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Obtain Cascade form realization, $$ y(n)-\frac{1}{4} y(n-1)-\frac{1}{8} y(n-2)=x(n)+3 x(n-1)+2 x(n-2) $$
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Solution:

$ y(n)-\frac{1}{4} y(n-1)-\frac{1}{8} y(n-2)=x(n)+3 x(n-1)+2 x(n-2)\\ $

Taking Z-transform,

$ Y(Z)-\frac{1}{4} Z^{-1} Y(Z)-\frac{1}{8} Z^{-2} Y(Z)\\ $

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

$ \frac{Y(Z)}{X(Z)}=\frac{1+3 Z^{-1}+2 Z^{-1}}{1-\frac{1}{4} Z^{-1}-\frac{1}{8} Z^{-2}}\\ $

$ =\frac{\left(1+Z^{-1}\right)\left(1+2 Z^{-1}\right)}{\left(1-\frac{1}{2} Z^{-1}\right)\left(1+\frac{1}{4} Z^{-1}\right)}\\ $

$ =H_1(Z) \cdot H_2(Z)\\ $

$ H_1(Z)=\frac{\left(1+Z^{-1}\right)}{\left(1-\frac{1}{2} Z^{-1}\right)}\\ $

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$ H_2(Z)=\frac{\left(1+2 Z^{-1}\right)}{\left(1+\frac{1}{4} Z^{-1}\right)}\\ $

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