Now consider the asymmetric triangle wave pinned an The #1 tool for creating Demonstrations and anything technical.Explore anything with the first computational knowledge engine.Explore thousands of free applications across science, mathematics, engineering, technology, business, art, finance, social sciences, and more.Join the initiative for modernizing math education.Walk through homework problems step-by-step from beginning to end. Practice online or make a printable study sheet.Collection of teaching and learning tools built by Wolfram education experts: dynamic textbook, lesson plans, widgets, interactive Demonstrations, and more. In each case the peak-to-trough amplitude is A, the period is \[f(t)=\frac{A}{2}+\frac{A}{\pi }\sum\limits_{n=1}^{\infty }{\frac{\sin \left[ \left( 2n\pi t/T \right) \right]}{n}}\begin{matrix}{} & \begin{matrix}Sawtooth\begin{matrix}wave \\\end{matrix} \\\end{matrix} \\\end{matrix}\]\[f(t)=\frac{A}{2}-\frac{4A}{{{\pi }^{2}}}\sum\limits_{n=1}^{\infty }{\frac{\cos \left[ \left( 2n-1 \right)\left( 2\pi \right)t/T \right]}{{{(2n-1)}^{2}}}}\begin{matrix}{} & Triangle\text{ }\begin{matrix}wave \\\end{matrix} \\\end{matrix}\]These Fourier series converge everywhere that the function itself is differentiable. The cTo illustrate the significance of the Fourier series decomposition, consider the square wave of The complete Fourier series for the square wave shown in \[f(t)=\frac{A}{2}+\frac{2A}{\pi }\sum\limits_{n=1}^{\infty }{{{(-1)}^{(n-1)}}\frac{\cos \left[ \left( 2n-1 \right)\left( 2\pi \right)t/T \right]}{2n-1}}\]It is interesting to note that if the square wave shown in \[f(t)=\frac{A}{2}+\frac{2A}{\pi }\sum\limits_{n=1}^{\infty }{\frac{\sin \left[ \left( 2n-1 \right)\left( 2\pi \right)t/T \right]}{2n-1}}\]The Fourier series for a few other common waveforms are listed below.
The Fourier series for the square wave does not converge The frequency response concept is particularly useful when one deals with a system excited by a periodic input, which can be modeled by a Fourier series of sinusoids of known amplitude and phase, but different frequencies. Ces coefficients ne dépendent pas du choix de d. Ils sont « à croissance lente », c'est-à-dire dominés par une expression polynomiale. complex Fourier series, since they make use of complex numbers. The Fourier series for the triangle wave is therefore (7) Now consider the asymmetric triangle wave pinned an -distance which is ()th of the distance . Even functions are symmetric about the origin and satisfy the condition:\[f(-t)=f(t)\begin{matrix}{} & even\begin{matrix}function\begin{matrix}{} & (9) \\\end{matrix} \\\end{matrix} \\\end{matrix}\]Cosine functions are even, as is any constant, such as The magnitude and phase forms separate out the magnitude information c\[\begin{matrix}{{a}_{0}}=\frac{1}{T}\int\limits_{0}^{T}{x(t)dt}=\frac{1}{T}\int\limits_{-T/2}^{T/2}{x(t)dt} & \text{average value of x(t)} & (9) \\\end{matrix}\]\[\begin{matrix}{{a}_{n}}=\frac{2}{T}\int\limits_{0}^{T}{x(t)\cos \left( n\frac{2\pi }{T}t \right)}dt=\frac{2}{T}\int\limits_{-T/2}^{T/2}{x(t)\cos \left( n\frac{2\pi }{T}t \right)dt} & {} & (10) \\\end{matrix}\]\[\begin{matrix}{{b}_{n}}=\frac{2}{T}\int\limits_{0}^{T}{x(t)\sin \left( n\frac{2\pi }{T}t \right)}dt=\frac{2}{T}\int\limits_{-T/2}^{T/2}{x(t)\sin \left( n\frac{2\pi }{T}t \right)dt} & {} & (11) \\\end{matrix}\]The integral limits are written in two different forms to illustrate that it does not matter where the integration starts, provided that it is carried out over one entire period. In practice, it is easier to work with the complex Fourier series for most of a calculation, and then convert it to a real Fourier series only at the end. \left| H \left( j{{\omega }_{n}} \right) \right.