Time-domain analysis of discrete-time systems - Continuous and Discrete Time Signals and Systems - page 457

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 = 0.5.ˆk.*(k>=0); % Input signal
>>A=[110.25]; % Coefficients with y[k]
>>B=[100]; %Coefficients with x[k]
>> Zi = filtic(B,A,[1 -2]); % Initial condition
>> Y = filter(B,A,X,Zi); % Calculate output

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

Y = [0.5000 -0.2500 0.3750 -0.1875 0.1563
-0.0781 0.0547].

To confirm if the M ATLAB code is correct, we also compute the values of
the output response in the range 0 ≤ k ≤ 5. We express y [ k + 2] + y [ k + 1] +
0 . 25 y [ k ]= x [ k + 2] as follows:

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

with ancillary conditions y [ − 1] = 1 and y [ − 2] =− 2. Solving the difference
equation iteratively yields

y [0] =− y [ − 1] − 0 . 25 y [ − 2] + x [0] =− 1 − 0 . 25( − 2) + 1 = 0 . 5 ,
y [1] =− y [0] − 0 . 25 y [ − 1] + x [1] =− 0 . 5 − 0 . 25(1) + 0 . 5 =− 0 . 25 ,
y [2] =− y [1] − 0 . 25 y [0] + x [2] =− ( − 0 . 25) − 0 . 25(0 . 5) + 0 . 25 = 0 . 375 ,
y [3] =− y [2] − 0 . 25 y [1] + x [3] =− 0 . 375 − 0 . 25( − 0 . 25) + 0 . 125
=− 0 . 1875 ,
y [4] =− y [3] − 0 . 25 y [2] + x [4] =− ( − 0 . 1875) − 0 . 25(0 . 375) + 0 . 0625
= 0 . 1563 ,

and
y [5] =− y [4] − 0 . 25 y [3] + x [2] =− 0 . 1563 − 0 . 25( − 0 . 1875) + 0 . 031 25
=− 0 . 0782 ,

which are the same as the values computed using M ATLAB .

The expressions for the initial conditions for the higher-order difference equa-
tions are more complex. Fortunately, most systems are causal with zero ancillary
conditions. The initial conditions Zi are zero in such cases.