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

Convolution in the time domain corresponds to plain polynomial multiplication in the $z$-transform domain, so this problem can also be solved by writing $X(z)$ and $H(z)$ as polynomials in $z^{-1}$ and multiplying them out.

From the given sequences:

\[ X(z) = 1 + 0.6\,z^{-1} \] \[ H(z) = 0.6\,z^{-1} + 0.8\,z^{-2} + 0.5\,z^{-3} + 0.3\,z^{-4} \]

The output transform is $Y(z) = X(z)H(z)$. Multiply term by term:

\[ Y(z) = (1)(0.6z^{-1}+0.8z^{-2}+0.5z^{-3}+0.3z^{-4}) + (0.6z^{-1})(0.6z^{-1}+0.8z^{-2}+0.5z^{-3}+0.3z^{-4}) \]

The first product gives $0.6z^{-1}+0.8z^{-2}+0.5z^{-3}+0.3z^{-4}$. The second product gives $0.36z^{-2}+0.48z^{-3}+0.30z^{-4}+0.18z^{-5}$.

Add the two, collecting terms of the same power of $z^{-1}$:

\[ Y(z) = 0.6z^{-1} + (0.8+0.36)z^{-2} + (0.5+0.48)z^{-3} + (0.3+0.30)z^{-4} + 0.18z^{-5} \] \[ Y(z) = 0.6z^{-1} + 1.16z^{-2} + 0.98z^{-3} + 0.60z^{-4} + 0.18z^{-5} \]

Reading off the coefficients directly gives the output sequence, since the coefficient of $z^{-n}$ is exactly $y[n]$:

\[ y[n] = [0,\ 0.6,\ 1.16,\ 0.98,\ 0.60,\ 0.18] \quad \text{for } n = 0,1,2,3,4,5 \]

Let's summarize:

• Multiplying $X(z)$ and $H(z)$ as polynomials is algebraically identical to convolving $x[n]$ and $h[n]$ in the time domain.
• Each coefficient of $z^{-n}$ in the product directly gives $y[n]$, with no separate sliding-sum step needed.
• The largest coefficient is $1.16$, at $n=2$, which rounds to $1.2$.

The maximum value of $y[n]$ is $1.2$.