0
$\begingroup$

I need some help about finding the Fourier Series coefficient of the given signal; $$ x(t) = \sin(10\pi t + \frac {\pi}{6} ) $$

I know that,

$$ a_{k} = \frac{1}{T}\int_{0}^{T} x(t)e^{-jkw_{0}t}dt $$

but substituting $\sin(10t + \frac {\pi}{6})$ in to that formula and taking the integral is really hard.

I'm pretty sure that something can be done by using the fact that

$$ x(t) = \sum_{k = -\infty}^{\infty} a_{k}e^{jkw_{0}t}dt $$ $$ \sin(t) = \frac {e^{jt} - e^{-jt}}{2j} $$

Need some help, thanks.

$\endgroup$
6
  • 1
    $\begingroup$ Write $x(t) = \sin (10t)\cos \frac{\pi}{6} + \sin \frac{\pi}{6} \cos (10t)$. $\endgroup$ – Daniel Fischer Nov 24 '13 at 15:35
  • $\begingroup$ What is $T$ here? $2 \pi$? $\endgroup$ – copper.hat Nov 24 '13 at 16:18
  • $\begingroup$ Is $x(t) = \sin (10t + \frac{\pi}{6})$ or $x(t) = \sin (10\pi t + \frac{\pi}{6})$? (You have both above.) $\endgroup$ – copper.hat Nov 24 '13 at 16:41
  • $\begingroup$ $ x(t) = \sin(10\pi t + \frac {\pi}{6} ) $ I have corrected it. $\endgroup$ – cecemelly Nov 24 '13 at 16:49
  • $\begingroup$ Still looks the same above. Are you sure that $T= \frac{1}{5 \pi}$? This $x$ is $\frac{1}{5}$-periodic. $\endgroup$ – copper.hat Nov 24 '13 at 16:56
2
$\begingroup$

Suppose $\hat{f}_k = \frac{1}{T} \int_0^T e^{-i k 2 \pi \frac{t}{T}} f(t) dt $, and $f$ is $T$-periodic. Let $\phi(t) = f(t-t_0)$, then you have $\hat{\phi}_k = \frac{1}{T} \int_0^T e^{-i k 2 \pi \frac{t}{T}} f(t-t_0)dt = \frac{1}{T} \int_0^T e^{-i k 2 \pi \frac{(t+t_0)}{T}} f(t)dt = e^{-i k 2 \pi \frac{t_0}{T}} \hat{f}_k$.

Assuming that $f(t) = \sin (10 \pi t)$ and $T= \frac{1}{5}$ we have $f(t) = \frac{1}{2i} (e^{i10 \pi t} - e^{-i10 \pi t}) = \frac{1}{2i} (e^{i2 \pi \frac{t}{T}} - e^{-i2 \pi \frac{t}{T}})$, and we can read off the Fourier coefficients as $\hat{f}_1 = \frac{1}{2 i}$, $\hat{f}_{-1} = -\frac{1}{2i}$ and all other $\hat{f}_k = 0$.

Since $x(t) = f(t+\frac{1}{60})$, the above formula (with $t_0 = -\frac{1}{60}$) gives $\hat{x}_1 = \frac{1}{2 i}e^{i \frac{\pi}{6}}$, $\hat{x}_{-1} = -\frac{1}{2 i}e^{-i \frac{\pi}{6}}$ and all other $\hat{x}_k = 0$.

Alternative: You could work directly from the expansion of $\sin ( 10 \pi t) = \frac{1}{2i} (e^{i2 \pi \frac{t}{T}} - e^{-i2 \pi \frac{t}{T}})$ to get $x(t) = \sin ( 10 \pi(t+\frac{1}{60})) = \frac{1}{2i} (e^{i2 \pi \frac{(t+\frac{1}{60})}{T}} - e^{-i2 \pi \frac{(t+\frac{1}{60})}{T}}) = \frac{1}{2i} (e^{i \frac{\pi}{6}} e^{i2 \pi \frac{t}{T}} - e^{-i \frac{\pi}{6}} e^{-i2 \pi \frac{t}{T}}) $, from which we can read off the coefficient of $t \to e^{i2 \pi \frac{t}{T}}$ to get $\hat{x}_1$, and similarly for $\hat{x}_{-1}$.

$\endgroup$
2
  • $\begingroup$ Thank you so much, your answer helped me to find the exact answer. $\endgroup$ – cecemelly Nov 24 '13 at 17:34
  • $\begingroup$ Delighted to help. $\endgroup$ – copper.hat Nov 24 '13 at 17:35
1
$\begingroup$

$$ x(t) = \sin(10\pi t + \frac {\pi}{6} ) $$ $$ x(t) = \frac {e^{j(10\pi t + \frac {\pi}{6})} - e^{-j(10\pi t + \frac {\pi}{6})}}{2j} $$ $$ x(t) = (\frac{1}{2j}e^{j\frac{\pi}{6}})e^{j10\pi t} - (\frac{1}{2j}e^{-j\frac{\pi}{6}})e^{-j10\pi t} $$ where $ T= \frac{1}{5}$ and $ w_{0} = 10\pi$

Therefore,

$$a_{0} = 0 $$ $$a_{1} = \frac{1}{2j} e^{j\frac{\pi}{6}} = \frac{1}{4} ({\frac{\sqrt3}{j} + 1})$$ $$a_{-1} = \frac{-1}{2j} e^{j\frac{-\pi}{6}} = \frac{1}{4} (1 - {\frac{\sqrt3}{j}})$$ and $$a_{k} = 0\ for \ |k|>1 $$

$\endgroup$
1
  • $\begingroup$ Looks good. ${}{}{}$ $\endgroup$ – copper.hat Nov 24 '13 at 17:34
0
$\begingroup$

Hint: there is no need for calculations. Just use the formula

$$ \sin(a+b) = \sin a cos b + \sin b \cos a ,$$

To get the Fourier series.

$\endgroup$
0

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.