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Chapter 5: Problem 5

Find the inverse \(Z\) transform of $$ F(z)=\frac{z^{2}(z-3)}{\left(z^{2}-2 z+1\right)(z-2)} $$

Short Answer

Expert verified
The inverse Z-transform is \(f(n) = 1 + 2n - 4(2^n)\).

Step by step solution

01

- Factorize the denominator

The given function is \[F(z)=\frac{z^{2}(z-3)}{(z^{2}-2z+1)(z-2)}.\] First, factorize the denominator. Notice that \(z^2 - 2z + 1 = (z-1)^2\). Thus, the function becomes \[F(z) = \frac{z^2(z-3)}{(z-1)^2 (z-2)}.\]
02

- Perform partial fraction decomposition

Express \(F(z)\) as a sum of partial fractions. We write:\[\frac{z^2(z-3)}{(z-1)^2 (z-2)} = \frac{A}{z-1} + \frac{B}{(z-1)^2} + \frac{C}{z-2}.\] Multiply both sides by the denominator \((z-1)^2 (z-2)\) to get:\[z^2(z-3) = A(z-1)(z-2) + B(z-2) + C(z-1)^2.\]
03

- Solve for coefficients A, B, and C

Set \(z = 1\) to solve for \(B\): \[1^2(1-3) = B(1-2),\] which simplifies to \(-2 = -B\) so \(B = 2\). Set \(z = 2\) to solve for \(C\): \[2^2(2-3) = C(2-1)^2\] which simplifies to \[-4 = C\], so \(C = -4\). Expand and simplify to solve for \(A\): Equate coefficients of like powers of \(z\) and solve: \[A(z-1)(z-2) = z^3 - 3z^2\] leads to \[A = 1\].
04

- Rewrite F(z)

Using the values of \(A, B,\) and \(C\), rewrite \(F(z)\) as\[F(z) = \frac{1}{z-1} + \frac{2}{(z-1)^2} - \frac{4}{z-2}.\]
05

- Find inverse Z-transform

Using standard inverse Z-transform pairs, find the inverse Z-transform of each term.For \(\frac{1}{z-1}\), the inverse Z-transform is \(1\). For \(\frac{2}{(z-1)^2}\), the inverse Z-transform is \(2n\). For \(\frac{-4}{z-2}\), the inverse Z-transform is \(-4(2^n)\).Thus, the inverse Z-transform of \(F(z)\) is \[f(n) = 1 + 2n - 4(2^n).\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Partial Fraction Decomposition
Partial fraction decomposition is a method used to break down complex rational expressions into simpler fractions that are easier to work with. In this problem, we start with the function \[F(z) = \frac{z^2(z-3)}{(z-1)^2 (z-2)}.\] The goal is to express this as a sum of simpler fractions, which helps in the next steps of finding the inverse Z-transform.

Here鈥檚 the process:
  • Write the given expression as a sum of partial fractions: \[\frac{A}{z-1} + \frac{B}{(z-1)^2} + \frac{C}{z-2}.\]
  • Multiply both sides by the denominator \((z-1)^2(z-2)\) to clear the fraction: \[z^2(z-3) = A(z-1)(z-2) + B(z-2) + C(z-1)^2.\]
Coefficients Solving
The next step in the partial fraction decomposition process is solving for the coefficients \(A\), \(B\), and \(C\).

Here鈥檚 a step-by-step approach to find these coefficients:
  • Start by setting \(z = 1\) to solve for \(B\): \[1^2(1-3) = B(1-2)\] which simplifies to \[-2 = -B\] and thus, \(B = 2\).
  • Set \(z = 2\) to solve for \(C\): \[2^2(2-3) = C(2-1)^2\] which simplifies to \[-4 = C\] and so, \(C = -4\).
  • Lastly, expand the left-hand side and compare coefficients of like powers of \(z\) to solve for \(A\): From \[A(z-1)(z-2) = z^3 - 3z^2, \] identify terms of \(z\) to find \[A = 1\].
Once solved, we recompose the original function using these coefficients. So, \[F(z) = \frac{1}{z-1} + \frac{2}{(z-1)^2} - \frac{4}{z-2}.\]
Inverse Z-Transform Pairs
The inverse Z-transform process utilizes pairs that map between the Z-domain and the time-domain representations. These pairs are essential for converting the decomposed function back to a form that can be understood as a time sequence. Here鈥檚 how we handle each term:
  • The term \(\frac{1}{z-1}\) corresponds to the inverse Z-transform of \(1\).
  • The term \(\frac{2}{(z-1)^2}\) corresponds to the inverse Z-transform of \(2n\).
  • The term \(\frac{-4}{z-2}\) corresponds to the inverse Z-transform of \(-4(2^n)\).
By applying these standard inverse Z-transform pairs, we get the final solution in the time-domain:

\[f(n) = 1 + 2n - 4(2^n).\]

This sequence is our desired inverse Z-transform, derived from the original function \(F(z)\). This method allows us to see how complex Z-domain expressions translate to simpler, interpretable time sequences.

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Most popular questions from this chapter

Use the derivative property to find the \(Z\) transform of \(f(n)=3^{n} n u(n-3)\).

Find the \(Z\) transforms of the sequence of values obtained when \(f(t)\) is sampled at regular intervals of \(t=T\) where (a) \(f(t)=\sinh t\) (b) \(f(t)=\cosh a t\) (c) \(f(t)=e^{-a t} \cosh b t\).

Show that the difference equation \(g(n+2)-g(n+1)-6 g(n)=0\) can be derived from the coupled difference equation $$ \begin{aligned} &f(n+1)=g(n) \\ &g(n+1)=g(n)+6 f(n) \end{aligned} $$ Find \(f(n)\) and \(g(n)\) given that \(f(1)=0\) and \(g(1)=1\).

Find the \(Z\) transform of each of the sequences with output: (a) \(f(n)=(-1)^{n} u(n)\) (b) \(f(n)=\left(4 n-2 a^{n}\right) u(n)\) (c) \(f(n)=(n-3) u(n)\) (d) \(f(n)=\left(5^{n+2}\right) u(n)\).

Given the sequence of \(n\) terms \(f(k)\) for \(0 \leq k \leq n-1\) with \(Z\) transform \(F_{n}(z)\), show that the \(Z\) transform of the sequence formed by continually repeating the terms \(f(k)\) is given as $$ F(z)=\frac{F_{n}(z)}{1-Z^{-n}} $$
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