Question

If \(f(t)\) is a periodic waveform with even symmetry, then its Fourier series expansion does not contain

• A. Sine terms
• B. Cosine terms
• C. Odd harmonics
• D. Even harmonics

Hint:

For Fourier Series: \[ \text{Even Signal} \Rightarrow \text{Only cosine terms} \] \[ \text{Odd Signal} \Rightarrow \text{Only sine terms} \] These symmetry properties greatly reduce calculations.

Answer 1

The Correct Option is A

Concept: The trigonometric Fourier series of a periodic signal is \[ f(t)=a_0+\sum_{n=1}^{\infty} \left( a_n\cos n\omega_0 t + b_n\sin n\omega_0 t \right). \] The symmetry properties of a signal greatly simplify the Fourier series. For an even function, \[ f(t)=f(-t). \]

Step 1: Examine the coefficient of sine terms.
The coefficient of sine terms is \[ b_n= \frac{2}{T} \int_{-T/2}^{T/2} f(t)\sin(n\omega_0 t)\,dt. \] Since \[ f(t) \] is even and \[ \sin(n\omega_0 t) \] is odd, their product is odd. \[ \text{Even}\times\text{Odd} = \text{Odd}. \]

Step 2: Use the property of odd functions.
The integral of an odd function over symmetric limits is zero. Therefore, \[ b_n=0. \] Hence all sine coefficients vanish.

Step 3: Write the Fourier series for an even function.
The Fourier series becomes \[ f(t) = a_0 + \sum_{n=1}^{\infty} a_n\cos(n\omega_0 t). \] Thus only cosine terms remain. \[ \boxed{\text{Sine terms are absent}} \]