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y(k) = y(k – 1) + a k

Pb. 2.12 Show that the response of the system:

y(k) = (1 – a)u(k) + a y(k – 1)

to a step signal of amplitude c; that is, u(k) = c for all positive k, is given by:

y(k) = c(1 – a ) for k = 0, 1, 2, …
k+1
where the initial condition y(–1) = 0.
Pb. 2.13 Given the first-order difference equation:

y(k) = u(k) + y(k – 1) for k = 0, 1, 2, …

with the input signal u(k) = k, and the initial condition y(–1) = 0. Verify that
its solution also satisfies the second-order difference equation

y(k) = 2y(k – 1) – y(k – 2) + 1

with the initial conditions y(0) = 0 and y(–1) = 0.
Pb. 2.14 Verify that the response of the system governed by the first-order
difference equation:

y(k) = bu(k) + a y(k – 1)

k
to the alternating input: u(k) = (–1) for k = 0, 1, 2, 3, … is given by:

yk() = b [(−1 ) + a k+1 ] for k = 0 , , , ,…1 2 3
k
1 + a
if the initial condition is: y(–1) = 0.
Pb. 2.15 The impulse response of a system is the output from this system
when excited by an input signal δ(k) that is zero everywhere, except at k = 0,
where it is equal to 1. Using this definition and the general form of the solu-
tion of a difference equation, write the output of a linear system described by:

y(k) – 3y(k – 1) – 4y(k – 2) = δ(k) + 2δ(k – 1)

The initial conditions are: y(–2) = y(–1) = 0.
 1
Answer: yk() =− (−1 ) + 6 ( ) 4 k    for k > 0
k

5
 5
© 2001 by CRC Press LLC