## 17Calculus - Laplace Transform Convolution

The concept of convolution is pretty straight-forward 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.

So what does that mean? Let's start backwards, with two functions multipled together in the s-domain. 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 t-domain, 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. $\displaystyle{ h(t) = \int_0^t{ f(t-x)g(x)~dx } }$ 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 $\displaystyle{ f(t) \ast g(t) = \int_0^t{ f(t-x)g(x)~dx } }$

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 s-domain, i.e. $$F(s)G(s)$$ IS $$F(s)$$ TIMES $$G(s)$$ but only in the s-domain, not in the t-domain.

So if we are working in the s-domain and we end up with two functions multipled together, we can use the convolution integral to convert back to the t-domain. We may even be able to evaluate the integral to determine our answer.

Let's watch a quick video clip getting the convolution result.

### blackpenredpen - general convolution [~15mins]

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

$$\displaystyle{ f(t) \ast g(t) = \int_0^t{ f(t-x)g(x)~dx } }$$

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.

### MIT OCW - Laplace Transforms and Convolution [10mins-28secs]

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.

### blackpenredpen - laplace transform of $$\sin(t)\cos(t)$$ vs laplace transform $$\sin(t) \ast \cos(t)$$ [10mins-28secs]

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 $$\displaystyle{ \mathcal{L}\{ h(t) \} = \frac{3}{(s-1)(s-4)} }$$. 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 $$\displaystyle{ \mathcal{L}\{ h(t) \} = \frac{3}{(s-1)(s-4)} }$$. First, use partial fractions and then use the convolution integral, showing that the answers are the same.

Solution

 1. Using partial fractions, we get $$\displaystyle{ \frac{3}{(s-1)(s-4)} = \frac{-1}{s-1} + \frac{1}{s-4} }$$ From the table of Laplace Transforms, we know $$\displaystyle{\mathcal{L}\left\{\frac{1}{s-a}\right\} = e^{at}}$$ So $$\displaystyle{ \mathcal{L}\left\{\frac{-1}{s-1}\right\} = -e^t }$$ and $$\displaystyle{ \mathcal{L}\left\{\frac{1}{s-4}\right\} = e^{4t} }$$ 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 $$\displaystyle{ F(s)=\frac{3}{s-1} }$$ and $$\displaystyle{ G(s)=\frac{1}{s-4} }$$ Using the table of Laplace Transforms, we get $$f(t)=3e^t$$ and $$g(t)=e^{4t}$$ Our convolution integral is $$\displaystyle{ h(t)=f(t)\ast g(t) = \int_0^t{ f(t-x)g(x)~dx } }$$ With our $$f(t)$$ and $$g(t)$$ we have $$\displaystyle{ h(t)=3\int_0^t{ e^{t-x}e^{4x}~dx } }$$ $$\displaystyle{ h(t)=3\int_0^t{ e^t e^{-x} e^{4x}~dx } }$$ $$\displaystyle{ h(t) = 3e^t \int_0^t{ e^{3x}~dx } }$$ $$\displaystyle{ h(t) = 3e^t \left[ \frac{e^{3x}}{3} \right]_0^t }$$ $$h(t) = e^t [ e^{3t} - e^0 ] = e^t(e^{3t}-1)$$ Final Answer from convolution: $$h(t) = e^t(e^{3t}-1)$$ The answers are equal.

$$h(t) = e^t(e^{3t}-1)$$

Proof

Okay, you should be asking yourself, why does this work? Here is a video with a proof.

### blackpenredpen - Proof of the Convolution Theorem [18mins-9secs]

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.

### Graphical Concept of Convolution Videos

videos by Michel vanBiezen

Convolution of Periodic Functions

Here is a video containing a summary of the Laplace Transforms of some periodic functions.

### Michel vanBiezen - Laplace Transform of Periodic Functions Summary [3mins-38secs]

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.

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

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

### 2236 video

video by Gregg Waterman

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

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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.

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

Problem Statement

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

Solution

### 2237 video

video by Gregg Waterman

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

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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

### 2238 video

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.

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

Problem Statement

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

Solution

### 2239 video

video by Gregg Waterman

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

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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.

$$\displaystyle{ \int_0^t{ h(t-\tau)e^{-2\tau}~d\tau } }$$

Problem Statement

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

Solution

### 2240 video

video by Gregg Waterman

$$\displaystyle{ \int_0^t{ h(t-\tau)e^{-2\tau}~d\tau } }$$

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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 $$v_o(t)=h(t)\ast v_s(t)$$.

$$v_o(t) = 20[e^{-t}-e^{-2t}]$$

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

### 2260 video

video by Michel vanBiezen

$$v_o(t) = 20[e^{-t}-e^{-2t}]$$

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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.

$$\displaystyle{ \frac{t\sin(t)}{2} }$$

Problem Statement

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

Solution

### 2241 video

video by blackpenredpen

$$\displaystyle{ \frac{t\sin(t)}{2} }$$

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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 $$\displaystyle{ \mathcal{L}[f(t)] = \frac{1}{1-e^{-sP}}\int_0^P{e^{-s\tau}f(\tau)~d\tau} }$$, 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 $$2a$$.

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

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 $$\displaystyle{ \mathcal{L}[f(t)] = \frac{1}{1-e^{-sP}}\int_0^P{e^{-s\tau}f(\tau)~d\tau} }$$, which is derived in practice problem 2257.

Solution

### 2258 video

video by Michel vanBiezen

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

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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. $$f(t+P)=f(t)$$.

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

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

### 3548 video

video by Michel vanBiezen

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

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You CAN Ace Calculus

### Trig Formulas

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 - double-angle formulas

$$\sin(2t) = 2\sin(t)\cos(t)$$

$$\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }$$

Set 4 - half-angle 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

 $$\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1-t^2}} }$$ $$\displaystyle{ \frac{d[\arccos(t)]}{dt} = -\frac{1}{\sqrt{1-t^2}} }$$ $$\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }$$ $$\displaystyle{ \frac{d[\arccot(t)]}{dt} = -\frac{1}{1+t^2} }$$ $$\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 -1}} }$$ $$\displaystyle{ \frac{d[\arccsc(t)]}{dt} = -\frac{1}{\abs{t}\sqrt{t^2 -1}} }$$

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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