Given a continuous-time system: 5500 x(t)[8(t – t) − 8(t+t)]dt − y(t) = 0.5 Explain what this system does. Is the system BIBO stable? Justify your answer. Is the system linear? Justify your answer. Is the system memoryless? Justify your answer. Is the system causal? Justify your answer. Is the system time variant? Justify your answer.
Given a continuous-time system: 5500 x(t)[8(t – t) − 8(t+t)]dt − y(t) = 0.5 Explain what this system does. Is the system BIBO stable? Justify your answer. Is the system linear? Justify your answer. Is the system memoryless? Justify your answer. Is the system causal? Justify your answer. Is the system time variant? Justify your answer.
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WHAT SYSTEM DOES ,PLEASE DO ALL( NEED NEAT HANDWRITTEN SOLUTION ONLY OTHERWISE DOWNVOTE).
![### Continuous-Time System Analysis
Given a continuous-time system:
\[ y(t) = 0.5 \int_{-\infty}^{\infty} x(\tau) [\delta(t - \tau) - \delta(t + \tau)] d\tau \]
#### Questions for Analysis:
1. **Explain what this system does.**
The given system's output \( y(t) \) is determined by convolving the input signal \( x(\tau) \) with the impulse response provided by the difference of two delta functions. Specifically, these delta functions are \( \delta(t - \tau) \), which represents a time shift, and \( \delta(t + \tau) \), which represents a time reflection and shift.
2. **Is the system BIBO stable? Justify your answer.**
**Bounded Input, Bounded Output (BIBO) Stability:** A system is BIBO stable if every bounded input produces a bounded output. To determine if this system is BIBO stable, we need to examine the impulse response:
\[ h(t) = 0.5[\delta(t) - \delta(t)] \]
Evaluating \( h(t) \):
\[ h(t) = 0.5[\delta(t) - \delta(t)] = 0 \]
Since the impulse response function \( h(t) \) is zero, the output \( y(t) \) will always be zero for any bounded input \( x(t) \). Therefore, the system is BIBO stable.
3. **Is the system linear? Justify your answer.**
**Linearity:** A system is linear if it satisfies the principles of superposition and scaling. The given system involves convolution, which is a linear operation. Therefore, the system is linear.
4. **Is the system memoryless? Justify your answer.**
**Memoryless:** A system is memoryless if the output at any time \( t \) depends only on the input at the same time \( t \). For convolution systems, the output depends on a range of input values. Hence, this system is not memoryless.
5. **Is the system causal? Justify your answer.**
**Causality:** A system is causal if the output at any time \( t \) depends only on input values at the present and past times (not](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F0aee0e3d-a57d-4304-8930-413dab086e20%2Fc02378d2-b0f6-484f-8608-f575e426de1f%2F0r988b_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Continuous-Time System Analysis
Given a continuous-time system:
\[ y(t) = 0.5 \int_{-\infty}^{\infty} x(\tau) [\delta(t - \tau) - \delta(t + \tau)] d\tau \]
#### Questions for Analysis:
1. **Explain what this system does.**
The given system's output \( y(t) \) is determined by convolving the input signal \( x(\tau) \) with the impulse response provided by the difference of two delta functions. Specifically, these delta functions are \( \delta(t - \tau) \), which represents a time shift, and \( \delta(t + \tau) \), which represents a time reflection and shift.
2. **Is the system BIBO stable? Justify your answer.**
**Bounded Input, Bounded Output (BIBO) Stability:** A system is BIBO stable if every bounded input produces a bounded output. To determine if this system is BIBO stable, we need to examine the impulse response:
\[ h(t) = 0.5[\delta(t) - \delta(t)] \]
Evaluating \( h(t) \):
\[ h(t) = 0.5[\delta(t) - \delta(t)] = 0 \]
Since the impulse response function \( h(t) \) is zero, the output \( y(t) \) will always be zero for any bounded input \( x(t) \). Therefore, the system is BIBO stable.
3. **Is the system linear? Justify your answer.**
**Linearity:** A system is linear if it satisfies the principles of superposition and scaling. The given system involves convolution, which is a linear operation. Therefore, the system is linear.
4. **Is the system memoryless? Justify your answer.**
**Memoryless:** A system is memoryless if the output at any time \( t \) depends only on the input at the same time \( t \). For convolution systems, the output depends on a range of input values. Hence, this system is not memoryless.
5. **Is the system causal? Justify your answer.**
**Causality:** A system is causal if the output at any time \( t \) depends only on input values at the present and past times (not
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