Question

The exponential Fourier series coefficient \( C_{-n} \) in terms of trigonometric Fourier coefficients is

• A. \( \frac{1}{2}(a_n + jb_n) \)
• B. \( \frac{1}{2}(a_n - jb_n) \)
• C. \( \frac{1}{2}(a_0 + jb_n) \)
• D. \( \frac{1}{2}(a_0 + a_n) \)

Hint:

\(C_0 = a_0\)
\(C_n = \frac{1}{2}(a_n - jb_n)\) for \(n \neq 0\)
\(C_{-n} = \frac{1}{2}(a_n + jb_n)\) (for \(n>0\))
For real signals, \(C_{-n} = C_n^*\).

Answer 1

The Correct Option is A

The trigonometric Fourier series is \(x(t) = a_0 + \sum_{n=1}^{\infty} (a_n \cos(n\omega_0 t) + b_n \sin(n\omega_0 t))\).

The exponential Fourier series is \(x(t) = \sum_{n=-\infty}^{\infty} C_n e^{jn\omega_0 t}\). 

Using Euler's formulas for \(\cos\) and \(\sin\):

\(\cos(n\omega_0 t) = \frac{e^{jn\omega_0 t} + e^{-jn\omega_0 t}}{2}\)

\(\sin(n\omega_0 t) = \frac{e^{jn\omega_0 t} - e^{-jn\omega_0 t}}{2j}\)

Substituting into the trigonometric series and rearranging terms: \(x(t) = a_0 + \sum_{n=1}^{\infty} \left[ \left(\frac{a_n}{2} + \frac{b_n}{2j}\right)e^{jn\omega_0 t} + \left(\frac{a_n}{2} - \frac{b_n}{2j}\right)e^{-jn\omega_0 t} \right]\)

Since \(1/j = -j\): \(x(t) = a_0 + \sum_{n=1}^{\infty} \left[ \frac{1}{2}(a_n - jb_n)e^{jn\omega_0 t} + \frac{1}{2}(a_n + jb_n)e^{-jn\omega_0 t} \right]\)

Comparing with the exponential series \(C_0 + \sum_{n=1}^{\infty} C_n e^{jn\omega_0 t} + \sum_{n=1}^{\infty} C_{-n} e^{-jn\omega_0 t}\):

\(C_0 = a_0\)

For \(n>0\): \(C_n = \frac{1}{2}(a_n - jb_n)\)

For \(n>0\): \(C_{-n} = \frac{1}{2}(a_n + jb_n)\)

\[ \boxed{\frac{1}{2}(a_n + jb_n)} \]