Dadas las siguientes ecuaciones en diferencias, (i) aplicar las fórmulas básicas y las propiedades de la transformada z para hallar la solución; (ii) bosquejar el gráfico de la solución; (iii) verificar la solución analítica comparándola con la solución iterativa en los primeros 5 instantes de muestreo.
- $y(k+1)-0.6y(k)=0,y(0)=1$
- $y(k+2)-4y(k+1)+3y(k)=0,y(0)=0,y(1)=1$
- $y(k+2)+0.1y(k+1)-0.02y(k)=0,y(0)=0,y(1)=-1$
- $y(k+2)-0.25y(k)=0,y(0)=1,y(1)=0$
- $y(k+2)-0.25y(k)=0,y(0)=0,y(1)=-1$
- $y(k)-0.25y(k-2)=0,y(0)=0,y(1)=-1$
- $y(k+3)+0.8y(k+2)=0,y(0)=1,y(1)=0,y(2)=1$
- $y(k+2)+0.4y(k+1)+0.04y(k)=0,y(0)=1,y(1)=0$
- $y(k+5) -0.2y(k+4) -0.35(k+3) =0,y(0)=1,y(1)=0,y(2)=-1,y(3)=0,y(4)=1$
- $y(k+2)-y(k+1)+y(k)=0,y(0)=1,y(1)=-1$
- $y(k+2)-y(k+1)+0.41y(k)=0,y(0)=0,y(1)=-1$
- $y(k+1)-0.6y(k)=2,y(0)=0$
- $y(k+1)-0.6y(k)=0.6^k,y(0)=-0.5$
- $y(k+1)-0.6y(k)=u_s(k-1),y(0)=1$
- $y(k+1)-0.6y(k)=u_s(k-2),y(0)=1$
- $y(k+1)-0.3y(k)=2,y(0)=0.5$
- $y(k+1)-0.3y(k)=k+3,y(0)=2$
- $y(k+1)-0.3y(k)=0.3^k,y(0)=0.3$
- $y(k+1)-0.3y(k)=\sin k,y(0)=0.3$
- $y(k+1)-0.3y(k)=u_s(k-2)-u_s(k-4),y(0)=2$
- $y(k+2)-0.4y(k+1)+0.04y(k)=1,y(0)=0,y(1)=0$
- $y(k+2)-0.4y(k+1)+0.04y(k)=u_s(k-1),y(0)=0,y(1)=0$
- $y(k+2)-0.4y(k+1)+0.04y(k)=k+1,y(0)=0,y(1)=0$
- $y(k+2)-0.4y(k+1)+0.04y(k)=0.2^k,y(0)=0,y(1)=-1$
- $y(k+2)-0.4y(k+1)+0.04y(k)=\sin 0.2k,y(0)=0,y(1)=-1$
- $y(k+2)-0.4y(k+1)+0.04y(k)=u_s(k-1),y(0)=y(1)=0$
- $y(k)-y(k-1)=0,y(0)=0.1$
- $y(k)-y(k-1)=u(k),u(k)=1,y(0)=0.1$
- $y(k)-y(k-1)=u(k),u(k)=-1,y(0)=0$
- $y(k)+y(k-1)=u(k),u(k)=(-1)^k,y(0)=1$
- $y(k)+y(k-1)=u_s(k-1)-u_s(k-2),y(0)=1$
- $y(k+1)+ay(k)=u(k+1)+au(k),u(k)=k,y(0)=0$
- $\left\{ \begin{array}{l} x(k+1) =y(k) \\ y(k+1) =1-0.4x(k)\\\end{array} \right., ~~x(0)=-1,y(0)=1$
- $y(k+2)+4y(k) =8\times 2^k u_s(k) ,y(0) =0,y(1) =0$
- $y(k+2)+4y(k)=8\times 2^k\sin (\pi k/2) u_s(k) ,y(0) =0,y(1) =0$
- $y(k+2)+4y(k) =2^{k-1} u_s(k-1) ,y(0) =y(1) =0$
Comentarios