From an engineering perspective, infinite sums can be easily calculated and coded up, so that From a pure math perspective, you can observe how the infinite sum varies with n, and get an idea of itsįrequency response via that analysis. It doesn't matter that the solution comes out to be an infinite sum. Series coefficients are found, the output can be quickly calculated. The beautiful thing about Fourier Series is that this method works for any periodic function, no matter how complicated. The output or solution voltage for R=1 Ohm, C=0.1 Farads, T=1s, is given in Figure 3: In this case, the output and input are equivalent, so that constant (non-varying) When n=0, then f=0, so the output voltage is easily found from equation. Since the electric circuit is linear, the total output voltage is given by the sum of all the components of the waveform: In equation, note that the frequency f has been substituted with n/T, because that is the frequency of the correspondingĬomplex exponential that the cn multiplies. Using equation in equation, the output voltageįor just this sinusoid (or complex exponential, they are basically the same), is: The coefficient that multiplies the complex exponential, with frequency given by f=n/T: To do this, let's choose a random component of the Fourier Series, say the nth component, corresponding to coefficient cn. Of a sum of sinusoidal functions, calculate the output via each one, and then sum up the solutions for each sinusoidal component. The facts in the proceeding paragraph mean that with Fourier Series, the solution is very simple. Finally, electric circuits are simple linear systems: this means that if an input voltage V1 producesĪn output X1, and an input voltage V2 produces an output X2, then when the input V1+V2 is applied, the output is X1+X2. And we know how to solve the circuit ofįigure 1 for any sinusoidal input. It just so happens that we know that any periodic function IS the sum of sinusoidal functions. If you don't think the answer has something to do with Fourier Series, you probably need to work on your reading comprehension skills. The question now is: how can we calculate the output voltage, Vo(t), when the input is not a sinusoidal function, but rather The output voltage can be easily found with some application of Ohm's Law (V=I*Z):įrom equation, we see that the output voltage can be easily calculated when the source voltage Vs is sinusoidal. If the source voltage has frequency f, then the impedance of the capacitor (Zc) is: For capacitors and inductors, the impedance is a complex number (meaning the voltage and current are out of phase),įrequency of the sinusoidal source voltage. The impedance is analogous to resistance: it is the ratio of the voltage across the capacitor to theĬurrent that flows through it. The capacitor has an impedance that is easily calculable. Have already been found on the complex coefficients page.Įlectric circuits like that of Figure 1 are easily solved in the source voltage is sinusoidal (sine or cosine function). The Fourier Series coefficients for this function The source voltage Vs(t) will be a periodic square wave shown in Figure 1. Side note: this simple circuit can be used as a low pass filter: high frequency Voltage across the capacitor, which we label as the output voltage. In Figure 1, there is a source voltage, Vs, in series with a resistor R, and a capacitor C. We will look at the circuit shown in Figure 1: On this page, an the Fourier Series is applied to a real world problem: determining the solution for an electric circuit. Re > 0 21 Σ 2lt - nT,) 811 - 12) = S Seelikud - ¿ Cariken a + jw in 1)! (a + jo)" 21 ? + G(no)8(w - nah), 2. - Fourier Series Example: Electric Circuit Fourier Series Application: Electric Circuits Previous: Derivation of Fourier Series Coefficients TABLE 5.1 Fourier Transform Properties Time Function Fourier Transform aF (W) + F2(W) Operation Linearity Time shift Time reversal afi(t) + bfz(t) f(t – to) f-t) F(we jest F(-) 0 Time scaling f(at) la -FC) ( Time transformation f(at - to) F(t) Duality Frequency shift Convolution 1 F e-jwto/a al a 21f(-w) F(w-wo) F (w)F2(0) f(t)ejant f(t)*f (t) Modulation (Multiplication) f(t)f (t) Fi(w)*F2() 27T Integration fludo f(T)dt i "(W) + F(0)8(w) Differentiation in time (jw)"F(W) d" dt" (-jt)"f(t) Differentiation in Frequency d" do" Symmetry f(t) real F(-w) = F"(w) Using the tables of Fourier Transform Pairs and Fourier Transform Properties, find the Fourier Transform of each of the following signals: a.
0 Comments
Leave a Reply. |
AuthorWrite something about yourself. No need to be fancy, just an overview. ArchivesCategories |