Fourier Series
Eigen stuff
Say we have an LTI system.
Given an input \(x(t) = \phi(t)\), If the output \(y(t) = k\ \phi(t)\), where \(k\) is a constant, then \(\phi(t)\) is an eigen-function.
The constant \(k\) is the eigen-value. It can be understood as some amplitude factor.
In the context of LTI signals, we know \(y(t) = x(t) * h(t)\).
A known eigenfunction for these LTI systems is \(\phi(t) = e^{j \omega t}\), such that:
\[\phi(t) * h(t) = H(\omega) \phi(t)\]Where
- \(\phi(t) = e^{j \omega t}\) is an eigen-function
- \(h(t)\) is the impulse response
- \(H(\omega) = \int_{-\infty}^{\infty} h(\tau) e^{-j \omega \tau} d\tau\) is the eigen-value, which is a function of the frequency \(\omega\)
We can use this to define the Fourier Series:
Fourier Series definition
\[x(t) = \sum_{k = -\infty}^{\infty} a_k e^{jk \omega t}\]
Where
- \(k\) is the counter (integer, real number)
- \(a_k\) is the constant of the k-th element
- \(\omega\) is the frequency (since this are sines and cosines in disguise)
How do we go from eigenfunctions to here? Of course I’m skipping a lot of stuff, but if you need a fuller version, go read this.
A general way of getting the constants is:
\[a_n = \frac{1}{T} \int_{-T/2}^{T/2} x(t) e^{-jn \omega t} dt\]
Where
- \(a_n\) is the n-th constant of the Fourier Series expansion of \(x(t)\)
- \(T\) is fundamental period
- \(x(t)\) is a signal
- \(n\) is the index of the Fourier Series expansion
- \(\omega\) is frequency
Reminder: \(T = \frac{2\pi}{\omega}\)
Discrete time
\[x[n] = \sum_{k = 0}^{N-1} a_k e^{jk\Omega n}\]
Where
- \(N\) is the period
- \(\Omega\) is frequency
\[a_k = \frac{1}{N} \sum_{n = 0}^{N-1} x[n] e^{-jk\Omega n}\]
Properties
Periodicity
\[x[n] = x[n + aN]\]Where
- \(N\) is the period
- \(a\) is some real int
Linearity - superposition
Time shifting
\[x[n-m] \rightarrow e^{-j (2\pi / N) km} c_k\]Where
- \(m\) is the shift
- \(c_k\) is the fourier series sum of \(x[n]\)
Problems
Type 1: Find \(a_k\) without the integral
Given:
- \(x(t)\), containing sin, cos, and constants
Find:
- The constants of its Fourier Series expansion \(a_k\)
Process
1. Rewrite all sines and cosines in their exponential form:
\[\cos{\omega t} = \frac{1}{2}(e^{j\omega t} + e^{-j\omega t})\] \[\sin{\omega t} = \frac{1}{2j}(e^{j\omega t} - e^{-j\omega t})\]2. Group the exponentials \(e^{jk\omega t}\) and simplify their coefficients
3. For each \(e^{jk\omega t}\), write the accompanying coefficients to \(a_k\)
Tips:
- Keep the expansion form in mind
- \(k\) MUST be of type int
- Consider the case \(k = 0\) separately (since \(e^0 = 1\), then \(a_0\) will be any left over constants)
- The sin exponential form is divided by a j, don’t forget it
- Remember \(e^{a+b} = e^a e^b\)
Type 2: Find \(a_k\) with the integral
Given:
- \(x(t)\), either a function or graph
Find:
- The constants of its Fourier Series expansion \(a_k\)
Process
1. If given a graph, rewrite as an equation and stating the limits.
- The graph may contain the period T as a variable or a number. If it is a number, make sure to replace it both in T and \(\omega\).
2. Consider the integral:
\[a_k = \frac{1}{T} \int_T x(t) e^{-jk \omega t} dt\]Tips:
- You should split the integral depending on the limits (if the functions is piecewise)
- If the case \(k = 0\) is problematic (undefined), re-evaluate the integral only for the case \(k = 0\), so that \(a_0 = \frac{1}{T} \int_T x(t) dt\)