1+τ-t) dτ = -1/6 (t-2)³

γ(t) =

{

• 0     t < -2
• 1/6 (t+2)³     -2 ≤ t < -1
• 1/3 (t³ + 3t + 2) + 1/6 (t² 6t + 6)     0 ≤ t < 1
• -1/6 (t-2)³     1 ≤ t < 2