Time-Domain Analysis and z Transform - Introduction to Digital Signal Processing and Filter Design - page 39

This is known as the convolution sum , denoted by a compact notation y(n) =

x(n) h(n) . The summation formula can be used to find the response due to any

input signal. So if we know the unit impulse response h(n) of the system, we
can find the output y(n) due to any input x(n) —therefore it is another model
for the discrete-time system. In contrast to the recursive algorithm, however,
note that the convolution sum cannot be used to find the response due to given
initial conditions. When and if the input signal is defined for −∞

∗

<n< ∞ or

∞ , obviously the lower index of summation is changed to −∞ .In
this case the convolution sum formula takes the general form

− M ≤ n<

∞

y(n) = x(k)h(n − k) (2.6)

k =−∞

For example, even though we know that h(n) = 0for −∞ <n< 0, if the input

sequence x(n) is defined for − M<n< ∞ ,thenwehavetousetheformula

= k =−∞ x(k)h(n y(n) − k) .If x(n) = 0for −∞ <n< 0, then we have to use

= k = 0 x(k)h(n

the formula y(n) k) .

To understand the procedure for implementing the summation formula, we
choose a graphical method in the following example. Remember that the recur-
sive algorithm cannot be used if the DT system is described by more than one
difference equation, and the convolution sum requires that we have the unit pulse
response of the system. We will find that these limitations are not present when
we use the z -transform method for analyzing the DT system performance in the
time domain.

−

Example 2.1

Given an h(n) and x(n) , we change the independent variable from n to k and
plot h(k) and x(k) as shown in Figure 2.4a,b. Note that the input sequence is
defined for − 2 ≤ k ≤ 5 but h(k) is a causal sequence defined for 0 ≤ k ≤ 4. Next

we do a time reversal and plot h( − k) in Figure 2.4c. When n ≥ 0, we obtain

h(n − k) by delaying (or shifting to the right) h( − k) by n samples; when n< 0,

the sequence h(

− k) is advanced (or shifted to the left). For every value of n ,

we have h(n − k) and x(k) and we multiply the samples of h(n − k) and x(k)

at each value of k and add the products.
For our example, we show the summation of the product when n =− 2in

Figure 2.4d, and show the summation of the product when n = 3inFigure2.4e.

The output y( − 2 ) has only one nonzero product = x( − 2 )h( 0 ) . But the output

sample y( 3 ) is equal to x( 0 )h( 3 ) + x( 1 )h( 2 ) + x( 2 )h( 1 ) + x( 3 )h( 0 ) .

But note that when n> 9, and n< − 2, the sequences h(n − k) and x(k) do

not have overlapping samples, and therefore y(n) = 0for n> 9and n< − 2.

Example 2.2

As another example, let us assume that the input sequence x(n) and also the
unit impulse response h(n) are given for 0 ≤ n< ∞ . Then output y(n) given