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For a signal u(sint); Fourier series is having
- a)Odd harmonics of sine only.
- b)Zero DC.
- c)DC and odd Harmonics of sine.
- d)All Harmonics of sine.
Correct answer is option 'C'. Can you explain this answer?
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Most Upvoted Answer
For a signal u(sint); Fourier series is havinga)Odd harmonics of sine ...
Fourier series representation of a signal u(sint)
The Fourier series representation of a periodic signal u(t) with period T is given by:
u(t) = a0 + ∑[an*cos(nωt) + bn*sin(nωt)]
Where a0 is the DC component, an and bn are the Fourier coefficients, and ω = 2π/T is the fundamental angular frequency.
In this case, the signal is u(sint), which means the input signal is a sine wave with angular frequency ω = sin(t). Let's analyze the Fourier series representation of this signal.
Odd harmonics of sine only
When we analyze the Fourier series of u(sint), we can observe that the cosine terms will become zero. This is because the product of sin(nωt) and cos(mωt) is zero for any positive integer values of n and m. Therefore, only the sine terms will remain in the Fourier series representation.
Zero DC component
The DC component, a0, represents the average value of the signal over one period. In the case of u(sint), the average value of the sine wave over one period is zero, as the positive and negative halves cancel each other out. Therefore, the DC component, a0, is zero.
DC and odd harmonics of sine
Based on the above analysis, we can conclude that the Fourier series representation of u(sint) will only have the odd harmonics of sine and a zero DC component. This implies that the Fourier coefficients bn will be non-zero for odd values of n, while the coefficients an and the DC component a0 will be zero.
Therefore, the correct answer is option 'C' - DC and odd harmonics of sine.
The Fourier series representation of a periodic signal u(t) with period T is given by:
u(t) = a0 + ∑[an*cos(nωt) + bn*sin(nωt)]
Where a0 is the DC component, an and bn are the Fourier coefficients, and ω = 2π/T is the fundamental angular frequency.
In this case, the signal is u(sint), which means the input signal is a sine wave with angular frequency ω = sin(t). Let's analyze the Fourier series representation of this signal.
Odd harmonics of sine only
When we analyze the Fourier series of u(sint), we can observe that the cosine terms will become zero. This is because the product of sin(nωt) and cos(mωt) is zero for any positive integer values of n and m. Therefore, only the sine terms will remain in the Fourier series representation.
Zero DC component
The DC component, a0, represents the average value of the signal over one period. In the case of u(sint), the average value of the sine wave over one period is zero, as the positive and negative halves cancel each other out. Therefore, the DC component, a0, is zero.
DC and odd harmonics of sine
Based on the above analysis, we can conclude that the Fourier series representation of u(sint) will only have the odd harmonics of sine and a zero DC component. This implies that the Fourier coefficients bn will be non-zero for odd values of n, while the coefficients an and the DC component a0 will be zero.
Therefore, the correct answer is option 'C' - DC and odd harmonics of sine.
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Community Answer
For a signal u(sint); Fourier series is havinga)Odd harmonics of sine ...
Given x(t) = u(sint)
Drawing waveform
We know for waveform like y(t) shown
Drawing waveform
We know for waveform like y(t) shown
DC component is zero and have sine odd harmonics.
Performing upward shift in y(t) by adding dc value of 0.5, then
x(t) = y(t) + 0.5
So, x(t) → DC + sine odd harmonics
Performing upward shift in y(t) by adding dc value of 0.5, then
x(t) = y(t) + 0.5
So, x(t) → DC + sine odd harmonics
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For a signal u(sint); Fourier series is havinga)Odd harmonics of sine only.b)Zero DC.c)DC and odd Harmonics of sine.d)All Harmonics of sine.Correct answer is option 'C'. Can you explain this answer?
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