Chapter 7: Problem 4
Find the convolution \(f(t) * g(t)\). $$ f(t)=t^{2}, g(t)=\cos t $$
Short Answer
Expert verified
The convolution is \(-t^3 \sin t - 2t + 2t \sin t\).
Step by step solution
01
Explanation of Convolution
The convolution of two functions \(f(t)\) and \(g(t)\) is given by the integral \((f * g)(t) = \int_{-fty}^{fty} f(\tau) g(t-\tau) \, d\tau\). The integration bounds are conventionally from \(-\infty\) to \(\infty\), but in practical problems, you can evaluate over the ranges where the functions are non-zero.
02
Setup the Convolution Integral
For our functions \(f(t) = t^2\) and \(g(t) = \cos(t)\), the convolution integral becomes: \((f * g)(t) = \int_{-fty}^{fty} \tau^2 \cos(t-\tau) \, d\tau\). Simplifying this, the actual integral we need to solve is \(\int_{0}^{t} \tau^2 \cos(t-\tau) \, d\tau\) because \(f(t)\) is defined from 0 to t.
03
Change of Variables
Substitute \(u = t - \tau\) so the differential \(d\tau = -du\) and the limits change accordingly: when \(\tau = 0, u = t\) and when \(\tau = t, u = 0\). Thus, the integral becomes \(- \int_{t}^{0} (t-u)^2 \cos u \, (-du)\), which simplifies to \(\int_{0}^{t} (t-u)^2 \cos u \, du\).
04
Expand the Integrand
Expand \((t-u)^2\) to \(t^2 - 2tu + u^2\) and rewrite the integral: \(\int_{0}^{t} (t^2 - 2tu + u^2) \cos u \, du = \int_{0}^{t} t^2 \cos u \, du - 2t\int_{0}^{t} u \cos u \, du + \int_{0}^{t} u^2 \cos u \, du\).
05
Integrate Term-by-Term
1. For \(\int_{0}^{t} t^2 \cos u \, du = t^2 \sin t\), because \(t^2\) is constant with respect to \(u\).2. Use integration by parts to solve \(- 2t \int_{0}^{t} u \cos u \, du\): let \(v = u\) and \(dw = \cos u \, du\), then \(dv = du\) and \(w = \sin u\). Solve it as \(-2t(u\sin u \bigg|_{0}^{t} - \int_{0}^{t} \sin u \, du) = -2t (t \sin t + \cos t - 1)\).3. Use integration by parts twice for \(\int_{0}^{t} u^2 \cos u \, du\), yielding \((u^2 \sin u + 2u \cos u - 2\sin u) \bigg|_{0}^{t}\), which simplifies to \((t^2\sin t + 2t \cos t - 2\sin t)\).
06
Combine the Solutions
Add all integrated parts together to get the final result: \(t^2 \sin t - 2t (t \sin t + \cos t - 1) + (t^2 \sin t + 2t \cos t - 2\sin t)\), which simplifies to \(-t^3 \sin t - 2t + 2t \sin t\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Integration by Parts
Integration by parts is a powerful technique used to solve certain integrals, especially when dealing with products of functions. It's based on the product rule for differentiation and is expressed as:\[ \int u \, dv = uv - \int v \, du \]Here's the strategy: you pick parts of the integrand to be "\(u\)" and "\(dv\)". You then differentiate \(u\) to get \(du\) and integrate \(dv\) to get \(v\).
- Choose \(u\) to be a function that simplifies when differentiated (often a polynomial).
- Choose \(dv\) such that it's easily integrable.
Change of Variables in Integration
Change of variables is a crucial method in integration where we simplify integrals by substituting a new variable. This is especially helpful when the integration is complicated due to the form of the function or limits.In practice, to perform a successful substitution:
- Select a substitution, like \(u = t - \tau\), that reduces complexity.
- Identify the differential relationship: if \(u = t - \tau\), then \(d\tau = -du\).
- Adjust the integration limits if definite: when \(\tau = 0\), \(u = t\); when \(\tau = t\), \(u = 0\).
Convolution Integral
The convolution integral is a mathematical way of combining two functions, providing a measure of how one function alters the shape and domain of another. In detail, the convolution \((f * g)(t)\) of two functions \(f(t)\) and \(g(t)\) is defined as:\[ (f * g)(t) = \int_{-\infty}^{\infty} f(\tau) g(t-\tau) \, d\tau \]This integral is key in systems analysis, used extensively in fields like signal processing and differential equations.
- It explores how one function broadens or contracts the other.
- The practical limits of integration depend on where the functions are non-zero.