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Chapter 8: Problem 22

\(f(t)=\delta(t-2)-2 \delta(t-1), 0 \leq t<3, f(t)=\) \(f(t-3)\) if \(t \geq 3\)

Short Answer

Expert verified
The function is defined as a piecewise repetition every 3 units, with \(f(t)=\delta(t-2)-2\delta(t-1)\) for \(0\leq t<3\) and \(f(t)=f(t-3)\) for \(t \geq 3\).

Step by step solution

01

- Understanding the Original Function

The given function is defined piecewise. For $$0 \leq t<3, f(t)=\delta(t-2)-2\delta(t-1)$$ For \(t \geq 3, f(t)=f(t-3)\).
02

- Piecewise Definition for Function between 0 and 3

For \(0 \leq t < 3\), the function is composed of two Dirac delta functions: $$f(t) = \delta(t-2) - 2\delta(t-1)$$.
03

- Applying the Piecewise Extension

For \(t \geq 3, f(t)=f(t-3)\).This means the original function will repeat every 3 units. For example, $$f(3) = f(3-3) = f(0),$$$$f(4) = f(4-3) = f(1).$$
04

- Combining Results

For a compact representation across the whole domain of \(t\): $$f(t) = \begin{cases} \delta(t-2) - 2\delta(t-1)\ & \text{if } 0 \leq t < 3, \ f(t-3) \ & \text{if } t \geq 3 \end{cases}$$

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Key Concepts

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

Periodic Functions
A function is periodic if it repeats its values in regular intervals or periods. Mathematically, a function \( f(t) \) is periodic with period \( T \) if for all \( t \):
  • \[ f(t + T) = f(t) \]

In the given exercise, for \( t \geq 3 \), the function repeats every 3 units. This means:
  • \[ f(t) = f(t-3) \].
  • For example, for \( t = 4 \): \[ f(4) = f(4-3) = f(1) \]

Understanding periodic functions helps in analyzing behaviors in cycles, which is essential in various fields such as signal processing, physics, and engineering.

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Most popular questions from this chapter

(Sawtooth Wave) If \(f(t)=t / a, 0 \leq t

Show that the convolution integral is associative by proving that \((f *(g * h))(t)=\) \(((f * g) * h)(t)\).

\(x^{\prime \prime}+4 x^{\prime}+13 x=f(t)\), $$ f(t)=\left\\{\begin{array}{l} 1,0 \leq t<1 \\ 2, t \geq 1 \end{array}, x(0)=x^{\prime}(0)=0\right. $$

\(x^{\prime \prime}+y^{\prime \prime}=x+t, y^{\prime}-x+x^{\prime}=0, x(0)=x^{\prime}(0)=\) \(y(0)=y^{\prime}(0)=0\)

\(y^{\prime \prime}+2 y^{\prime}+y=2 t e^{-t}-e^{-t}, y(0)=1, y^{\prime}(0)=\) \(-1\)
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