Eserciziario di Elaborazione dei Segnali Inf Musicale Pedersini

es. 4.2

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

Y(z)=0.6 z^-1Y(z)-0.08 z^-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}

zero=1

pole: z²-0.6z+0.08=0

z₁=0.4   z₂=0.2

H(e^i2π1/4)=H(e^iπ/2)=H(i)=^1-i/_{1-0.6i+0.08i²}=¹⁺ⁱ/_1+0.6i-0.08=¹⁺ⁱ/_0.92+0.6i

|H(i)|=^√2/_{√0.8(1.2)^1/2}=^√2/_√1.2808=1.287

∠H(i)=-π/4   arctan(^0.6/_0.82)=-0.2075 rad (11.882°)

(1-z^-1)/(1-0.6z^-1+0.08z^-2)=^1-z-1/_{(1-0.6z^-1)(1-0.2z^-1)}=^A/_(1-0.6z^-1) + ^B/_(1-0.2z^-1)

A-0.2z^-1A+B-0.2z^-1B=z^-1(0.2A+0.2B)+A+B=1-z^-1

&lion;  A+B=1A=1-B0.2A+0.2B=10.2-0.2B+0.4B=1B=0.8/0.2=B=4A=-3

1.¹/_1-0.2z^-1-3.¹/_1-0.6z^-1=(¹/_1-0.2z^-1 - ¹/_1-0.6z^-1)=^{(0.2)m-3.(0.6)m}u(n)

es. 4.2

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

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

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

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

• zer1 = 1

• pol1 = z² - 0.6z + 0.08 = 0

z₁ = 0.4 z₂ = 0.2

H(e^i2π₄) = H(e^iπ₂) = H(i) = ^{1 - i}/_{1 - 0.6 i + 0.08/ⁱ²} = ^{1 + i}/_{1 + 0.6i - 0.08} = ^{1 + i}/_{0.92 + 0.6i}

|H(i)| = ^√2/_{√0.8(1 + 2)} = ^√2/_√1.2064 = 1.287

∠H(i) = ^π/₄ - arctan (^0.6/_0.82) = -2.0735 rad (11.883°)

• ^{1 - z-1}/_{1 - 0.6z^-1 + 0.08z^-2} = ^{1 - z-1}/_{(1 - 0.6z^-1)(1 - 0.2z^-1)} = ^A/_{(1 - 0.6z^-1)} + ^B/_{(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 A = 1 - B} {0.2A + 0.2B = 1 0.2 - 0.2B + 0.2B = 1 {B = 0.8/0.2 = 4 A = -3}

4•¹/_{1 - 0.2z^-1} - 3•¹/_{1 - 0.6z^-1} = (^(0.2)n - 3(0.6)ⁿ) u(n)

x(n) ← ─ ◻ + ◻ → y(n) ↑ │0.6 ↑ │ └───┤ z^-1 │ -0.08 ─────→◻

S(t) = 5t, ⁻₁/2 ≤ t < 1/2

con T = 2s

S(t) = ∑_m=0^∞ C_m e^−i2πmt

C_m = 1/2 ₋₁∫¹ 5t e^−i2πmt dt

0Hz ⇒ _m/2 = 0    m = 0

0.5Hz ⇒ m/2 = 1/2    m = 1

1Hz ⇒ m/2 = 1    m = 2

C₀ = 1/2 ₋₁∫¹ 5t e^−i2π0t dt = 1/2 ₋₁∫¹ 5t dt = 5/2 ¹[_1/2 t²]₋₁ = 5/2[1/2 − 1/2] = 0

C_ⁿ = 1/2 ₋₁∫¹ 5t e^{−i2π1⁄₂t} dt = 1/2 ₋₁∫¹ 5t e^−iπt dt = 5/2 ₋₁∫¹ t e^−iπt dt

= 5/2{ ^f[_{t/ iπ} e^−iπt]₋₁ − ₋₁∫¹ e^−iπt dt} = 5/2 { ¹⁄_iπ e^−iπ em −¹⁄_iπ e^iπ − [⁻¹⁄_iπ e −iπt]

= 5/2 {¹⁄_iπ (e^−iπ + e^iπ)] + 1/ (e^−iπ − e^iπ)} = 5/2⁄² − 2/iπ = 5/iπ − 5/iπ =

C_² =¹⁄_{2 −1}∫¹ 5t e^{−i2π2⁄₂t} dt = 5/2 ₋₁∫¹ t e−i2πt dt

= 5/2{[^t/ _i2π⁄s>e_ͨ