The Convolution Sum for Discrete-Time LTI Systems

The Convolution Sum for Discrete-Time LTI Systems Andrew W. H. House 01 June 004 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(k)δ(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 x(k)δ(n 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): ax 1 (n) + bx (n) L ay 1 (n) + by (n) Time Invariance (TI): x(n n o ) L y(n n o ) We can use the property of linearity to distribute the system L over our input. [ ] y(n) = L x(k)δ(n k) = x(k)l [δ(n 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)] ENGI 784 Discrete-Time Systems and Signals 1

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 which gives us the following. x(n) L y(n) y(n) = 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) = x(k)h(n k) 1.1 Comments on the DT Convolution Sum 1. A system s impulse response, h(n), completely characterizes the behaviour of the system.. 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(k)h(n 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. CT y(t) = x(t) h(t) = x(τ)h(t τ)dτ ENGI 784 Discrete-Time Systems and Signals

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. 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.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) 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. s(n) = x(n) h(n) = x(k)h(n k) Thus the step response is as follows, found by substituting our actual signals into the general convolution sum. s(n) = u(k) ( ) n k 1 u(n k) Let s look at this step response in smaller ranges to see what happens. First, consider the case where n < 0. ENGI 784 Discrete-Time Systems and Signals 3

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 =. 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). s(n) = u(k) ( ) n k 1 u(n 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. ENGI 784 Discrete-Time Systems and Signals 4

s(n) = = = = = ( ) n k 1 u(k) u(n k) n ( ) n k 1 1 1 k=0 We can pull out any terms only in n since that is not the summation variable. n ( ) n ( ) k 1 1 k=0 ( ) n n ( 1 1 k=0 ( ) n n 1 k k=0 ) k Now we have a form consistent with a geometric series. We can use that to solve. n Recall k = 1 n+1 = n+1 1 1 k=0 So we have s(n) as follows. ( ) n 1 ( s(n) = n+1 1 ) ( ) n 1 = ( n 1) ( ) ( n ( ) n 1 1 = 1) ( ) n ( ) n ( ) n 1 1 1 = 1 ( ) n 1 s(n) = We can visualize this, say for n =, as shown below. Note how the system output comes from the overlap of the input signal and the shifted and flipped impulse response. ENGI 784 Discrete-Time Systems and Signals 5

So, overall, we have the following step response. [ ( ) n ] 1 s(n) = u(n) 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.: 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. ENGI 784 Discrete-Time Systems and Signals 6

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 ] [ 1 = u(n) ( ) ] n 4 1 u(n 4). Now consider a different system, with impulse response h (n) = δ(n) ( 1 )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. ENGI 784 Discrete-Time Systems and Signals 7

s (n) = u(n) h (n) [ ( ) n 1 = u(n) δ(n) u(n)] ( ) n 1 = u(n) δ(n) u(n) u(n) We can use the sifting property on the first term and the second term is just the step response from the first example. s (n) = u(n) s(n) [ ( ) n ] 1 = u(n) u(n) factoring out u(n) ( [ ( ) n ]) 1 = u(n) [ ( ) n ] 1 = u(n) ( ) n 1 s (n) = u(n) 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.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. ENGI 784 Discrete-Time Systems and Signals 8

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) = for 4 n 5 0 for n > 5 Let x(n) = u(n 4). ENGI 784 Discrete-Time Systems and Signals 9

We want to determine y(n) = x(n) h(n) = = = x(k)h(n k) u(k 4)h(n k) We can use u(k 4) to change the summation limits, but it doesn t help much. 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). ENGI 784 Discrete-Time Systems and Signals 10

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) = 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) =. This is shown below. ENGI 784 Discrete-Time Systems and Signals 11

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. ENGI 784 Discrete-Time Systems and Signals 1

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(k)h(n k) = h(k)x(n k) 3. 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 is the same as is the same as is the same as x(n) h(n) g(n) y(n) x(n) h(n) g(n) y(n) x(n) g(n) h(n) y(n) x(n) g(n) h(n) y(n) and so the systems in series can be reordered. ENGI 784 Discrete-Time Systems and Signals 13

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. is the same as x(n) h(n) + g(n) y(n) 3.4 Identity We have previously established that δ(n) is the identity with respect to discrete-time convolution. Recall x(n) = x(k)δ(n k) = x(n) δ(n) So x(n) δ(n) = x(n). This concept is quite easily extended, so x(n) δ(n n o ) = x(n n o ) for n o x(n n o ) δ(n n 1 ) = x(n (n o + n 1 )) for n o, n 1 Z. Z and Example 3.1: Convolution Properties 1. Consider the following interconnection of systems. ENGI 784 Discrete-Time Systems and Signals 14

The system impulse response of the above overall system can be determined from the properties of DT convolution. h overall (n) = h 1 (n) [(h (n) h 3 (n)) + h 4 (n)] So we get the following overall system output. y(n) = x(n) h overall (n) = x(n) h 1 (n) [(h (n) h 3 (n)) + h 4 (n)] which can be further reducded if desired as = x(n) h 1 (n) h (n) h 3 (n) + x(n) h 1 (n) h 4 (n). 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). h overall (n) = [h 1 (n) + h (n)] [(h 3 (n) [h 4 (n) + h 5 (n)]) + h 6 (n)] h 7 (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. ENGI 784 Discrete-Time Systems and Signals 15