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The original problem is from the Problem Set 7 of MIT OpenCourseware: Find the Fourier series coefficients for $$ x(t)=sin(10\pi t+\frac{\pi}{6}) $$ What I did is to rewrite it in exponential form $\frac{1}{2j}e^{j\frac{\pi}{6}}e^{j10\pi t}-\frac{1}{2j}e^{-j\frac{\pi}{6}}e^{-j10\pi t}$, and take $\omega_0=10\pi$ as the fundamental frequency. The non-zero coefficients I got are $a_1=\frac{1}{2j}e^{j\frac{\pi}{6}}$ and $a_{-1}=-\frac{1}{2j}e^{-j\frac{\pi}{6}}$.

However, the solution takes $\omega_0=2\pi$, which gives $a_5=\frac{1}{2j}e^{j\frac{\pi}{6}}$ and $a_{-5}=-\frac{1}{2j}e^{-j\frac{\pi}{6}}$.

I can't find a proper explanation for this problem. What should I take as fundamental frequency? Can the fundamental frequency be arbitary number for continuous time Fourier series?

Here is the link of the problem set: Problem set 7

The link of the solution: Solution

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  • $\begingroup$ Can you link to the original document with that exercise? $\endgroup$ – Matt L. Mar 8 '20 at 8:34
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    $\begingroup$ Yes, I have edited the post and put the related links there. I am asking about P7.3(a) in the document. $\endgroup$ – DXer Mar 8 '20 at 8:57
  • $\begingroup$ Then why don't we consider $\omega_0=\pi$ or $\pi/2$ or any other submultiple of $\pi$ and work appropriately? There seems to be a hole here, it does not make sense. Your solution looks good to me. $\endgroup$ – GKH Mar 8 '20 at 10:48
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Your solution is correct. Obviously, any periodic function with period $T$ is also periodic with period $kT$, $k\in\mathbb{Z}^+$. In the given example, the smallest possible period is $T=1/5$, which is equivalent to $\omega_0=10\pi$. There is no obvious reason why one should choose $T^\prime=kT$ and $\omega_0^{\prime}=\omega_0/k$, with $k>1$. Nevertheless, such a choice, even though unmotivated, is equally correct, because the resulting Fourier series are identical:

$$\begin{align}x(t)&=\sum_nc_ne^{jn\omega_0t}\\&=\sum_nc_n^{\prime}e^{jn\omega_0^{\prime}t}\\&=\sum_nc_n^{\prime}e^{j\frac{n}{k}\omega_0t}\\&=\sum_mc_{km}^{\prime}e^{jm\omega_0t}\end{align}\tag{1}$$

With $\omega_0^{\prime}=\omega_0/k$ we get

$$c^{\prime}_n=\begin{cases}c_{n/k},&n=mk\\0,&\textrm{otherwise}\end{cases}$$

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Fourier series is used to obtain any function using the weighted summation of sinusoidal functions. The above sinusoidal function is also obtained using weighted sums of different sinusoidal frequencies with the fundamental frequency being ω0=2π and its harmonics. You can find a lot of detail about Fourier series and how its coefficients are calculated here: https://en.wikipedia.org/wiki/Fourier_series

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  • $\begingroup$ But based on fundamental frequency $\omega_0=\frac{2\pi}{T_0}$, I still can't get $2\pi$ if $T_0$ is the fundamental period 1/5 $\endgroup$ – DXer Mar 8 '20 at 5:42
  • $\begingroup$ The Fourier series uses sine functions with fundamental frequency = 2π and its harmonics to obtain the sine function with the frequency 10π. So a5 is the 5th coefficient of Fourier series using frequency 10π to obtain the sin(10πt+π/6) function. $\endgroup$ – DSP Novice Mar 8 '20 at 5:56
  • $\begingroup$ Since the given solution says that they considered 2pi to be fundamental frequency, I tailored my answer. Your solution looks right as well. $\endgroup$ – DSP Novice Mar 9 '20 at 2:22

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