Exercise · ZW-CONVOLUTION-004 · ★☆☆ Basic
If \(F=\mathcal L\{f\}\) and \(G=\mathcal L\{g\}\), what is \(\mathcal L\{f*g\}\)?
Choose one option.
Show answer
- C. \(F(s)\,G(s)\)
Explanation
The convolution theorem says the transform of the convolution is the product of the transforms; the transform of the ordinary product \(fg\) is not.
Review the concept: Convolution
The convolution \(f*g\) of two functions \(f\) and \(g\), defined for \(t\ge0\), is the function of \(t\) that integrates from \(0\) to \(t\) the product of \(f(\tau)\) and a reflected, shifted copy of \(g\).
\[(f*g)(t)=\int_0^t f(\tau)\,g(t-\tau)\,d\tau\]
More practice problems: convolution
- ZW-CONVOLUTION-001 What is (1*1)(t)?
- ZW-CONVOLUTION-002 What is (t*1)(t)?
- ZW-CONVOLUTION-003 What is (e^t*1)(t)?
- ZW-CONVOLUTION-005 Use the convolution theorem: what is mathcal L\1*t\?
- ZW-CONVOLUTION-006 What is mathcal L^-1\(1)/(s(s+1))\, viewed as 1*e^-t?
- ZW-CONVOLUTION-007 What is mathcal L^-1\(1)/(s^2(s-1))\?
- ZW-CONVOLUTION-008 What is (sin t*1)(t)?
- ZW-CONVOLUTION-009 Select the true statement about convolution.