Consider the (causal) IIR filter described by the difference equation y[n] – 0.25y[n – 1] = x[n] + 0.5x[n – 1] (i) Determine the system function of this filter (including its region of convergence). Is the filter stable? (ii) Using z-transforms, determine the response y[n] of this filter to the two-sided input sequence x[n] = 3.0u[–n - 1] + 2.0(0.75)"u[n] (iii) Determine the impulse response of a noncausal LTI system that satisfies the same difference equation.

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**Consider a Causal IIR Filter**

A causal Infinite Impulse Response (IIR) filter is characterized by the following difference equation:

\[ y[n] - 0.25y[n-1] = x[n] + 0.5x[n-1] \]

To analyze this filter, we will address the following points:

### (i) System Function and Stability

**Determine the system function (including its region of convergence) and assess the filter's stability.**

**Solution:**
1. **System Function (H(z))**: Using the z-transform of the difference equation, we can express the system function \( H(z) \) as:
   
   \[ Y(z) - 0.25Y(z)z^{-1} = X(z) + 0.5X(z)z^{-1} \]

   Rearrange this to solve for \( H(z) \):

   \[ Y(z) (1 - 0.25z^{-1}) = X(z) (1 + 0.5z^{-1}) \]

   \[ H(z) = \frac{Y(z)}{X(z)} = \frac{1 + 0.5z^{-1}}{1 - 0.25z^{-1}} \]

   Writing in standard form:

   \[ H(z) = \frac{1 + 0.5z^{-1}}{1 - 0.25z^{-1}} \]
   
2. **Region of Convergence (ROC)**: For causal systems, the ROC is \( |z| > 0.25 \).

3. **Stability**: A system is stable if the ROC includes the unit circle \( |z| = 1 \). Here, the ROC \( |z| > 0.25 \) does indeed include the unit circle, hence the filter is stable.

### (ii) Filter Response to a Given Input Sequence

**Using z-transforms, determine the response \( y[n] \) of this filter to the two-sided input sequence:**

\[ x[n] = 3.0u[-n - 1] + 2.0(0.75)^n u[n] \]

**Solution:**
1. The given input sequence consists of two terms:

   - \( 3.0u[-n - 1] \): A step function starting at \( n = -1
Transcribed Image Text:**Consider a Causal IIR Filter** A causal Infinite Impulse Response (IIR) filter is characterized by the following difference equation: \[ y[n] - 0.25y[n-1] = x[n] + 0.5x[n-1] \] To analyze this filter, we will address the following points: ### (i) System Function and Stability **Determine the system function (including its region of convergence) and assess the filter's stability.** **Solution:** 1. **System Function (H(z))**: Using the z-transform of the difference equation, we can express the system function \( H(z) \) as: \[ Y(z) - 0.25Y(z)z^{-1} = X(z) + 0.5X(z)z^{-1} \] Rearrange this to solve for \( H(z) \): \[ Y(z) (1 - 0.25z^{-1}) = X(z) (1 + 0.5z^{-1}) \] \[ H(z) = \frac{Y(z)}{X(z)} = \frac{1 + 0.5z^{-1}}{1 - 0.25z^{-1}} \] Writing in standard form: \[ H(z) = \frac{1 + 0.5z^{-1}}{1 - 0.25z^{-1}} \] 2. **Region of Convergence (ROC)**: For causal systems, the ROC is \( |z| > 0.25 \). 3. **Stability**: A system is stable if the ROC includes the unit circle \( |z| = 1 \). Here, the ROC \( |z| > 0.25 \) does indeed include the unit circle, hence the filter is stable. ### (ii) Filter Response to a Given Input Sequence **Using z-transforms, determine the response \( y[n] \) of this filter to the two-sided input sequence:** \[ x[n] = 3.0u[-n - 1] + 2.0(0.75)^n u[n] \] **Solution:** 1. The given input sequence consists of two terms: - \( 3.0u[-n - 1] \): A step function starting at \( n = -1
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