Fourier series in continuous time domain while representing $a_k$ in rectangular form $$ a_k = B_k + jC_k$$ But when using the value of $a_k$ in the main equation: $$ x(t) = a_0 + 2\sum^{+\infty}_{k\ =\ 1} [B_k\cos k\omega_0t - jC_k\sin k\omega_0t]$$

I want to ask from where $-$(minus) sign comes as there was no minus sign main Equation $$ x(t) = a_0 + 2\sum^{+\infty}_{k\ =\ 1} 2\mathrm{Re}\left[ e^j(k \omega_0t +\theta_k)\right]$$

  • $\begingroup$ The two equations for $x(t)$ cannot be equal. The first contains an imaginary component. The second is purely real-valued. That, and it would be nice to know the relationship between $a_k$ and $\theta_k$. $\endgroup$ – Peter K. Oct 5 '15 at 14:27
  • $\begingroup$ That depends on how you exactly define $\theta_k$. Something like this should work $\theta_k = atan(\frac{-C_k}{B_k})$ $\endgroup$ – Hilmar Oct 5 '15 at 15:06

I'm afraid that all the formulas in your question are inaccurate. From what I understand, you're interested in the representation of real-valued signals $x(t)$:

$$x(t)=a_0+2\Re\left\{\sum_{k=1}^{\infty}a_ke^{jk\omega_0t}\right\}= a_0+2\sum_{k=1}^{\infty}\Re\left\{a_ke^{jk\omega_0t}\right\}\tag{1}$$

With $a_k=B_k+jC_k$ and $e^{jk\omega_0t}=\cos(k\omega_0t)+j\sin(k\omega_0t)$ you get


which you can substitute into Eq. $(1)$ to get the equation in your question (but without the $j$!). The minus sign in $(2)$ appears simply because $j\cdot j=-1$.

  • $\begingroup$ Yup in understand now, I have posted a question. please answer me if you have time $\endgroup$ – Aadnan Farooq A Oct 6 '15 at 7:50

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