Math Problem Statement

  1. Compute the Convolution of 𝑓 ( 𝑑 ) = 𝑒 βˆ’ 𝑑 f(t)=e βˆ’t and 𝑔 ( 𝑑 ) = 𝑑 g(t)=t.

Solution

To compute the convolution of the functions f(t)=eβˆ’tf(t) = e^{-t} and g(t)=tg(t) = t, we use the formula for the convolution of two functions f(t)f(t) and g(t)g(t) defined as:

(fβˆ—g)(t)=∫0tf(Ο„)g(tβˆ’Ο„) dΟ„(f * g)(t) = \int_0^t f(\tau) g(t - \tau) \, d\tau

Step-by-Step Computation

We have:

  • f(t)=eβˆ’tf(t) = e^{-t}
  • g(t)=tg(t) = t

The convolution integral becomes:

(fβˆ—g)(t)=∫0teβˆ’Ο„(tβˆ’Ο„) dΟ„(f * g)(t) = \int_0^t e^{-\tau} (t - \tau) \, d\tau

Now, expand (tβˆ’Ο„)(t - \tau) inside the integral:

(fβˆ—g)(t)=∫0teβˆ’Ο„t dΟ„βˆ’βˆ«0teβˆ’Ο„Ο„β€‰dΟ„(f * g)(t) = \int_0^t e^{-\tau} t \, d\tau - \int_0^t e^{-\tau} \tau \, d\tau

We will now compute each term individually.

1. First Integral: ∫0teβˆ’Ο„t dΟ„\int_0^t e^{-\tau} t \, d\tau

Since tt is a constant with respect to Ο„\tau, we can factor it out:

t∫0teβˆ’Ο„β€‰dΟ„t \int_0^t e^{-\tau} \, d\tau

The integral of eβˆ’Ο„e^{-\tau} is:

∫eβˆ’Ο„β€‰dΟ„=βˆ’eβˆ’Ο„\int e^{-\tau} \, d\tau = -e^{-\tau}

Evaluating from 0 to tt:

t[βˆ’eβˆ’Ο„]0t=t(βˆ’eβˆ’t+e0)=t(1βˆ’eβˆ’t)t \left[-e^{-\tau}\right]_0^t = t \left( -e^{-t} + e^{0} \right) = t \left( 1 - e^{-t} \right)

2. Second Integral: ∫0teβˆ’Ο„Ο„β€‰dΟ„\int_0^t e^{-\tau} \tau \, d\tau

This is a standard integral, and we can solve it using integration by parts.

Let:

  • u=Ο„u = \tau and dv=eβˆ’Ο„dΟ„dv = e^{-\tau} d\tau
  • du=dΟ„du = d\tau and v=βˆ’eβˆ’Ο„v = -e^{-\tau}

Using the integration by parts formula ∫udv=uvβˆ’βˆ«vdu\int u dv = uv - \int v du:

∫0teβˆ’Ο„Ο„β€‰dΟ„=[βˆ’Ο„eβˆ’Ο„]0t+∫0teβˆ’Ο„β€‰dΟ„\int_0^t e^{-\tau} \tau \, d\tau = \left[ -\tau e^{-\tau} \right]_0^t + \int_0^t e^{-\tau} \, d\tau

First, evaluate βˆ’Ο„eβˆ’Ο„-\tau e^{-\tau} from 0 to tt:

βˆ’Ο„eβˆ’Ο„βˆ£0t=βˆ’teβˆ’t+0=βˆ’teβˆ’t-\tau e^{-\tau} \Big|_0^t = -t e^{-t} + 0 = -t e^{-t}

Now, compute the remaining integral:

∫0teβˆ’Ο„β€‰dΟ„=[βˆ’eβˆ’Ο„]0t=1βˆ’eβˆ’t\int_0^t e^{-\tau} \, d\tau = \left[-e^{-\tau} \right]_0^t = 1 - e^{-t}

So, the second integral is:

βˆ’teβˆ’t+(1βˆ’eβˆ’t)-t e^{-t} + (1 - e^{-t})

Combine the results

Now, combining the results of both integrals:

(fβˆ—g)(t)=t(1βˆ’eβˆ’t)βˆ’(βˆ’teβˆ’t+1βˆ’eβˆ’t)(f * g)(t) = t(1 - e^{-t}) - \left( -t e^{-t} + 1 - e^{-t} \right)

Simplifying:

(fβˆ—g)(t)=t(1βˆ’eβˆ’t)+teβˆ’tβˆ’1+eβˆ’t(f * g)(t) = t(1 - e^{-t}) + t e^{-t} - 1 + e^{-t}

(fβˆ—g)(t)=tβˆ’teβˆ’t+teβˆ’tβˆ’1+eβˆ’t(f * g)(t) = t - te^{-t} + t e^{-t} - 1 + e^{-t}

The terms βˆ’teβˆ’t- te^{-t} and teβˆ’tt e^{-t} cancel out, leaving:

(fβˆ—g)(t)=tβˆ’1+eβˆ’t(f * g)(t) = t - 1 + e^{-t}

Final Result

(fβˆ—g)(t)=tβˆ’1+eβˆ’t(f * g)(t) = t - 1 + e^{-t}


Would you like more details on any part of the process? Or do you have any questions?

Here are some related questions:

  1. How would you compute the convolution for different types of functions, like polynomials or sinusoidal functions?
  2. What is the role of the convolution integral in signal processing?
  3. How do you compute the inverse Laplace transform of a convolution?
  4. Can we compute the convolution of two functions without using integration?
  5. What is the relationship between convolution and the Fourier transform?

Tip: When performing convolution, always check if you can simplify the integral by factoring constants out of the integral before evaluating it.

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Math Problem Analysis

Mathematical Concepts

Convolution
Integration
Exponential Functions
Integration by Parts

Formulas

Convolution formula: (f * g)(t) = ∫_0^t f(Ο„) g(t - Ο„) dΟ„
Integration by parts formula: ∫ u dv = uv - ∫ v du

Theorems

Convolution Theorem
Integration by Parts Theorem

Suitable Grade Level

Grades 11-12