Convolution
Today’s Objectives:
Students will be able to:
a) Use convolution to solve
transient SDOF problems
Jean Marie Constant Duhamel (1797-1872)
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The response of a SDOF system to a unit impulse at
time 0. (Approximately)
Impulse represented by narrow pulse
Response of System
This is called the “impulse response function” and is given by:
x(t ) =
1
m ωd
e −ζ ωn t sin (ωd t ) = g ( t )
We can scale this if need be. The response to an impulse F at time zero.
x(t ) =
F − ζ ωn t
e
sin (ωd t ) = F g ( t )
m ωd
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The response to an impulse F at time t > 0.
x(t ) =
F −ζ ωn ( t −τ )
e
sin (ωd t − τ ) = F g ( t − τ )
m ωd
Response is same – just shifted in time!
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Now we use the principle of superposition
Here is a set of three impulses.
Each has its own response.
The total response is equal to
x ( t ) = F1 g ( t ) + F2 g ( t − τ 2 ) + F1 g ( t − τ 3 )
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Duhamel or Convolution Integral
The response of a SDOF system to a sequence of impulses is just a similar sum.
x(t ) =
∑ Fi g ( t − τ i )
i
By subdividing finely enough, we can model the arbitrary function as a sum of
impulses, and by examining the limit of the sum as the width of the impulses shrinks to
zero, we get an integral.
t
∫0
F (τ ) g ( t − τ ) dτ =
1
m ωd
t
∫0
F (τ ) e
−ζ ωn ( t −τ )
F(t)
x(t ) =
t
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sin (ωd ( t − τ ) ) dτ
Base Motion Excitation.
The problem we are solving looks like (remember z = x - y = relative motion)
m &&
z + c z& + k z = − m &&
y
In other words, the forcing function, F, is replaced by − m &&
y
Consequently, our convolution integral is now
z (t ) = −
1
t
ωd
∫0
−ζ ω t −τ
&&
y (τ ) e n ( ) sin (ωd ( t − τ ) ) dτ
So if the base acceleration is specified, we can evaluate this in much the same way that
we did with the specified force load. Just don’t forget the minus sign and the removal
of that “m” in the denominator out front.
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Online demonstration of convolution
http://www.jhu.edu/~signals/convolve/index.html
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Example
A SDOF system has the parameters m = 100 kg, c = 50 N-s/m and k = 1200 N/m. The
system is subjected to the chain of three pulses shown. Calculate the maximum value of
the response. Determine the time at which this maximum occurred.
Let’s look at a Maple solution and a Matlab solution
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Maple Solution to example
Click Here for Maple
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Matlab results
Function to define G
Main routine
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Function to
define F
Comparison of results
Maple results
Matlab results
Response
0.3
x (m)
0.2
0.1
0
-0.1
0
1
2
3
4
time (s)
5
6
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7
8
In-class exercise
The water tank shown has a mass of 2400 lb. It is mounted on a 20 ft high beamcolumn whose bending stiffness, EI is 500 x 106 lb-in2. The base experiences a shock
type acceleration over a duration of 2.0 sec which varies as shown: Assume ζ = 0.04
0.3 g
2.0 sec
-0.3 g
Using convolution in either Matlab or Maple, calculate the maximum
displacement. Use a Simulink model to check these results.
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