The Convolution Sum for DT LTI Systems
The Convolution Sum for Discrete-Time LTI Systems
Andrew W. H. House
01 June 2004
1
The Basics of the Convolution Sum
Consider a DT LTI system, L.
x(n) −→ L −→ y(n)
DT convolution is based on an earlier result where we showed that any signal x(n) can be
expressed as a sum of impulses.
x(n) =
∞
X
x(k)δ(n − k)
k=−∞
So let us consider x(n) written in this form to be our input to the LTI system.
" ∞
#
X
x(k)δ(n − k)
y(n) = L [x(n)] = L
k=−∞
This looks like our general linear form with a scalar x(k) and a signal in n, δ(n − k). Recall
that for an LTI system:
• Linearity (L): ax1 (n) + bx2 (n) −→ L −→ ay1 (n) + by2 (n)
• Time Invariance (TI): x(n − no ) −→ L −→ y(n − no )
We can use the property of linearity to distribute the system L over our input.
" ∞
#
∞
X
X
y(n) = L
x(k)δ(n − k) =
x(k)L [δ(n − k)]
k=−∞
k=−∞
So now we wonder, what is L [δ(n − k)]? Well, we can figure it out. Suppose we know how
L acts on one impulse δ(n), and we call it
h(n) = L [δ(n)]
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then by time invariance we get our answer.
h(n − k) = L [δ(n − k)]
δ(n − k) −→ L −→ h(n − k)
This means that if we know one input-output pair for this system, namely
δ(n) −→ L −→ h(n)
then we can infer
x(n) −→ L −→ y(n)
which gives us the following.
y(n) =
∞
X
x(k)h(n − k)
k=−∞
This is the convolution sum for DT LTI systems.
The convolution sum for x(n) and h(n) is usually written as shown here.
y(n) = x(n) ∗ h(n) =
∞
X
x(k)h(n − k)
k=−∞
1.1
Comments on the DT Convolution Sum
1. A system’s impulse response, h(n), completely characterizes the behaviour of the system.
2. The impulse response h(n) can be generated directly from δ(n) through L, since δ(n) is
an actual signal in DT. Thus, we can actually find the impulse response experimentally.
3. Compare convolution in DT and CT.
• DT convolution has an output variable n and a dummy variable k which causes
a shift and flip.
DT y(n) = x(n) ∗ h(n) =
∞
X
x(k)h(n − k)
k=−∞
• CT convolution has one signal in terms of the dummy variable, and the other
shifted and flipped on the dummy variable, but centred on the output variable.
Z∞
CT y(t) = x(t) ∗ h(t) =
x(τ )h(t − τ )dτ
−∞
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4. CT convolution is a model of behaviour of CT systems.
DT convolution is a model of behaviour of DT systems, but also an algorithm we can
use to implement the system, since it is computable. This is another reason why we
prefer DT signals and systems.
2
Examples of Convolution
Convolution is best understood when seen in action. Let’s look at a couple of examples,
one using signals and impulse responses defined functionally, and another with an impulse
response defined point-wise.
Example 2.1: DT Convolution: Step Response
Say we are given the following signal x(n) and system impulse response h(n).
n
1
x(n) = u(n) and h(n) =
u(n)
2
We wish to find the step response s(n) of the system (i.e. the response of the
system to the unit step input x(n) = u(n). This is shown below.
∞
X
s(n) = x(n) ∗ h(n) =
x(k)h(n − k)
k=−∞
Thus the step response is as follows, found by substituting our actual signals into
the general convolution sum.
∞
X
n−k
1
s(n) =
u(k)
u(n − k)
2
k=−∞
Let’s look at this step response in smaller ranges to see what happens.
• First, consider the case where n < 0.
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Here, s(n) = 0. This is because u(n − k) (and the associated exponential)
will be starting at a point less than 0 in the k domain, and will extend to
−∞, whereas u(k) starts at 0 and extends to +∞. We can visualize this,
say for a value of n = −2.
Notice that there is no non-zero overlap of x(k) and h(n − k). Since they
are multiplied together, the zero part of one signal cancels out the non-zero
part of the other, and vice versa. Thus, s(n) = 0 for n < 0.
• The more interesting case is when n ≥ 0.
Recall the convolution sum we are using to determin s(n).
∞
X
n−k
1
u(n − k)
s(n) =
u(k)
2
k=−∞
Note that u(k) means we know the summation will be 0 for all values of
k < 0, so we can change the lower limit of the summation to 0. Similarly,
the u(n − k) term means that the summation for all values of k > n will
be 0, since that unit step is flipped and extends toward −∞. So, we can
change the upper limit of the summation to n. In the range 0 ≤ k ≤ n,
both of the unit steps will have a value of 1. This is shown below.
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∞
X
n−k
1
u(n − k)
s(n) =
u(k)
2
k=−∞
n−k
n
X
1
1·
·1
=
2
k=0
We can pull out any terms only in n
since that is not the summation variable.
n n −k
X
1
1
=
2
2
k=0
n X
n −k
1
1
=
2
2
k=0
n X
n
1
2k
=
2
k=0
Now we have a form consistent with a geometric series. We can use that to
solve.
n
X
1 − 2n+1
Recall
2k =
= 2n+1 − 1
1
−
2
k=0
So we have s(n) as follows.
s(n) =
=
=
=
s(n) =
n
1
2n+1 − 1
2
n
1
(2 · 2n − 1)
2
!
n
−n
1
1
2·
−1
2
2
−n n
n
1
1
1
2·
−1·
2
2
2
n
1
2−
2
We can visualize this, say for n = 2, as shown below. Note how the system
output comes from the overlap of the input signal and the shifted and flipped
impulse response.
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So, overall, we have the following step response.
n 1
s(n) = 2 −
u(n)
2
The u(n) comes from our first case above since s(n) = 0 for n < 0, and obviously
the other part comes from the expression found in the second case above.
Now consider some variations on this first example.
Example 2.2: Variations on the Step Response
1. How would the system in the previous example react if the input was x(n) =
u(n) − u(n − 4)?
We could work this through mathematically, using the convolution sum,
but that is not necessary in this case. Remember, our system is LTI. (If the
system is not LTI, there is no valid impulse response). Since the system is
LTI, we can break down its response.
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y(n) = L[x(n)]
= L[u(n) − u(n − 4)]
= L[u(n)] − L[u(n − 4)]
In the prior example we determined s(n) = L[u(n)] and since the system is
time invariant, we also know that s(n − 4) = L[u(n − 4)].
So, overall, we have the following system output y(n).
y(n) = s(n) − s(n − 4)
"
n n−4 #
1
1
= 2−
u(n) − 2 −
u(n − 4)
2
2
2. Now consider a different system, with impulse response h2 (n) = 2δ(n) −
( 12 )n u(n). We want the step response of this LTI system.
Since this system is LTI, and since the second term of the expression is the
system considered in our first example, we can solve as follows without using
the convolution sum.
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s2 (n) = u(n) ∗ h2 (n)
n
1
u(n)
= u(n) ∗ 2δ(n) −
2
n
1
= u(n) ∗ 2δ(n) − u(n) ∗
u(n)
2
We can use the sifting property on the first term
and the second term is just the step response
from the first example.
s2 (n) = 2u(n) − s(n)
n 1
= 2u(n) − 2 −
u(n)
2
factoring out u(n)
n 1
u(n)
=
2− 2−
2
n 1
= 2−2−
u(n)
2
n
1
s2 (n) =
u(n)
2
These variations have shown how we can deal with more complex systems as
combinations of simpler systems that are often already known.
Now let’s consider an example which is not so nicely mathematically defined.
Example 2.3: Graphical Convolution
Even though the convolution sum is nicely defined, sometimes it can’t be nicely
worked out. Thus, it is also useful to understand the concept of how the convolution sum works.
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In the convolution sum, the impulse response is written as h(n−k), meaning that
in the k domain, the impulse response is shifted by n and flipped around that
point. We can visualize the convolution operation as that shifted-and-flipped
impulse response sliding along the k axis from −∞ to ∞ as the summation
occurs. Whenever there is some non-zero overlap between this shifted-and-flipped
impulse response and the input signal, the system output will be non-zero (unless
the non-zero overlaps cancel each other).
Let’s consider a specific example.
We are given the impulse response shown below.

0
for n < 0



1
for 0 ≤ n ≤ 3
h(n) =
−2
for 4 ≤ n ≤ 5



0
for n > 5
Let x(n) = u(n − 4).
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We want to determine
y(n) = x(n) ∗ h(n) =
=
∞
X
k=−∞
∞
X
x(k)h(n − k)
u(k − 4)h(n − k)
k=−∞
We can use u(k − 4) to change the summation
limits, but it doesn’t help much.
∞
X
=
1 · h(n − k)
k=4
but we have no convenient functional representation of h(n) to allow us to solve.
Consider the solution in a piece-wise fashion.
For n < 4, y(n) = 0 since there is only zero overlap between the two signals.
This is illustrated below.
For n ≥ 4, we need to visualize what is happening in order to determine the
value of y(n). In the figure below, we can see how h(n − k) slides along the k
axis and overlaps with x(k).
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So we have to determine the value of y(n) for specific values of n. We start at
the first point of non-zero overlap, when n = 4.
When n = 4, y(4) = 1 since only one point of the two signals overlaps, and
1 · 1 = 1. This is shown in the figure below.
When n = 5, y(5) = 2 since it is the sum of the two overlapping points. This is
shown in the figure below.
Similarly, y(6) = 3 and y(7) = 4.
Note that when n = 8, we have a negative overlap, and so y(8) = 2. This is
shown below.
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For the case where n ≥ 9, y(n) = 1 since it is summing over the entire length of
the impulse response. This is shown in the figure below.
Thus, we can plot our overall y(n) as shown here.
Thus, we have evaulated the convolution some graphically by taking advantage
of this shifting and flipping behaviour.
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3
Basic Properties of DT Convolution
Discrete-time convolution has several useful properties that allows us to solve systems more
easily.
3.1
Commutativity
Convolution is a commutative operation, meaning signals can be convolved in any order.
x(n) ∗ h(n) = h(n) ∗ x(n)
This quite naturally is true of the convolution sums themselves, as well.
∞
X
x(k)h(n − k) =
k=−∞
3.2
∞
X
h(k)x(n − k)
k=−∞
Associativity
Convolution is associative, meaning that convolution operations in series can be done in any
order.
(x(n) ∗ h(n)) ∗ g(n) = x(n) ∗ (h(n) ∗ g(n))
This is significant because it means systems in series can be reordered.
Thus we have
x(n) −→ h(n) −→ g(n) −→ y(n)
is the same as
x(n) −→ h(n) ∗ g(n) −→ y(n)
is the same as
x(n) −→ g(n) ∗ h(n) −→ y(n)
is the same as
x(n) −→ g(n) −→ h(n) −→ y(n)
and so the systems in series can be reordered.
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3.3
Distributivity
Convolution is distributive over addition.
x(n) ∗ [h(n) + g(n)] = x(n) ∗ h(n) + x(n) ∗ g(n)
This is significant to all parallel connections because it means the following two arrangements
are equivalent.
x(n) −→ h(n) + g(n) −→ y(n)
is the same as
3.4
Identity
We have previously established that δ(n) is the identity with respect to discrete-time convolution.
∞
X
Recall x(n) =
x(k)δ(n − k) = x(n) ∗ δ(n)
k=−∞
So x(n) ∗ δ(n) = x(n).
This concept is quite easily extended, so x(n) ∗ δ(n − no ) = x(n − no ) for no ∈ Z and
x(n − no ) ∗ δ(n − n1 ) = x(n − (no + n1 )) for no , n1 ∈ Z.
Example 3.1: Convolution Properties
1. Consider the following interconnection of systems.
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The system impulse response of the above overall system can be determined
from the properties of DT convolution.
hoverall (n) = h1 (n) ∗ [(h2 (n) ∗ h3 (n)) + h4 (n)]
So we get the following overall system output.
y(n) = x(n) ∗ hoverall (n)
= x(n) ∗ h1 (n) ∗ [(h2 (n) ∗ h3 (n)) + h4 (n)]
which can be further reducded if desired as
= x(n) ∗ h1 (n) ∗ h2 (n) ∗ h3 (n) + x(n) ∗ h1 (n) ∗ h4 (n)
2. Consider the following interconnection of systems.
For the above systems, we can apply the same principles as in the first part
of the example. Once we have found the overall impulse response, we can
find the output just as with any other impulse response (albeit with a lot
more tedium).
hoverall (n) = [h1 (n) + h2 (n)] ∗ [(h3 (n) ∗ [h4 (n) + h5 (n)]) + h6 (n)] ∗ h7 (n)
As we have seen, it is possible to build more complex systems from interconnections of simpler ones, and still have an overall impulse response to represent the
system.
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