Kurt's Assignment: Difference between revisions
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=Common Synthesizer Waveforms= |
=Common Synthesizer Waveforms= |
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Many synthesizers employ a variety of waveforms to produce varied sounds. The most common waveform is the sine wave. However, in additive synthesis, multiple waveforms can be added together to create a different waveform with different characteristics. The basis for this form of synthesis is the Fourier series: |
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<math>\begin{align} |
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==Square Wave== |
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x(t) &= x(t+T) = a_0 + \sum_{n=1}^\infty a_n \cos(n\omega_0t) + b_n \sin(n\omega_0t)\\ |
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a_0 &= \frac{1}{T}\int_0^T f(t)\, dt\\ |
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a_n &= \frac{2}{T}\int_0^T f(t)\cos(n\omega_0t)\, dt\\ |
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b_n &= \frac{2}{T}\int_0^T f(t)\sin(n\omega_0t)\, dt\\ |
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\end{align} |
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</math> |
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<math>x(t) = x(t+T) = a_0 + \sum_{n=1}^\infty a_n \cos(n\omega_0t) + b_n \sin(n\omega_0t)</math> |
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The four basic waveforms are Sine Waves, Square Waves, Triangle Waves, and Sawtooth Waves. |
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===Square Wave=== |
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By inspection of the waveform, the DC component of the wave will be 0. Also, since the waveform is odd, a<sub>n</sub> will be 0. Here is the proof: |
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<math>\begin{align} |
<math>\begin{align} |
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a_0 &= \frac{1}{T}\int_0^T |
a_0 &= \frac{1}{T}\int_0^{\frac{1}{2}T} H\, dt + \frac{1}{T}\int_{\frac{1}{2}T}^T -H\, dt\\ |
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&=\frac{1}{T}\ |
&=\frac{1}{T}\left[Ht\right]\bigg|_{t=0}^{\frac{1}{2}T} - \frac{1}{T}\left[Ht\right]\bigg|_{t={\frac{1}{2}T}}^T\\ |
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&=\frac{1}{T}[ |
&=\frac{1}{T}H\frac{1}{2}T-0 - \left[\frac{1}{T}HT - \frac{1}{T}H\frac{1}{2}T\right]\\ |
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&=\frac{H}{2} - \left[H-\frac{1}{2}H\right]\\ |
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&=\frac{H}{2}-\frac{H}{2}\\ |
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&=0 |
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\end{align}</math> |
\end{align}</math> |
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TODO: finish |
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<math>\begin{align} |
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a_n &= \frac{2}{T}\int_0^{\frac{1}{2}T} H\cos(n\omega_0t)\, dt + \frac{2}{T}\int_{\frac{1}{2}T}^T -H\cos(n\omega_0t)\, dt\\ |
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&=\frac{2}{T}\left[\frac{H}{n\omega_0}\sin(n\omega_0t)\right]_0^{\frac{1}{2}T} + \frac{2}{T}\left[\frac{-H}{n\omega_0}\sin\left(n\omega_0t\right)\right]_{\frac{1}{2}T}^T\\ |
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&=\frac{2}{T}\left[\frac{H}{n\frac{2\pi}{T}}\sin\left(n\frac{2\pi}{T}\frac{1}{2}T\right)-0\right] + \frac{2}{T}\left[-0+\frac{H}{n\frac{2\pi}{T}}\sin\left(n\frac{2pi}{T}\frac{1}{2}T\right)\right]\\ |
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&=\frac{2}{T}\left[\frac{TH}{2\pi n}\underbrace{\sin(n\pi)}_\text{0}\right] + \frac{2}{T}\left[\frac{TH}{2\pi n}\underbrace{\sin(n\pi)}_\text{0}\right]\\ |
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&=0 |
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\end{align}</math> |
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This just leaves the sine component of the waveform found below. |
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<math>\begin{align} |
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b_n &= \frac{2}{T}\int_0^{\frac{1}{2}T} H\sin(n\omega_0t)\, dt + \frac{2}{T}\int_{\frac{1}{2}T}^T -H\sin(n\omega_0t)\, dt\\ |
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&=\frac{2}{T}\left[\frac{-H}{n\omega_0}\cos(n\omega_0t)\right]_0^{\frac{1}{2}T} + \frac{2}{T}\left[\frac{H}{n\omega_0}\cos(n\omega_0t)\right]_{\frac{1}{2}T}^T\\ |
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&=\frac{2}{T}\left[-\frac{H}{n\frac{2\pi}{T}}\cos(n\frac{2\pi}{T}\frac{1}{2}T)+\frac{H}{n\frac{2\pi}{T}}\right] + \frac{2}{T}\left[\frac{H}{n\frac{2\pi}{T}}\cos(n\frac{2\pi}{T}T)-\frac{H}{n\frac{2\pi}{T}}\cos(n\frac{2\pi}{T}\frac{1}{2}T)\right]\\ |
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&=\frac{2}{T}\left[-\frac{TH}{2\pi n}\cos(n\pi)+\frac{TH}{2\pi n}\right] + \frac{2}{T}\left[\frac{TH}{2\pi n}\underbrace{\cos(2 \pi n)}_\text{1}-\frac{TH}{2\pi n}\cos(n\pi)\right]\\ |
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&=-\frac{H}{\pi n}\cos(n\pi)+\frac{H}{\pi n} + \frac{H}{\pi n}-\frac{H}{\pi n}\cos(n\pi)\\ |
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&=\frac{2H}{\pi n}-\frac{2H}{\pi n}\cos(n\pi)\\ |
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\end{align}</math> |
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Finally, resulting in the Fourier series for a Square Wave. |
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<math>\text{Square Wave Fourier Series: }x(t) = x(t+T) = \sum_{n=1}^\infty \left(\frac{2H}{\pi n}-\frac{2H}{\pi n}\cos(n\pi)\right) \sin(n\omega_0t) </math> |
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===Triangle Wave=== |
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Like the Square wave, the DC component of the Triangle Wave is 0 by inspection. Also, since the triangle wave is odd, it is made up only by sine components. |
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<math>\begin{align} |
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a_0 &= \frac{1}{T}\int_{-\frac{1}{4}T}^{\frac{1}{4}T}\left(\frac{4H}{T}t \right)\,dt + \frac{1}{T}\int_{\frac{1}{4}T}^{\frac{3}{4}T}\left(-\frac{4H}{T}t+2H\right) \,dt\\ |
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&= \frac{1}{T}\left[\frac{1}{2}\frac{4H}{T}t^2\right]_{-\frac{1}{4}T}^{\frac{1}{4}T} + \frac{1}{T}\left[-\frac{1}{2}\frac{4H}{T}t^2+2Ht\right]_{\frac{1}{4}T}^{\frac{3}{4}T}\\ |
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&= \frac{1}{T}\left[\frac{2H}{T}\left(\frac{1}{4}T\right)^2-\frac{2H}{T}\left(-\frac{1}{4}T\right)^2\right] + \frac{1}{T}\left[-\frac{2H}{T}\left(\frac{3}{4}T\right)^2+2H\left(\frac{3}{4}T\right)-\left(-\frac{2H}{T}\left(\frac{1}{4}T\right)^2+2H\left(\frac{1}{4}T\right)\right)\right]\\ |
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&= \frac{1}{T}\left[\underbrace{\frac{2H}{T}\frac{1}{16}T^2-\frac{2H}{T}\frac{1}{16}T^2}_\text{0}\right] + \frac{1}{T}\left[-\frac{2H}{T}\frac{9}{16}T^2+\frac{3}{2}HT+\frac{2H}{T}\frac{1}{16}T^2-\frac{1}{2}HT\right]\\ |
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&= 0 + \frac{1}{T}\left[-\frac{18H}{16T}T^2+\frac{3}{2}HT+\frac{2H}{16T}T^2-\frac{1}{2}HT\right]\\ |
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&= \frac{1}{T}\left[\underbrace{-\frac{18}{16}HT+\frac{24}{16}HT+\frac{2}{16}HT-\frac{8}{16}HT}_\text{0}\right]\\ |
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&= 0 |
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\end{align}</math> |
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<math>\begin{align} |
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a_n &= \frac{2}{T}\int_{-\frac{1}{4}T}^{\frac{1}{4}T}\left(\frac{4H}{T}t\right)\cos\left(n\omega_0t\right) \,dt + \frac{2}{T}\int_{\frac{1}{4}T}^{\frac{3}{4}T}\left(-\frac{4H}{T}t+2H\right)\cos\left(n\omega_0t\right)\,dt\\ |
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&= \frac{8H}{T^2}\int_{-\frac{1}{4}T}^{\frac{1}{4}T}t\,\cos\left(n\omega_0t\right) \,dt + \frac{4H}{T}\int_{\frac{1}{4}T}^{\frac{3}{4}T}\left(-\frac{2}{T}t+1\right)\cos\left(n\omega_0t\right)\,dt\\ |
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&= \frac{8H}{T^2}\left[\frac{t}{n\omega_0}\sin(n\omega_0 t)+\frac{1}{n^2\omega_0 ^2}\cos(n\omega_0 t)\right]_{-\frac{1}{4}T}^{\frac{1}{4}T} + \frac{4H}{T}\left[\frac{-\frac{2}{T}t+1}{n\omega_0} \sin(n\omega_0 t) - \frac{2}{Tn^2\omega_0^2} \cos(n\omega_0 t)\right]_{-\frac{1}{4}T}^{\frac{3}{4}T}\\ |
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&= \frac{8H}{T^2}\left[\left[\frac{\frac{1}{4}T}{n\frac{2\pi}{T}}\sin\left(n\frac{2\pi}{T}\frac{1}{4}T\right)+\frac{1}{n^2\frac{4\pi^2}{T^2}}\cos\left(n\frac{2\pi}{T}\frac{1}{4}T\right)\right] - \left[\frac{-\frac{1}{4}T}{n\frac{2\pi}{T}}\sin\left(n\frac{2\pi}{T}\left(-\frac{1}{4}T\right)\right)+\frac{1}{n^2\frac{4\pi^2}{T^2}}\cos\left(n\frac{2\pi}{T}\left(-\frac{1}{4}T\right)\right)\right]\right] + \frac{4H}{T}\left[\left[\frac{-\frac{2}{T}\left(\frac{3}{4}T\right)+1}{n\frac{2\pi}{T}} \sin\left(n\frac{2\pi}{T}\frac{3}{4}T\right) - \frac{2}{Tn^2\frac{4\pi^2}{T^2}}\cos\left(n\frac{2\pi}{T}\frac{3}{4}T\right)\right] - \left[\frac{-\frac{2}{T}\left(-\frac{1}{4}T\right)+1}{n\frac{2\pi}{T}} \sin\left(n\frac{2\pi}{T}\left(-\frac{1}{4}T\right)\right) - \frac{2}{Tn^2\frac{4\pi^2}{T^2}}\cos\left(n\frac{2\pi}{T}\left(-\frac{1}{4}T\right)\right)\right]\right]\\ |
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&= \frac{8H}{T^2}\left[\left[\frac{T^2}{8\pi n}\sin\left(\frac{1}{2}n\pi\right)+\frac{T^2}{n^2 4\pi^2}\cos\left(\frac{1}{2}n\pi\right)\right] - \left[-\frac{T^2}{8\pi n}\sin\left(-\frac{1}{2}n\pi\right)+\frac{T^2}{n^2 4 \pi^2}\cos\left(-\frac{1}{2}n\pi\right)\right]\right] + \frac{4H}{T} \left[\left[-\frac{T}{4\pi n}\sin\left(\frac{3}{2}n\pi\right)-\frac{T}{2\pi^2 n^2}\cos\left(\frac{3}{2}n\pi\right)\right] - \left[\frac{3T}{4\pi n}\sin\left(-\frac{1}{2}n \pi\right)-\frac{T}{2\pi^2 n^2}\cos\left(-\frac{1}{2}n\pi\right)\right]\right]\\ |
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&= \frac{8H}{T^2}\left[\frac{T^2}{8\pi n}\sin\left(\frac{1}{2}n\pi\right)-\frac{T^2}{8\pi n}\sin\left(-\frac{1}{2}n\pi\right)+\underbrace{\frac{T^2}{n^2 4\pi^2}\cos\left(\frac{1}{2}n\pi\right)-\frac{T^2}{n^2 4\pi^2}\cos\left(-\frac{1}{2}n\pi\right)}_\text{0}\right] + \frac{4H}{T}\left[-\frac{T}{4\pi n}\sin\left(\frac{3}{2}n\pi\right) - \frac{3T}{4\pi n}\sin\left(-\frac{1}{2}n\pi\right) - \underbrace{\frac{T}{2\pi^2 n^2} \cos\left(\frac{3}{2}n\pi\right) + \frac{T}{2\pi^2 n^2} \cos\left(-\frac{1}{2}n\pi\right)}_\text{0}\right]\\ |
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&= \frac{8H}{T^2}\left[\frac{2T^2}{8\pi n}\sin\left(\frac{1}{2}n\pi\right)\right] + \text{integrate other side}\\ |
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&= \frac{2H}{\pi n}\sin\left(\frac{1}{2}n\pi\right) + \text{integrate other side} |
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\end{align}</math> |
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<math>\begin{align} |
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test function &= |
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\end{align}</math> |
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===Sawtooth Wave=== |
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==OCTAVE Scripts to Plot Fourier Series== |
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====Square Wave==== |
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clf; %Clear Figure |
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t=0:.01:10; %Limits of the graph |
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T=2*pi %Definition of the period |
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M=100 %Number of iterations to undergo |
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sum1=0; %Initialize the sum to <br> |
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%----------FOURIER SERIES----------% |
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for m=1:1:M, %For m=1, increment by 1 until you get to M |
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if(m!=0) |
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sum1 = sum1 + ((2/(pi*m))-(2/(pi*m))*cos(m*pi))*sin(m*2*pi/T*t); |
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end |
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end<br> |
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%---------------PLOT---------------% |
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plot(t,real(sum1),'b-') |
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title('Fourier Series Representation of a Wave') |
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xlabel('time (seconds)') |
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ylabel('Function') |
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grid on;<br> |
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legend(num2str(M) ' terms'); |
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print("squarewave.png","-dpng") % Prints the plot to a png file called squarewave.png |
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TODO: |
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*finish the triangle wave derivation |
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*start sawtooth wave derivation |
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*put pictures in |
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*implement triangle wave Fourier Series in OCTAVE |
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*implement sawtooth wave Fourier Series in OCTAVE |
Latest revision as of 23:11, 1 November 2010
Common Synthesizer Waveforms
Many synthesizers employ a variety of waveforms to produce varied sounds. The most common waveform is the sine wave. However, in additive synthesis, multiple waveforms can be added together to create a different waveform with different characteristics. The basis for this form of synthesis is the Fourier series:
The four basic waveforms are Sine Waves, Square Waves, Triangle Waves, and Sawtooth Waves.
Square Wave
By inspection of the waveform, the DC component of the wave will be 0. Also, since the waveform is odd, an will be 0. Here is the proof:
This just leaves the sine component of the waveform found below.
Finally, resulting in the Fourier series for a Square Wave.
Triangle Wave
Like the Square wave, the DC component of the Triangle Wave is 0 by inspection. Also, since the triangle wave is odd, it is made up only by sine components.
Sawtooth Wave
OCTAVE Scripts to Plot Fourier Series
Square Wave
clf; %Clear Figure t=0:.01:10; %Limits of the graph T=2*pi %Definition of the period M=100 %Number of iterations to undergo sum1=0; %Initialize the sum to
%----------FOURIER SERIES----------% for m=1:1:M, %For m=1, increment by 1 until you get to M if(m!=0) sum1 = sum1 + ((2/(pi*m))-(2/(pi*m))*cos(m*pi))*sin(m*2*pi/T*t); end end
%---------------PLOT---------------% plot(t,real(sum1),'b-') title('Fourier Series Representation of a Wave') xlabel('time (seconds)') ylabel('Function') grid on;
legend(num2str(M) ' terms'); print("squarewave.png","-dpng") % Prints the plot to a png file called squarewave.png
TODO:
- finish the triangle wave derivation
- start sawtooth wave derivation
- put pictures in
- implement triangle wave Fourier Series in OCTAVE
- implement sawtooth wave Fourier Series in OCTAVE