Question

5.3-16 A professor recently received an unexpected $10 (a futile bribe attached to a test). Being the savvy investor that she is, the professor decides to invest the $10 into a savings account that earns 0.5% interest compounded monthly (6.17% APY). Furthermore, she decides to supplement this initial investment with an additional $5 deposit made every month, beginning the month immediately following her initial investment.
(a) Model the professor's savings account as a constant coefficient linear difference equation. Designate yln] as the account balance at month n, where n corresponds to the first month that interest is awarded (and that her $5 deposits begin).
(b) Determine a closed-form solution for y[n] That is, you should express yIn] as a function only of n.
(c) If we consider the professor's bank account as a system, what is the system impulse response h[n]? What is the system transfer function Hz]?
(d) Explain this fact: if the input to the professor's bank account is the everlasting exponential xn] 1 is not y[n] I"H[I]-HII]. 1, then the output

Answer 1

Answer:

a) y (n + 1) = 1.005 y(n) + 5U n

y (n + 1) - 1.005 y(n) = 5U (n)

b) Z^-1(Z(y0)=y(n) = [1010(1.005)^n - 1000(1)^n] U(n)

c) h(n) = (1.005)^n U(n - 1) + 10(1.005)^n U(n)

Explanation:

Her bank account can be modeled as:

y (n + 1) = y (n) + 0.5% y(n) + $5

y (n + 1) = 1.005 y(n) + 5U n

Given that y (0) = $10

y (n + 1) - 1.005 y(n) = 5U (n)

Apply Z transform on both sides

= ZY ((Z) - Z(y0) - 1.005) Z = 5 U (Z)

U(Z) = Z {U(n)} = Z/ Z - 1

Y(Z) [Z- 1.005] = Z y(0) + 5Z/ Z - 1

= 10Z/ Z - 1.005 + 5Z/(Z - 1) (Z - 1.005)

Y(Z) = 10Z/ Z - 1.005 + 1000Z/ Z - 1.005 + 1000Z/ Z - 1

= 1010Z/Z- 1.005 - 1000Z/Z-1

Apply inverse Z transform

Z^-1(Z(y0)) = y(n) = [1010(1.005)^n - 1000(1)^n] U(n)

Impulse response in output when input f(n) = S(n)

That is,

y(n + 1)= 1. 005y (n) + 8n

y(n + 1) - 1.005y (n) = 8n

Apply Z transform

ZY (Z) - Z(y0) - 1.005y(Z) = 1

HZ (Z - 1.005) = 1 + 10Z [Therefore y(Z) = H(Z)]

H(Z) = 1/ Z - 1.005 + 10Z/Z - 1. 005

Apply inverse laplace transform

= h(n) = (1.005)^n U(n - 1) + 10(1.005)^n U(n)