Time-domain analysis of discrete-time systems - Continuous and Discrete Time Signals and Systems - page 456

For an n -order difference equation, the input variable yinitial is set to

yinitial = [ y [ − 1] , y [ − 2] ,..., y [ − n ]] .

To illustrate the usage of the built-in function filter , let us repeat

Example 10.1 using M ATLAB .

Example 10.17
The DT sequence x [ k ] = 2 ku [ k ] is applied at the input of an LTID system
described by the following difference equation:

y [ k + 1] − 0 . 4 y [ k ] = x [ k ] ,

with the ancillary condition y [ − 1] = 4 . Compute the output response y [ k ]of
the LTID system for 0 ≤ k ≤ 50 using M ATLAB .

Solution
The M ATLAB code used to solve the difference equation is listed below. The
explanation follows each instruction in the form of comments.

>> k = [0:50]; % time index k = [-1, 0, 1,
% ...50]
>> X = 2*k.*(k>=1); % Input signal
>>A=[1-0.4]; % Coefficients with y[k]
>>B=[01]; %Coefficients with x[k]
>> Zi = filtic(B,A,4); % Initial condition
>> Y = filter(B,A,X,Zi); % Calculate output

The output response is stored in the vector Y . Printing the first six values of the
output response yields

Y = [1.6 0.6400 2.2560 4.9024 7.9610 11.1844],

which corresponds to the values of the output response y [ k ] for the duration
0 ≤ k ≤ 5. Comparing with the numerical solution obtained in Example 10.1,
we observe that the two results are identical.

Next we proceed with a second-order difference equation.

Example 10.18
The DT sequence x [ k ] = 0.5 k u [ k ] is applied at the input of an LTID system
described by the following second-order difference equation:

y [ k + 2] + y [ k + 1] + 0 . 25 y [ k ] = x [ k + 2] ,

with ancillary conditions y [ − 1] = 1 and y [ − 2] =− 2. Compute the output
response y [ k ] of the LTID system for 0 ≤ k ≤ 50 using M ATLAB .