Question

If the Z-transform of a finite-duration discrete-time signal \( x[n] \) is \( X(z) \), then the Z-transform of the signal \( y[n] = x[2n] \) is:

• A. \( Y(z) = X(z^2) \)
• B. \( Y(z) = \frac{1}{2}\left[ X(z^{-1/2}) + X(-z^{-1/2}) \right] \)
• C. \( Y(z) = \frac{1}{2}\left[ X(z^{1/2}) + X(-z^{1/2}) \right] \)
• D. \( Y(z) = \frac{1}{2}\left[ X(z^2) + X(-z^2) \right] \)

Hint:

Z-Transform:} The Z-transform of a sequence \( x[n] \) is: \[ X(z) = \sum_{n=0}^{N-1} x[n]z^{-n}. \] Fractional Substitution:} - Substituting \( z^{1/2} \) or \( -z^{1/2} \) into \( X(z) \) helps analyze fractional intervals. - For \( x(-z^{1/2}) \), alternating signs are introduced by \((-1)^n\). Averaging for Downsampling:} Adding \( X(z^{1/2}) \) and \( X(-z^{1/2}) \) isolates even powers: \[ Y(z) = \frac{1}{2} \left[X(z^{1/2}) + X(-z^{1/2})\right]. \] Result:} For \( y[n] = x[2n] \), the Z-transform is: \[ Y(z) = 1 + 3z^{-1}. \]

Answer 1

The Correct Option is C

Let \( x[n] \leftrightarrow X(z) \), where \( X(z) \) represents the Z-transform of the sequence \( x[n] \). \[ X(z) = \sum_{n=0}^{3} x[n]z^{-n}, \] where the summation limits are from \( 0 \) to \( 3 \) because \( x[n] \) has 4 elements. Substituting the values \( x[n] = \{1, 2, 3, 1\} \), we obtain: \[ X(z) = 1 + 2z^{-1} + 3z^{-2} + z^{-3}. \] \subsection*{Substituting \( z^{1/2} \) into \( X(z) \)} To analyze the sequence at a fractional power of \( z \), substitute \( z^{1/2} \) into \( X(z) \): \[ X(z^{1/2}) = 1 + 2z^{-1/2} + 3z^{-1} + z^{-3/2}. \] \subsection*{Substituting \( -z^{1/2} \) into \( X(z) \)} Next, substitute \( -z^{1/2} \) into \( X(z) \), accounting for the alternating signs due to powers of \((-1)\): \[ \begin{aligned} x(-z^{1/2}) &= \left[1 + 2(-1)^{-1}z^{-1/2} + 3(-1)^{-2}z^{-1} + 1(-1)^{-3}z^{-3/2}\right]
&= \left[1 + (-2)z^{-1/2} + 3z^{-1} + (-1)z^{-3/2}\right]
&= \left[1 - 2z^{-1/2} + 3z^{-1} - z^{-3/2}\right]. \end{aligned} \] This is labeled as: \[ x(-z^{1/2}) = \left[1 - 2z^{-1/2} + 3z^{-1} - z^{-3/2}\right]. \tag{iii} \] \subsection*{Averaging \( X(z^{1/2}) \) and \( X(-z^{1/2}) \)} Adding \( X(z^{1/2}) \) and \( X(-z^{1/2}) \) isolates terms where even powers of \( z^{-1/2} \) are present: \[ x(z^{1/2}) + x(-z^{1/2}) = \left[2 + 6z^{-1}\right]. \] Taking the average: \[ \frac{1}{2}\left[x(z^{1/2}) + x(-z^{1/2})\right] = \left[1 + 3z^{-1}\right]. \tag{iv} \] \subsection*{Sequence Representation} The original sequence is: \[ x[n] = \{1, 2, 3, 1\}. \] If we downsample by keeping only the even indices (\(n = 2k\)): \[ y[n] = x[2n] = \{1, 3\}. \] The Z-transform of the downsampled sequence is: \[ y(z) = 1 + 3z^{-1}. \tag{v} \] \subsection*{Connection Between \( Y(z) \) and Averaging} The relationship between the Z-transform of the downsampled sequence \( Y(z) \) and the original sequence's fractional analysis is: \[ Y(z) = \frac{1}{2}\left[x(z^{1/2}) + x(-z^{1/2})\right]. \] This demonstrates that the downsampling operation is mathematically related to the averaged contributions of the sequence at fractional powers of \( z \).