Differential Equations
Solving for Impulse Response
Linear systems are often described using differential
equations. For example:
dy
d2y
+
5
+ 6y = f (t)
dt2
dt
where f(t) is the input to the system and y(t) is the
output.
We cannot solve for the impulse response directly so
we solve for the step response and then differentiate
it to get the impulse response.
Differential
Equation
solve
We know how to solve for y given a specific input f .
Step response
We now cover an alternative approach:
differentiate
Differential
Equation
solve
Impulse response
Impulse response
Any input
Any input
convolution
convolution
Corresponding
Output
Corresponding
Output
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Motivation: Convolution
Solving for Step Response
If we know the response of a linear system to a step
input, we can calculate the impulse response and hence
we can find the response to any input by convolution.
If we want to find the step response of
dy
+ 5y = f (t)
dt
where f is the input and y is the output. It would be
nice if we could put f (t) = H(t) and solve. Unfortunately we don’t know of a way to do this directly. So
we
Suppose we want to know how a car’s suspension responds to lots of different types of road surface.
We measure how the suspension responds to a step
input (or calculate the step response from a theoretical model of the system).
We can then find the impulse response and use convolution to find the car’s behaviour for any road surface profile.
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1. set f (t) = 1, and solve for just t ≥ 0
2. set the boundary condition y(0) = 0 (also ẏ(0) =
0 for second order equations) to imply that f (t)
was zero for all t < 0.
We thus have a complete solution because y = 0 for
t < 0, and we have found y for all t ≥ 0.
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Boundary Condition Justification
Step Response Example
Prove that y = 0 at t = 0 by contradiction.
We know that y(t) = 0 for all t < 0. Therefore the
only way for y to equal something other than zero at
t = 0 is if there is a step discontinuity in y at t = 0 .
Assume that y has a step of height h at t = 0 . If y
has a step discontinuity at t = 0 then dy
dt must have
a delta function at t = 0.
Step 1: set f (t) = 1, and solve for just t ≥ 0.
dy
+ 5y = 1
dt
Complimentary function: ẏ + 5y = 0 ⇒ y = Ae−5t
So we have:
• f (t) is a step function so |f (t)| ≤ 1 for all t.
• |y| ≤ h at t = 0.
1
Particular Integral: try y = λ (a const) ⇒ y = 5
1
General Solution: y = Ae−5t + 5
• dy
dt → ∞ at t = 0.
Step 2: set the boundary condition y = 0 at t = 0
Which violates the original equation at t = 0.
dy
= f (t) − 5y
dt
As the RHS is finite but the LHS is infinite. Therefore
y must be continuous at t = 0, and we can use the
initial condition y(0) = 0.
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1 = 0 ⇒ A = −1
y(0) = 0 ⇒ A + 5
5
−5t for t ≥ 0.
So step response is y(t) = 1
5 1−e
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Find the Impulse Response
dy
d2y
+
13
+ 12y = f (t)
dt2
dt
Step → Impulse Response
Impulse
Response
g(t)
integrate
differentiate
1. Find the General Solution with f (t) = 1
Step
Response
Complimentary function is y = Ae−12t + Be−t
1
Particular integral is y = 12
1 + Ae−12t + Be−t
General solution is y = 12
1
−5t
Step response is y(t) = 5 1 − e
for t ≥ 0.
2. Set boundary conditions y(0) = ẏ(0) = 0 to get
the step response.
1
12 + A + B = 0
Impulse response g(t) is given by:
−12A − B = 0
1 and B = − 1
⇒ A = 132
11


0, t < 0



g(t) =

1
d

−5t

= e−5t, t ≥ 0
1−e

−12t
−t
e
1 +e
Thus Step Response is y = 12
132 − 11
dt 5
3. Differentiate the step response to get the impulse
response.
g(t) =
23
e−t − e−12t
dy
=
dt
11
, (t > 0)
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Using the Impulse Response
If we have a system input composed of impulses,
f (t) = 3δ(t − 1) + 4δ(t − 2)
we can find the corresponding system output using
superposition.
More General Input
Suppose our input is composed of lots of delta functions:
f (t) =
n
y(t) = 3g(t − 1) + 4g(t − 2)

= 3
e−(t−1) − e−12(t−1) 
11
+4 
e−(t−2) − e−12(t−2)
11
pn δ(t − qn )
Then the corresponding system output will be


X
y(t) =
X
n

pn g(t − qn )

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Section 2: Summary
Differential Equation
ay + by + cy + d = f(t)
Section 3
solve
ay + by + cy + d = 1
with boundary conditions
y(0) = 0 and y(0) = 0
Convolution
Step response
In this section we derive the convolution integral and
show its use in some examples.
differentiate
Impulse response
Any input
convolution
Corresponding
Output
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Convolution
Our goal is to calculate the output, y(t) of a linear system using the input, f (t), and the impulse response
of the system, g(t).
An impulse at time t = 0 produces the impulse response.
δ (t)
k δ(t)
t
t
An impulse delayed to time t = τ produces a delayed
impulse response starting at time τ .
δ(t−τ ) Linear
System
g(t−τ )
τ
τ
Linear
System
k g(t)
t
g(t)
Linear
System
t
A scaled impulse at time t = 0 produces a scaled
impulse response.
t
An impulse that has been scaled by k and delayed to
time t = τ produces an impulse response scaled by
k and starting at time τ .
k δ(t−τ ) Linear
System
τ
t
k g(t−τ )
τ
t
t
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Consider the input, f (t) to be made up of a sequence
of strips of width ∆τ . Each of these strips is similar
to a delta function and thus leads to a system output of an appropriately scaled and delayed impulse
response.
δ (t−τ ) f(τ ) ∆ τ
f(t)
leads to response
g(t−τ ) f(τ ) ∆τ
y(t) =
−∞
g(t − τ )f (τ )dτ
• Treat t as a constant when evaluating the integral.
The integration variable is τ .
∆τ
t
τ
The response of the system, y(t) is thus the sum of
these delayed, scaled impulse responses. (Provided
g(t) = 0 for t < 0.)
y(t) ≈
Z t
X
g(t − τ )f (τ )∆τ
• t is time as it relates to the output of the system
y(t).
• τ is time as it relates to the input of the system
f (τ ).
All
slices
Let the width of the slices tend to zero. The sum turns
into an integral called the convolution integral.
y(t) =
Z t
−∞
g(t − τ )f (τ )dτ
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Convolution Example 2
Convolution Example 1
For the same system (g(t) = e−5t , t ≥ 0), find the
output for input
Consider a system with impulse response
g(t) =
(
0
,t<0
−5t
e
,t≥0
f (t) =
Find the output for input f (t) = H(t) (step function).


 0, t < 0
v, 0 < t < k


0, t > k
f(t)
v
k
t
Using the convolution integral, the answer is given by
y(t) =
=
Z t
−∞
Z t
g(t − τ )f (τ )dτ
y(t) =
e−5(t−τ )H(τ )dτ
−∞
Z t
e−5(t−τ )dτ
=
0
1 −5(t−τ ) t
=
e
5
0
1
=
1 − e−5t
5
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Z t
g(t − τ )f (τ )dτ
−∞
 R
t

g(t − τ ) × 0 dτ,
t<0


−∞





R0




−∞ Rg(t − τ ) × 0 dτ



+ 0t g(t − τ ) v dτ,
0<t<k
=



R0




−∞ Rg(t − τ ) × 0 dτ




+ 0k g(t − τ ) v dτ


R


+ kt g(t − τ ) × 0 dτ, t > k
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Case (a): t < 0
Rt
−∞ g(t − τ ) × 0 dτ = 0 so y(t) = 0 for all t < 0.
Convolution Example 3
Case (b): 0 < t < k
y(t) =
Z t
0
g(t − τ ) v dτ =
=
Z t
0
vh
e−5(t−τ ) v dτ
e−5(t−τ )
it
0
5
v
=
1 − e−5t
5
For the same system (g(t) = e−5t , t ≥ 0), find the
output for input
f (t) =
(
0,
t<0
sin(ωt), t > 0
f(t)
t
Case (c): t > k
y(t) =
Z k
0
g(t − τ ) v dτ =
=
=
Z k
0
vh
−5(t−τ )
e
Using the convolution integral, the answer is given by
v dτ
i
−5(t−τ ) k
e
0
5
v 5k
e − 1 e−5t
5
y(t)
(a)
(b)
k (c)
y(t) =
Z t
g(t − τ )f (τ )dτ
−∞
 R
t

t<0

−∞ g(t − τ ) × 0 dτ,



R0
=


−∞ Rg(t − τ ) × 0 dτ



+ 0t g(t − τ ) sin(ωτ ) dτ, 0 < t
t
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Convolution Summary
Differential Equation
ay + by + cy + d = f(t)
Case (a): t < 0
Rt
−∞ g(t − τ ) × 0 dτ = 0 so y(t) = 0 for all t < 0.
solve
ay + by + cy + d = 1
with boundary conditions
y(0) = 0 and y(0) = 0
Case (b): 0 < t
y(t) =
=
Z t
0
Z t
g(t − τ ) sin(ωτ ) dτ
e−5(t−τ ) sin(ωτ ) dτ
0
Z t
= Im
e−5(t−τ )eiωτ dτ
Step response
 0

t 


(5+iω)τ


e
−5t


= Im e


5 + iω


0
(
)
eiωt − e−5t
differentiate
Impulse response: g(t)
= Im
5 + iω
5 sin(ωt) − ω cos(ωt) + ωe−5t
=
25 + ω 2
Any
Input: f(t)
convolution
y(t) =
37
Z t
−∞
Corresponding
Output: y(t)
g(t − τ ) f (τ ) dτ
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Complete Example
Find the impulse response of
d2y
dy
+
3
+ 2y = f (t)
dt2
dt
hence find the output when the input f (t) = H(t)e−t .
3. Differentiate the step response to get the impulse
response.
g(t) =
dy
= e−t − e−2t
dt
1. Find the General Solution with f (t) = 1
4. Use the convolution integral to find the output for
the required input.
Complimentary function is y = Ae−t + Be−2t
Particular integral is y = 1
2
The required input is f (t) = e−t , t > 0.
1 + Ae−t + Be−2t
General solution is y = 2
y(t) =
2. Set boundary conditions y(0) = ẏ(0) = 0 to get
the step response.
1
2+A+B =0
−A − 2B = 0
⇒ A = −1 and B = 1
2
=
=
=
Z t
−∞
Z t Z0t
0
h
g(t − τ )f (τ )dτ
e−(t−τ ) − e−2(t−τ ) e−τ dτ
e−t − eτ −2t dτ
i
τ −2t t
τe − e
0
−t
− 1) e + e−2t
−t
= (t
−2t
−t + e
Thus Step Response is y = 1
2−e
2
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Section 3: Summary
Section 4
Convolution integral (memorise this):
f (t) =
input
g(t) =
impulse response
y(t) =
y(t) =
Evaluating Convolution Integrals
output
Z t
−∞
g(t − τ ) f (τ ) dτ
Way to find the output of a linear system, described
by a differential equation, for an arbitrary input:
• Find general solution to equation for input = 1.
• Set boundary conditions y(0) = ẏ(0) = 0 to get
the step response.
A way of rearranging the convolution integral is described and illustrated.
The differences between convolution in time and space
are discussed and the concept of causality is introduced.
The section ends with an example of spatial convolution.
• Differentiate to get the impulse response.
• Use convolution integral together with the impulse
response to find the output for any desired input.
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