Question

7. An LTI system is described by the difference equation y[n] + 0.5y[n - 1] = x[n] - 0.25x[n - 1] with initial conditions y[-1] = 2. (a) (4 pts) Write down the system transfer function $H(z)$. (b) (2 pts) Sketch the pole-zero diagram of this system. (c) (4 pts) Compute the system output y[n] for the zero-input case (i.e., x[n] = 0 for n ? 0). (d) (5 pts) Compute the system output y[n] for the input x[n] = (\frac{3}{4})^n u[n] (assuming zero initial conditions).

          7. An LTI system is described by the difference equation
y[n] + 0.5y[n - 1] = x[n] - 0.25x[n - 1]
with initial conditions y[-1] = 2.
(a) (4 pts) Write down the system transfer function $H(z)$.
(b) (2 pts) Sketch the pole-zero diagram of this system.
(c) (4 pts) Compute the system output y[n] for the zero-input case (i.e., x[n] = 0 for
n ? 0).
(d) (5 pts) Compute the system output y[n] for the input x[n] = (\frac{3}{4})^n u[n] (assuming
zero initial conditions).

7. An LTI system is described by the difference equation
y[n] + 0.5y[n - 1] = x[n] - 0.25x[n - 1]
with initial conditions y[-1] = 2.
(a) (4 pts) Write down the system transfer function H(z).
(b) (2 pts) Sketch the pole-zero diagram of this system.
(c) (4 pts) Compute the system output y[n] for the zero-input case (i.e., x[n] = 0 for
n ? 0).
(d) (5 pts) Compute the system output y[n] for the input x[n] = ((3)/(4))^n u[n] (assuming
zero initial conditions).

Added by Joseph J.

University Physics with Modern Physics

Hugh D. Young 14th Edition

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7. An LTI system is described by the difference equation y[n] + 0.5y[n-1] = x[n] - 0.25x[n-1] with initial conditions y[-1] = 2 (a) (4 pts) Write down the system transfer function H(z) (b) (2 pts) Sketch the pole-zero diagram of this system. n â‰¥ 0 (zero initial conditions)