Question#1: The discrete time signal shown in problem 1.32(d) (Above). x(n) = 5

Question

Question#1:

The discrete time signal shown in problem 1.32(d) (Above). x(n) = 5 sin (Pi n). What is the value of Pi?

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Question#2:

A system has the following input/output relation: y(n) = x(n) + .5 y(n -1). If the input x(n) is the sinusoidal signal shown in problem 1.32(d) (Above), the steady-state output will have the following form: y(n) = A sin(Bn+C). Determine the value of A.

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Question#3:

A system has the following input/output relation: y(n) = x(n) + .5 y(n -1). If the input x(n) is the sinusoidal signal shown in problem 1.32(d) (Above), the steady-state output will have the following form: y(n) = A sin(Bn+C). Determine the value of B.

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Question#4:

Please see problem 1.32(d) (Above). If the x(n) shown in the problem represents the sampled version of an analog signal x(t) and the sampling rate was 10000 samples per second, what is the corresponding analog frequency, i.e., if x(t) is expressed as x(t) = sin (Bt), what is the value of B? (The units of B are radians per second.). (Assume that the sampling rate used to obtain x(n) from x(t) was sufficiently fast to represent the true frequency of x(t).).

Explanation / Answer

Question#1

pi = 3.14

Question#2:

y(n) = x(n) + .5 y(n -1)

let homogeneous solution is Yh(n) = c^n

                                                     Y(n) = (0.5)^n

for particular solution ::: let Yp(n) = Asin (Pi n)

then ::     Asin (Pi n)   = 5 sin (Pi n) + A sin (Pi (n-1))

                  A   =5/2

hence particular solution is   :: Yp(n) = 2.5*sin(pi*n)

Y(n) = Yh(n) + Yp(n)

         = (0.5)^n + 2.5*sin(pi*n)

for stedy state ::: Y(n) = 0+ 2.5*sin(pi*n)

                    hence   A = 2.5

Question#3:

B = pi

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