Questions: LTI system starts at rest (no stored values) and has an impulse response h[n] = 1.25δ[n] − 0.5δ[n − 1] + δ[n − 3] and an input signal x[n] = 10 cos(0.25πn + 15o) (u[n − 1] − u[n − 3)) Find a closed form (analytic expression) for the output of the sys

Solution Steps

We need to find the output \( y[n] \) of a Linear Time-Invariant (LTI) system given its impulse response \( h[n] \) and input signal \( x[n] \). The impulse response is: \[ h[n] = 1.25\delta[n] - 0.5\delta[n-1] + \delta[n-3] \] The input signal is: \[ x[n] = 10 \cos(0.25\pi n + 15^\circ) (u[n-1] - u[n-3]) \]

The input signal \( x[n] \) can be simplified by considering the unit step functions \( u[n-1] \) and \( u[n-3] \): \[ x[n] = 10 \cos(0.25\pi n + 15^\circ) \text{ for } 1 \leq n < 3 \] \[ x[n] = 0 \text{ otherwise} \]

The output \( y[n] \) of the LTI system is given by the convolution of \( x[n] \) and \( h[n] \): \[ y[n] = x[n] * h[n] = \sum_{k=-\infty}^{\infty} x[k] h[n-k] \]

Since \( x[n] \) is non-zero only for \( 1 \leq n < 3 \), we only need to consider these values in the convolution sum: \[ y[n] = \sum_{k=1}^{2} x[k] h[n-k] \]

We compute \( y[n] \) for each \( n \) by substituting the values of \( x[k] \) and \( h[n-k] \):

For \( n = 1 \): \[ y[1] = x[1]h[0] + x[0]h[1] = 10 \cos(0.25\pi \cdot 1 + 15^\circ) \cdot 1.25 = 10 \cos(0.25\pi + 15^\circ) \cdot 1.25 \]

For \( n = 2 \): \[ y[2] = x[2]h[0] + x[1]h[1] = 10 \cos(0.25\pi \cdot 2 + 15^\circ) \cdot 1.25 - 10 \cos(0.25\pi + 15^\circ) \cdot 0.5 \]

For \( n = 3 \): \[ y[3] = x[2]h[1] + x[1]h[2] + x[0]h[3] = 10 \cos(0.25\pi \cdot 2 + 15^\circ) \cdot (-0.5) + 10 \cos(0.25\pi + 15^\circ) \cdot 0 + 0 \cdot 1 \]

For \( n = 4 \): \[ y[4] = x[2]h[2] + x[1]h[3] = 10 \cos(0.25\pi \cdot 2 + 15^\circ) \cdot 0 + 10 \cos(0.25\pi + 15^\circ) \cdot 1 \]

Final Answer