TINA_Advanced_Topics.pdf

In tina fourier analysis must be always performed in

Info icon This preview shows pages 51–55. Sign up to view the full content.

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
Image of page 51

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

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)
Image of page 52
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).
Image of page 53

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

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}
Image of page 54
Image of page 55
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern