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An LTI system has the relationship y[n] = ∑ 𝑥[𝑘]𝑔[𝑛 − 2𝑘]∞𝑘= −∞ , where g[n] = u[n] –u[n-4]. Determine y[n] if a) x[n] = δ[n-1], b) x[n] = δ[n-2].

Question

An LTI system has the relationship

y[n]=k=x[k]g[n2k] y[n] = \sum_{k = -\infty}^{\infty} x[k] g[n - 2k]
where
g[n]=u[n]u[n4] g[n] = u[n] - u[n-4]
Determine y[n] if
a) x[n]=δ[n1] x[n] = \delta[n-1]
b) x[n]=δ[n2] x[n] = \delta[n-2]

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Solution

To determine y[n] for the given LTI system, we need to substitute the given values of x[n] into the equation y[n] = ∑ 𝑥[𝑘]𝑔[𝑛 − 2𝑘]∞𝑘= −∞ and simplify the expression.

a) For x[n] = δ[n-1], we substitute this value into the equation:

y[n] = ∑ δ[k-1] * g[n - 2k] from k = -∞ to ∞

Since δ[k-1] is non-zero only when k = 1, the summation simplifies to:

y[n] = δ[1-1] * g[n - 2(1)] = δ[0] * g[n - 2] = g[n - 2]

Now, let's substitute the expression for g[n]:

y[n] = (u[n] - u[n-4])[n - 2]

Since u[n] is the unit step function, it is equal to 1 for n >= 0 and 0 for n < 0. Similarly, u[n-4] is equal to 1 for n >= 4 and 0 for n < 4.

Therefore, we can simplify the expression further:

y[n] = (1 - 1)[n - 2] = 0 for n < 2 y[n] = (1 - 0)[n - 2] = n - 2 for 2 <= n < 4 y[n] = (0 - 0)[n - 2] = 0 for n >= 4

b) For x[n] = δ[n-2], we follow the same steps as above:

y[n] = ∑ δ[k-2] * g[n - 2k] from k = -∞ to ∞

Since δ[k-2] is non-zero only when k = 2, the summation simplifies to:

y[n] = δ[2-2] * g[n - 2(2)] = δ[0] * g[n - 4] = g[n - 4]

Substituting the expression for g[n]:

y[n] = (u[n] - u[n-4])[n - 4]

Using the properties of the unit step function, we can simplify the expression:

y[n] = (1 - 1)[n - 4] = 0 for n < 4 y[n] = (1 - 0)[n - 4] = n - 4 for n >= 4

Therefore, the values of y[n] for the given LTI system are:

  • For x[n] = δ[n-1]: y[n] = 0 for n < 2, y[n] = n - 2 for 2 <= n < 4, and y[n] = 0 for n >= 4.
  • For x[n] = δ[n-2]: y[n] = 0 for n < 4, and y[n] = n - 4 for n >= 4.

This problem has been solved

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