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I've already wrote about that trouble (link here), but I don't understand where I've made a mistake.

Full description of the task is as follows:

Z-transform of sequence {x(k)} describe by the equation. $$ X(z) = \frac{2.5 - 3.15z^{-1} + 1.2z^{-2}}{1 - 2.3z^{-1} + 1.2z^{-2}} $$

Calculate of samples x(-2), x(-1), x(0), x(1) and x(2) double-side of sequence {x(k)}, which correspond to this z-transform.
Note - Double-side sequence is sequence to consist samples before zero and after zero.

Step 1

I calculate of the integer part of the expression (just divided the numerator by denominator),

$$ X(z) = \frac{2.5 - 3.15z^{-1} + 1.2z^{-2}}{1 - 2.3z^{-1} + 1.2z^{-2}}\qquad(1.1) $$ transform to, $$ X(z) = 2.5 + \frac{2.6z^{-1} - 1.8z^{-2}}{1 - 2.3z^{-1} + 1.2z^{-2}}\qquad(1.2) $$

Step 2

I transformed denominator like $$\frac{(z – p1)(z – p2)}{z^{-2}}\qquad(2.1)$$. $$ 1 - 2.3z^{-1} + 1.2z^{-2} = (1 - 1.5z^{-1})(1 - 0.8z^{-1})\qquad(2.2) $$

Step 3

I transformed of the expression from: $$ X(z) = 2.5 + \frac{2.6z^{-1} - 1.8z^{-2}}{1 - 2.3z^{-1} + 1.2z^{-2}}\qquad(3.1) $$ to $$ X(z) = 2.5 + X{_1}(z)\qquad(3.2) $$ and then $$X{_1}(z)\qquad(3.3)$$ transform to $$ X{_1}(z) = \frac{A}{1 - 1.5z^{-1}} + \frac{B}{1 - 0.8z^{-1}}\qquad(3.4) $$

Step 4

Calculate of coefficient A and B

Calculate of A

$$ (1 - 1.5z^{-1})X{_1}(z) = A + \frac{B(1 - 1.5z^{-1})}{1 - 0.8z^{-1}},\qquad(4.1)\qquad if\qquad z=1.5 $$ hence $$ A = \frac{2.6z^{-1} - 1.8z^{-2}}{1 - 0.8z^{-1}},\qquad(4.2)\qquad if\qquad z = 1.5 $$ so $$ A = 2 $$

Calculate B

$$ (1 - 0.8z^{-1})X{_1}(z) = \frac{A(1 - 0.8z^{-1})}{1 - 1.5z^{-1}},\qquad(4.3)\qquad if\qquad z=0.8 $$ hence $$ B = \frac{2.6z^{-1} - 1.8z^{-2}}{1 - 1.5z^{-1}},\qquad(4.4)\qquad if\qquad z = 0.8 $$ so $$ B = -0.5 $$

Step 5

$$ X{_1}(z) = \frac{2}{1 - 1.5z^{-1}} + \frac{-0.5}{1 - 0.8z^{-1}}\qquad(5.1) $$ and $$ X(z) = 2.5 + \frac{2}{1 - 1.5z^{-1}} + \frac{-0.5}{1 - 0.8z^{-1}}\qquad(5.2) $$ Sequence of {x(k)} has the following representation $$ x(k) = 2(1.5)^{k}, if k<0; x(k) = 2.5, if k=0; x(k) = -0.5(0.8)^{k}, if k>0; $$ And after that we can calculate Sequence of {x(k)} - x(-2), x( 1), x(0), x(1) и x(2) x(-2) = -0.88889; x(-1) = -1.33333; x(0) = 2.50; x(1) = -0.40; x(2) = -0.32.

Could you check my solve and say where I made mistake. Thank you very much!

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  • $\begingroup$ +1 for showing your work. $\endgroup$ – Ahmad Bazzi Dec 28 '18 at 13:04
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$$X_1(z) = \frac{2.6z^{-1} - 1.8z^{-2}}{1 - 2.3z^{-1} + 1.2z^{-2}} =\frac{A}{1 - 1.5z^{-1}} + \frac{B}{1 - 0.8z^{-1}} \tag{1}$$

Mistake is that $A \neq 2$ and $B \neq -0.5$.

Well then, what are they ? There aren't any $A,B$ where the above equation is true. Why? Take the limit $z \rightarrow \infty$. The term $\frac{A}{1 - 1.5z^{-1}} + \frac{B}{1 - 0.8z^{-1}}$ factors to maximally $z^{-1}$ in the numerator whereas $\frac{2.6z^{-1} - 1.8z^{-2}}{1 - 2.3z^{-1} + 1.2z^{-2}} $ contains a $z^{-2}$ term!!!!. So a contradiction !! The correct decomposition is as follows: $$X_1(z) = \frac{2.6z^{-1} - 1.8z^{-2}}{1 - 2.3z^{-1} + 1.2z^{-2}} =\frac{A}{1 - 1.5z^{-1}} + \frac{B+Cz^{-1}}{1 - 0.8z^{-1}}$$

Taking a common denominator in $\frac{A}{1 - 1.5z^{-1}} + \frac{B}{1 - 0.8z^{-1}}$, equation $(1)$ is now equivalent to $$2.6z^{-1} - 1.8z^{-2} =A+B + (C -1.5B - 0.8)z^{-1} -1.5Cz^{-1}$$ By identification, we get the following system of equations $$\begin{bmatrix} 1.0000 & 1.0000 & 0\\ 0 & -1.5000 & 1.0000\\ 0 & 0 & -1.5000 \end{bmatrix} \begin{bmatrix} A\\B\\C \end{bmatrix} = \begin{bmatrix} 0\\ 3.4000\\ -1.8000 \end{bmatrix} $$ which will give you \begin{align} A &= 1.4667\\ B &= -1.4667 \\ C &= 1.2 \end{align} Can you continue ?

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There are two mistakes in your calculations. The first one has been pointed out in Ahmad Bazzi's answer. However, I would suggest a different expansion of the given function $X(z)$, which requires the introduction of only two unknown constants:

$$X(z)=2.5+\frac{Az^{-1}}{1-1.5z^{-1}}+\frac{Bz^{-1}}{1-0.8z^{-1}}\tag{1}$$

I'm sure you can determine these two constants yourself.

The second mistake is the inverse transform of the basic function

$$G(z)=\frac{1}{1-\alpha z^{-1}}\tag{2}$$

For $|z|>\alpha$ the inverse $\mathcal{Z}$-transform of $(2)$ is

$$g[n]=\alpha^nu[n]\tag{3}$$

where $u[n]$ is the unit step sequence.

For $|z|<\alpha$, the inverse $\mathcal{Z}$-transform of $(2)$ is

$$g[n]=-\alpha^nu[-n-1]\tag{4}$$

From $(2)-(4)$, the inverse $\mathcal{Z}$-transform of $(1)$ is

$$x[n]=2.5\delta[n]-A(1.5)^{n-1}u[-n]+B(0.8)^{n-1}u[n-1]\tag{5}$$

For $n=0$ you have contributions from two terms:

$$x[0]=2.5-A(1.5)^{-1}\tag{6}$$


Note that another possible expansion of $X(z)$ is

$$X(z)=1+\frac{C}{1-1.5z^{-1}}+\frac{D}{1-0.8z^{-1}}$$

resulting in

$$x[n]=\delta[n]-C(1.5)^nu[-n-1]+D(0.8)^nu[n]$$

which of course leads to the same result as above. This approach was taken in this answer to your previous question.

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Correct solution of this task is:

$$ X(z) = 2.5 + X{_1}(z)\qquad(3.2) $$ and then $$X{_1}(z)\qquad(3.3)$$ transform to $$ X{_1}(z) = \frac{A}{1 - 1.5z^{-1}} + \frac{B}{1 - 0.8z^{-1}}\qquad(3.4) $$ where A = 2 and B = -0.5

$$X(z)=2.5+\frac{A}{1-1.5z^{-1}}+\frac{B}{1-0.8z^{-1}}$$

resulting in

$$x[n]=2.5\delta[n]-A(1.5)^nu[-n-1]+B(0.8)^nu[n]$$

Correct double-side of sequence {x(k)} is:

x(-2) = -0.88889; x(-1) = -1.33333; x(0) = 0.50; x(1) = -0.40; x(2) = -0.32.

Thank you very much everyone!

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