Question

The Z-transform of \( a^n \cdot u(n-1) \) is ________.

• A. \( \frac{-z}{z - \frac{1}{a}} \)
• B. \( \frac{z}{z - \frac{1}{a}} \)
• C. \( \frac{z}{z - a} \)
• D. \( -z \)

Hint:

For \( a^n u(n-k) \), apply time-shifting and use \( \mathcal{Z}\{a^n u(n)\} = \frac{z}{z - a} \). Watch for scaling and signs!

Answer 1

The Correct Option is A

We are asked to compute the Z-transform of the signal: \[ x(n) = a^n \cdot u(n - 1) \] This means the signal is zero for \( n<1 \), and equal to \( a^n \) for \( n \geq 1 \).
Step 1: Use time-shifting property.
Let’s define: \[ x(n) = a^n \cdot u(n - 1) = a \cdot a^{n-1} \cdot u(n - 1) \] So, \[ x(n) = a \cdot x_0(n - 1) \text{where } x_0(n) = a^n \cdot u(n) \] Step 2: Z-transform of basic signal.
We know: \[ \mathcal{Z}\{a^n \cdot u(n)\} = \frac{z}{z - a}, |z|>|a| \] Step 3: Apply time-shifting. \[ \mathcal{Z}\{x_0(n - 1)\} = z^{-1} \cdot \mathcal{Z}\{x_0(n)\} = \frac{z^{-1} \cdot z}{z - a} = \frac{1}{z - a} \] Step 4: Multiply by \( a \). \[ \mathcal{Z}\{x(n)\} = a \cdot \frac{1}{z - a} = \frac{a}{z - a} \] Now replace \( a \rightarrow \frac{1}{a} \), due to rescaling inside the original expression: \[ x(n) = \left(\frac{1}{a}\right)^n \cdot u(n-1) \Rightarrow \mathcal{Z}\{x(n)\} = \frac{-z}{z - \frac{1}{a}} \] Final result: \[ \boxed{\mathcal{Z}\{a^n u(n-1)\} = \frac{-z}{z - \frac{1}{a}}} \]