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Let x(t) be a periodic signal with time period T, Let y(t) = x(t – to) + x(t + to) for some to. The fourier series coefficients of y(t) are denoted by bk. If bk = 0 for all odd K. Then to can be equal to
- a)T/8
- b)T/4
- c)T/2
- d)2T
Correct answer is option 'B'. Can you explain this answer?
Most Upvoted Answer
Let x(t) be a periodic signal with time period T, Let y(t) = x(t &ndas...
Understanding the Problem
The signal y(t) is defined as y(t) = x(t - to) + x(t + to). The Fourier series coefficients of y(t) are denoted by bk, and it is given that bk = 0 for all odd k. This means that y(t) only contains even harmonics.
Significance of bk = 0 for Odd k
- A signal with only even harmonics is symmetric with respect to time.
- This symmetry indicates that y(t) must exhibit an even function property, which is a key condition for having only even Fourier coefficients.
Time Shifts and Periodicity
- The periodic signal x(t) has a period T.
- The shifts of to in y(t) affect the symmetry of the signal.
Determining the Value of to
- The crucial point is that for y(t) to maintain the even symmetry, the shifts (to) must relate to the period T.
- Analyzing the conditions, we find that if to = T/4, then the shifts will ensure that x(t) appears symmetrically around t=0.
Verification of Option B (T/4)
- If to = T/4:
- Then y(t) = x(t - T/4) + x(t + T/4).
- This results in a symmetric configuration around t = 0, ensuring that the contributions to the Fourier coefficients from odd k cancel out.
Concluding the Explanation
- Therefore, the only value of to that maintains the even symmetry in the Fourier series representation of y(t) is T/4.
- This confirms that the correct answer is option 'B'.
This understanding not only illustrates the relationship between periodic signals and their Fourier series but also highlights the importance of symmetry in signal analysis.
The signal y(t) is defined as y(t) = x(t - to) + x(t + to). The Fourier series coefficients of y(t) are denoted by bk, and it is given that bk = 0 for all odd k. This means that y(t) only contains even harmonics.
Significance of bk = 0 for Odd k
- A signal with only even harmonics is symmetric with respect to time.
- This symmetry indicates that y(t) must exhibit an even function property, which is a key condition for having only even Fourier coefficients.
Time Shifts and Periodicity
- The periodic signal x(t) has a period T.
- The shifts of to in y(t) affect the symmetry of the signal.
Determining the Value of to
- The crucial point is that for y(t) to maintain the even symmetry, the shifts (to) must relate to the period T.
- Analyzing the conditions, we find that if to = T/4, then the shifts will ensure that x(t) appears symmetrically around t=0.
Verification of Option B (T/4)
- If to = T/4:
- Then y(t) = x(t - T/4) + x(t + T/4).
- This results in a symmetric configuration around t = 0, ensuring that the contributions to the Fourier coefficients from odd k cancel out.
Concluding the Explanation
- Therefore, the only value of to that maintains the even symmetry in the Fourier series representation of y(t) is T/4.
- This confirms that the correct answer is option 'B'.
This understanding not only illustrates the relationship between periodic signals and their Fourier series but also highlights the importance of symmetry in signal analysis.
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Let x(t) be a periodic signal with time period T, Let y(t) = x(t – to) + x(t + to) for some to. The fourier series coefficients of y(t) are denoted by bk. If bk = 0 for all odd K. Then to can be equal toa)T/8b)T/4c)T/2d)2TCorrect answer is option 'B'. Can you explain this answer?
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