On this page, an the Fourier Series is applied to a real world
problem: determining the solution for an electric circuit. Particularly, we will
look at the circuit shown in Figure 1:
Figure 1. A series R-C circuit.
In Figure 1, there is a source voltage, Vs, in series with a
resistor R, and a capacitor C. We are interested in finding the voltage across
the capacitor, which we label as the output voltage. Side note: this simple
circuit can be used as a low pass filter: high frequency noise can be elminated.
The source voltage Vs(t) will be a periodic square wave shown in Figure 1. The Fourier Series coefficients for this function have already been found on the complex coefficients page.
Figure 2. A periodic square waveform.
Electric circuits like that of Figure 1 are easily solved in the
source voltage is sinusoidal (sine or cosine function). When this happens, the
capacitor has an impedance that is easily calculable. The impedance is analogous
to resistance: it is the ratio of the voltage across the capacitor to the
current that flows through it. For capacitors and inductors, the impedance is a
complex number (meaning the voltage and current are out of phase), that depends
on the frequency of the sinusoidal source voltage. If the source voltage has
frequency f, then the impedance of the capacitor (Zc) is:
The output voltage can be easily found with some application of Ohm's Law (V=I*Z):
From equation [2], we see that the output voltage can be easily calculated when the source voltage Vs is sinusoidal.
The question now is: how can we calculate the output voltage, Vo(t), when the input is not a sinusoidal function, but rather any periodic function f(t)?
If you don't think the answer has something to do with Fourier Series, you probably need to work on your reading comprehension skills. It just so happens that we know that any periodic function IS the sum of sinusoidal functions. And we know how to solve the circuit of Figure 1 for any sinusoidal input. Finally, electric circuits are simple linear systems: this means that if an input voltage V1 produces an output X1, and an input voltage V2 produces an output X2, then when the input V1+V2 is applied, the output is X1+X2.
The facts in the proceeding paragraph mean that with Fourier Series, the solution is very simple. We rewrite the square wave in terms of a sum of sinusoidal functions, calculate the output via each one, and then sum up the solutions for each sinusoidal component.
To do this, let's choose a random component of the Fourier Series, say the nth component, corresponding to coefficient cn. This is the coefficient that multiplies the complex exponential, with frequency given by f=n/T:
Using equation [3] in equation [2], the output voltage for just this sinusoid (or complex exponential, they are basically the same), is:
In equation [4], note that the frequency f has been substituted with n/T, because that is the frequency of the corresponding complex exponential that the cn multiplies. The cn values were already calculated on the square wave page.
Since the electric circuit is linear, the total output voltage is given by the sum of all the components of the waveform:
When n=0, then f=0, so the output voltage is easily found from equation [2]. In this case, the output and input are equivalent, so that constant (non-varying) term passes directly to the output.
The output or solution voltage for R=1 Ohm, C=0.1 Farads, T=1s, is given in Figure 3:
Figure 3. Output Voltage, Vo(t).
The beautiful thing about Fourier Series is that this method works
for any periodic function, no matter how complicated. Once the Fourier Series
coefficients are found, the output can be quickly calculated. It doesn't matter
that the solution comes out to be an infinite sum - this is still very
enlightening. From a pure math perspective, you can observe how the infinite sum
varies with n, and get an idea of its frequency response via that analysis. From
an engineering perspective, infinite sums can be easily calculated and coded up,
so that the solution can be found quickly.
This is the power of the Fourier Series. Many, many problems in engineering and physics can be solved analytically for the case of a pure sinusoid input function. By using Fourier Series, the solution for all periodic functions can be quickly found. Hence, Fourier Series is a very useful tool.
The source voltage Vs(t) will be a periodic square wave shown in Figure 1. The Fourier Series coefficients for this function have already been found on the complex coefficients page.
[Equation 1] |
---|
The output voltage can be easily found with some application of Ohm's Law (V=I*Z):
[Equation 2] |
---|
From equation [2], we see that the output voltage can be easily calculated when the source voltage Vs is sinusoidal.
The question now is: how can we calculate the output voltage, Vo(t), when the input is not a sinusoidal function, but rather any periodic function f(t)?
If you don't think the answer has something to do with Fourier Series, you probably need to work on your reading comprehension skills. It just so happens that we know that any periodic function IS the sum of sinusoidal functions. And we know how to solve the circuit of Figure 1 for any sinusoidal input. Finally, electric circuits are simple linear systems: this means that if an input voltage V1 produces an output X1, and an input voltage V2 produces an output X2, then when the input V1+V2 is applied, the output is X1+X2.
The facts in the proceeding paragraph mean that with Fourier Series, the solution is very simple. We rewrite the square wave in terms of a sum of sinusoidal functions, calculate the output via each one, and then sum up the solutions for each sinusoidal component.
To do this, let's choose a random component of the Fourier Series, say the nth component, corresponding to coefficient cn. This is the coefficient that multiplies the complex exponential, with frequency given by f=n/T:
[Equation 3] |
---|
Using equation [3] in equation [2], the output voltage for just this sinusoid (or complex exponential, they are basically the same), is:
[Equation 4] |
---|
In equation [4], note that the frequency f has been substituted with n/T, because that is the frequency of the corresponding complex exponential that the cn multiplies. The cn values were already calculated on the square wave page.
Since the electric circuit is linear, the total output voltage is given by the sum of all the components of the waveform:
[Equation 5] |
---|
When n=0, then f=0, so the output voltage is easily found from equation [2]. In this case, the output and input are equivalent, so that constant (non-varying) term passes directly to the output.
The output or solution voltage for R=1 Ohm, C=0.1 Farads, T=1s, is given in Figure 3:
This is the power of the Fourier Series. Many, many problems in engineering and physics can be solved analytically for the case of a pure sinusoid input function. By using Fourier Series, the solution for all periodic functions can be quickly found. Hence, Fourier Series is a very useful tool.
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