Question

A discrete linear time invariant system has an impulse response function given by \[ h[n] = \delta[n] + \frac{1}{2}\delta[n - 1] + \frac{1}{3}\delta[n - 2]. \] For input signal \( x[n] = \delta[n] + \delta[n - 1] \), which of the following option(s) is/are CORRECT for the output signal \( y[n] \)?

• A. \( y[2] = \frac{11}{6} \)
• B. \( y[2] = \frac{5}{6} \)
• C. \( y[k] = 0 \) for all \( k \geq 3 \)
• D. \( y[k] = 0 \) for all \( k \geq 4 \)

Hint:

In convolution, when computing the output, remember that the delta function \( \delta[n] \) is non-zero only when \( n = 0 \). This simplifies the calculation by limiting the non-zero contributions to only certain values of \( n \).

Answer 1

The Correct Option is B, D

The output signal \( y[n] \) of a discrete linear time-invariant (LTI) system is obtained by convolving the input signal \( x[n] \) with the impulse response function \( h[n] \). \[ y[n] = x[n] * h[n] \] where \( * \) denotes convolution. Given that the input signal \( x[n] = \delta[n] + \delta[n - 1] \), the output \( y[n] \) can be computed as: 

Using the linearity of convolution and the properties of the delta function, we get: \[ y[n] = \delta[n] * \delta[n] + \delta[n] * \frac{1}{2}\delta[n - 1] + \delta[n] * \frac{1}{3}\delta[n - 2] + \delta[n - 1] * \delta[n] + \delta[n - 1] * \frac{1}{2}\delta[n - 1] + \delta[n - 1] * \frac{1}{3}\delta[n - 2]. \] Simplifying each term: \[ y[n] = \delta[n] + \frac{1}{2} \delta[n - 1] + \frac{1}{3} \delta[n - 2] + \delta[n - 1] + \frac{1}{2} \delta[n - 2] + \frac{1}{3} \delta[n - 3]. \] Now, we compute \( y[n] \) for different values of \( n \): - For \( y[2] \): \[ y[2] = \delta[2] + \frac{1}{2} \delta[1] + \frac{1}{3} \delta[0] + \delta[1] + \frac{1}{2} \delta[0] + \frac{1}{3} \delta[-1]. \] Since \( \delta[n] \) is 1 only when \( n = 0 \) and 0 otherwise, we get: \[ y[2] = \frac{1}{2} + \frac{1}{2} = \frac{5}{6}. \] So, the correct answer for \( y[2] \) is \( \frac{5}{6} \), which corresponds to option (B). - For \( y[k] \), where \( k \geq 3 \), the terms involving \( \delta[n] \) for \( n \geq 3 \) will be zero because \( \delta[n] \) is non-zero only when \( n = 0 \). Hence, for \( k \geq 3 \), \( y[k] = 0 \). Therefore, the correct answer for \( y[k] \) when \( k \geq 4 \) is \( y[k] = 0 \), which corresponds to option (D).