Question

Bestimmen Sie die Partialbruchzerlegung von (a) \(F(z)=\frac{3 z-5}{z^{2}-4 z+3}\), (b) \(G(z)=\frac{2}{z\left(z^{2}+4\right)}\) und daraus mit Hilfe der Tabelle der LAPLACE-Transformierten diejenigen Funktionen, deren LAPLACE-Transformierten \(F(z)\) bzw. \(G(z)\) sind.

Short answer

(a) -2e^t + 5e^{3t}, (b) 0.5 + 0.25\sin(2t)

Step-by-step solution

Factor the Denominator of (a)

For the function \( F(z) = \frac{3z-5}{z^2-4z+3} \), we first need to factor the denominator. The quadratic equation \( z^2 - 4z + 3 \) can be factored as \((z-1)(z-3)\). This gives us simpler linear factors to work with in the partial fraction decomposition.

Set Up Partial Fraction Decomposition for (a)

Express \( F(z) = \frac{3z-5}{(z-1)(z-3)} \) as a sum of partial fractions: \( \frac{A}{z-1} + \frac{B}{z-3} \). We need to find the constants \( A \) and \( B \).

Solve for Constants in (a)

Use the partial fractions setup: \( 3z - 5 = A(z-3) + B(z-1) \). Solve for \( A \) and \( B \) by plugging in suitable values for \( z \) or comparing coefficients. For \( z=1 \), \( A = -2 \), and for \( z=3 \), \( B = 5 \).

Write Full Partial Fraction for (a)

Substitute \( A \) and \( B \) back into the formula: \( F(z) = \frac{-2}{z-1} + \frac{5}{z-3} \). This is the partial fraction decomposition of \( F(z) \).

Factor the Denominator of (b)

For \( G(z) = \frac{2}{z(z^2+4)} \), note that the denominator \( z(z^2+4) \) cannot be factored further using real factors.

Set Up Partial Fraction Decomposition for (b)

For \( G(z) \), set it up as \( \frac{A}{z} + \frac{Bz+C}{z^2+4} \). This accounts for the linear and quadratic terms.

Solve for Constants in (b)

Set \( 2 = A(z^2+4) + (Bz+C)z \) and solve for \( A \), \( B \), and \( C \) by comparing coefficients or plugging in convenient values for \( z \): solving yields \( A = 0.5 \), \( B = 0 \), \( C = 0.5 \).

Write Full Partial Fraction for (b)

Substitute \( A \), \( B \), \( C \) into the formula: \( G(z) = \frac{0.5}{z} + \frac{0.5}{z^2+4} \). This is the partial fraction decomposition of \( G(z) \).

Use Laplace Transform Table for (a)

Using the Laplace transform table, the term \( \frac{1}{z-1} \) corresponds to \( e^t \), and \( \frac{1}{z-3} \) corresponds to \( e^{3t} \). Thus, \( F(z) \) corresponds to the inverse Laplace transform \( -2e^t + 5e^{3t} \).

Use Laplace Transform Table for (b)

For \( G(z) \), \( \frac{1}{z} \) corresponds to \( 1 \), and \( \frac{1}{z^2+4} \) corresponds to \( \frac{1}{2} \sin(2t) \). Therefore, \( G(z) \) corresponds to the inverse Laplace transform \( 0.5 + 0.5 \frac{1}{2} \sin(2t) \).

Key concepts

Laplace Transform

Laplace transform is a powerful integral transform method used in engineering and physics to transform a time-domain function into a complex frequency-domain function. This transformation simplifies the process of solving linear differential equations, making it a valuable tool in control theory and systems analysis.

The Laplace transform of a function \( f(t) \) is defined by the integral:\[L\{f(t)\} = F(s) = \int_{0}^{\infty} e^{-st} f(t) \, dt\]where \( s \) is a complex number, and \( F(s) \) is the transformed function.

The transform effectively converts differential equations into algebraic equations for easier manipulation. Once transformed, solutions can be determined and then converted back to the time domain through the inverse Laplace transform. Using Laplace tables, one can find corresponding transforms for standard functions, facilitating quick solutions in problem-solving environments.

Quadratic Factoring

Quadratic factoring is a mathematical process that involves expressing a quadratic equation as a product of its linear factors. This is essential in the context of partial fraction decomposition, especially when dealing with functions of the form \( ax^2 + bx + c \).

For example, consider the quadratic equation \( z^2 - 4z + 3 \). To factor it, we need to find two numbers whose product is the constant term (3) and whose sum is the linear coefficient (-4). These numbers are -1 and -3. Thus, the factors are \( (z-1)(z-3) \).

Factoring simplifies the polynomial expressions, making them suitable for decomposition. This step is crucial because partial fraction decomposition relies on these factors to separate complex fractions into simpler, more manageable parts.

• Identify the coefficient of the squared term, the linear term, and constant values.
• Determine potential factors of the constant term that, when summed, match the linear coefficient.
• Reconstruct the expression as the product of linear binomials.

Understanding and applying quadratic factoring is an indispensable skill in algebraic manipulation for solving equations and performing integral operations.

Inverse Laplace Transform

The inverse Laplace transform is the process of converting a frequency-domain function, back into the time-domain representation. This is vital for interpreting solutions to differential equations in forms that describe real-world phenomena like circuits or harmonic oscillations.

To find the inverse Laplace transform, we often reference a table of Laplace transforms, matching terms from the frequency domain back to their time-domain equivalents. For example:

• A term like \( \frac{1}{s-a} \) pairs with \( e^{at} \) in the time domain.
• Terms like \( \frac{1}{s^2+\omega^2} \) correspond to \( \frac{1}{\omega} \sin(\omega t) \).

The steps involve identifying recognizable patterns from the Laplace table and applying inverse relationships correctly.

In the context of our given functions, the inverse Laplace transforms enable us to express decomposed partial fractions back to the original functions, articulated in terms of real-time behavior. This conversion ensures that mathematical solutions align with physical or theoretical models in practical applications.