Question

The figure below shows the impulse response of a linear time-invariant (LTI) system, \(h[n] = [0, 0.6, 0.8, 0.5, 0.3, 0]\) for \(n = 0\) to \(5\). For an input \(x[n] = [1, 0.6, 0, 0, 0, 0]\) for \(n = 0\) to \(5\), the maximum value of its output \(y[n]\) is . (Round off to one decimal place)

Hint:

Since \(x[n]\) has only two nonzero samples, \(y[n]=h[n]+0.6\,h[n-1]\); compute this for each n and find the largest value.

Answer 1

Correct Answer: 1.2

Step 1: Recall the convolution formula.
For an LTI system, the output is the convolution of the input and the impulse response:
\[ y[n] = x[n] * h[n] = \sum_{k} x[k]\, h[n-k] \]
Since \(x[n]\) is nonzero only at \(n = 0\) (\(x[0] = 1\)) and \(n = 1\) (\(x[1] = 0.6\)), the sum collapses to just two terms:
\[ y[n] = x[0]\, h[n] + x[1]\, h[n-1] = h[n] + 0.6\, h[n-1] \]

Step 2: List the impulse response values.
From the given sequence, \(h[0] = 0\), \(h[1] = 0.6\), \(h[2] = 0.8\), \(h[3] = 0.5\), \(h[4] = 0.3\), \(h[5] = 0\), and \(h[n] = 0\) outside this range.

Step 3: Compute y[n] for each n.
\[ y[0] = h[0] + 0.6\,h[-1] = 0 + 0 = 0 \]
\[ y[1] = h[1] + 0.6\,h[0] = 0.6 + 0.6(0) = 0.6 \]
\[ y[2] = h[2] + 0.6\,h[1] = 0.8 + 0.6(0.6) = 0.8 + 0.36 = 1.16 \]
\[ y[3] = h[3] + 0.6\,h[2] = 0.5 + 0.6(0.8) = 0.5 + 0.48 = 0.98 \]
\[ y[4] = h[4] + 0.6\,h[3] = 0.3 + 0.6(0.5) = 0.3 + 0.30 = 0.60 \]
\[ y[5] = h[5] + 0.6\,h[4] = 0 + 0.6(0.3) = 0.18 \]

Step 4: Find the maximum.
The computed sequence is \(y[n] = [0, 0.6, 1.16, 0.98, 0.60, 0.18]\). The largest value occurs at \(n=2\): \(y[2] = 1.16\). Rounded to one decimal place, this is \(1.2\).

Final Answer:
The maximum value of \(y[n]\) is \(1.2\). \[ \boxed{y_{max} = 1.2} \]