barcode lib ssrs + OSx(0) + O.~X(-I) = 0.75 - 0.25 + 0.5 = 1 in Software

Making Code 128 Code Set B in Software + OSx(0) + O.~X(-I) = 0.75 - 0.25 + 0.5 = 1

+ OSx(0) + O.~X(-I) = 0.75 - 0.25 + 0.5 = 1
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+ 0 . 5 ~ ( 1+ OSx(0) = 1 - 0.75 + 0.25 + 0.5 = 1 )
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Now, substituting these derived initial conditions into Eq. (1.25), we have y(O) = 0.5 Al + A2 = I y( I) = 0.25 A ,ei"I3 ~
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Writing this pair of equations in the two unknowns A I and
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in matrix form,
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and solving, we find
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Substituting into Eq. (1.25) and simplifying, we find, after a bit of algebra.
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An important observation to make about this solution is that. because the difference equation has real coefficients, the roots of the characteristic polynomial are in complex-conjugate pairs. This ensures that the unit sample response is real. With a real-valued input x(n), the response must be real and, therefore, it follows that A2 will be the complex conjugate of A 1 :
A second-order recursive system is described by the LCCDE
(a) Find the unit sample response h(n) of this system.
(h) Find the system's response to the input x ( n ) = u(n) - u(n - 10) with zero initial conditions.
CHAP. I]
SIGNALS AND SYSTEMS
(c) Find the-system's response to the input x ( n ) = ( ; ) " u ( n ) with zero initial conditions.
( a ) To find the unit sample response, we must solve the difference equation with x ( n ) = S ( n ) and initial rest conditions. The characteristic equation is
Therefore, the homogeneous solution is
Because the particular solution is zero when the system input is a unit sample, Eq. ( 1 . 2 6 ) represents the total solution. To find the constants A l and A2, we must derive the initial conditions at n = 0 and n = I. With initial rest conditions, y ( - I) = y ( - 2 ) = 0, it follows that
We may now write two equations in the two unknowns A1 and A2 by evaluating Eq. ( 1 . 2 6 ) at n = 0 and n = I as follows:
Solving for A l and A2, we find
A,=-2 A2=3
Thus,
y(n) = -2(i)"
+ 3($)"
and the unit sample response is
h(n)= [-2(;)"
+3(a)"]u(n)
(b) To find the response of the system to x ( n ) = u ( n ) - u ( n - l o ) , we may proceed in one of two ways. First, we may perform the convolution of h ( n ) with x ( n ) :
Alternatively, noting that the input is a sum of two steps, we may find the step response of the system, s ( n ) , and then using linearity. write the response as
Using this approach, we see from part ( a ) that the step response for n
0 is
Evaluating the sums using the geometric series. we find
Thus, the desired solution is
y ( n ) = s ( n ) - s ( n - 10) = [ 2 ( f ) " - ( : ) " ] u ( n ) - [ 2 ( f )
11-10
( i ) n - l o ] u ( n- 10)
SIGNALS AND SYSTEMS
[CHAP. I
(c) With x(n) = (f)"u(n), note that .r(n) has the same form as one of the terms in the homogeneous solution. Therefore, the particular solution will not be of the form y,(n) = C(:)" as indicated in Table 1-2. If we were to substitute this particular solution into the difference equation, we would find that no value of C would work. As is the case when a root of the characteristic equation is of second order, the particular solution has the form
yp(n) = cn(!)'' Substituting this into the difference equation, we have
Dividing through by ( f )I1,we have
Solving for C , we find that C = -2. Thus, the total solution is
We now must solve for the constants A1 and A> As we did in part (a),with zero initial conditions we find that y(0) = I and y ( l ) = $. Therefore, evaluating Eq. (1.27) at 11 = 0 and n = 1, we obtain the following two equations in the two unknowns A l and A2:
Solving for A1 and A z , we find that A1 = 4 and A2 = 3. Thus, the total solution becomes
A $100,000 mortgage is to be paid off in equal monthly payments of d dollars. Interest, compounded monthly, is charged at the rate of 10 percent per annum on the unpaid balance [e.g., after the first month . the total debt equals ($100,000 ~ $ l 0 0 , 0 0 0 ) ] Determine the amount of the payment, d, so that the mortgage is paid off in 30 years, and find the total amount of payments that are made over the 30-year period.
The total unpaid balance at the end of the nth month. in the absence of any additional loans or payments, is equal to the unpaid balance in the previous month plus the interest charged on the unpaid balance for the previous month. Therefore, with y(n) the balance at the end of the nth month we have
is the interest charged on the unpaid balance. In addition, the balance must be adjusted by the net where B = amount of money leaving the bank into your pocket, which is simply the amount borrowed in the nth month minus the amount paid to the bank in the nth month. Thus
where xb(n) is the amount borrowed in the nth month and xp(n) is the amount paid in the nth month. Combining terms, we have y(n) - vy(r - 1) = xh(n) - x,,(n) = x(n) and x(n) is the net amount of money in the nth month that leaves the bank. Because where v = I B = 1 a principal of p dollars is borrowed during month zero, and payments of d dollars begin with month I, the driving function, x(n), is x(n) = .wh(n)- .u,,(n) = p8(n) - du(n - 1) and the difference equation for y(n) becomes
+ 9,
CHAP. I]
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