The concept of convolution is pretty straightforward but it takes some thinking to understand it clearly. You can think of it this way. Functions that are convolved are combined in a different way using an integral.
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So what does that mean? Let's start backwards, with two functions multipled together in the sdomain. So we have some function \(h(t)\) whose Laplace Transform is the product of two functions. \[ \mathcal{L}\{ h(t) \} = F(s)G(s)\] So, your first inclination is to say that \(h(t)\) is just the product of \(f(t)\) and \(g(t)\) and, although that is a logical conclusion, it is wrong. Remember the product rule from basics derivatives, i.e. \((fg)' = f'g+g'f.\) We can't just take the derivative of each of them and multiply them together, i.e. \((fg)' \neq f'g'.\) Just like the product rule, we can't say that the functions are just multiplied together in the tdomain, i.e. \[ \mathcal{L}\{ f(t)g(t) \} \neq F(s)G(s)\] However, there is a relationship between \(f(t), g(t), F(s), G(s)\) expressed this way. \[\displaystyle{ h(t) = \int_0^t{ f(tx)g(x)~dx } }\] The integral on the right of the equal sign is called the convolution of \(f\) and \(g\) and we write this as \(f \ast g\). So now we can write the Laplace Transform of a pair of convolved functions as \[ \mathcal{L}\{ f(t) \ast g(t) \} = F(s)G(s)\] where \[\displaystyle{ f(t) \ast g(t) = \int_0^t{ f(tx)g(x)~dx } }\]
Now, we have to be careful here, \(f(t) \ast g(t)\) is NOT \(f(t)\) TIMES \(g(t)\) and the \(\ast\) notation does NOT mean composite either. This is a unique new way of combining two functions. The reason it is so important is because of what happens in the sdomain, i.e. \(F(s)G(s)\) IS \(F(s)\) TIMES \(G(s)\) but only in the sdomain, not in the tdomain.
So if we are working in the sdomain and we end up with two functions multipled together, we can use the convolution integral to convert back to the tdomain. We may even be able to evaluate the integral to determine our answer.
Let's watch a quick video clip getting the convolution result.
video by blackpenredpen 

Before we go on, let's put everything together that we have so far. The equations are summarized in the following table.
The Convolution Integral 

\(\displaystyle{ f(t) \ast g(t) = \int_0^t{ f(tx)g(x)~dx } }\) 
The Laplace Transform 
\( \mathcal{L}\{ f(t) \ast g(t) \} = F(s)G(s)\) 
Here is a video that explains this again. It is always good to hear the same concepts from different sources, multiple times.
video by MIT OCW 

Okay, you should be convinced by now that \(f(t)g(t) \neq f(t) \ast g(t)\). However, this next video shows an example that will really help you cement that in your mind.
video by blackpenredpen 

Convolution Properties
Convolution has the following properties, which work as you would expect.
Constant Property (c is a constant) 
\( cf \ast g = f \ast cg = c(f \ast g) \) 
Commutative Property 
\(f \ast g = g \ast f\) 
Associative Property 
\( (f \ast g) \ast h = f \ast (g \ast h) \) 
Distributive Property 
\( f \ast (g+h) = f \ast g + f \ast h \) 
Before we go on, let's do an example. Try it on your own before looking at the solution.
Find \(h(t)\) where \(\displaystyle{ \mathcal{L}\{ h(t) \} = \frac{3}{(s1)(s4)} }\). First, use partial fractions and then use the convolution integral, showing that the answers are the same.
\( h(t) = e^t(e^{3t}1) \)
Find \(h(t)\) where \(\displaystyle{ \mathcal{L}\{ h(t) \} = \frac{3}{(s1)(s4)} }\). First, use partial fractions and then use the convolution integral, showing that the answers are the same.
Solution 

1. Using partial fractions, we get 
\(\displaystyle{ \frac{3}{(s1)(s4)} = \frac{1}{s1} + \frac{1}{s4} }\) 
From the table of Laplace Transforms, we know \(\displaystyle{\mathcal{L}\left\{\frac{1}{sa}\right\} = e^{at}}\) 
So \(\displaystyle{ \mathcal{L}\left\{\frac{1}{s1}\right\} = e^t }\) and \(\displaystyle{ \mathcal{L}\left\{\frac{1}{s4}\right\} = e^{4t} }\) 
Putting the results together, we have \(h(t)=e^{4t}e^{t}=e^t(e^{3t}1)\) 
Final Answer from partial fractions: \(h(t)=e^t(e^{3t}1)\) 
2. To use the convolution integral, we need to assign \(F(s)\) and \(G(s)\). 
Let \(\displaystyle{ F(s)=\frac{3}{s1} }\) and \(\displaystyle{ G(s)=\frac{1}{s4} }\) 
Using the table of Laplace Transforms, we get 
\(f(t)=3e^t\) and \(g(t)=e^{4t}\) 
Our convolution integral is 
\(\displaystyle{ h(t)=f(t)\ast g(t) = \int_0^t{ f(tx)g(x)~dx } }\) 
With our \(f(t)\) and \(g(t)\) we have 
\(\displaystyle{ h(t)=3\int_0^t{ e^{tx}e^{4x}~dx } }\) 
\(\displaystyle{ h(t)=3\int_0^t{ e^t e^{x} e^{4x}~dx } }\) 
\(\displaystyle{ h(t) = 3e^t \int_0^t{ e^{3x}~dx } }\) 
\(\displaystyle{ h(t) = 3e^t \left[ \frac{e^{3x}}{3} \right]_0^t }\) 
\( h(t) = e^t [ e^{3t}  e^0 ] = e^t(e^{3t}1) \) 
Final Answer from convolution: \( h(t) = e^t(e^{3t}1) \) 
The answers are equal. 
Final Answer 

\( h(t) = e^t(e^{3t}1) \) 
Proof
Okay, you should be asking yourself, why does this work? Here is a video with a proof.
video by blackpenredpen 

Graphical Concept of Convolution
To understand what is going on with the convolution integral in a graphical sense, here is a set of videos that explains it and shows several examples.
videos by Michel vanBiezen 

Convolution of Periodic Functions
Here is a video containing a summary of the Laplace Transforms of some periodic functions.
video by Michel vanBiezen 

Here are some practice problems to help you get familiar with the convolution integral. After that, you can use the convolution integral to solve differential equations.
Practice
Unless otherwise instructed, solve these problems using the convolution integral.
Note: Although we use \(x\) on this page as our variable of integration, some of these practice problem solutions use the greek letter tau \(\tau\) for the same purpose.
Basic
Evaluate \( \mathcal{L}\{ \int_0^t{ 3(t\tau)^2 \sin(2\tau)d\tau }\} \)
Problem Statement 

Evaluate \( \mathcal{L}\{ \int_0^t{ 3(t\tau)^2 \sin(2\tau)d\tau }\} \)Evaluate \( \mathcal{L}\{ \int_0^t{ 3(t\tau)^2 \sin(2\tau)d\tau }\} \) using the convolution integral.
Final Answer 

\(\displaystyle{ \frac{12}{s^3(s^2+4)} }\)
Problem Statement 

Evaluate \( \mathcal{L}\{ \int_0^t{ 3(t\tau)^2 \sin(2\tau)d\tau }\} \)Evaluate \( \mathcal{L}\{ \int_0^t{ 3(t\tau)^2 \sin(2\tau)d\tau }\} \) using the convolution integral.
Solution 

video by Gregg Waterman 

Final Answer 

\(\displaystyle{ \frac{12}{s^3(s^2+4)} }\)
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Evaluate \( \mathcal{L}\{ \int_0^t{ \tau e^{2(t\tau)}~d\tau } \} \)
Problem Statement 

Evaluate \( \mathcal{L}\{ \int_0^t{ \tau e^{2(t\tau)}~d\tau } \} \) using the convolution integral.
Final Answer 

\(\displaystyle{ \frac{1}{s^2(s+2)} }\)
Problem Statement 

Evaluate \( \mathcal{L}\{ \int_0^t{ \tau e^{2(t\tau)}~d\tau } \} \) using the convolution integral.
Solution 

video by Gregg Waterman 

Final Answer 

\(\displaystyle{ \frac{1}{s^2(s+2)} }\)
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\(\displaystyle{ \frac{5}{2}\int_0^t{ (t\tau)^2 e^{4\tau} ~ d\tau } }\)
Problem Statement 

\(\displaystyle{ \frac{5}{2}\int_0^t{ (t\tau)^2 e^{4\tau} ~ d\tau } }\)
Solution 

video by Gregg Waterman 

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Write \(\displaystyle{ \mathcal{L}^{1}\left\{ \frac{3s}{(s^2+16)^2} \right\} }\) as a convolution integral.
Problem Statement 

Write \(\displaystyle{ \mathcal{L}^{1}\left\{ \frac{3s}{(s^2+16)^2} \right\} }\) as a convolution integral.
Final Answer 

\(\displaystyle{ \int_0^t{ \cos[4(t\tau)]\sin(4\tau)~d\tau } }\)
Problem Statement 

Write \(\displaystyle{ \mathcal{L}^{1}\left\{ \frac{3s}{(s^2+16)^2} \right\} }\) as a convolution integral.
Solution 

video by Gregg Waterman 

Final Answer 

\(\displaystyle{ \int_0^t{ \cos[4(t\tau)]\sin(4\tau)~d\tau } }\)
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Write \(\displaystyle{ \mathcal{L}^{1}\left\{ \frac{H(s)}{s+2} \right\} }\) as a convolution integral.
Problem Statement 

Write \(\displaystyle{ \mathcal{L}^{1}\left\{ \frac{H(s)}{s+2} \right\} }\) as a convolution integral.
Final Answer 

\(\displaystyle{ \int_0^t{ h(t\tau)e^{2\tau}~d\tau } }\)
Problem Statement 

Write \(\displaystyle{ \mathcal{L}^{1}\left\{ \frac{H(s)}{s+2} \right\} }\) as a convolution integral.
Solution 

video by Gregg Waterman 

Final Answer 

\(\displaystyle{ \int_0^t{ h(t\tau)e^{2\tau}~d\tau } }\)
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For a certain system, the impulse response is \(h(t)=2e^{2t}\) and the source voltage is \(v_s(t)=10e^{t}\). Determine the output response \(v_o(t)=h(t)\ast v_s(t)\).
Problem Statement 

For a certain system, the impulse response is \(h(t)=2e^{2t}\) and the source voltage is \(v_s(t)=10e^{t}\). Determine the output response \(v_o(t)=h(t)\ast v_s(t)\).
Final Answer 

\( v_o(t) = 20[e^{t}e^{2t}] \)
Problem Statement 

For a certain system, the impulse response is \(h(t)=2e^{2t}\) and the source voltage is \(v_s(t)=10e^{t}\). Determine the output response \(v_o(t)=h(t)\ast v_s(t)\).
Solution 

video by Michel vanBiezen 

Final Answer 

\( v_o(t) = 20[e^{t}e^{2t}] \)
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Intermediate
Evaluate \(\displaystyle{ \mathcal{L}^{1}\left\{ \frac{s}{(s^2+1)^2} \right\} }\) using the convolution integral.
Problem Statement 

Evaluate \(\displaystyle{ \mathcal{L}^{1}\left\{ \frac{s}{(s^2+1)^2} \right\} }\) using the convolution integral.
Final Answer 

\(\displaystyle{ \frac{t\sin(t)}{2} }\)
Problem Statement 

Evaluate \(\displaystyle{ \mathcal{L}^{1}\left\{ \frac{s}{(s^2+1)^2} \right\} }\) using the convolution integral.
Solution 

video by blackpenredpen 

Final Answer 

\(\displaystyle{ \frac{t\sin(t)}{2} }\)
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Calculate the Laplace Transform of the periodic function which is a constant pulse of height \(k\) and width \(a\). The period is \(2a\).
Problem Statement 

Calculate the Laplace Transform of the periodic function which is a constant pulse of height \(k\) and width \(a\). The period is \(2a\).
Hint 

The Laplace Transform of a general periodic function \(f(t)\) of period P is \(\displaystyle{ \mathcal{L}[f(t)] = \frac{1}{1e^{sP}}\int_0^P{e^{s\tau}f(\tau)~d\tau} }\), which is derived in practice problem 2257.
Problem Statement 

Calculate the Laplace Transform of the periodic function which is a constant pulse of height \(k\) and width \(a\). The period is \(2a\).
Final Answer 

\(\displaystyle{ \mathcal{L}[f(t)] = \frac{k(1e^{as})}{s(1e^{2as})} }\)
Problem Statement 

Calculate the Laplace Transform of the periodic function which is a constant pulse of height \(k\) and width \(a\). The period is \(2a\).
Hint 

The Laplace Transform of a general periodic function \(f(t)\) of period P is \(\displaystyle{ \mathcal{L}[f(t)] = \frac{1}{1e^{sP}}\int_0^P{e^{s\tau}f(\tau)~d\tau} }\), which is derived in practice problem 2257.
Solution 

video by Michel vanBiezen 

Final Answer 

\(\displaystyle{ \mathcal{L}[f(t)] = \frac{k(1e^{as})}{s(1e^{2as})} }\)
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Advanced
Find the Laplace Transform of a periodic function \(f(t)\) with period \(P\), i.e. \(f(t+P)=f(t)\).
Problem Statement 

Find the Laplace Transform of a periodic function \(f(t)\) with period \(P\), i.e. \(f(t+P)=f(t)\).
Hint 

Your answer will contain a definite integral since you are not given a specific function \(f(t)\).
Problem Statement 

Find the Laplace Transform of a periodic function \(f(t)\) with period \(P\), i.e. \(f(t+P)=f(t)\).
Final Answer 

\(\displaystyle{ \mathcal{L}[f(t)] = \frac{1}{1e^{sP}}\int_0^P{ e^{s\tau} f(\tau) ~d\tau } }\)
Problem Statement 

Find the Laplace Transform of a periodic function \(f(t)\) with period \(P\), i.e. \(f(t+P)=f(t)\).
Hint 

Your answer will contain a definite integral since you are not given a specific function \(f(t)\).
Solution 

video by Michel vanBiezen 

Final Answer 

\(\displaystyle{ \mathcal{L}[f(t)] = \frac{1}{1e^{sP}}\int_0^P{ e^{s\tau} f(\tau) ~d\tau } }\)
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You CAN Ace Calculus
external links you may find helpful 

Paul's Online Math Notes  Differential Equations  Convolution 
The Unit Circle
The Unit Circle [wikipedia]
Basic Trig Identities
Set 1  basic identities  

\(\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }\) 
\(\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }\) 
\(\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }\) 
\(\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }\) 
Set 2  squared identities  

\( \sin^2t + \cos^2t = 1\) 
\( 1 + \tan^2t = \sec^2t\) 
\( 1 + \cot^2t = \csc^2t\) 
Set 3  doubleangle formulas  

\( \sin(2t) = 2\sin(t)\cos(t)\) 
\(\displaystyle{ \cos(2t) = \cos^2(t)  \sin^2(t) }\) 
Set 4  halfangle formulas  

\(\displaystyle{ \sin^2(t) = \frac{1\cos(2t)}{2} }\) 
\(\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }\) 
Trig Derivatives
\(\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }\) 
\(\displaystyle{ \frac{d[\cos(t)]}{dt} = \sin(t) }\)  
\(\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }\) 
\(\displaystyle{ \frac{d[\cot(t)]}{dt} = \csc^2(t) }\)  
\(\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }\) 
\(\displaystyle{ \frac{d[\csc(t)]}{dt} = \csc(t)\cot(t) }\) 
Inverse Trig Derivatives
\(\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1t^2}} }\) 
\(\displaystyle{ \frac{d[\arccos(t)]}{dt} = \frac{1}{\sqrt{1t^2}} }\)  
\(\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }\) 
\(\displaystyle{ \frac{d[\arccot(t)]}{dt} = \frac{1}{1+t^2} }\)  
\(\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 1}} }\) 
\(\displaystyle{ \frac{d[\arccsc(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 1}} }\) 
Trig Integrals
\(\int{\sin(x)~dx} = \cos(x)+C\) 
\(\int{\cos(x)~dx} = \sin(x)+C\)  
\(\int{\tan(x)~dx} = \ln\abs{\cos(x)}+C\) 
\(\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C\)  
\(\int{\sec(x)~dx} = \) \( \ln\abs{\sec(x)+\tan(x)}+C\) 
\(\int{\csc(x)~dx} = \) \( \ln\abs{\csc(x)+\cot(x)}+C\) 
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Practice Instructions
Unless otherwise instructed, solve these problems using the convolution integral.
Note: Although we use \(x\) on this page as our variable of integration, some of these practice problem solutions use the greek letter tau \(\tau\) for the same purpose.