# Fourier Series coefficients/Trigonometric functions

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.

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

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 $\hat{x}_{-1}$.

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

$$x(t) = \sin(10\pi t + \frac {\pi}{6} )$$ $$x(t) = \frac {e^{j(10\pi t + \frac {\pi}{6})} - e^{-j(10\pi t + \frac {\pi}{6})}}{2j}$$ $$x(t) = (\frac{1}{2j}e^{j\frac{\pi}{6}})e^{j10\pi t} - (\frac{1}{2j}e^{-j\frac{\pi}{6}})e^{-j10\pi t}$$ 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 $$a_{k} = 0\ for \ |k|>1$$

• Looks good. ${}{}{}$ – copper.hat Nov 24 '13 at 17:34

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.