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Problem 8.03PP
The one-sided z-transform is defined as 
00
F ® = ! ] / ( * » ■ * ■
0
(a) Show that the one-sided transform of f(k+ ^) is Z{ f(k + 1)} = zF(z) - z f ̂ ) .
(b) Use the one-sided transform to solve for the transforms of the Fibonacci numbers generated 
by the difference equation u(k+2) = u(k+^)+ u(k). Let u(0) = u(^) = [Hint: You will need to find a 
general expression for the transform of f(k + 2) in terms of the transform of f[k).]
(c) Compute the pole locations of the transform of the Fibonacci numbers.
(c) Compute the pole locations of the transform of the Fibonacci numbers.
(d) Compute the inverse transform of the Fibonacci numbers.
(e) Show that, if u(k) represents the kth Fibonacci number, then the ratio u(k + 1 )/u(k) will
approach • This is the golden ratio valued so highly by the Greeks.
Step-by-step solution
w
step 1 of 5
Find the z-transform o f / (^+ 1)
z { / { k + i ) ] = ± / { k + i ) z - ^
Z { / ( * + l ) } = 2 / ( . / ) z ^ " ‘. where i + l = J
Z [ / { k + \ ) } = z Z / O ) z - ' - : ^ f ( 0)
0
Thus, we get|z(/(A + l)) =zF(s) - ^ ( 0)|
0 >)
Step 2 of 5
We know that,
a ( i+ 2 ) - a ( i+ l ) - a ( j t ) = 0
■We have Z [ / ( i + 2)] = I V (z) - 2 V (0) - JeT (1)
Obtain the z transform and simplify further.
z^U (z) - z^u (0) - za (1) - [z U (z) - za (0)] - U {z ) = 0
{ z ^ - z - l)C /(^ ) = z)a(O ) + za (l)
Step 3 of 5
Since. a (0 ) = a ( l ) an d a (l) = l ,
W e g e t y ( 2 ) = - j ^
Thus, we get U(2) =
z ^ - z - 1
(c)
- z - 1
Find the poles ofthe system U(z) =
2
z^ - z - 1 = 0
1 ± V5
s 1.618, and ot̂ = -0 .618 
Thus, the poles are = 1.618, and -0 .618 |
Step 4 of 5
We know that U (z) = - j ---------^
We can rewrite this as Z/ (z) = ----- ^ ^ .
z z
By the long division rule, we get the quotient as l- fz “* -Fz"® +z~^ +... 
Hence, a (* ) = U , 2,3,5.............
Step 5 of 5
Apply the principle o f partial fraction to U (z) = ----------.
z^ — z — 1
U (z) =
( 1 - 0 0 - ^ " ) 
a ^ -O i . O j- c i i
--------- I=r -------------E7(z)= ^ ----- T ----- T
 ̂ ̂ 1 - q z '^ l - O jZ ‘ ^
Finding the invers e z transform of ̂ ( z ) :
u [k ) = - o f + — S — o f
O i- O i O i-C ^
Thus, we get
(«)
S i n c e , I