LTI system accepts the input x[n] = 2"u[n] nd produces the output y[n] = c8[n] + (-1)"u[n] here c # 0 is a real constant. O Determine the system function H(z) and its region of convergence, expressing your answer in terms of c. i) If the input x1[n] = (-4)", ne Z results in zero output at all times, determine the value of c. Also, determine the output Y2[n] produced by the input x2[n] = 3", n E Z (It is not necessary to invert z-transforms in this problem).
LTI system accepts the input x[n] = 2"u[n] nd produces the output y[n] = c8[n] + (-1)"u[n] here c # 0 is a real constant. O Determine the system function H(z) and its region of convergence, expressing your answer in terms of c. i) If the input x1[n] = (-4)", ne Z results in zero output at all times, determine the value of c. Also, determine the output Y2[n] produced by the input x2[n] = 3", n E Z (It is not necessary to invert z-transforms in this problem).
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![### Problem 4: LTI System Analysis
#### Problem Statement
An LTI (Linear Time-Invariant) system accepts the input:
\[ x[n] = 2^n u[n] \]
and produces the output:
\[ y[n] = c\delta[n] + (-1)^n u[n] \]
where \( c \neq 0 \) is a real constant.
##### Tasks:
1. **Determine the system function \( H(z) \) and its region of convergence, expressing your answer in terms of \( c \).**
2. **If the input**
\[ x_1[n] = (-4)^n, \quad n \in \mathbb{Z} \]
results in zero output at all times, determine the value of \( c \). Also, determine the output \( y_2[n] \) produced by the input:
\[ x_2[n] = 3^n, \quad n \in \mathbb{Z} \]
##### Notes:
- It is not necessary to invert \( z \)-transforms in this problem.
#### Step-by-Step Solution
**(i) Determining the System Function \( H(z) \)**
1. Using the input-output relationship of the LTI system, the output in the z-domain \( Y(z) \) can be expressed as:
\[ Y(z) = H(z)X(z) \]
2. Given the input \( x[n] = 2^n u[n] \), taking the z-transform:
\[ X(z) = \frac{1}{1 - 2z^{-1}} \quad \text{for } |z| > 2 \]
3. Given the output \( y[n] = c\delta[n] + (-1)^n u[n] \), the z-transform of \( y[n] \) is:
\[ Y(z) = c + \frac{1}{1 + z^{-1}} \quad \text{for } |z| > 1 \]
4. Therefore, the system function \( H(z) \) is:
\[ H(z) = \frac{Y(z)}{X(z)} = \left( c + \frac{1}{1 + z^{-1}} \right) \left( 1 - 2z^{-1} \right) \]
**Region of Convergence:**
\[ |z| > 2 \text](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd07e8233-789e-4760-ba71-e70869af7b54%2F77c18f62-474d-4106-9344-59f8dd953abd%2Fbf2udo_processed.png&w=3840&q=75)
Transcribed Image Text:### Problem 4: LTI System Analysis
#### Problem Statement
An LTI (Linear Time-Invariant) system accepts the input:
\[ x[n] = 2^n u[n] \]
and produces the output:
\[ y[n] = c\delta[n] + (-1)^n u[n] \]
where \( c \neq 0 \) is a real constant.
##### Tasks:
1. **Determine the system function \( H(z) \) and its region of convergence, expressing your answer in terms of \( c \).**
2. **If the input**
\[ x_1[n] = (-4)^n, \quad n \in \mathbb{Z} \]
results in zero output at all times, determine the value of \( c \). Also, determine the output \( y_2[n] \) produced by the input:
\[ x_2[n] = 3^n, \quad n \in \mathbb{Z} \]
##### Notes:
- It is not necessary to invert \( z \)-transforms in this problem.
#### Step-by-Step Solution
**(i) Determining the System Function \( H(z) \)**
1. Using the input-output relationship of the LTI system, the output in the z-domain \( Y(z) \) can be expressed as:
\[ Y(z) = H(z)X(z) \]
2. Given the input \( x[n] = 2^n u[n] \), taking the z-transform:
\[ X(z) = \frac{1}{1 - 2z^{-1}} \quad \text{for } |z| > 2 \]
3. Given the output \( y[n] = c\delta[n] + (-1)^n u[n] \), the z-transform of \( y[n] \) is:
\[ Y(z) = c + \frac{1}{1 + z^{-1}} \quad \text{for } |z| > 1 \]
4. Therefore, the system function \( H(z) \) is:
\[ H(z) = \frac{Y(z)}{X(z)} = \left( c + \frac{1}{1 + z^{-1}} \right) \left( 1 - 2z^{-1} \right) \]
**Region of Convergence:**
\[ |z| > 2 \text
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