0
$\begingroup$

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 $$ x(t) = a_0 + 2\sum^{+\infty}_{k\ =\ 1} 2\mathrm{Re}\left[ e^j(k \omega_0t +\theta_k)\right]$$

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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

$$\Re\left\{a_ke^{jk\omega_0t}\right\}=B_k\cos(k\omega_0t)-C_k\sin(k\omega_0t)\tag{2}$$

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 $j\cdot j=-1$.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.