Fourier series - by Ray Betz

Fourier Series

If

1. $x(t)=x(t+T)$
2. Dirichlet conditions are satisfied

then we can write

${\mathbf {x}}(t)=\sum _{k=-\infty }^{\infty }\alpha _{k}e^{\frac {j2\pi kt}{T}}$

The above equation is called the complex fourier series. Given $x(t)$, we may determine $\alpha _{k}$ by taking the inner product of $\alpha _{k}$ with $x(t)$. Let us assume a solution for $\alpha _{k}$ of the form $e^{\frac {j2\pi nt}{T}}$. Now we take the inner product of $\alpha _{k}$ with $x(t)$. $<\alpha _{k}|x(t)>=<e^{\frac {j2\pi nt}{T}}|\sum _{k=-\infty }^{\infty }\alpha _{k}e^{\frac {j2\pi kt}{T}}>$ $=\int _{-{\frac {T}{2}}}^{\frac {T}{2}}x(t)e^{\frac {-j2\pi nt}{T}}dt$ $=\int _{-{\frac {T}{2}}}^{\frac {T}{2}}\sum _{k=-\infty }^{\infty }\alpha _{k}e^{\frac {j2\pi kt}{T}}e^{\frac {-j2\pi nt}{T}}dt$ $=\sum _{k=-\infty }^{\infty }\alpha _{k}\int _{-{\frac {T}{2}}}^{\frac {T}{2}}e^{\frac {j2\pi (k-n)t}{T}}dt$

If $k=n$ then,

$\sum _{k=-\infty }^{\infty }\alpha _{k}\int _{-{\frac {T}{2}}}^{\frac {T}{2}}e^{\frac {j2\pi (k-n)t}{T}}dt=\int _{-{\frac {T}{2}}}^{\frac {T}{2}}1dt=T$

If $k\neq n$ then,

$\sum _{k=-\infty }^{\infty }\alpha _{k}\int _{-{\frac {T}{2}}}^{\frac {T}{2}}e^{\frac {j2\pi (k-n)t}{T}}dt=0$

We can simplify the above two conclusion into one equation.

$\sum _{k=-\infty }^{\infty }\alpha _{k}\int _{-{\frac {T}{2}}}^{\frac {T}{2}}e^{\frac {j2\pi (k-n)t}{T}}dt=\sum _{k=-\infty }^{\infty }T\delta _{k,n}\alpha _{k}=T\alpha _{n}$

So, we may conclude $\alpha _{n}={\frac {1}{T}}\int _{-{\frac {T}{2}}}^{\frac {T}{2}}x(t)e^{\frac {-j2\pi nt}{T}}dt$

Orthogonal Functions

The function $y_{n}(t)$ and $y_{m}(t)$ are orthogonal on $(a,b)$ if and only if $<y_{n}(t)|y_{m}(t)>=\int _{a}^{b}y_{n}^{*}(t)y_{m}(t)dt=0$.

The set of functions are orthonormal if and only if $<y_{n}(t)|y_{m}(t)>=\int _{a}^{b}y_{n}^{*}(t)y_{m}(t)dt=\delta _{m,n}$.

Linear Systems

I may come back to this latter...

Fourier Series (indepth)

I would like to take a closer look at $\alpha _{k}$ in the Fourier Series. Hopefully this will provide a better understanding of $\alpha _{k}$.

We will seperate x(t) into three parts; where $\alpha _{k}$ is negative, zero, and positive. ${\mathbf {x}}(t)=\sum _{k=-\infty }^{\infty }\alpha _{k}e^{\frac {j2\pi kt}{T}}=\sum _{k=-\infty }^{-1}\alpha _{k}e^{\frac {j2\pi kt}{T}}+\alpha _{0}+\sum _{k=1}^{\infty }\alpha _{k}e^{\frac {j2\pi kt}{T}}$

Now, by substituting $n=-k$ into the summation where $k$ is negative and substituting $n=k$ into the summation where $k$ is positive we get: $\sum _{k=1}^{\infty }\alpha _{-n}e^{\frac {-j2\pi nt}{T}}+\alpha _{0}+\sum _{k=1}^{\infty }\alpha _{n}e^{\frac {j2\pi nt}{T}}$

Recall that $\alpha _{n}={\frac {1}{T}}\int _{-{\frac {T}{2}}}^{\frac {T}{2}}x(u)e^{\frac {-j2\pi nt}{T}}dt$

If $x(t)$ is real, then $\alpha _{n}^{*}=\alpha _{-n}$. Let us assume that $x(t)$ is real.

$x(t)=\alpha _{0}+\sum _{n=1}^{\infty }(\alpha _{n}e^{\frac {j2\pi nt}{T}}+\alpha _{n}^{*}e^{\frac {-j2\pi nt}{T}})$

Recall that $y+y^{*}=2Re(y)$ Here is further clarification on this property

So, we may write:

$x(t)=\alpha _{0}+\sum _{n=1}^{\infty }2Re(\alpha _{n}e^{\frac {j2\pi nt}{T}})$

Fourier Transform

Fourier transforms emerge because we want to be able to make Fourier expressions of non-periodic functions. We can take the limit of those non-periodic functions to get a fourier expression for the function.

Remember that: $x(t)=x(t+T)=\sum _{k=-\infty }^{\infty }\alpha _{k}e^{\frac {j2\pi kt}{T}}=\sum _{k=-\infty }^{\infty }1/T\int _{-{\frac {T}{2}}}^{\frac {T}{2}}x(u)e^{\frac {-j2\pi ku}{T}}due^{\frac {j2\pi kt}{T}}$

So, $\lim _{x\to \infty }x(t)=\int _{-\infty }^{\infty }(\int _{-\infty }^{\infty }x(u)e^{-j2\pi fu}du)e^{j2\pi ft}df$

From the above limit we define $x(t)$ and $X(f)$.

$x(t)={\mathcal {F}}^{-1}[X(f)]=\int _{-\infty }^{\infty }X(f)e^{j2\pi ft}df$

$X(f)={\mathcal {F}}^{-1}[x(t)]=\int _{-\infty }^{\infty }x(t)e^{j2\pi ft}dt$

We can take the derivitive of $x(t)$ and then put in terms of the reverse fourier transform.

${\frac {dx}{dt}}=\int _{-\infty }^{\infty }j2\pi fX(f)e^{j2\pi ft}df={\mathcal {F}}^{-1}[j2\pi fX(f)]$