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### Transcript of The Convolution Sum for Discrete-Time LTI ahouse/engi7824/course_notes_7824_part6.pdfThe Convolution...

• 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 beexpressed as a sum of impulses.

x(n) =

k=

x(k)(n k)

So let us consider x(n) written in this form to be our input to the LTI system.

y(n) = L [x(n)] = L

[

k=

x(k)(n k)

]

This looks like our general linear form with a scalar x(k) and a signal in n, (n k). Recallthat 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.

y(n) = L

[

k=

x(k)(n k)

]=

k=

x(k)L [(n k)]

So now we wonder, what is L [(n k)]? Well, we can figure it out. Suppose we know howL acts on one impulse (n), and we call it

h(n) = L [(n)]

ENGI 7824 Discrete-Time Systems and Signals 1

• The Convolution Sum for DT LTI Systems

then by time invariance we get our answer.

h(n k) = L [(n k)]

(n k) L h(n k)

This means that if we know one input-output pair for this system, namely

(n) L h(n)

then we can inferx(n) L y(n)

which gives us the following.

y(n) =

k=

x(k)h(n 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) =

k=

x(k)h(n k)

1.1 Comments on the DT Convolution Sum

1. A systems impulse response, h(n), completely characterizes the behaviour of the sys-tem.

2. The impulse response h(n) can be generated directly from (n) through L, since (n) isan 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 causesa shift and flip.

DT y(n) = x(n) h(n) =

k=

x(k)h(n k)

CT convolution has one signal in terms of the dummy variable, and the othershifted and flipped on the dummy variable, but centred on the output variable.

CT y(t) = x(t) h(t) =

x()h(t )d

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• The Convolution Sum for DT LTI Systems

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 canuse to implement the system, since it is computable. This is another reason why weprefer DT signals and systems.

2 Examples of Convolution

Convolution is best understood when seen in action. Lets look at a couple of examples,one using signals and impulse responses defined functionally, and another with an impulseresponse 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).

x(n) = u(n) and h(n) =

(1

2

)nu(n)

We wish to find the step response s(n) of the system (i.e. the response of thesystem to the unit step input x(n) = u(n). This is shown below.

s(n) = x(n) h(n) =

k=

x(k)h(n k)

Thus the step response is as follows, found by substituting our actual signals intothe general convolution sum.

s(n) =

k=

u(k)

(1

2

)nku(n k)

Lets look at this step response in smaller ranges to see what happens.

First, consider the case where n < 0.

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• The Convolution Sum for DT LTI Systems

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 theyare multiplied together, the zero part of one signal cancels out the non-zeropart 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).

s(n) =

k=

u(k)

(1

2

)nku(n k)

Note that u(k) means we know the summation will be 0 for all values ofk < 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 willbe 0, since that unit step is flipped and extends toward . So, we canchange 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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• The Convolution Sum for DT LTI Systems

s(n) =

k=

u(k)

(1

2

)nku(n k)

=n

k=0

1 (

1

2

)nk 1

We can pull out any terms only in n

since that is not the summation variable.

=n

k=0

(1

2

)n(1

2

)k=

(1

2

)n nk=0

(1

2

)k=

(1

2

)n nk=0

2k

Now we have a form consistent with a geometric series. We can use that tosolve.

Recalln

k=0

2k =1 2n+1

1 2= 2n+1 1

So we have s(n) as follows.

s(n) =

(1

2

)n (2n+1 1

)=

(1

2

)n(2 2n 1)

=

(1

2

)n(2 (

1

2

)n 1

)

= 2 (

1

2

)n(1

2

)n 1

(1

2

)ns(n) = 2

(1

2

)nWe can visualize this, say for n = 2, as shown below. Note how the systemoutput comes from the overlap of the input signal and the shifted and flippedimpulse response.

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• The Convolution Sum for DT LTI Systems

So, overall, we have the following step response.

s(n) =

[2

(1

2

)n]u(n)

The u(n) comes from our first case above since s(n) = 0 for n < 0, and obviouslythe 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 thesystem is not LTI, there is no valid impulse response). Since the system isLTI, we can break down its response.

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• The Convolution Sum for DT LTI Systems

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 istime 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)

=

[2

(1

2

)n]u(n)

[2

(1

2

)n4]u(n 4)

2. Now consider a different system, with impulse response h2(n) = 2(n) (1

2)nu(n). We want the step response of this LTI system.

Since this system is LTI, and since the second term of the expression is thesystem considered in our first example, we can solve as follows without usingthe convolution sum.

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• The Convolution Sum for DT LTI Systems

s2(n) = u(n) h2(n)

= u(n) [2(n)

(1

2

)nu(n)

]= u(n) 2(n) u(n)

(1

2

)nu(n)

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)

= 2u(n)[2

(1

2

)n]u(n)

factoring out u(n)

=

(2

[2

(1

2

)n])u(n)

=

[2 2

(1

2

)n]u(n)

s2(n) =

(1

2

)nu(n)

These variations have shown how we can deal with more complex systems ascombinations of simpler systems that are often already known.

Now lets 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 cant be nicelyworked out. Thus, it is also useful to understand the concept of how the convo-lution sum works.

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• The Convolution Sum for DT LTI Systems

In the convolution sum, the impulse response is written as h(nk), meaning thatin the k domain, the impulse response is shifted by n and flipped around thatpoint. We can visualize the convolution operation as that shifted-and-flippedimpulse response sliding along the k axis from to as the summationoccurs. Whenever there is some non-zero overlap between this shifted-and-flippedimpulse response and the input signal, the system output will be non-zero (unlessthe non-zero overlaps cancel each other).

Lets consider a specific example.

We are given the impulse response shown below.

h(n) =

0 for n < 01 for 0 n 32 for 4 n 50 for n > 5

Let x(n) = u(n 4).

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• The Convolution Sum for DT LTI Systems

We want to determine

y(n) = x(n) h(n) =

k=

x(k)h(n k)

=

k=

u(k 4)h(n k)

We can use u(k 4) to change the summationlimits, but it doesnt help much.

=

k=4

1 h(n k)

but we have no convenient functional representation of h(n) to allow us to solve.