Convolution

Ch 6.6:  The Convolution Integral   Sometimes it is possible to write a Laplace transform H(s) as H(s) = F(s)G(s), where F(s) and G(s) are the transforms of known functions f and g, respectively. In this case we might expect H(s) to be the transform of the product of f and g. That is, does H(s) = F(s)G(s) = L{f }L{g} = L{f g}? On the next slide we give an example that shows that this equality does not hold, and hence the Laplace transform cannot in general be commuted with ordinary multiplication. In this section we examine the convolution of f and g, which can be viewed as a generalized product, and one for which the Laplace transform does commute.

Example 1   Let f (t) = 1 and g(t) = sin(t). Recall that the Laplace Transforms of f and g are 1 1 L{ f (t )} = L{ 1 } = , L{ g (t ) } = L{ sin t } = 2 s s +1

Thus L{ f (t ) g (t )} = L{ sin t } =

and

1 s2 +1

1 L{ f (t ) } L{ g (t ) } = 2 s s +1

(

)

Therefore for these functions it follows that L{ f (t ) g (t )} ≠ L{ f (t ) } L{ g (t ) }

Theorem 6.6.1 Suppose F(s) = L{f (t)} and G(s) = L{g(t)} both exist for s > a ≥ 0. Then H(s) = F(s)G(s) = L{h(t)} for s > a, where t

t

0

0

h(t ) = ∫ f (t − τ ) g (τ )dτ = ∫ f (t ) g (t − τ )dτ

The function h(t) is known as the convolution of f and g and the integrals above are known as convolution integrals. Note that the equality of the two convolution integrals can be seen by making the substitution u = t - τ. The convolution integral defines a “generalized product” and can be written as h(t) = ( f *g)(t). See text for more details.

Theorem 6.6.1 Proof Outline ∞

F ( s )G ( s ) = ∫ e

− su

0

∞

f (u )du ∫ e − sτ g (τ )dτ 0

∞

∞

0

0

= ∫ g (τ )dτ ∫ e − s (τ +u ) f (u )du

∫

∞

=∫

∞

=∫

∞

=

0

0

0

∞

g (τ )dτ ∫ e − st f (t − τ )dt τ

∞

∫τ

e − st g (τ ) f (t − τ )dtdτ

t

e − st f (t − τ )g (τ )dτ dt

∫

0

t  = ∫ e ∫ f (t − τ ) g (τ )dτ  dt  0  0 = L{ h(t )} ∞

− st

(t = τ + u )

Example 2 Find the Laplace Transform of the function h given below. t

h(t ) = ∫ (t − τ ) sin 2τdτ 0

Solution: Note that f (t) = t and g(t) = sin2t, with 1 F ( s ) = L{ f (t )} = L{t} = 2 s 2 G ( s ) = L{g (t )} = L{sin 2t} = 2 s +4 Thus by Theorem 6.6.1, L{ h(t )} = H ( s) = F ( s )G ( s ) =

(

2

s2 s2 + 4

)

Example 3: Find Inverse Transform

(1 of 2)

Find the inverse Laplace Transform of H(s), given below. H (s) =

2 s 2 ( s − 2)

Solution: Let F(s) = 2/s2 and G(s) = 1/(s - 2), with f (t ) = L−1{ F ( s)} = 2t

g (t ) = L−1{ G ( s)} = e 2t

Thus by Theorem 6.6.1, L { H ( s )} = h(t ) = 2 ∫ (t − τ )e 2τ dτ −1

t

0

Example 3: Solution h(t)

(2 of 2)

We can integrate to simplify h(t), as follows. t

t

t

h(t ) = 2 ∫ (t − τ )e dτ = 2 t ∫ e dτ − 2∫ τ e 2τ dτ 2τ

0

2τ

0

0

t 2τ t  = te − τe − ∫ e 2τ dτ  0 0   0 1   = t e 2t − 1 − t e 2t − e 2t − 1  2   1 1 = te 2t − t − t e 2t + e 2t − 2 2 1 1 = e 2t − t − 2 2 2τ t

[

]

[

]

Example 4: Initial Value Problem

(1 of 4)

Find the solution to the initial value problem y′′ + 4 y = g (t ), y (0) = 3, y′(0) = −1

Solution: L{ y′′} + 4 L{ y} = L{g (t )}

or

[ s L{ y} − sy(0) − y′(0)] + 4L{ y} = G(s) 2

Letting Y(s) = L{y}, and substituting in initial conditions,

(s

Thus

2

+ 4 ) Y ( s ) = 3s − 1 + G ( s )

Y (s) =

3s − 1 G ( s ) + 2 2 s +4 s +4

Example 4: Solution

(2 of 4)

We have 3s − 1 G ( s ) Y (s) = 2 + 2 s +4 s +4  s  1 2  1 2  = 3 2 −  2 +  2 G ( s)    s + 4 2 s + 4 2 s + 4

Thus 1 1 t y (t ) = 3 cos 2t − sin 2t + ∫ sin 2(t − τ ) g (τ )dτ 2 2 0

Note that if g(t) is given, then the convolution integral can be evaluated.

y′′ + 4 y = g (t ), y (0) = 3, y′(0) = −1

Example 4: Laplace Transform of Solution

(3 of 4)

Recall that the Laplace Transform of the solution y is Y (s) =

3s − 1 G ( s ) + 2 = Φ(s) + Ψ (s) 2 s +4 s +4

Note Φ (s) depends only on system coefficients and initial conditions, while Ψ (s) depends only on system coefficients and forcing function g(t). Further, φ(t) = L-1{Φ (s)} solves the homogeneous IVP y′′ + 4 y = 0, y (0) = 3, y′(0) = −1 while ψ(t) = L-1{Ψ (s)} solves the nonhomogeneous IVP y′′ + 4 y = g (t ), y (0) = 0, y′(0) = 0

Example 4: Transfer Function

(4 of 4)

Examining Ψ (s) more closely, Ψ ( s) =

G (s) 1 = H ( s ) G ( s ), where H ( s ) = s2 + 4 s2 + 4

The function H(s) is known as the transfer function, and depends only on system coefficients. The function G(s) depends only on external excitation g(t) applied to system. If G(s) = 1, then g(t) = δ(t) and hence h(t) = L-1{H(s)} solves the nonhomogeneous initial value problem y′′ + 4 y = δ (t ), y (0) = 0, y′(0) = 0 Thus h(t) is response of system to unit impulse applied at t = 0, and hence h(t) is called the impulse response of system.

Input-Output Problem

(1 of 3)

Consider the general initial value problem ay′′ + by′ + cy = g (t ), y (0) = y0 , y′(0) = y0′

This IVP is often called an input-output problem. The coefficients a, b, c describe properties of physical system, and g(t) is the input to system. The values y0 and y0' describe initial state, and solution y is the output at time t. Using the Laplace transform, we obtain

[

]

a s 2Y ( s ) − sy (0) − y′(0) + b[ sY ( s ) − y (0)] + cY ( s ) = G ( s )

or

(as + b) y0 + ay0′ G ( s) Y (s) = + = Φ(s) + Ψ (s) 2 2 as + bs + c as + bs + c

ay′′ + by′ + cy = g (t ), y (0) = y0 , y′(0) = y0′

Laplace Transform of Solution

(2 of 3)

We have Y (s) =

(as + b) y0 + ay0′ G ( s) + = Φ(s) + Ψ (s) 2 2 as + bs + c as + bs + c

As before, Φ (s) depends only on system coefficients and initial conditions, while Ψ (s) depends only on system coefficients and forcing function g(t). Further, φ(t) = L-1{Φ (s)} solves the homogeneous IVP ay′′ + by′ + cy = 0, y (0) = y0 , y′(0) = y0′

while ψ(t) = L-1{Ψ (s)} solves the nonhomogeneous IVP ay′′ + by′ + cy = g (t ), y (0) = 0, y′(0) = 0

Transfer Function

(3 of 3)

Examining Ψ (s) more closely, G(s) 1 Ψ (s) = 2 = H ( s )G ( s ), where H ( s ) = 2 as + bs + c as + bs + c

As before, H(s) is the transfer function, and depends only on system coefficients, while G(s) depends only on external excitation g(t) applied to system. Thus if G(s) = 1, then g(t) = δ(t) and hence h(t) = L-1{H(s)} solves the nonhomogeneous IVP ay′′ + by′ + cy = δ (t ), y (0) = 0, y′(0) = 0 Thus h(t) is response of system to unit impulse applied at t = 0, and hence h(t) is called the impulse response of system, with ψ (t ) = L { H ( s )G ( s )} = ∫ h(t − τ ) g (τ )dτ −1

t

0