11.3-8 Solve y[k] + 2y[k - 1] + 2y[k - 2] = f[k - 1] + 2f[k - 2] with y[0] = 0, y[1] = 1, and f[k] = e^{k}u[k].
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First, let's rewrite the given equation in terms of the unit step function u[k]: y[k] + 2y[k-1] + 2y[k-2] = f[k-1] + 2f[k-2] Since f[k] = e*u[k], we can substitute it into the equation: y[k] + 2y[k-1] + 2y[k-2] = e*u[k-1] + 2e*u[k-2] Show more…
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