# Frequencyhz 000 250g 500g 750g 1000g amplitude vs 000

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Frequency[Hz] 0.00 2.50G 5.00G 7.50G 10.00G Amplitude [Vs] 0.00 250.00p 500.00p 750.00p 1.00n Amplitude spectrum generated by Tina's Interpreter You can easily check, that the above spectrum provided by the closed formula and the spectrum provided by Tina’s Fourier Spectrum command are exactly the same. It’s rather easy to compare these curves using the clipboard. First select the curve which was generated by the Interpreter, then copy this curve onto the clipboard then in the diagram window go back to the previous page and paste the curve back. The curves will be identical, so we do not show here both.

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Example : Transform of a sine wave (fourex4.sch) The Fourier transform of a sine wave results in a Dirac-delta function, that is an infinitely high spike of zero width at the frequency of a sine wave. We can not reproduce that with DFT. We can actually examine sine waves of finite length. The longer the sine wave is, the taller and narrower the spectrum becomes. Let us first examine a 3ms long sine wave with 1kHz frequency. Time [s] 0.00 1.00m 2.00m 3.00m Voltage [V] -1.00 -500.00m 0.00 500.00m 1.00 T=3m If you run Fourier spectrum on this curve you will get the following result. Frequency [Hz] 0 1k 2k 3k 4k 5k Amplitude [Vs] 0.00 500.00u 1.00m 1.50m 2.00m T 2 =1.5m
Although the peak value of this spectrum is correct the shape of the spectrum is quite rough. To get a better spectrum change the simulation time to 21ms and carry out the analysis again. At FFT the frequency resolution is inversely proportional to the duration of the signal. Time [s] 0.00 1.00m 2.00m 3.00m 4.00m 5.00m Voltage [V] -1.00 -500.00m 0.00 500.00m 1.00 Note that the actual simulation time in this example was 21ms to get a better spectrum. (Not shown on the figure.) Frequency [Hz] 0 1k 2k 3k 4k 5k Amplitude [Vs] 0.00 500.00u 1.00m 1.50m 2.00m

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Let us check this result using the theory of Fourier transform: F } sin ) ( { 0 t t d ω , where > < = = t 0 if , 1 T or t 0 t if , 0 ) ( ) ( ) ( T T t t t d ε ε = = = = + T t j T t j t j t j t j T t j T dt e dt e j dt e j e e t d dt te t d j F 0 ) ( 0 ) ( 0 0 0 0 0 2 1 2 ) ( sin ) ( ) ( ω ω ω ω ω ω ω ω ω ω ) ( 2 ) sin( ) ( 2 ) sin( 0 0 2 ) ( 0 0 2 ) ( 0 0 ω ω ω ω ω ω ω ω ω ω ω ω + + + j T e j T e T j T j Note, that the main amplitude of the spectrum at 0 ω ω = and at 0 ω ω = is 2 T . Let us draw this function with Tina’s Interpreter (fourex3.ipr). In this example we will use a simplified expression, because this part is significant only regarding the amplitude of the spectrum. The following simplified function calculates the amplitude spectrum. Function SinusAmplSpe(f); Begin w := 2 * pi * f; w0 := 2 * pi * 1k; T := 3m; SinusAmplSpe := Abs(Sin((w - w0) * T / 2) / (w - w0) – Sin((w + w0) * T / 2) / (w + w0)); End; Draw(SinusAmplSpe(Frequency), AmplSpe) To try out the above example invoke the Interpreter from the Tools menu, type in the above text or read in fourex3.ipr by selecting the Open command from the File menu. If you press the Run button after a short time the following function appears Frequency[Hz] 0.00 1.00k 2.00k 3.00k 4.00k 5.00k Amplitude [Vs] 0.00 500.00u 1.00m 1.50m 2.00m Amplitude spectrum calculated by Tina's Interpreter
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