PRACTICAL MATLAB® FOR ENGINEERS PRACTICAL MATLAB

464 Practical MATLAB® Applications for Engineers

The impulse response would then be given by

hn hn h()

1

2


 (1)()







If the system is causal, then h(n) = 0, for n < 0. Then h(−1) = 0 , and the recursive
difference equation can be evaluated numerically as follows:

N = 0h(0) = 1

N = 1 h(1) − 1/2(1) = 0 h(1) = 1/2 = (1/ 2 )^1

N = 2 h(2) − 1/2(1/2) = 0 h(2) = 1/4 = (1/2)2

N = 3 h(3) − 1/2(1/4) = 0 h(3) = 1/2 = (1/2)3

N = 4 h(4) − 1/2(1/8) = 0 h(4) = 1/16 = (1/2)^4

Observe that the coeffi cients of h(n) are (1/2)n, for n > 0, or

h(n)

1

2

u(n)

1

2

()

















∑ 

nk

k

nk

0

R.5.14 Note that the examples presented in R.5.12 and R.5.13 given by the following dif-
ference equations present identical impulse responses. If the impulse responses are
identical, then the two given systems are equivalent.
a. g(n) = f(n) + ( 1 / 2 ) f(n – 1) + ( 1 / 2 )^2 f(n – 2) + ... + ( 1 / 2 )k f(n – k)
b. g(n) – ( 1 / 2 ) g(n – 1) = f(n)
Observe that in the fi rst equation, the output depends only on the input sequence,
whereas in the second equation, the output depends on its previous output scaled
(g(n)), delayed by one time unit, as well as by the input sequence.

R.5.15 A memoryless system is a system where the output depends only on the input
sequence, and a memory system is a system where the output depends on the
previous outputs as well as input.

R.5.16 Recall that any arbitrary discrete sequence can be represented as a superposition of
delayed weighted impulses (see Chapter 1 of this book), which is given by

fn fk n k f n f n f f

n

()

()( ) ( )( ) ( )( ) () ()

 



∑ 


 
... 22 1101(()

()( ) ()( ) ()( )

n

fn fn fknk





1

2233


...

R.5.17 Let f (n) → g(n). Recall that “→” denotes transforms to. Then

(n) → h(n)

FIGURE 5.6
System block diagram of R.5.13.

f(n) + g(n)

+

1/2

Z−^1

∑