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If a periodic signal xp(t) has the Fourier coefficient X(k), and when the signal is delayed by 2sec its 5th harmonic gets delayed in phase by 0.1 radian, the fundamental frequency of the signal in milli rad/sec is
- a)10
- b)15
- c)14
- d)7
Correct answer is option 'A'. Can you explain this answer?
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Most Upvoted Answer
If a periodic signal xp(t) has the Fourier coefficient X(k), and when...
When the signal is delayed by 2 sec kth harmonic gets phase delayed by kw02
It is given that 5w02 = 0.1
10w0 = 0.1
w0 = 0.01 = 10 milli rad/sec
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If a periodic signal xp(t) has the Fourier coefficient X(k), and when...
Fundamental Frequency and Harmonics
The fundamental frequency of a periodic signal is the lowest frequency component present in the signal. It is denoted by f0 and is equal to the reciprocal of the period of the signal. The harmonics of a signal are integer multiples of the fundamental frequency.
Given Information
In this question, we are given that a periodic signal xp(t) has the Fourier coefficient X(k). When the signal is delayed by 2 seconds, its 5th harmonic gets delayed in phase by 0.1 radian.
Phase Delay Calculation
To calculate the phase delay, we need to first determine the angular frequency of the 5th harmonic before and after the delay.
The angular frequency of the 5th harmonic before the delay is given by:
ω5 = 5 * 2πf0
After the delay, the angular frequency of the 5th harmonic is given by:
ω5' = 5 * 2πf0'
where f0' is the fundamental frequency after the delay.
We are given that the phase delay of the 5th harmonic is 0.1 radian. Therefore, we can write the following equation:
ω5' = ω5 + 0.1
Fundamental Frequency Calculation
To find the fundamental frequency after the delay, we need to relate the original and delayed signals. Since the delay is 2 seconds, we can write:
f0' = 1 / (T + 2)
where T is the period of the original signal.
Using the relation between angular frequency and frequency, we can write:
ω5' = 5 * 2πf0' = 5 * 2π / (T + 2)
Now, substituting the value of ω5' from the earlier equation, we get:
5 * 2π / (T + 2) = 5 * 2πf0 + 0.1
Simplifying the equation, we have:
1 / (T + 2) = f0 + 0.1 / (5f0)
Further simplifying and rearranging the equation, we get:
5f0(T + 2) = 1 + 0.1(T + 2)
Expanding and rearranging the equation, we finally get:
4.9T + 0.2 = 0
Solving for T, we find T = -0.2 / 4.9 ≈ -0.0408 seconds.
Since the period T cannot be negative, this solution is not feasible. Therefore, there must be an error in the question or the given answer options.
The fundamental frequency of a periodic signal is the lowest frequency component present in the signal. It is denoted by f0 and is equal to the reciprocal of the period of the signal. The harmonics of a signal are integer multiples of the fundamental frequency.
Given Information
In this question, we are given that a periodic signal xp(t) has the Fourier coefficient X(k). When the signal is delayed by 2 seconds, its 5th harmonic gets delayed in phase by 0.1 radian.
Phase Delay Calculation
To calculate the phase delay, we need to first determine the angular frequency of the 5th harmonic before and after the delay.
The angular frequency of the 5th harmonic before the delay is given by:
ω5 = 5 * 2πf0
After the delay, the angular frequency of the 5th harmonic is given by:
ω5' = 5 * 2πf0'
where f0' is the fundamental frequency after the delay.
We are given that the phase delay of the 5th harmonic is 0.1 radian. Therefore, we can write the following equation:
ω5' = ω5 + 0.1
Fundamental Frequency Calculation
To find the fundamental frequency after the delay, we need to relate the original and delayed signals. Since the delay is 2 seconds, we can write:
f0' = 1 / (T + 2)
where T is the period of the original signal.
Using the relation between angular frequency and frequency, we can write:
ω5' = 5 * 2πf0' = 5 * 2π / (T + 2)
Now, substituting the value of ω5' from the earlier equation, we get:
5 * 2π / (T + 2) = 5 * 2πf0 + 0.1
Simplifying the equation, we have:
1 / (T + 2) = f0 + 0.1 / (5f0)
Further simplifying and rearranging the equation, we get:
5f0(T + 2) = 1 + 0.1(T + 2)
Expanding and rearranging the equation, we finally get:
4.9T + 0.2 = 0
Solving for T, we find T = -0.2 / 4.9 ≈ -0.0408 seconds.
Since the period T cannot be negative, this solution is not feasible. Therefore, there must be an error in the question or the given answer options.
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