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Question Number 151204 by kalenis last updated on 19/Aug/21

	Answered by JDamian last updated on 19/Aug/21
	$(1) Z{y[k+2]+3y[k+1]+2y[k]}=Z{x[k]} z^2 Y(z)+3zY(z)+2Y(z)=X(z) (z^2 +3z+2)Y(z)=X(z) H(z)=((Y(z))/(X(z)))=(1/(z^2 +3z+2)) = (1/(z+1))−(1/(z+2)) ∀∣z∣<1 (2) H(z)= (1/(z+1))−(1/(z+2)) ∀∣z∣<1 h[k]=Z^(−1) {H(z)}=Z^(−1) {(1/(z+1))}−Z^(−1) {(1/(z+2))}= =u[k+1]−u[k+2]$
	$(1)$ $Z {y [k + 2] + 3 y [k + 1] + 2 y [k]} = Z {x [k]}$ $z^{2} Y (z) + 3 z Y (z) + 2 Y (z) = X (z)$ $(z^{2} + 3 z + 2) Y (z) = X (z)$ $H (z) = \frac{Y (z)}{X (z)} = \frac{1}{z^{2} + 3 z + 2} = \frac{1}{z + 1} - \frac{1}{z + 2} \forall ∣ z ∣< 1$ $(2)$ $H (z) = \frac{1}{z + 1} - \frac{1}{z + 2} \forall ∣ z ∣< 1$ $h [k] = Z^{- 1} {H (z)} = Z^{- 1} {\frac{1}{z + 1}} - Z^{- 1} {\frac{1}{z + 2}} =$ $= u [k + 1] - u [k + 2]$
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