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.
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](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd07e8233-789e-4760-ba71-e70869af7b54%2F1fcf2879-61e1-43df-bfb6-f6cf7f2e3d73%2Ffufhq1u_processed.png&w=3840&q=75)
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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