Assume a system function of an LTI system is5 z?H(z)(22 – z – 6)The input to the system x(n) = (-2)"u(n). Assuming y(n) and Y(z) are the Z-transform pair, and y(n) isthe output of the above system:a) determine Y(z) assuming ROC of H(z) is |z|>3b) Determine y(n) by inverting Y(z) using the method of residuals (assuming ROC of H(z) is |z|>3)c) What is the ROC of a stable system given by H(z) ?d) What is the response y(n) of a stable system given by H(z)?

“Since you have posted a question with multiple sub-parts, we will solve the first three sub-parts for you. To get the remaining sub-part solved please repost the complete question and mention the sub-parts to be solved.”The system function of an LTI system is, H(z)=5z2(z2-z-6)The input to the system is given by,x(n) = (-2)nu(n)Part (a) - Given input is,x(n)=(-2)nu(n)Applying z-transform,x(n) → X(z)X(z) =zz+2Considering, ROC of H(z) is z>3.Now, the output of the system is,Y(z) = H(z)·X(z)Y(z) =5z2(z2-z-6)·zz+2Y(z) =5z2(z2-3z+2z-6)·zz+2Y(z) =5z2z(z-3)+2(z-3)·zz+2Y(z) =5z2(z-3)+(z+2)·zz+2Y(z) =5z3(z-3)+(z+2)2Part (b) - For y(n),Using the method of residuals,Y(z) = 5z3(z-3)+(z+2)2Y(z)z = 5z2(z-3)+(z+2)2using partial fraction,5z2(z-3)+(z+2)2 = A(z-3)+B(z+2)+C(z+2)2After solving for A, B and C we get,A=1.8B=3.2C=-4So, Y(z)z=1.8(z-3)+3.2(z+2)-4(z+2)2Y(z) = 1.8z(z-3)+3.2z(z+2)-4z(z+2)2Applying inverse laplace transform,y(n) = 1.8(3)nu(n)+3.2(-2)nu(n)-4(-2)nu(n) Part (c) - H(z) = 5z2(z-3)+(z+2)Number of poles are 2.i.e. P=3, - 2We know, for a stable system the ROC must contain the unit circle.i.e. z<1Here, Both the poles are outside the unit circle.Therefore, ROC must be z <3 and z<2 and the common area is z<2.