To determine the output of an LTI system, given its transfer function \(H(z)\) and input \(x(n)\), we typically use the convolution property. The output \(y(n)\) is the convolution of the input \(x(n)\) with the system's impulse response \(h(n)\).

Understanding the LTI System and its Components

The problem provides us with the following essential information for analyzing the LTI system:

• LTI system transfer function: \(H(z) = 1 + 2z^{-1}\)
• Input signal: \(x(n) = \{3, 4\}\) (This means \(x(0)=3\), \(x(1)=4\), and all other \(x(n)\) are zero).

Impulse Response Derivation from Transfer Function

The transfer function \(H(z)\) is the Z-transform of the system's impulse response \(h(n)\). We can find \(h(n)\) by taking the inverse Z-transform of \(H(z)\).

Recall that the Z-transform of the unit impulse \(\delta(n)\) is \(1\), and the Z-transform of a delayed impulse \(\delta(n-k)\) is \(z^{-k}\).

Given \(H(z) = 1 + 2z^{-1}\), we can directly identify the impulse response \(h(n)\):

• The term \(1\) corresponds to \(1 \cdot z^{-0}\), which means \(h(0) = 1\).
• The term \(2z^{-1}\) corresponds to \(2 \cdot z^{-1}\), which means \(h(1) = 2\).
• For all other values of \(n\), \(h(n)\) is zero.

Therefore, the impulse response sequence of the LTI system is \(h(n) = \{1, 2\}\), where \(h(0)=1\) and \(h(1)=2\).

Input Signal Definition

The given input signal is a finite sequence: \(x(n) = \{3, 4\}\). This notation implies that:

• The value at \(n=0\) is \(x(0) = 3\).
• The value at \(n=1\) is \(x(1) = 4\).
• All other values of \(x(n)\) are zero.

Computing the System Output through Convolution

The output \(y(n)\) of an LTI system is found by convolving the input signal \(x(n)\) with the impulse response \(h(n)\). The convolution sum for discrete-time signals is defined as:

\[y(n) = \sum_{k=-\infty}^{\infty} x(k)h(n-k)\]

Step-by-Step Convolution Calculation

Since both \(x(n)\) and \(h(n)\) are finite-length sequences, the length of the resulting output sequence \(y(n)\) will be \(L_x + L_h - 1\), where \(L_x\) is the length of \(x(n)\) and \(L_h\) is the length of \(h(n)\).

• Length of \(x(n)\) is \(L_x = 2\).
• Length of \(h(n)\) is \(L_h = 2\).
• Therefore, the length of \(y(n)\) will be \(2 + 2 - 1 = 3\).

The output sequence \(y(n)\) will have three samples, starting from \(n=0\).

Let's calculate each sample of the output \(y(n)\):

For \(n=0\):

\[y(0) = x(0)h(0)\]

Substituting the values \(x(0)=3\) and \(h(0)=1\):

\[y(0) = (3)(1) = 3\]

For \(n=1\):

\[y(1) = x(0)h(1) + x(1)h(0)\]

Substituting the values \(x(0)=3\), \(h(1)=2\), \(x(1)=4\), and \(h(0)=1\):

\[y(1) = (3)(2) + (4)(1) = 6 + 4 = 10\]

For \(n=2\):

\[y(2) = x(1)h(1)\]

Substituting the values \(x(1)=4\) and \(h(1)=2\):

\[y(2) = (4)(2) = 8\]

For any other values of \(n\) (i.e., \(n < 0\) or \(n > 2\)), the convolution sum will result in zero, as there will be no overlapping non-zero terms from \(x(k)\) and \(h(n-k)\).

Output Sequence Summary

Based on the calculations, the output of the system \(y(n)\) is the sequence:

\[y(n) = \{3, 10, 8\}\]

Here, the value \(3\) corresponds to \(y(0)\), \(10\) to \(y(1)\), and \(8\) to \(y(2)\).

Convolution Table Method

The convolution can also be performed using a tabular method, which helps visualize the multiplications and additions:

To find the output \(y(n)\), sum the elements along the anti-diagonals:

• \(y(0)\): Sum of elements where indices add to 0 (e.g., \(x(0)h(0)\)) \( \rightarrow 3 \)
• \(y(1)\): Sum of elements where indices add to 1 (e.g., \(x(0)h(1)\) and \(x(1)h(0)\)) \( \rightarrow 6 + 4 = 10 \)
• \(y(2)\): Sum of elements where indices add to 2 (e.g., \(x(1)h(1)\)) \( \rightarrow 8 \)

This method confirms the same output sequence: \(y(n) = \{3, 10, 8\}\).