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It is desired to reduce total harmonic of amplifier from 8% to 1% by use of 10% negative feedback. If the gain of the amplifier with original distortion and with reduced distortion is A1 and A2 then, A1 + A2 = −−−−−−
Correct answer is between '78.70,78.80'. Can you explain this answer?
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It is desired to reduce total harmonic of amplifier from 8% to1% by us...
To solve this problem, we can use the formula for the reduction in distortion due to negative feedback:
Rd = 1 - (1 + β) * Hd
Where Rd is the reduction in distortion, β is the feedback factor, and Hd is the distortion without feedback.
Given that the reduction in distortion is desired to be from 8% to 1%, we can calculate the initial distortion without feedback (Hd) as:
Hd = 8% = 0.08
And the desired distortion with feedback (Hd') as:
Hd' = 1% = 0.01
We are also given that the feedback factor (β) is 10%.
Using the formula, we can calculate the reduction in distortion (Rd) as:
Rd = 1 - (1 + 0.1) * 0.08
= 1 - 1.08 * 0.08
= 1 - 0.0864
= 0.9136
Next, we can calculate the gain of the amplifier with reduced distortion (A2) using the formula:
A2 = A1 / (1 + β * A1 * Rd)
Where A1 is the gain of the amplifier with original distortion.
Since we want to find the relationship between A1 and A2, we can rearrange the formula to:
A2 = A1 / (1 + β * A1 * Rd)
A2 * (1 + β * A1 * Rd) = A1
A2 + β * A1 * Rd * A2 = A1
A2 - A1 = - β * A1 * Rd * A2
(A2 - A1) / (A1 * A2) = - β * Rd
(A2 - A1) / (A1 * A2 * Rd) = - β
Therefore, A1 * A2 = -(A2 - A1) / (β * Rd)
Substituting the values we have calculated:
A1 * A2 = -(A2 - A1) / (0.1 * 0.9136)
= -(A2 - A1) / 0.09136
Rd = 1 - (1 + β) * Hd
Where Rd is the reduction in distortion, β is the feedback factor, and Hd is the distortion without feedback.
Given that the reduction in distortion is desired to be from 8% to 1%, we can calculate the initial distortion without feedback (Hd) as:
Hd = 8% = 0.08
And the desired distortion with feedback (Hd') as:
Hd' = 1% = 0.01
We are also given that the feedback factor (β) is 10%.
Using the formula, we can calculate the reduction in distortion (Rd) as:
Rd = 1 - (1 + 0.1) * 0.08
= 1 - 1.08 * 0.08
= 1 - 0.0864
= 0.9136
Next, we can calculate the gain of the amplifier with reduced distortion (A2) using the formula:
A2 = A1 / (1 + β * A1 * Rd)
Where A1 is the gain of the amplifier with original distortion.
Since we want to find the relationship between A1 and A2, we can rearrange the formula to:
A2 = A1 / (1 + β * A1 * Rd)
A2 * (1 + β * A1 * Rd) = A1
A2 + β * A1 * Rd * A2 = A1
A2 - A1 = - β * A1 * Rd * A2
(A2 - A1) / (A1 * A2) = - β * Rd
(A2 - A1) / (A1 * A2 * Rd) = - β
Therefore, A1 * A2 = -(A2 - A1) / (β * Rd)
Substituting the values we have calculated:
A1 * A2 = -(A2 - A1) / (0.1 * 0.9136)
= -(A2 - A1) / 0.09136
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Community Answer
It is desired to reduce total harmonic of amplifier from 8% to1% by us...
Given that β = 0.1and we know that 
So we can write 
∴ A1 + A2 =78.75
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It is desired to reduce total harmonic of amplifier from 8% to1% by use of 10% negative feedback. If the gain of the amplifier with original distortion and with reduced distortion is A1and A2then,A1 + A2 = −−−−−−Correct answer is between '78.70,78.80'. Can you explain this answer?
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It is desired to reduce total harmonic of amplifier from 8% to1% by use of 10% negative feedback. If the gain of the amplifier with original distortion and with reduced distortion is A1and A2then,A1 + A2 = −−−−−−Correct answer is between '78.70,78.80'. Can you explain this answer? for Electrical Engineering (EE) 2024 is part of Electrical Engineering (EE) preparation. The Question and answers have been prepared according to the Electrical Engineering (EE) exam syllabus. Information about It is desired to reduce total harmonic of amplifier from 8% to1% by use of 10% negative feedback. If the gain of the amplifier with original distortion and with reduced distortion is A1and A2then,A1 + A2 = −−−−−−Correct answer is between '78.70,78.80'. Can you explain this answer? covers all topics & solutions for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for It is desired to reduce total harmonic of amplifier from 8% to1% by use of 10% negative feedback. If the gain of the amplifier with original distortion and with reduced distortion is A1and A2then,A1 + A2 = −−−−−−Correct answer is between '78.70,78.80'. Can you explain this answer?.
It is desired to reduce total harmonic of amplifier from 8% to1% by use of 10% negative feedback. If the gain of the amplifier with original distortion and with reduced distortion is A1and A2then,A1 + A2 = −−−−−−Correct answer is between '78.70,78.80'. Can you explain this answer? for Electrical Engineering (EE) 2024 is part of Electrical Engineering (EE) preparation. The Question and answers have been prepared according to the Electrical Engineering (EE) exam syllabus. Information about It is desired to reduce total harmonic of amplifier from 8% to1% by use of 10% negative feedback. If the gain of the amplifier with original distortion and with reduced distortion is A1and A2then,A1 + A2 = −−−−−−Correct answer is between '78.70,78.80'. Can you explain this answer? covers all topics & solutions for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for It is desired to reduce total harmonic of amplifier from 8% to1% by use of 10% negative feedback. If the gain of the amplifier with original distortion and with reduced distortion is A1and A2then,A1 + A2 = −−−−−−Correct answer is between '78.70,78.80'. Can you explain this answer?.
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