a) Determine the inverse z-transform of X(z) = 3/(1 - 0.2z^(-1)) + 2/(1 - 0.7z^(-1)) if: i. ROC: |z| > 1 ii. ROC: |z| < 0.7 iii. ROC: 0.7 < |z| < 1 b) A linear time-invariant syste... a) Determine the inverse z-transform of X(z) = 3/(1 - 0.2z^(-1)) + 2/(1 - 0.7z^(-1)) if: i. ROC: |z| > 1 ii. ROC: |z| < 0.7 iii. ROC: 0.7 < |z| < 1 b) A linear time-invariant system is characterized by the system function H(z) = 1/(1 - 0.8z^(-1)) + 3/(1 - 4z^(-1)). Specify the ROC of H(z) and determine h(n) for the following conditions: i. The system is stable ii. The system is causal iii. The system is anticausal. Read more

Steps to Solve

1. Inverse Z-Transform for ROC: |z| > 1

To find the inverse z-transform of

$$ X(z) = \frac{3}{1 - 0.2z^{-1}} + \frac{2}{1 - 0.7z^{-1}} $$

for $|z| > 1$, we can use the property that the inverse z-transform of $\frac{1}{1 - az^{-1}}$ is $a^n u[n]$.

So,

• For the first term: The inverse z-transform is $3(0.2)^n u[n]$.

• For the second term: The inverse z-transform is $2(0.7)^n u[n]$.

Thus, the total inverse z-transform is:

$$ h[n] = 3(0.2)^n u[n] + 2(0.7)^n u[n] $$

2. Inverse Z-Transform for ROC: |z| < 0.7

For $|z| < 0.7$, we need to consider the inverse z-transform for the case $|z| < 1$. The inverse transforms remain the same, with the signs of $n$ being negative (anti-causal). Thus, we have:

• For the first term (anti-causal): The inverse z-transform is $3(0.2)^{-n-1} u[-n-1]$.

• For the second term: The inverse z-transform is $2(0.7)^{-n-1} u[-n-1]$.

So the resulting inverse z-transform is:

$$ h[n] = 3(0.2)^{-n-1} u[-n-1] + 2(0.7)^{-n-1} u[-n-1] $$

3. Inverse Z-Transform for ROC: 0.7 < |z| < 1

For $0.7 < |z| < 1$, we have a mixed case where both causal and anti-causal parts will occur:

• The causal part (for the first term): $3(0.2)^n u[n]$.

• The anti-causal part (for the second term): $2(0.7)^{-n-1} u[-n-1]$.

Thus, the total inverse z-transform is:

$$ h[n] = 3(0.2)^n u[n] + 2(0.7)^{-n-1} u[-n-1] $$

4. Region of Convergence for H(z)

Given the system function

$$ H(z) = \frac{1}{1 - 0.8z^{-1}} + \frac{3}{1 - 4z^{-1}} $$

The ROCs are determined as follows, considering poles:

• For $\frac{1}{1 - 0.8z^{-1}}$, the pole is at z = 0.8.
• For $\frac{3}{1 - 4z^{-1}}$, the pole is at z = 4.

5. Stability Analysis

• Stable: The system is stable if the ROC includes the outer region of the outermost pole, i.e. $|z| > 4$.

• Causal: The system is causal if the ROC is outside the outermost pole and includes $|z| > 4$.

• Anticausal: The system is anti-causal if the ROC is inside the innermost pole, i.e. $|z| < 0.8$.

6. Determining h(n)

Analyze h(n) based on the above ROCs:

• For stability (ROC: $|z| > 4$):

$$ h[n] = 0.8^n u[n] + 4^n u[n] $$

• For causality:

$$ h[n] = 0.8^n u[n] + 0 \quad \text{(for $|z| > 0.8$)} $$

• For anticausality:

$$ h[n] = 0 + 4^{-n} u[-n-1] \quad \text{(for $|z| < 0.8$)} $$