Question

Find the z-transform of x[n] = $\delta$[n] + $\delta$[n + 1]:

• A. 1 + z$^{-1}$
• B. z + 1
• C. 1 + z
• D. z$^{-1}$

Hint:

For any finite-length sequence, you can find the Z-transform directly by inspection:
\[ X(z) = \dots + x[-1]z^1 + x[0]z^0 + x[1]z^{-1} + \dots \] Here, $x[0]=1$ and $x[-1]=1$. Substituting these values gives:
\[ X(z) = 1 \cdot z^1 + 1 \cdot z^0 = z + 1 \] This method is fast and highly reliable.

Answer 1

The Correct Option is C

Step 1: Understanding the Problem:
This question asks for the Z-transform of a discrete-time sequence composed of two unit impulse components: one at the origin and one advanced by one sample.

Step 2: Key Formula or Approach:
The definition of the unilateral/bilateral Z-transform is:
\[ X(z) = \sum_{n=-\infty}^{\infty} x[n] z^{-n} \]
The Z-transform of a unit impulse $\delta[n]$ is:
\[ \mathcal{Z}\{\delta[n]\} = 1 \]
The time-shifting property of the Z-transform states:
\[ \mathcal{Z}\{x[n - n_0]\} = z^{-n_0} X(z) \]

Step 3: Detailed Explanation:

• Given signal:
\[ x[n] = \delta[n] + \delta[n+1] \]
• By utilizing the linearity property of the Z-transform, we can compute the transform of each term individually:
\[ X(z) = \mathcal{Z}\{\delta[n]\} + \mathcal{Z}\{\delta[n+1]\} \]
• The first term is a standard impulse at $n = 0$:
\[ \mathcal{Z}\{\delta[n]\} = 1 \]
• For the second term, we apply the time-shifting property where the shift is $n_0 = -1$:
\[ \mathcal{Z}\{\delta[n - (-1)]\} = z^{-(-1)} \mathcal{Z}\{\delta[n]\} = z^1 \times 1 = z \]
• Adding the two terms together:
\[ X(z) = 1 + z \]
• This finite-duration sequence has an ROC that encompasses the entire z-plane, except for $z = \infty$.

Step 4: Final Answer
The Z-transform of the given signal is $1 + z$, which corresponds to option (C).