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What are the abscissae of convergence for the following functions? (a) \(\mathrm{e}^{5 t}\) (b) \(\mathrm{e}^{-3 t}\) (c) \(\sin 2 t\) (d) \(\sinh 3 t\) (e) \(\cosh 2 t\) (f) \(t^{4}\) (g) \(\mathrm{e}^{-5 t}+t^{2}\) (h) \(3 \cos 2 t-t^{3}\) (i) \(3 \mathrm{e}^{2 t}-2 \mathrm{e}^{-2 t}+\sin 2 t\) (j) \(\sinh 3 t+\sin 3 t\)

Short Answer

Expert verified
(a) 5, (b) -3, (c) 0, (d) 3, (e) 2, (f) 0, (g) 0, (h) 0, (i) 2, (j) 3.

Step by step solution

01

Understand the Concept of Abscissa of Convergence

The abscissa of convergence is a concept used in complex analysis related to power series and Laplace transforms. For a function, it is linked to the convergence of the Laplace transform \(\mathcal{L}\{f(t)\}(s) = \int_0^\infty e^{-st} f(t) \, dt\). The abscissa of convergence \(\sigma_c\) is the real part of \(s\) beyond which the integral converges.
02

Analyze Given Functions

We need to identify each given function's behavior to determine its abscissa of convergence. Most functions can be cast into an exponential form to easily identify convergence criteria.
03

Determine Abscissa for \(\mathrm{e}^{5 t}\)

For \(\mathrm{e}^{5t}\), the Laplace transform is \(\int_0^\infty e^{(-s+5)t} \, dt\). It converges if \(s > 5\), so the abscissa is \(5\).
04

Determine Abscissa for \(\mathrm{e}^{-3 t}\)

For \(\mathrm{e}^{-3t}\), the transform is \(\int_0^\infty e^{(-s-3)t} \, dt\). It converges if \(s > -3\), making the abscissa \(-3\).
05

Determine Abscissa for \(\sin 2t\)

For \(\sin 2t\), since it is periodic, consider it as part of \(e^{i(2t)}\). The Laplace transform converges for \(s > 0\), so the abscissa is \(0\).
06

Determine Abscissa for \(\sinh 3t\)

The hyperbolic sine can be expressed as \(\frac{e^{3t} - e^{-3t}}{2}\). The more dominant term is \(e^{3t}\), thus \(s > 3\), giving an abscissa of \(3\).
07

Determine Abscissa for \(\cosh 2t\)

Similarly, \(\cosh 2t = \frac{e^{2t} + e^{-2t}}{2}\), dominated by \(e^{2t}\). It converges for \(s > 2\), hence the abscissa is \(2\).
08

Determine Abscissa for \(t^4\)

A polynomial like \(t^4\) converges for all \(s > 0\). Therefore, the abscissa is \(0\).
09

Determine Abscissa for \(\mathrm{e}^{-5t} + t^2\)

With \(\mathrm{e}^{-5t}\) requiring \(s > -5\) and \(t^2\) requiring \(s > 0\), \(t^2\) dictates that \(s > 0\). Thus, the abscissa is \(0\).
10

Determine Abscissa for \(3 \cos 2t - t^3\)

\(3 \cos 2t\) requires \(s > 0\) (being periodic), and \(-t^3\) requires \(s > 0\) (being polynomial). Abscissa is determined by the largest, which is \(0\).
11

Determine Abscissa for \(3e^{2t} - 2e^{-2t} + \sin 2t\)

The function \(3e^{2t}\) requiring \(s > 2\) has the largest abscissa demand compared to \(-2e^{-2t}\) which requires \(s > -2\) and \(\sin 2t\) which requires \(s > 0\). Thus, the abscissa is \(2\).
12

Determine Abscissa for \(\sinh 3t + \sin 3t\)

\(\sinh 3t\) has an abscissa requirement of \(s > 3\) and \(\sin 3t\) requires \(s > 0\). The dominant term \(s > 3\) dictates the abscissa, so it is \(3\).

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Key Concepts

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

Laplace transform
The Laplace transform is a powerful mathematical tool used to transform functions of time into functions of complex frequency. This transformation makes solving differential equations easier, among other applications in engineering and physics. The Laplace transform of a function \( f(t) \) is denoted as \( \mathcal{L}\{f(t)\}(s) \) and is defined by the integral:
  • \( \mathcal{L}\{f(t)\}(s) = \int_0^\infty e^{-st} f(t) \, dt \)
In this integral:
  • \( s \) is a complex number \( s = \sigma + i\omega \).
  • \( \sigma \) affects the rate of convergence.
The Laplace transform allows us to work in the frequency domain, where linear operations become easier to handle. It simplifies the handling of linear time-invariant systems. Understanding the abscissa of convergence \( \sigma_c \) is essential as it indicates where the Laplace transform of a function converges.
Complex analysis
Complex analysis is a branch of mathematics focusing on functions of complex numbers. It extends calculus to complex-valued functions, offering powerful tools to solve problems involving complex numbers.
  • A key focus is on analyticity, which means a function is complex differentiable.
  • Complex analysis tools include concepts like contour integration and the Cauchy-Riemann equations.
  • Analytic functions lead to properties like infinite differentiation and obey the Cauchy integral theorem.
These properties make complex analysis very useful in physics, engineering, and number theory. In the context of Laplace transforms, complex analysis helps understand the behavior of functions in the complex plane, especially around their abscissa of convergence, determining the conditions under which the transformation converges.
Exponential functions
Exponential functions are a crucial component of mathematics and various applied fields. They are among the class of functions that grow (or decay) at a rate proportional to their value:
  • General form of an exponential function: \( f(t) = ae^{kt} \), where \( a \) is a constant, and \( k \) determines the rate of growth.
  • If \( k > 0 \), the function shows exponential growth.
  • If \( k < 0 \), the function shows exponential decay.
Exponential functions are encountered frequently in the context of Laplace transforms, as they appear in the transformation kernels \( e^{-st} \) and in many practical functions \( f(t) \). They have continuous derivatives, making them ideal candidates for such operations. Understanding their properties, especially in relation to the Laplace domain, is vital for calculating the abscissae of convergence, since these tell us where the combined exponential terms allow the integral to converge.

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Most popular questions from this chapter

Use Laplace transforms to find the value of \(x\) given that $$ \begin{aligned} &4 \frac{\mathrm{d} x}{\mathrm{~d} t}+6 x+y=2 \sin 2 t \\ &\frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}}+x-\frac{\mathrm{d} y}{\mathrm{~d} t}=3 \mathrm{e}^{-2 t} \end{aligned} $$ and that \(x=2\) and \(\mathrm{d} x / \mathrm{d} t=-2\) when \(t=0\).

Find \(\mathscr{L}^{-1}\\{F(s)\\}\) when \(F(s)\) is given by (a) \(\frac{1}{(s+3)(s+7)}\) (b) \(\frac{s+5}{(s+1)(s-3)}\) (c) \(\frac{s-1}{s^{2}(s+3)}\) (d) \(\frac{2 s+6}{s^{2}+4}\) (e) \(\frac{1}{s^{2}\left(s^{2}+16\right)}\) (f) \(\frac{s+8}{s^{2}+4 s+5}\) (g) \(\frac{s+1}{s^{2}\left(s^{2}+4 s+8\right)}\) (h) \(\frac{4 s}{(s-1)(s+1)^{2}}\) (i) \(\frac{s+7}{s^{2}+2 s+5}\) (j) \(\frac{3 s^{2}-7 s+5}{(s-1)(s-2)(s-3)}\) (k) \(\frac{5 s-7}{(s+3)\left(s^{2}+2\right)}\) (1) \(\frac{s}{(s-1)\left(s^{2}+2 s+2\right)}\) (m) \(\frac{s-1}{s^{2}+2 s+5}\) (n) \(\frac{s-1}{(s-2)(s-3)(s-4)}\) (o) \(\frac{3 s}{(s-1)\left(s^{2}-4\right)}\) (p) \(\frac{36}{s\left(s^{2}+1\right)\left(s^{2}+9\right)}\) (q) \(\frac{2 s^{2}+4 s+9}{(s+2)\left(s^{2}+3 s+3\right)}\) (r) \(\frac{1}{(s+1)(s+2)\left(s^{2}+2 s+10\right)}\)

Solve the differential equation $$ \frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}}+2 \frac{\mathrm{d} x}{\mathrm{~d} t}+2 x=\cos t $$ subject to the initial conditions \(x=x_{0}\) and \(\mathrm{d} x / \mathrm{d} t=x_{1}\) at \(t=0\). Identify the steady-state and transient solutions. Find the amplitude and phase shift of the steady-state solution.

Using Laplace transform methods, solve for \(t \geqslant 0\) the following differential equations, subject to the specified initial conditions. (a) \(\frac{\mathrm{d} x}{\mathrm{~d} t}+3 x=\mathrm{e}^{-2 t} \quad\) subject to \(x=2\) at \(t=0\) (b) \(3 \frac{\mathrm{d} x}{\mathrm{~d} t}-4 x=\sin 2 t \quad\) subject to \(x=\frac{1}{3}\) at \(t=0\) (c) \(\frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}}+2 \frac{\mathrm{d} x}{\mathrm{~d} t}+5 x=1\) subject to \(x=0\) and \(\frac{\mathrm{d} x}{\mathrm{~d} t}=0\) at \(t=0\) (d) \(\frac{\mathrm{d}^{2} y}{\mathrm{~d} t^{2}}+2 \frac{\mathrm{d} y}{\mathrm{~d} t}+y=4 \cos 2 t\) subject to \(y=0\) and \(\frac{\mathrm{d} y}{\mathrm{~d} t}=2\) at \(t=0\) (e) \(\frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}}-3 \frac{\mathrm{d} x}{\mathrm{~d} t}+2 x=2 \mathrm{e}^{-4 t}\) subject to \(x=0\) and \(\frac{\mathrm{d} x}{\mathrm{~d} t}=1\) at \(t=0\) (f) \(\frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}}+4 \frac{\mathrm{d} x}{\mathrm{~d} t}+5 x=3 \mathrm{e}^{-2 t}\) subject to \(x=4\) and \(\frac{\mathrm{d} x}{\mathrm{~d} t}=-7\) at \(t=0\) (g) \(\frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}}+\frac{\mathrm{d} x}{\mathrm{~d} t}-2 x=5 \mathrm{e}^{-t} \sin t\) subject to \(x=1\) and \(\frac{\mathrm{d} x}{\mathrm{~d} t}=0\) at \(t=0\) (h) \(\frac{\mathrm{d}^{2} y}{\mathrm{~d} t^{2}}+2 \frac{\mathrm{d} y}{\mathrm{~d} t}+3 y=3 t\) subject to \(y=0\) and \(\frac{\mathrm{d} y}{\mathrm{~d} t}=1\) at \(t=0\) (i) \(\frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}}+4 \frac{\mathrm{d} x}{\mathrm{~d} t}+4 x=t^{2}+\mathrm{e}^{-2 t}\) subject to \(x=\frac{1}{2}\) and \(\frac{\mathrm{d} x}{\mathrm{~d} t}=0\) at \(t=0\) (j) \(9 \frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}}+12 \frac{\mathrm{d} x}{\mathrm{~d} t}+5 x=1\) subject to \(x=0\) and \(\frac{\mathrm{d} x}{\mathrm{~d} t}=0\) at \(t=0\) (k) \(\frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}}+8 \frac{\mathrm{d} x}{\mathrm{~d} t}+16 x=16 \sin 4 t\) subject to \(x=-\frac{1}{2}\) and \(\frac{\mathrm{d} x}{\mathrm{~d} t}=1\) at \(t=0\) (1) \(9 \frac{\mathrm{d}^{2} y}{\mathrm{~d} t^{2}}+12 \frac{\mathrm{d} y}{\mathrm{~d} t}+4 y=\mathrm{e}^{-t}\) subject to \(y=1\) and \(\frac{\mathrm{d} y}{\mathrm{~d} t}=1\) at \(t=0\)(m) \(\frac{\mathrm{d}^{3} x}{\mathrm{~d} t^{3}}-2 \frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}}-\frac{\mathrm{d} x}{\mathrm{~d} t}+2 x=2+t\) (n) \(\frac{\mathrm{d}^{3} x}{\mathrm{~d} t^{3}}+\frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}}+\frac{\mathrm{d} x}{\mathrm{~d} t}+x=\cos 3 t\) subject to \(x=0, \frac{\mathrm{d} x}{\mathrm{~d} t}=1\) and \(\frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}}=0\) at \(t=0\) subject to \(x=0, \frac{\mathrm{d} x}{\mathrm{~d} t}=1\) and \(\frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}}=1\) at \(t=0\)

(a) Find the inverse Laplace transform of $$ \frac{1}{(s+1)(s+2)\left(s^{2}+2 s+2\right)} $$ (b) A voltage source \(V \mathrm{e}^{-t} \sin t\) is applied across a series \(L C R\) circuit with \(L=1, R=3\) and \(C=\frac{1}{2}\). Show that the current \(i(t)\) in the circuit satisfies the differential equation $$ \frac{\mathrm{d}^{2} i}{\mathrm{~d} t^{2}}+3 \frac{\mathrm{d} i}{\mathrm{~d} t}+2 i=V \mathrm{e}^{-t} \sin t $$ Find the current \(i(t)\) in the circuit at time \(t \geqslant 0\) if \(i(t)\) satisfies the initial conditions \(i(0)=1\) and \((\mathrm{d} i / \mathrm{d} t)(0)=2\)

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