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Is the system h(t) = exp(-jwt) stable?
- a)Yes
- b)No
- c)Can’t say
- d)None of the mentioned
Correct answer is option 'C'. Can you explain this answer?
Verified Answer
Is the system h(t) = exp(-jwt) stable?a)Yesb)Noc)Can’t sayd)None...
If w is a complex number with Im(w) < 0, we could have an unstable situation as well. Hence, we cannot conclude [no constraints on w given].
Most Upvoted Answer
Is the system h(t) = exp(-jwt) stable?a)Yesb)Noc)Can’t sayd)None...
The given system h(t) = exp(-jwt) is a complex exponential function. In order to determine its stability, we need to analyze the behavior of its magnitude as t goes to infinity.
The magnitude of exp(-jwt) is given by |exp(-jwt)| = |cos(wt) - j*sin(wt)| = sqrt(cos^2(wt) + sin^2(wt)) = 1 for all values of t.
Since the magnitude of h(t) is constant and equal to 1, it does not grow or decay as t goes to infinity. Therefore, the system h(t) = exp(-jwt) is stable.
The correct answer is a) Yes.
The magnitude of exp(-jwt) is given by |exp(-jwt)| = |cos(wt) - j*sin(wt)| = sqrt(cos^2(wt) + sin^2(wt)) = 1 for all values of t.
Since the magnitude of h(t) is constant and equal to 1, it does not grow or decay as t goes to infinity. Therefore, the system h(t) = exp(-jwt) is stable.
The correct answer is a) Yes.
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