Let's solve the given system step by step as requested. ### **Given:** \[ H(z) = \frac{z - \frac{1}{2}}{z^2 + \frac{3}{4}z + \frac{1}{8}} \] --- ## **a. Find the linear difference equation for the discrete time system** ### **Step 1: Write the system in terms of \( Y(z) \) and \( X(z) \)** Since \( H(z) = \frac{Y(z)}{X(z)} \): \[ Y(z) = H(z)X(z) \] \[ Y(z) (z^2 + \frac{3}{4}z + \frac{1}{8}) = X(z) (z - \frac{1}{2}) \] --- ### **Step 2: Write in time domain using inverse Z-transform** Recall: - \( zY(z) \leftrightarrow y[k+1] \) - \( z^2Y(z) \leftrightarrow y[k+2] \) - \( zX(z) \leftrightarrow x[k+1] \) - etc. So, \[ z^2 Y(z) + \frac{3}{4}z Y(z) + \frac{1}{8}Y(z) = zX(z) - \frac{1}{2}X(z) \] Taking inverse Z-transform: \[ y[k+2] + \frac{3}{4}y[k+1] + \frac{1}{8}y[k] = x[k+1] - \frac{1}{2}x[k] \] **Final Linear Difference Equation:** \[ \boxed{ y[k+2] + \frac{3}{4}y[k+1] + \frac{1}{8}y[k] = x[k+1] - \frac{1}{2}x[k] } \] --- ## **b. Determine \( h[k] \), the unit impulse response of the system** ### **Step 1: Express \( H(z) \) for inverse Z-transform** Let’s factor and/or use partial fraction expansion for easier inversion. \[ H(z) = \frac{z - \frac{1}{2}}{z^2 + \frac{3}{4}z + \frac{1}{8}} \] Let’s factor the denominator: \[ z^2 + \frac{3}{4}z + \frac{1}{8} \] Use quadratic formula: \( z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) \( a = 1, b = \frac{3}{4}, c = \frac{1}{8} \) \[ z = \frac{-\frac{3}{4} \pm \sqrt{\left(\frac{3}{4}\right)^2 - 4 \cdot 1 \cdot \frac{1}{8}}}{2} = \frac{-\frac{3}{4} \pm \sqrt{\frac{9}{16} - \frac{1}{2}}}{2} = \frac{-\frac{3}{4} \pm \sqrt{\frac{9}{16} - \frac{8}{16}}}{2} = \frac{-\frac{3}{4} \pm \sqrt{\frac{1}{16}}}{2} = \frac{-\frac{3}{4} \pm \frac{1}{4}}{2} \] Roots: \[ z_1 = \frac{-\frac{3}{4} + \frac{1}{4}}{2} = \frac{-\frac{2}{4}}{2} = -\frac{1}{4} \] \[ z_2 = \frac{-\frac{3}{4} - \frac{1}{4}}{2} = \frac{-1}{2} \] So, \[ z^2 + \frac{3}{4}z + \frac{1}{8} = (z + \frac{1}{4})(z + \frac{1}{2}) \] Rewrite \( H(z) \): \[ H(z) = \frac{z - \frac{1}{2}}{(z + \frac{1}{4})(z + \frac{1}{2})} \] --- ### **Step 2: Partial Fraction Expansion** Let’s express: \[ \frac{z - \frac{1}{2}}{(z + \frac{1}{4})(z + \frac{1}{2})} = \frac{A}{z + \frac{1}{4}} + \frac{B}{z + \frac{1}{2}} \] Multiply both sides by the denominator: \[ z - \frac{1}{2} = A(z + \frac{1}{2}) + B(z + \frac{1}{4}) \] Solve for A, B: Set \( z = -\frac{1}{2} \): \[ (-\frac{1}{2}) - \frac{1}{2} = A(0) + B(-\frac{1}{2} + \frac{1}{4}) \implies -1 = B(-\frac{1}{4}) \implies B = 4 \] Set \( z = -\frac{1}{4} \): \[ (-\frac{1}{4}) - \frac{1}{2} = A(-\frac{1}{4} + \frac{1}{2}) + B(0) \implies -\frac{3}{4} = A(\frac{1}{4}) \implies A = -3 \] So, \[ H(z) = \frac{-3}{z + \frac{1}{4}} + \frac{4}{z + \frac{1}{2}} \] --- ### **Step 3: Find \( h[k] \) using inverse Z-transform** Recall: \[ \mathcal{Z}^{-1}\left(\frac{1}{z - a}\right) = a^k u[k] \] But we have \( z + a \). So, \[ \frac{1}{z + a} = \frac{1}{z - (-a)} \rightarrow (-a)^k u[k] \] So, \[ h[k] = -3 \cdot \left(-\frac{1}{4}\right)^k + 4 \cdot \left(-\frac{1}{2}\right)^k, \quad k \geq 0 \] --- **Final Answer for impulse response:** \[ \boxed{ h[k] = -3 \left(-\frac{1}{4}\right)^k + 4 \left(-\frac{1}{2}\right)^k, \quad k \geq 0 } \] --- ## **c. The frequency response function \( H(e^{j\Omega}) \)** ### **Step 1: Substitute \( z = e^{j\Omega} \) in \( H(z) \)** \[ H(e^{j\Omega}) = \frac{e^{j\Omega} - \frac{1}{2}}{(e^{j\Omega})^2 + \frac{3}{4}e^{j\Omega} + \frac{1}{8}} \] --- ### **Step 2: Write the final answer** \[ \boxed{ H(e^{j\Omega}) = \frac{e^{j\Omega} - \frac{1}{2}}{e^{2j\Omega} + \frac{3}{4}e^{j\Omega} + \frac{1}{8}} } \] --- ## **Summary of Answers** ### **a. Linear Difference Equation:** \[ y[k+2] + \frac{3}{4}y[k+1] + \frac{1}{8}y[k] = x[k+1] - \frac{1}{2}x[k] \] ### **b. Impulse Response:** \[ h[k] = -3 \left(-\frac{1}{4}\right)^k + 4 \left(-\frac{1}{2}\right)^k, \quad k \geq 0 \] ### **c. Frequency Response:** \[ H(e^{j\Omega}) = \frac{e^{j\Omega} - \frac{1}{2}}{e^{2j\Omega} + \frac{3}{4}e^{j\Omega} + \frac{1}{8}} \]

Let's analyze and solve the given problem step by step, providing clear explanations at each stage. --- ### **Given Transfer Function:** \[ H(z) = \frac{z - \frac{1}{2}}{z^2 + \frac{3}{4}z + \frac{1}{8}} \] --- ## **a. Derive the linear difference equation** ### **Step 1: Express in terms of \(Y(z)\) and \(X(z)\)** Since \(H(z) = \frac{Y(z)}{X(z)}\), multiplying both sides by \(X(z)\) gives: \[ Y(z) = H(z) \times X(z) \] which leads to: \[ Y(z) \times \left(z^2 + \frac{3}{4}z + \frac{1}{8}\right) = X(z) \times \left(z - \frac{1}{2}\right) \] ### **Step 2: Convert to the time domain** Recall the properties of the Z-transform: - \(zY(z) \leftrightarrow y[k+1]\) - \(z^2 Y(z) \leftrightarrow y[k+2]\) - \(z X(z) \leftrightarrow x[k+1]\) - \(X(z) \leftrightarrow x[k]\) Applying the inverse Z-transform: \[ y[k+2] + \frac{3}{4} y[k+1] + \frac{1}{8} y[k] = x[k+1] - \frac{1}{2} x[k] \] **Final difference equation:** \[ \boxed{ y[k+2] + \frac{3}{4} y[k+1] + \frac{1}{8} y[k] = x[k+1] - \frac{1}{2} x[k] } \] This relation describes how current and past outputs relate to current and past inputs. --- ## **b. Find the impulse response \(h[k]\)** ### **Step 1: Factor the denominator** First, factor the quadratic denominator: \[ z^2 + \frac{3}{4}z + \frac{1}{8} \] Compute roots using the quadratic formula: \[ z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \(a=1\), \(b=\frac{3}{4}\), \(c=\frac{1}{8}\): \[ z = \frac{-\frac{3}{4} \pm \sqrt{\left(\frac{3}{4}\right)^2 - 4 \times 1 \times \frac{1}{8}}}{2} \] Calculate discriminant: \[ \left(\frac{3}{4}\right)^2 = \frac{9}{16} \] \[ 4 \times \frac{1}{8} = \frac{1}{2} = \frac{8}{16} \] So, \[ z = \frac{-\frac{3}{4} \pm \sqrt{\frac{9}{16} - \frac{8}{16}}}{2} = \frac{-\frac{3}{4} \pm \sqrt{\frac{1}{16}}}{2} \] which simplifies to: \[ z = \frac{-\frac{3}{4} \pm \frac{1}{4}}{2} \] Calculating each root: - For the plus sign: \[ z_1 = \frac{-\frac{3}{4} + \frac{1}{4}}{2} = \frac{-\frac{2}{4}}{2} = -\frac{1}{4} \] - For the minus sign: \[ z_2 = \frac{-\frac{3}{4} - \frac{1}{4}}{2} = \frac{-1}{2} \quad \] Thus, the denominator factors as: \[ (z + \frac{1}{4})(z + \frac{1}{2}) \] ### **Step 2: Partial fraction expansion** Express \(H(z)\) as: \[ H(z) = \frac{z - \frac{1}{2}}{(z + \frac{1}{4})(z + \frac{1}{2})} \] Set: \[ \frac{z - \frac{1}{2}}{(z + \frac{1}{4})(z + \frac{1}{2})} = \frac{A}{z + \frac{1}{4}} + \frac{B}{z + \frac{1}{2}} \] Multiply through both sides by the denominator: \[ z - \frac{1}{2} = A(z + \frac{1}{2}) + B(z + \frac{1}{4}) \] Now, solve for \(A\) and \(B\): - To find \(A\): Set \(z = -\frac{1}{2}\): \[ -\frac{1}{2} - \frac{1}{2} = A(0) + B(-\frac{1}{2} + \frac{1}{4}) \Rightarrow -1 = B(-\frac{1}{4}) \Rightarrow B = 4 \] - To find \(A\): Set \(z = -\frac{1}{4}\): \[ -\frac{1}{4} - \frac{1}{2} = A(-\frac{1}{4} + \frac{1}{2}) + B(0) \Rightarrow -\frac{3}{4} = A(\frac{1}{4}) \Rightarrow A = -3 \] ### **Step 3: Write the inverse Z-transform for \(h[k]\)** Recall the inverse Z-transform: \[ \mathcal{Z}^{-1}\left(\frac{1}{z + a}\right) = (-a)^k u[k] \] Thus, \[ h[k] = A \times (-\frac{1}{4})^k + B \times (-\frac{1}{2})^k \] Substitute the values: \[ h[k] = -3 \times \left(-\frac{1}{4}\right)^k + 4 \times \left(-\frac{1}{2}\right)^k, \quad k \geq 0 \] --- ### **Final expression for impulse response:** \[ \boxed{ h[k] = -3 \left(-\frac{1}{4}\right)^k + 4 \left(-\frac{1}{2}\right)^k, \quad k \geq 0 } \] --- ## **c. Find the frequency response \(H(e^{j\Omega})\)** ### **Step 1: Substitute \(z = e^{j\Omega}\) into \(H(z)\)** \[ H(e^{j\Omega}) = \frac{e^{j\Omega} - \frac{1}{2}}{(e^{j\Omega})^2 + \frac{3}{4} e^{j\Omega} + \frac{1}{8}} \] This expression describes the system's response at each frequency \(\Omega\). --- ## **Summary of Results:** - **Difference equation:** \[ \boxed{ y[k+2] + \frac{3}{4} y[k+1] + \frac{1}{8} y[k] = x[k+1] - \frac{1}{2} x[k] } \] - **Impulse response:** \[ \boxed{ h[k] = -3 \left(-\frac{1}{4}\right)^k + 4 \left(-\frac{1}{2}\right)^k, \quad k \geq 0 } \] - **Frequency response:** \[ \boxed{ H(e^{j\Omega}) = \frac{e^{j\Omega} - \frac{1}{2}}{e^{2j\Omega} + \frac{3}{4}e^{j\Omega} + \frac{1}{8}} } \] This completes the detailed step-by-step solution with explanations.

Let's carefully analyze and solve the given problem step by step, providing clear explanations at each stage. --- ### **Given Transfer Function:** \[ H(z) = \frac{z - \frac{1}{2}}{z^2 + \frac{3}{4}z + \frac{1}{8}} \] --- ## **a. Derive the linear difference equation** ### **Step 1: Express \(Y(z)\) in terms of \(X(z)\)** Since the transfer function relates input and output in the Z-domain as \(H(z) = \frac{Y(z)}{X(z)}\), multiplying both sides by \(X(z)\) gives: \[ Y(z) = H(z) \times X(z) \] which leads to: \[ Y(z) \times \left(z^2 + \frac{3}{4}z + \frac{1}{8}\right) = X(z) \times \left(z - \frac{1}{2}\right) \] ### **Step 2: Convert to the time domain** Recall the properties of the Z-transform: - \(zY(z) \leftrightarrow y[k+1]\) - \(z^2 Y(z) \leftrightarrow y[k+2]\) - \(z X(z) \leftrightarrow x[k+1]\) - \(X(z) \leftrightarrow x[k]\) Applying the inverse Z-transform, each term corresponds to: \[ \boxed{ y[k+2] + \frac{3}{4} y[k+1] + \frac{1}{8} y[k] = x[k+1] - \frac{1}{2} x[k] } \] This is the **linear difference equation** governing the system. It relates current and past outputs to current and past inputs. --- ## **b. Find the impulse response \(h[k]\)** The impulse response is the system output when the input is a unit impulse \(\delta[k]\). In the Z-domain, this corresponds to evaluating \(H(z)\) directly, as \(H(z)\) is the Z-transform of \(h[k]\). ### **Step 1: Factor the denominator** First, factor the quadratic denominator: \[ z^2 + \frac{3}{4}z + \frac{1}{8} \] Calculate roots using the quadratic formula: \[ z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \(a=1\), \(b=\frac{3}{4}\), \(c=\frac{1}{8}\). Compute discriminant: \[ \left(\frac{3}{4}\right)^2 - 4 \times 1 \times \frac{1}{8} = \frac{9}{16} - \frac{1}{2} = \frac{9}{16} - \frac{8}{16} = \frac{1}{16} \] Square root: \[ \sqrt{\frac{1}{16}} = \frac{1}{4} \] Calculate roots: \[ z = \frac{-\frac{3}{4} \pm \frac{1}{4}}{2} \] - For the plus sign: \[ z_1 = \frac{-\frac{3}{4} + \frac{1}{4}}{2} = \frac{-\frac{2}{4}}{2} = -\frac{1}{4} \] - For the minus sign: \[ z_2 = \frac{-\frac{3}{4} - \frac{1}{4}}{2} = \frac{-1}{2} \] Thus, the denominator factors as: \[ (z + \frac{1}{4})(z + \frac{1}{2}) \] ### **Step 2: Partial fraction expansion** Express \(H(z)\) as a sum: \[ H(z) = \frac{z - \frac{1}{2}}{(z + \frac{1}{4})(z + \frac{1}{2})} = \frac{A}{z + \frac{1}{4}} + \frac{B}{z + \frac{1}{2}} \] Multiply both sides by the denominator to solve for \(A\) and \(B\): \[ z - \frac{1}{2} = A(z + \frac{1}{2}) + B(z + \frac{1}{4}) \] To find \(A\), set \(z = -\frac{1}{2}\): \[ -\frac{1}{2} - \frac{1}{2} = A(0) + B(-\frac{1}{2} + \frac{1}{4}) \Rightarrow -1 = B(-\frac{1}{4}) \Rightarrow B = 4 \] To find \(A\), set \(z = -\frac{1}{4}\): \[ -\frac{1}{4} - \frac{1}{2} = A(-\frac{1}{4} + \frac{1}{2}) + B(0) \Rightarrow -\frac{3}{4} = A(\frac{1}{4}) \Rightarrow A = -3 \] ### **Step 3: Inverse Z-transform** Recall that: \[ \mathcal{Z}^{-1}\left( \frac{1}{z + a} \right) = (-a)^k u[k] \] Thus, \[ h[k] = A \times \left(-\frac{1}{4}\right)^k + B \times \left(-\frac{1}{2}\right)^k = -3 \times \left(-\frac{1}{4}\right)^k + 4 \times \left(-\frac{1}{2}\right)^k \] for \(k \ge 0\). **Final expression for the impulse response:** \[ \boxed{ h[k] = -3 \left(-\frac{1}{4}\right)^k + 4 \left(-\frac{1}{2}\right)^k, \quad k \ge 0 } \] --- ## **c. Find the frequency response \(H(e^{j\Omega})\)** ### **Step 1: Substituting \(z = e^{j\Omega}\)** The frequency response is obtained by replacing \(z\) with \(e^{j\Omega}\): \[ H(e^{j\Omega}) = \frac{e^{j\Omega} - \frac{1}{2}}{(e^{j\Omega})^2 + \frac{3}{4} e^{j\Omega} + \frac{1}{8}} \] This expression describes how the system reacts at each normalized frequency \(\Omega\). --- ## **Summary** - **Difference Equation:** \[ \boxed{ y[k+2] + \frac{3}{4} y[k+1] + \frac{1}{8} y[k] = x[k+1] - \frac{1}{2} x[k] } \] - **Impulse Response:** \[ \boxed{ h[k] = -3 \left(-\frac{1}{4}\right)^k + 4 \left(-\frac{1}{2}\right)^k, \quad k \ge 0 } \] - **Frequency Response:** \[ \boxed{ H(e^{j\Omega}) = \frac{e^{j\Omega} - \frac{1}{2}}{e^{2j\Omega} + \frac{3}{4} e^{j\Omega} + \frac{1}{8}} } \] This completes the detailed step-by-step solution with explanations for each part.

Certainly! Here's a detailed, step-by-step explanation for each part of the question, written in a clear and concise manner suitable for Chegg guidelines: --- **Given Transfer Function:** \[ H(z) = \frac{z - \frac{1}{2}}{z^2 + \frac{3}{4}z + \frac{1}{8}} \] --- ### **a) Derive the linear difference equation** **Explanation:** The transfer function \( H(z) \) relates the Z-transforms of the output \( y[k] \) and input \( x[k] \). Specifically, \( H(z) = \frac{Y(z)}{X(z)} \). To find the difference equation that describes how the current and past outputs relate to the current and past inputs, we rearrange this relation: \[ Y(z) = H(z) \times X(z) \] Multiplying both sides by the denominator of \( H(z) \) gives: \[ Y(z) \times \left( z^2 + \frac{3}{4}z + \frac{1}{8} \right) = X(z) \times \left( z - \frac{1}{2} \right) \] Applying inverse Z-transform properties—where powers of \( z \) correspond to shifts in time—results in a difference equation: \[ y[k+2] + \frac{3}{4} y[k+1] + \frac{1}{8} y[k] = x[k+1] - \frac{1}{2} x[k] \] This equation shows how the output at future times relates to past outputs and inputs, capturing the system's behavior. --- ### **b) Find the impulse response \( h[k] \)** **Explanation:** The impulse response \( h[k] \) is the output of the system when the input is a unit impulse \( \delta[k] \). It characterizes the system's inherent response. To find \( h[k] \), we analyze the transfer function \( H(z) \) directly: 1. **Factor the denominator polynomial** to find its roots, which helps in partial fraction decomposition. The roots are found using the quadratic formula, yielding two roots: \( -\frac{1}{4} \) and \( -\frac{1}{2} \). 2. **Express \( H(z) \) as a sum of simpler fractions**: \[ H(z) = \frac{z - \frac{1}{2}}{(z + \frac{1}{4})(z + \frac{1}{2})} = \frac{A}{z + \frac{1}{4}} + \frac{B}{z + \frac{1}{2}} \] 3. **Solve for constants \( A \) and \( B \)** by substituting suitable values of \( z \). This yields \( A = -3 \) and \( B = 4 \). 4. **Inverse Z-transform** each term: recall that \( \frac{1}{z + a} \) transforms to \( (-a)^k u[k] \). Applying this, the impulse response becomes: \[ h[k] = -3 \left(-\frac{1}{4}\right)^k + 4 \left(-\frac{1}{2}\right)^k \] This formula gives the system's response at each discrete time \( k \). --- ### **c) Find the frequency response \( H(e^{j\Omega}) \)** **Explanation:** The frequency response describes how the system processes signals at different frequencies. To find it, substitute \( z = e^{j\Omega} \) into the transfer function: \[ H(e^{j\Omega}) = \frac{e^{j\Omega} - \frac{1}{2}}{(e^{j\Omega})^2 + \frac{3}{4} e^{j\Omega} + \frac{1}{8}} \] This expression allows us to analyze the magnitude and phase of the system's response across the frequency spectrum \( \Omega \), which is useful for understanding filtering characteristics or stability. --- **Summary:** - The **difference equation** describes the system's recursive behavior in the time domain. - The **impulse response** provides insight into the system's inherent dynamics. - The **frequency response** reveals how the system behaves at various frequencies. This step-by-step explanation combines mathematical derivations with conceptual understanding, aligning with Chegg guidelines for clarity and completeness.