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A stable real linear time-invariant system with single pole at p, has a transfer function
H(s) = (s2+100)/s−p with a dc gain of 5. The smallest positive frequency, in rad/s, at unity gain is closest to:
- a)8.84
- b)122.87
- c)78.13
- d)11.08
Correct answer is option 'A'. Can you explain this answer?
Most Upvoted Answer
A stable real linear time-invariant system with single pole at p, has...
To find the smallest positive frequency at unity gain, we first need to determine the frequency response of the given transfer function.
The transfer function H(s) is given as H(s) = (s^2 + 100)/(s - p).
To find the frequency response, we substitute s = jω (where j is the imaginary unit and ω is the angular frequency) into the transfer function.
H(jω) = ((jω)^2 + 100)/((jω) - p)
Simplifying further,
H(jω) = (-ω^2 + 100)/(jω - p)
Next, we need to calculate the magnitude of the frequency response.
|H(jω)| = |(-ω^2 + 100)/(jω - p)|
|H(jω)| = sqrt((-ω^2 + 100)^2 + (jω - p)^2)
At unity gain, |H(jω)| = 1.
Therefore,
sqrt((-ω^2 + 100)^2 + (jω - p)^2) = 1
Squaring both sides of the equation, we get
(-ω^2 + 100)^2 + (jω - p)^2 = 1
Expanding and simplifying,
ω^4 - 200ω^2 + 10000 + ω^2 - 2jωp + p^2 = 1
ω^4 - 199ω^2 + 10000 - 2jωp + p^2 = 1
Rearranging the terms,
ω^4 - 199ω^2 + (p^2 - 999) - 2jωp = 0
Since the system is stable, the imaginary part of the frequency response should be zero.
Therefore,
-2jωp = 0
This implies that either ω = 0 or p = 0.
However, we are looking for the smallest positive frequency, so ω = 0 is not valid.
Therefore, p = 0.
Substituting p = 0 into the transfer function,
H(s) = (s^2 + 100)/s
At DC (s = 0), the transfer function has a gain of 5.
Therefore,
H(0) = (0^2 + 100)/0 = 100/0
This is undefined.
Hence, the given transfer function does not have a stable real linear time-invariant system with a single pole at p.
Therefore, the given question is incorrect and does not have a correct answer.
The transfer function H(s) is given as H(s) = (s^2 + 100)/(s - p).
To find the frequency response, we substitute s = jω (where j is the imaginary unit and ω is the angular frequency) into the transfer function.
H(jω) = ((jω)^2 + 100)/((jω) - p)
Simplifying further,
H(jω) = (-ω^2 + 100)/(jω - p)
Next, we need to calculate the magnitude of the frequency response.
|H(jω)| = |(-ω^2 + 100)/(jω - p)|
|H(jω)| = sqrt((-ω^2 + 100)^2 + (jω - p)^2)
At unity gain, |H(jω)| = 1.
Therefore,
sqrt((-ω^2 + 100)^2 + (jω - p)^2) = 1
Squaring both sides of the equation, we get
(-ω^2 + 100)^2 + (jω - p)^2 = 1
Expanding and simplifying,
ω^4 - 200ω^2 + 10000 + ω^2 - 2jωp + p^2 = 1
ω^4 - 199ω^2 + 10000 - 2jωp + p^2 = 1
Rearranging the terms,
ω^4 - 199ω^2 + (p^2 - 999) - 2jωp = 0
Since the system is stable, the imaginary part of the frequency response should be zero.
Therefore,
-2jωp = 0
This implies that either ω = 0 or p = 0.
However, we are looking for the smallest positive frequency, so ω = 0 is not valid.
Therefore, p = 0.
Substituting p = 0 into the transfer function,
H(s) = (s^2 + 100)/s
At DC (s = 0), the transfer function has a gain of 5.
Therefore,
H(0) = (0^2 + 100)/0 = 100/0
This is undefined.
Hence, the given transfer function does not have a stable real linear time-invariant system with a single pole at p.
Therefore, the given question is incorrect and does not have a correct answer.
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Community Answer
A stable real linear time-invariant system with single pole at p, has...
H(s) = (s2+100)/s−p
D.C. gain = 5
D.C. gain = H{s}|s = 0
= (s2+100)/s−p|s=0 = 5
100/−p = 5 ⇒ p= −20
H(s) = (s2+100)/(s+20)
For frequency domain s = jω
Frequency at unity gain
(100 − w2)2 = (400 + w2)
Solving this equation
ω=8.84 rad/sec and 11.08 radisec.
Smallest positive frequency = 8.84 rad/sec.
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A stable real linear time-invariant system with single pole at p, has a transfer functionH(s) = (s2+100)/s−p with a dc gain of 5. The smallest positive frequency, in rad/s, at unity gain is closest to:a)8.84b)122.87c)78.13d)11.08Correct answer is option 'A'. Can you explain this answer? for Electronics and Communication Engineering (ECE) 2025 is part of Electronics and Communication Engineering (ECE) preparation. The Question and answers have been prepared according to the Electronics and Communication Engineering (ECE) exam syllabus. Information about A stable real linear time-invariant system with single pole at p, has a transfer functionH(s) = (s2+100)/s−p with a dc gain of 5. The smallest positive frequency, in rad/s, at unity gain is closest to:a)8.84b)122.87c)78.13d)11.08Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for Electronics and Communication Engineering (ECE) 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A stable real linear time-invariant system with single pole at p, has a transfer functionH(s) = (s2+100)/s−p with a dc gain of 5. The smallest positive frequency, in rad/s, at unity gain is closest to:a)8.84b)122.87c)78.13d)11.08Correct answer is option 'A'. Can you explain this answer?.
A stable real linear time-invariant system with single pole at p, has a transfer functionH(s) = (s2+100)/s−p with a dc gain of 5. The smallest positive frequency, in rad/s, at unity gain is closest to:a)8.84b)122.87c)78.13d)11.08Correct answer is option 'A'. Can you explain this answer? for Electronics and Communication Engineering (ECE) 2025 is part of Electronics and Communication Engineering (ECE) preparation. The Question and answers have been prepared according to the Electronics and Communication Engineering (ECE) exam syllabus. Information about A stable real linear time-invariant system with single pole at p, has a transfer functionH(s) = (s2+100)/s−p with a dc gain of 5. The smallest positive frequency, in rad/s, at unity gain is closest to:a)8.84b)122.87c)78.13d)11.08Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for Electronics and Communication Engineering (ECE) 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A stable real linear time-invariant system with single pole at p, has a transfer functionH(s) = (s2+100)/s−p with a dc gain of 5. The smallest positive frequency, in rad/s, at unity gain is closest to:a)8.84b)122.87c)78.13d)11.08Correct answer is option 'A'. Can you explain this answer?.
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