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barcode generator for ssrs Solving for X(z), we have in Software
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Solving this quadratic equation for r ( m ) , we find
r(m)= 2 However, because ~ ( n> 0, it follows that r ( m ) is the positive root, which is ) Finally, because
then
(c) For the first property, we may simply substitute the closedform expression for the Fibonacci sequence into the equation, and verify that it is true. For the second property, from the definition of the Fibonacci sequence we have which we may rewrite as
x2(n
+ 2)  x 2 ( n + I ) = x ( n ) [ x ( n )+ 2x(n + I)] However, note that
+ 3) = x ( n + I) + x ( n + 2 ) = x ( n ) + 2 x ( n + I) Substituting this into the previous equation, we have the desired property. 4.38
A savings account pays interest at the rate of 5 percent per year with interest compounded monthly.
(a) If $50 is deposited into the account every month for 60 months, find the balance in the account at the end of the 60 months. Assume that the money is deposited on the first day of the month s o that, at the end of the month, an entire month's interest has been earned.
(b) If no deposits are made for the next 60 months, find the account balance at the end of the next 60month period. CHAP. 41 (c) THE ZTRANSFORM
Instead of being compounded monthly, suppose that the bank offers to compound the interest daily. Compute the account balance at the end of 60 months and 120 months and compare your balances with those obtained when the interest is compounded monthly. (a) The savirigs account balance at the beginning of the nth month is equal to the balance in the previous month plus the amount deposited in the nth month plus the interest earned on the balance from the previous month. Therefore, if y(n) is the balance at the beginning of the nth month, where j3 is the interest earned on the account, and x(n) is the amount deposited into the savings account in the nth month. Taking ztransforms, and solving for Y (z), we have where v = I
+ j3. With $50 deposits beginning with month number zero, x(n) = 50u(n), and
Y ( z ) = 50 1 (1  vz')(I
z1) Performing a partial fraction expansion of Y(z), we have
Taking the inverse ztransform, we have
at the end of 60 months. after earning I month's interest, but prior to making With v = I + B, and j3 = the next deposit, the balance is (b) With no deposits for the next 60 months, the balance at the end of the first 60 months simply grows as y(n) = y(60). vn"' Therefore.
> 60 y( 120) = 4,379.42 (c) With the interest compounded daily, let us compute the effective monthly interest rate. Assuming a balance of $ I at the beginning of the month, the difference equation that describes the daily balance, w(n), is where j3 =
g. ztransforms as we did in part (a), the solution to this difference equation is Using
where v = 1 + B . Assuming that a month is 30 days long, for I month's interest we have w(30) = v30 = 1.004 175 Using v = 1.004175 in Eq. (4.8), we have THE zTRANSFORM The deterministic autocorrelation sequence corresponding to a sequence x ( n ) is defined as [CHAR 4
( a ) Express r , ( n ) as the convolution of two sequences, and find the ztransform of r,(n) in terms of the ztransform of x ( n ) . (b) If x ( n ) = a n u ( n ) ,where la1 < I , find the autocorrelation sequence, r,(n), and its ztransform. (a) From the definition of the deterministic autocorrelation, we see that r,(n) is the convolution of x(n) with x(n), rx(n) = x(n) * x(n) Therefore, using the timereversal property of the ztransform, it follows that If the region of convergence of X(z) is R,, the region of convergence of R,(z) will be the intersection of the regions R, and 1/R,,. Therefore, if this intersection is to be nonempty, R, must include the unit circle.

