Question

-) Solve: un+2 - 3un+1 +2un = 2^n , given u0 = 0, u1 = 1 by using Z-transforms.​

Answer 1

Step-by-step explanation:

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Answer 2

Answer:

$u_{n} =2^{n}n$

Step-by-step explanation:

Step 1:

Given, $u_{n+2} -3u_{n+1}+2u_{n} = 2^{n}$

This can be written as $u(n+2)-3u(n+1)+2u(n)=2^{n}$

Taking the Z-transform of both sides, we get

$Z{\{u(n+2)\}} -3Z{\{u(n+1)\}} + 2Z{\{u(n)\}} = Z \{{2^{n}}\}$

Step 2:

$[z^{2} F(z)-z^{2}u(0)-zu(1)]-3[zF(z)-zu(0)]+2F(z)=\frac{z}{z-2}$

Putting values $u_{0}=0,u_{1}=1$ in the above equation,

$\implies$$[z^{2} F(z)-0-z]-3[zF(z)-0]+2F(z)=\frac{z}{z-2}$

$\implies$$F(z)(z^{2} +3z+2)=\frac{z}{z-2}+z$

Step 3:

Factorizing $z^{2} -3z+2$ = $(z-1)(z-2)$,

$\implies$   $F(z)(z-1)(z-2)=\frac{z}{z-2}+z$

$\implies F(z)(z-1)(z-2)=\frac{z+z^{2}-2z }{z-2}$

$\implies F(z)(z-1)(z-2)=\frac{z^{2}-z }{z-2}$

$\implies F(z)(z-1)(z-2)=\frac{z(z-1)}{z-2}$

$\implies \frac{ F(z)}{z} =\frac{(z-1)}{(z-2)(z-1)(z-2)}$  

 $\implies \frac{ F(z)}{z} =\frac{1}{(z-2)(z-2)}$  

Step 4:

Solving the RHS by partial fraction method,

$\frac{F(z)}{z} =\frac{1}{(z-2)(z-2)} = \frac{A}{z-2} +\frac{B }{(z-2)^{2}}$           ...(i)

$\implies 1=A(z-2)+B\\\implies 1=Az-2A+B\\\\\therefore Az = 0$and        $-2A+B=1$

$\implies$ $A=0$                                      $B=1$

Substituting values of A and B in eq (i),

$\frac{F(z)}{z} =\frac{1 }{(z-2)^{2}}$

$\implies$ $F(z) =\frac{z }{(z-2)^{2}}$

Taking inverse Z transformation,

$u_{n} =2^{n}n$

Hence, $u_{n} =2^{n}n$

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