# In tina fourier analysis must be always performed in

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well. In Tina Fourier analysis must be always performed in conjunction with the transient analysis. This means, that first you always have to carry out a transient analysis on the circuit. When the diagram window appears you have to select the curve you want to process using the cursor and from the Process menu choose Fourier series or Fourier spectrum . In the first part of this booklet we will investigate Fourier series and later on Fourier spectrum. Fourier series One type of spectral analysis is called Fourier or harmonic decomposition. It’s based on the theory, that any periodic function can be expressed as the sum, or series of sinusoidal functions. If a periodic function is expressed this way, each component in the series must be periodic over the same interval as the original function. The components are integer multiples or harmonics of the original function’s fundamental (base) frequency. This decomposition is described by the following formulas: ) sin cos ( ) ( 0 t k B t k A t f k k k ω ω + = = , or in complex form f t C e k k jk t ( ) = =−∞ ω , where A B k k , , and C k are the Fourier coefficients. Let us see this decomposition in case of some practical waveforms: Let us demonstrate the above theorem with Tina in case of a square wave of f=1kHz frequency and 1V amplitude up to three sine waves. The easiest way is to put together a simple circuit (fourex1.sch). The circuit consists of three sine wave generators connected in series and the reference square wave generator. The frequencies of the sine wave generators are kHz f 1 1 = , kHz f f 3 3 1 2 = = and kHz f f 5 5 1 3 = = , however the amplitudes are , 2732 . 1 4 1 = = π A 4244 . 0 3 1 1 2 = = A A and 2546 . 0 5 1 1 3 = = A A according to the above formula. Triangle wave Square wave f t t t t ( ) sin sin sin ... = + 8 3 5 2 1 3 1 5 2 2 π ω ω ω f t t t t ( ) sin sin sin ... = + + + 4 3 5 1 3 1 5 π ω ω ω

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The result of the transient analysis can be seen on the following figure. Time [s] 0.00 250.00u 500.00u 750.00u 1.00m Output -2.00 -1.00 0.00 1.00 2.00 f 1+2+3 = 4 π [sin( ω t)+ 1 3 sin(3 ω t)+ 1 5 sin(5 ω t))] f 1+2 = 4 π [sin( ω t)+ 1 3 sin(3 ω t))] f 1 = 4 π sin( ω t)
Note that the above figure was entirely made by Tina using its Equation Editor in the Diagram Window. For example the ) sin( 4 1 t f ω π = is described in the following way in the Equation Editor’s text mode: \i(f,1)=\f(4,\s(p))sin(\s(w)t) as it can be seen on the next figure. If you press the View button you can preview the formula in the Equation Editor. Press the Edit button to return into the text mode if you want to make further changes. Finally press the Copy button to place the formula in the Diagram Window. Obviously if you connected more than 3 generators in series, you would have achieved a better match. There is an easy way to demonstrate this statement. Using Tina’s Interpreter you can easily write a function that calculates the sum of any number of sine waves (fourex1.ipr).

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Function Square(t, n); {Sum of n sine waves, n: number of sine waves, t: time} Begin f := 1k; w := 2 * pi * f; x := 0; For i := 0 To n - 1 Do x := x + 1 / (2 * i + 1) * sin((2 * i + 1) * w * t); Square := 4 / pi * x; End; {Sum of 20 sine waves}
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