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What are the steady state values of the signals, 1-exp(-t), and 1-k*exp(-k*t)?
- a)1, k
- b)1, 1/k
- c)k, k
- d)1, 1
Correct answer is option 'D'. Can you explain this answer?
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What are the steady state values of the signals, 1-exp(-t), and 1-k*ex...
Consider limit at t tending to infinity, we obtain 1 for both cases.
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What are the steady state values of the signals, 1-exp(-t), and 1-k*ex...
Steady State Values of the Signals
The steady state value of a signal is the value it approaches as time approaches infinity. In this case, we are given two signals: 1 - exp(-t) and 1 - k*exp(-k*t). We need to find their steady state values.
Signal 1: 1 - exp(-t)
To find the steady state value of this signal, we need to determine its behavior as t approaches infinity. Let's analyze the signal:
- As t approaches infinity, the exponential term exp(-t) approaches 0.
- Therefore, the signal becomes 1 - 0 = 1.
Hence, the steady state value of the signal 1 - exp(-t) is 1.
Signal 2: 1 - k*exp(-k*t)
Now, let's analyze the second signal to find its steady state value:
- As t approaches infinity, the exponential term exp(-k*t) approaches 0 if k > 0.
- However, if k = 0, the exponential term is always equal to 1, regardless of the value of t.
- Therefore, the signal becomes 1 - k*0 = 1.
Hence, the steady state value of the signal 1 - k*exp(-k*t) is 1, regardless of the value of k.
Conclusion
Therefore, the steady state values of the signals 1 - exp(-t) and 1 - k*exp(-k*t) are both 1. This means that as time approaches infinity, both signals converge to a value of 1.
The steady state value of a signal is the value it approaches as time approaches infinity. In this case, we are given two signals: 1 - exp(-t) and 1 - k*exp(-k*t). We need to find their steady state values.
Signal 1: 1 - exp(-t)
To find the steady state value of this signal, we need to determine its behavior as t approaches infinity. Let's analyze the signal:
- As t approaches infinity, the exponential term exp(-t) approaches 0.
- Therefore, the signal becomes 1 - 0 = 1.
Hence, the steady state value of the signal 1 - exp(-t) is 1.
Signal 2: 1 - k*exp(-k*t)
Now, let's analyze the second signal to find its steady state value:
- As t approaches infinity, the exponential term exp(-k*t) approaches 0 if k > 0.
- However, if k = 0, the exponential term is always equal to 1, regardless of the value of t.
- Therefore, the signal becomes 1 - k*0 = 1.
Hence, the steady state value of the signal 1 - k*exp(-k*t) is 1, regardless of the value of k.
Conclusion
Therefore, the steady state values of the signals 1 - exp(-t) and 1 - k*exp(-k*t) are both 1. This means that as time approaches infinity, both signals converge to a value of 1.
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