Question

The amplitude and phase spectrum of exponential Fourier Series about vertical axis respectively, is

• A. Symmetrical, symmetrical
• B. Symmetrical, antisymmetrical
• C. Antisymmetrical, antisymmetrical
• D. Antisymmetrical, symmetrical

Hint:

For real signals, Fourier coefficients exhibit conjugate symmetry: \(C_{-n} = C_n^*\).
This implies amplitude spectrum \(|C_n|\) is even.
This implies phase spectrum \(\angle C_n\) is odd.

Answer 1

The Correct Option is B

For a real-valued signal \(x(t)\), the exponential Fourier series coefficients \(C_n\) satisfy the conjugate symmetry property: \(C_{-n} = C_n^*\), where \(C_n^*\) is the complex conjugate of \(C_n\). Let \(C_n = |C_n|e^{j\theta_n}\), where \(|C_n|\) is the amplitude and \(\theta_n = \angle C_n\) is the phase. Then \(C_n^* = |C_n|e^{-j\theta_n}\). Since \(C_{-n} = C_n^*\): Amplitude spectrum: \(|C_{-n}| = |C_n^*| = |C_n|\). This means the amplitude spectrum is an even function (symmetrical about the vertical axis \(n=0\)). Phase spectrum: \(\angle C_{-n} = \angle C_n^* = -\theta_n = -\angle C_n\). This means the phase spectrum is an odd function (antisymmetrical about the vertical axis \(n=0\)). Thus, the amplitude spectrum is symmetrical, and the phase spectrum is antisymmetrical. \[ \boxed{\text{Symmetrical, antisymmetrical}} \]