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]](https://www.tinkutara.com/question/Q151240.png)
$$\left(\mathrm{1}\right) \\ $$$$\mathrm{Z}\left\{{y}\left[{k}+\mathrm{2}\right]+\mathrm{3}{y}\left[{k}+\mathrm{1}\right]+\mathrm{2}{y}\left[{k}\right]\right\}=\mathrm{Z}\left\{{x}\left[{k}\right]\right\} \\ $$$${z}^{\mathrm{2}} {Y}\left({z}\right)+\mathrm{3}{zY}\left({z}\right)+\mathrm{2}{Y}\left({z}\right)={X}\left({z}\right) \\ $$$$\left({z}^{\mathrm{2}} +\mathrm{3}{z}+\mathrm{2}\right){Y}\left({z}\right)={X}\left({z}\right) \\ $$$${H}\left({z}\right)=\frac{{Y}\left({z}\right)}{{X}\left({z}\right)}=\frac{\mathrm{1}}{{z}^{\mathrm{2}} +\mathrm{3}{z}+\mathrm{2}}\:=\:\frac{\mathrm{1}}{{z}+\mathrm{1}}−\frac{\mathrm{1}}{{z}+\mathrm{2}}\:\:\forall\mid{z}\mid<\mathrm{1} \\ $$$$\left(\mathrm{2}\right) \\ $$$${H}\left({z}\right)=\:\frac{\mathrm{1}}{{z}+\mathrm{1}}−\frac{\mathrm{1}}{{z}+\mathrm{2}}\:\:\:\:\:\:\forall\mid{z}\mid<\mathrm{1} \\ $$$${h}\left[{k}\right]=\mathrm{Z}^{−\mathrm{1}} \left\{{H}\left({z}\right)\right\}=\mathrm{Z}^{−\mathrm{1}} \left\{\frac{\mathrm{1}}{{z}+\mathrm{1}}\right\}−\mathrm{Z}^{−\mathrm{1}} \left\{\frac{\mathrm{1}}{{z}+\mathrm{2}}\right\}= \\ $$$$={u}\left[{k}+\mathrm{1}\right]−{u}\left[{k}+\mathrm{2}\right] \\ $$