VIPSolutionsDetermine if the LTI system with the following impulse response is memoryless, causal, and stable. Note that ∙ denotes multiplication. ℎ [ � � ] = 5 ∙ � � [ � � + 4 ] ― 5 ∙ � � [ � � ― 2 ] ( where u [ n ] is the unit - step function ) ( i ) First sketch h [ n ] . Please clearly label the axes. ( ii ) Is this system memoryless? Please justify your answer. ( iii ) Is this system causal? Please justify your answer. ( iv ) Is this system stable? Please indicate the appropriate numerical value, and show your work.
Question:
Determine if the LTI system with the following impulse response is memoryless, causal, and stable. Note that ∙ denotes multiplication. ℎ [ � � ] = 5 ∙ � � [ � � + 4 ] ― 5 ∙ � � [ � � ― 2 ] ( where u [ n ] is the unit - step function ) ( i ) First sketch h [ n ] . Please clearly label the axes. ( ii ) Is this system memoryless? Please justify your answer. ( iii ) Is this system causal? Please justify your answer. ( iv ) Is this system stable? Please indicate the appropriate numerical value, and show your work.
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Asked by: Prakash
Created at: 2025-11-01 05:04:26
Asked by: Prakash
Created at: 2025-11-01 05:04:26
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# Discrete-Time LTI System: Impulse Response Analysis
Given:
\[
h[n] = 5 \cdot u[n + 4] - 5 \cdot u[n - 2]
\]
where \(u[n]\) is the unit step function.
---
## (i) Sketch \( h[n] \)
### Step 1: Understand the Impulse Response
- \(u[n+4]\): Unit step "turns on" at \(n = -4\).
- \(u[n-2]\): Unit step "turns on" at \(n = 2\).
Thus,
- For \(n < -4\): Both unit steps are , \(h[n] = \).
- For \(-4 \leq n < 2\): \(u[n+4]=1\), \(u[n-2]=\), \(h[n]=5\).
- For \(n \geq 2\): Both are 1, so \(h[n]=5-5=\).
### Step 2: Piecewise Representation
\[
h[n] =
\begin{cases}
5, & -4 \leq n < 2 \\
, & \text{otherwise}
\end{cases}
\]
### Step 3: Sketch
```plaintext
^
| ******
| * *
| 5 ---*----*----*----*----*----*----*----*----*----*----*----*----*-->
| -4 -3 -2 -1 1 2 3 4 5 n
|
+-------------------------------------------------------------->
```
- **y-axis:** \(h[n]\)
- **x-axis:** \(n\)
- \(h[n]=5\) for \(n = -4, -3, -2, -1, , 1\)
- \(h[n]=\) elsewhere
---
## (ii) Is the System Memoryless?
**Definition:**
A system is memoryless if its output at time \(n\) depends only on the input at time \(n\).
**Impulse Response Criterion:**
For memorylessness, \(h[n]\) must be nonzero only at \(n=\).
**Here:**
\(h[n]\) is nonzero for multiple values of \(n\) (\(-4 \leq n < 2\)), not just at \(n=\).
**Conclusion:**
**The system is _not memoryless_.**
---
## (iii) Is the System Causal?
**Definition:**
A system is causal if \(h[n]=\) for all \(n<\).
**Here:**
\(h[n]=5\) for \(n = -4, -3, -2, -1\), i.e., for \(n < \).
**Conclusion:**
**The system is _not causal_.**
---
## (iv) Is the System Stable?
**Definition:**
A system is BIBO stable if \(\sum_{n=-\infty}^{\infty} |h[n]| < \infty\).
**Computation:**
- Nonzero \(h[n]=5\) for \(n = -4, -3, -2, -1, , 1\) (6 values).
\[
\sum_{n=-\infty}^{\infty} |h[n]| = 6 \times 5 = 30 < \infty
\]
**Conclusion:**
**The system is _stable_.**
**Numerical value:** \(\sum |h[n]| = 30\)
---
## Summary Table
| Property | Answer | Justification |
|---------------|----------|------------------------------------------------|
| Memoryless | No | \(h[n] \neq \) for \(n \neq \) |
| Causal | No | \(h[n] \neq \) for \(n < \) |
| Stable | Yes | \(\sum |h[n]| = 30 < \infty\) |
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