2.9b.2 Continuous-Time Fourier Series

Continuous-Time Fourier Series (CTFS)

The Continuous-Time Fourier Series (CTFS) is a mathematical method used to represent a periodic continuous-time signal as a sum of sinusoidal components or, equivalently, complex exponentials. The CTFS decomposes an uncountably infinite signal into a series of countably infinite harmonics $k{\omega }_0$ that represent the signal’s frequency content. The Fourier series for a $T$-periodic continuous-time signal $x:\mathbb{R}\to \mathbb{C}$ is
$x(t)={\sum }_{k=-\infty }^{\infty }{X}_k{e}^{jk{\omega }_{0}t}⟺X_k=\frac{1}{T}{\int }_{-\frac{T}{2}}^{\frac{T}{2}}x(t)e^{-jk{\omega }_{0}t}dt.$

Intuition

By decomposing $x(t)$ into a linear combination of its Fourier coefficients $X[k]$ with its basis vectors ${\varphi }_k(t)=e^{j\omega kt},$ we quickly get our synthesis equation
$x(t)={\sum }_{k=-\infty }^{\infty }X[k]e^{j{\omega }_{0}kt},$
where the right hand side is summed over infinity because although $x(t)$ is $p$-periodic, it is a continuous signal and will hold an infinite number of coefficients $X[k]$ over $k\in [0,p)$.

Since a CTFS Fourier basis ${\varphi }_k$ is periodic on $t$ and aperiodic on $k$ (as shown in the properties section below), the synthesis equation can be thought of decomposing an uncountably infinite signal $x(t)$ into a countably infinite set of frequencies $k{\omega }_0$.

We must proceed with a continuous-time “projection.” We take another basis ${\varphi }_{\ell }$ and see that
$⟨x,{\varphi }_{\ell }⟩=X[\ell ]⟨{\varphi }_{\ell },{\varphi }_{\ell }⟩⟺X[\ell ]=\frac{⟨x,{\varphi }_{\ell }⟩}{⟨{\varphi }_{\ell },{\varphi }_{\ell }⟩}.$
Since we are working in continuous-time (and therefore both $x(t)$ and ${\varphi }_k(t)$ are continuous and $p$-periodic), we define

\langle \phi_\ell, \phi_\ell \rangle &= \int^{\frac{p}{2}}_{-\frac{p}{2}}\phi_\ell(t) \phi_\ell\textasteriskcentered(t)dt \\ &= \int^{\frac{p}{2}}_{-\frac{p}{2}}e^{j\omega_0 \ell t} \cdot e^{-j\omega_0 \ell t}dt \\ &= \int^{\frac{p}{2}}_{-\frac{p}{2}}dt \\ &= \frac{p}{2} - \left(-\frac{p}{2} \right) \\ &= p. \end{align}

Note that we take the interval $[-\frac{p}{2},\frac{p}{2}]$ in order to avoid the intuitive interval $[0,p]=[0,0]$.

Similarly,

\langle x, \phi_\ell \rangle = \int^{\frac{p}{2}}_{-\frac{p}{2}} x(t) \phi_\ell\textasteriskcentered(t). \end{align}$$ Therefore, our analysis equation becomes $$X[k] = \frac{1}{p}\int^{\frac{p}{2}}_{-\frac{p}{2}}x(t)e^{-jk\omega_0t}.$$ # Properties #### (Basis $t$-Periodicity) $\phi_k(t+p)=\phi_k(t)$ We find that $$\phi_k(t+p)=e^{j\omega kt} \cdot e^{j\omega k p}=e^{j\omega kt} \cdot 1 = \phi_k(t),$$ so $\phi_k$ is periodic on $t$. Therefore, the continuous spectrum is periodic on time. #### (No Basis $k$-Periodicity) $\phi_{k+p}(t) \neq \phi_k(t)$ We also look at $\phi_k$'s periodicity on $k$: $$\phi_{k + p}(t)=e^{j\omega kt} \cdot e^{j\omega pt}=e^{j\omega kt} \cdot e^{j2\pi t},$$ which against intuition is not periodic on $k$, as $e^{j2\pi t}$ can hold any value $t \in \mathbb{R}$ and may not be a multiple of $2\pi$. Therefore, the continuous spectrum is aperiodic on frequency. #### (*Linearity*) #### (*Time Shift*) # Examples >[!example] A CTFS Puzzle > >Consider a periodic, continuous-time sinusoidal signal $x(t)$ with the following properties: >1. $x(t)=x(t+4)$ for all $t \in \mathbb{R}$. >2. For $k=1,2,3$, there exists $t \in \mathbb{R}$, such that $x(t) \neq x(t+k)$. >3. $X_{k}=X\textasteriskcentered_{-k}$ for all $k \in \mathbb{N}$. >4. $X_{k}=0$ for $k > 2$. >5. The average value of $x(t)$ over one period is 4. >6. $\int^{1}_{-3}|x(t)|^2dt=128$. >7. $x(t)=x(-t)$. >8. $x(0)=12$. > >We proceed in the following manner: >1. Tells us that $x(t)$ has a period $p=\frac{4}{\ell}$. >2. Verifies that $\ell=1$ (i.e. the fundamental period of $x$ is $p=4$). >3. Tells us that the signal is real, since the coefficients' conjugates are symmetric. >4. Since $k \in [0, 3]$, this means that $X_{3}=X_{-3}=0$. >5. This can be translated into $\frac{1}{4}\int^{2}_{-2}x(t)dt=4$, and implies that $X_{0}=4$ (since by definition $X_{0}=\frac{1}{T}\int^{\frac{T}{2}}_{-\frac{T}{2}}x(t)e^{j 0 \omega_{0} t}$. >6. Using Parseval's theorem, $\frac{1}{4}\int^{1}_{-3}|x(t)|^2dt=\sum^{\infty}_{k=-\infty}|X_{k}|^2=32$. Since $X_{0}=4$ and $X_{3}=X_{-3}=0$, $|X_{1}|^2+|X_{-1}|^2+|X_{2}|^2+|X_{-2}|^2=16$. Since $X_{1}=X_{-1}$ and $X_{2}=X_{-2}$, $|X_{1}|^2+|X_{2}|^2=8$. >7. This means that the function is even. Therefore, since it is sinusoidal, it may only contain cosine functions (and perhaps a coefficient). >8. We can formulate $x(t)=X_{-2}e^{-j\pi t}+X_{-1}e^{-j\frac{\pi}{2}t}+X_{0}+X_{1}e^{j\frac{\pi}{2}t}+X_{2}e^{j\pi t}$. For $x(0)$, this formulation equals $x(0)=X_{0}+2X_{1}+2X_{2}=12$. Since $X_{0}=4$, $X_{1}+X_{2}=4$. > >We now need to simply find $X_{1}$ and $X_{2}$. We compare the two equations $X_{1}+X_{2}=4$ (obtained in 8.) and $|X_{1}|^2+|X_{2}|^2=8$ (obtained in 6.). We find that $X_{1}=X_{2}=2$. Therefore, we get: >$$\begin{align} x(t)&=2e^{-j\pi t}+2e^{-j\frac{\pi}{2}t}+4+2e^{j\frac{\pi}{2}t}+2e^{j\pi t} \\ &= 4\cos(\frac{\pi}{2}t)+4\cos(\pi t)+4. \end{align}$$ >[!example] Continuous-Time Trigonometric Fourier Series > >