Tutorials - Digital Signal Processing: An Introduction with MATLAB and Applications - page 278

If 1 B 4b (i.e.,

p

Case 2:

1 4b is pure imaginary), then

\2 ) b\1 :

\1 ) 1 j p

z p 4b 1

Hence, combining Case 1 and Case 2 we get 0 \ b \ 1.

MATLAB: try b = 0.8; r = roots([1 - 1 b]); a = abs(r)

Tutorial 43

Q: Find the values of b for which the system shown below is stable, knowing that
b is real.

x ( n )

y ( n )

z −1

b / 2 b / 3

Solution: It is better to define some auxiliary points on such diagrams. In this
question we define r(n), hence the other side of z -1 will be r(n - 1).

r ( n )

y ( n )

x ( n )

z −1

b / 2 b / 3

r ( n − 1 )

y ð n Þ¼ r ð n Þþð b = 3 Þ r ð n 1 Þ ð 1 Þ

r ð n Þ¼ x ð n Þþð b = 2 Þ r ð n 1 Þ ð 2 Þ

It is not easy to find the I/O relationship from the time domain equations. Hence,
take the z-transform of ( 1 ) and ( 2 ) as follows:

R ð z Þ

Y ð z Þ¼ R ð z Þþ b

3 z 1 R ð z Þ¼ 1 þ b 3 z 1

ð 3 Þ

R ð z Þ

R ð z Þ¼ X ð z Þþ b

2 z 1 R ð z Þ) X ð z Þ¼ 1 b 2 z 1

ð 4 Þ