Time-domain analysis of discrete-time systems - Continuous and Discrete Time Signals and Systems - page 437

Table 10.1. Convolution of x [ k ] and h [ k ] using the sliding tape method for Example 10.8

m ... − 5 − 4 − 3 − 2 − 101234567 ... y [ k ]

x [ m ] 222

h [ m ] 12345

h [ − m ] 5 4 3 2 1
h [ − 1 − m ] 5 4 3 2 1

− 1

h [0 − m ] 5 4 3

h [1 − m ] 5 4 321 1 6

h [2 − m ] 5 4321 2 2

h [3 − m ] 54321 3 8

h [4 − m ] 54321 4 4

h [5 − m ] 54321 5 8

h [6 − m ] 54321 6 0

h [7 − m ] 54321 7 0

h [ m ] for different values of m . Following the steps involved in convolution,
we generate the values for the sequence h [ k − m ] and store the value in a row.
To generate the values of h [ k − m ], we first form the function h [ − m ], which
is obtained by time-inverting h [ m ]. The result is illustrated in the fourth row
of Table 10.1. The time-reversed function h [ − m ] is used to generate h [ k − m ]
by right-shifting h [ − m ]by k time units. For example, the fifth row contains
the values of the function h [ − 1 − m ] = h [ − ( m + 1)]. Similarly, rows (6)-(13)
contain the values of the function h [ k − m ] = h [ − ( m − k )] for the range 0
≤ k ≤ 7. In order to calculate y [ k ] for a fixed value of k , we multiply the entries
in the row containing x [ m ] by the corresponding entries contained in the row
for h [ k − m ] and then evaluate the summation:

∞

y [ k ] = x [ m ] h [ k − m ] .

m =−∞

For k =− 1, we note that the non-zero entries of x [ m ] and h [ k − m ]donot
overlap. Therefore, y [ k ] = 0 for k =− 1. Since there is also no overlap for
k < − 1, the output y [ k ] = 0 for k ≤− 1.
The aforementioned multiplication process is repeated for different values
of k .For k = 0, we note that the only overlap between the non-zero values of
x [ m ] and h [ − m ] occurs for m = 0. The output response is therefore given by

y [0] = 2 1 = 2 .

These values of time instant k = 0 and the output response y [0] = 2 are stored
in the last two columns of row (6), corresponding to the entries of h [0 − m ]
in Table 10.1. Similarly, for k = 1, we observe that the overlap between the
non-zero values of x [ m ] and h [1 − m ] occurs for m = 0 and 1. The output
response is given by

y [0] = 2 2 + 2 1 = 6