Discussion
\(F(t)\) is a periodic square wave function as shown. It takes only two values, 4 and 0, and stays at each of these values for 1 second before changing. What is the constant term in the Fourier series expansion of \(F(t)\)?
Answer: B
The constant term is 2.
The constant term of a Fourier series is the average value of the function over one period. Here the period is 2 seconds: the function is 4 for one second and 0 for the next.
\[ \begin{aligned} a_{0} &= \frac{1}{2}\int_{0}^{2} F(t)\,dt \\ &= \frac{4 \times 1 + 0 \times 1}{2} = 2 \end{aligned} \]
4 is the peak value, not the mean; it would be the answer only if the wave stayed at 4 all the time. 1 and 3 are not the mean of this waveform, since the wave spends equal time at 4 and at 0. A symmetric square wave between 0 and 4 always averages to the midpoint, 2.
Post your comment
Please log in or register to post a comment.
Examination Categories
- SSC Exams
- Banking Exams
- Teaching Exams
- Civil Services Exams
- Railways Exams
- Engineering Recruitment Exams
- Defence Exams
- State Government Exams
- Police Exams
- Insurance Exams
- Nursing Exams
- Other Government Exams
- NRA CET
- PG Entrance Exams
- Campus Placement
- MBA Entrance Exams
- CUET and UG Entrance
- Judiciary Exams
- Regulatory Body Exams
- Software Development