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A Fourier-tétel states that any periodic waveform can be synthesized by adding appropriately weighted sine and cosine terms of various frequencies. The theorem is well-covered in other textbooks, so we will only summarize the results and show some examples.

Let our periodic function be f (t) = f (t ±nT) where T is the time of one period and n is an integer number.

w₀= 2p/ T the fundamental angular frequency.

Valami által Fourier-tétel, the periodic function can be written as the following sum:

ahol

A_n és B_n are the Fourier-együtthatók és az összeg a Fourier sorozat.

Another form, probably a bit more practical:

ahol

A₀ = C₀ is the DC or average value, A₁, B₁ és C₁ are the fundamental components, and the others are the harmonic terms.

While only a few terms may be required to approximate some waveforms, others will require many terms.

Generally, the more terms included, the better the approximation, but for waveforms containing steps, such as rectangular impulses, the Gibbs-jelenség comes into play. As the number of terms increases, the overshoot becomes concentrated in an ever smaller period of time.

An egyenletes funkció f(t) = f(-t) (axis symmetry) requires only cosine terms.

An páratlan funkció f(t) = – f(-t) (point symmetry) requires only sine terms.

A hullámforma tükör vagy félhullámú szimmetria csak páratlan harmonics in its Fourier representation.

Here we will not deal with the Fourier series expansion, but will only use a given sum of sines and cosines as an excitation for a circuit.

In the earlier chapters of this book, we dealt with sinusoidal excitation. If the circuit is linear, the szuperpozíció tétel is valid. For a network with nonsinusoidal periodic excitation, superposition allows us to calculate the currents and voltages due to each Fourier sinusoid term one at a time. When all are calculated, we finally summarize the harmonic components of the response.

It is a bit complicated to determine the different terms of the periodic voltages and currents and, in fact, it may yield an overload of information. In practice, we would like to simply make measurements. We can measure the different harmonic terms using a harmonikus analizátor, spectrum analyzer, wave analyzer or Fourier analyzer. Mindezek complicated and probably yield more data than needed. Sometimes it is sufficient to describe a periodic signal only by its average values. But there are several kinds of average measurements.

ÁTLAGOS ÉRTÉKEK

Egyszerű átlag or DC term was seen in the Fourier representation as A₀

This average can be measured with instruments such as the Deprez DC instruments.

Hatékony érték or rms (root mean square) has the following definition:

This is the most important average value because the heat dissipated in resistors is proportional to the effective value. Many digital and some analog voltmeters can measure the effective value of voltages and currents.

Abszolút átlag

This average is no longer important; earlier instruments measured this form of average.

If we know the Fourier representation of a voltage or current waveform, we can also calculate the average values as follows:

Egyszerű átlag or DC term was seen in the Fourier representation as A₀ = C₀

Hatékony érték or rms (root mean square) is, after integrating the Fourier series of the voltage:

A klirr faktor is a very important ratio of the average values:

It is the ratio of the effective value of the higher harmonic terms to the effective value of the fundamental harmonic:

There seems to be a contradiction here–we solve network in terms of harmonic components, but we measure average quantities.

Mutassuk be a módszert egyszerű példákkal:

Example 1

Find the time function and the effective (rms) value of the voltage v_C(T)

ha R = 5 ohm, C = 10 mF és v (t) = (100 + 200 cos (w₀t) + 30 cos (3 w₀t - 90 °)) V, ahol az alapvető szögfrekvencia w₀= 30 krad / s.

Try using the superposition theorem to solve the problem.

The first step is to find the transfer function as a function of the frequency. For simplicity, use the substitution: s = j w

Now substitute the component values and s = j k w₀ahol k = 0; 1; 3 ebben a példában és w₀= 30 krad/s. V, A, ohm, mF és Mrad / s egységek:

It is helpful to use a table to organise the steps of the numerical solution:

We can summarise the steps of the superposition solution in another table. As we have already seen, to find the complex peak value of a component, we should multiply the complex peak value of the component of the excitation by the value of the complex transfer function:

And finally we can give the time function knowing the complex peak values of the components:

v_C(t) = 100 + 110 cos(w₀t - 56.3°) + 6.51 cos (3w₀t - 167.5°) V

The rms (effective) value of the voltage is:

As you can see, TINA’s measuring instrument measures this rms value.

Példa 2

Find the time function and the effective (rms) value of the current i(t)

ha R = 5 ohm, C = 10 mF és v (t) = (100 + 200 cos (w₀t) + 30 cos (3w₀t - 90 °)) V ahol az alapvető szögfrekvencia w₀= 30 krad / s.

Try to solve the problem using the superposition theorem.

The steps of the solution are similar to Example 1, but the transfer function is different.

Now substitute the numerical values and s = j k w_0,ahol k = 0; 1; 3 ebben a példában.

V, A, ohm, mF és Mrad / s egységek:

It’s helpful to use a table during the numerical solution:

We can summarise the steps of the superposition in another table. As we have already seen, to find the peak value of a component, we should multiply the complex peak value of that component of the excitation by the value of the complex transfer function. Use the complex peak values of the components of the excitation:

And finally, knowing the complex peak values of the components we can state the time function:

i (t) = 32.4 cos (w₀t + 33.7°) + 5.85 cos (3w₀t - 77.5°) [A]

The rms value of the current:

You can often do a sanity check for part of the solution. For example, a capacitor can have a DC voltage but not a DC current.

Példa 3

Obtain the time function of the voltage V_ab if R₁= 12 ohm, R₂ = 14 ohm, L = 25 mH, and

C = 200 mF. The generator voltage is v(t)=(50 + 80 cos(w₀t) + 30 cos (2 w₀t + 60 °)) V, ahol az alapvető frekvencia f₀ = 50 Hz.

The first step is to find the transfer function:

Substituting numerical values in V, A, ohm, mH, mF, kHz units:

A két táblázat egyesítése:

k V Sk		V ABK
0 50		50
1 80		79.3 e-j66.3
2 30 ej60		29.7 e-j44.7

Finally the time function:

v_ab(t) = 50 + 79.3 cos (w₁t - 66.3°) + 29.7 cos (2w₁t - 44.7°) [V]

and the rms value:

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