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The Fourier series of a real periodic function has only
P. cosine terms if it is even
Q. sine terms if it is even
R. cosine terms if it is odd
S. sine terms if it is odd
Which of the above statements are correct?
- a)P and S
- b)P and R
- c)Q and S
- d)Q and R
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
The Fourier series of a real periodic function has onlyP. cosine terms...
Because sine function is odd and cosine is even function.
Most Upvoted Answer
The Fourier series of a real periodic function has onlyP. cosine terms...
Fourier Series of Real Periodic Function
Fourier series is a mathematical technique to represent a periodic function as a sum of sine and cosine terms. The Fourier series of a real periodic function f(x) is given by:
f(x) = a0/2 + Σ(an*cos(nωx) + bn*sin(nωx))
where ω = 2π/T is the angular frequency, T is the period of the function, and an and bn are the Fourier coefficients given by:
an = (2/T) ∫f(x)cos(nωx)dx
bn = (2/T) ∫f(x)sin(nωx)dx
Even and Odd Functions
A function f(x) is said to be even if f(-x) = f(x) for all x, i.e., it is symmetric about the y-axis. Examples of even functions are cos(x), x², |x|, etc.
A function f(x) is said to be odd if f(-x) = -f(x) for all x, i.e., it is symmetric about the origin. Examples of odd functions are sin(x), x³, etc.
Cosine and Sine Terms in Fourier Series
If a function f(x) is even, then its Fourier series contains only cosine terms and no sine terms. This is because the integral of an odd function over a symmetric interval is zero, and hence the bn coefficients vanish.
Similarly, if a function f(x) is odd, then its Fourier series contains only sine terms and no cosine terms. This is because the integral of an even function over a symmetric interval is zero, and hence the an coefficients vanish.
Answer
From the above discussion, we can conclude that:
- The Fourier series of a real periodic function has only cosine terms if it is even
- The Fourier series of a real periodic function has only sine terms if it is odd
Therefore, the correct answer is option 'A', i.e., the Fourier series of a real periodic function has only cosine terms if it is even and sine terms if it is odd.
Fourier series is a mathematical technique to represent a periodic function as a sum of sine and cosine terms. The Fourier series of a real periodic function f(x) is given by:
f(x) = a0/2 + Σ(an*cos(nωx) + bn*sin(nωx))
where ω = 2π/T is the angular frequency, T is the period of the function, and an and bn are the Fourier coefficients given by:
an = (2/T) ∫f(x)cos(nωx)dx
bn = (2/T) ∫f(x)sin(nωx)dx
Even and Odd Functions
A function f(x) is said to be even if f(-x) = f(x) for all x, i.e., it is symmetric about the y-axis. Examples of even functions are cos(x), x², |x|, etc.
A function f(x) is said to be odd if f(-x) = -f(x) for all x, i.e., it is symmetric about the origin. Examples of odd functions are sin(x), x³, etc.
Cosine and Sine Terms in Fourier Series
If a function f(x) is even, then its Fourier series contains only cosine terms and no sine terms. This is because the integral of an odd function over a symmetric interval is zero, and hence the bn coefficients vanish.
Similarly, if a function f(x) is odd, then its Fourier series contains only sine terms and no cosine terms. This is because the integral of an even function over a symmetric interval is zero, and hence the an coefficients vanish.
Answer
From the above discussion, we can conclude that:
- The Fourier series of a real periodic function has only cosine terms if it is even
- The Fourier series of a real periodic function has only sine terms if it is odd
Therefore, the correct answer is option 'A', i.e., the Fourier series of a real periodic function has only cosine terms if it is even and sine terms if it is odd.
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The Fourier series of a real periodic function has onlyP. cosine terms if it is even Q. sine terms if it is even R. cosine terms if it is odd S. sine terms if it is oddWhich of the above statements are correct?a)P and S b)P and R c)Q and S d)Q and RCorrect answer is option 'A'. Can you explain this answer?
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