Zero-input response of a second-order circuit-EEWORLD

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Any circuit described by a second-order differential equation is called a second-order circuit. A second-order circuit contains two independent energy storage elements. This section takes a series circuit as an example to discuss the zero-input response of a second-order circuit.

Figure 8-7-1

Figure 8-7-1 is a series circuit. When , assuming that the capacitor C has been charged, the initial voltage is , and the inductor L is in the zero initial state, that is , at the moment, the switch S is closed. Find the zero input response , and .

Select the reference direction of each voltage and current as shown in Figure 8-7-1. After the switch S is closed, write the differential equation describing the circuit according to Kirchhoff's voltage law:

(Formula 8-7-1)

There are two unknown variables i and in (Equation 8-7-1) . Substituting into the above equation and eliminating it , we get:

Right now:

(Formula 8-7-2)

You can also get:

(Formula 8-8-3)

(Equation 8-7-2) and (Equation 8-7-3) are completely consistent in form, both are linear constant coefficient second-order homogeneous differential equations. You can choose any one of them to solve, and now choose (Equation 8-7-2). To solve the second-order differential equation, two initial conditions are required to determine the integral constant.

According to the route switching rules:

The characteristic equation is:

The characteristic roots are:

(Formula 8-7-4)

The characteristic roots are only related to the circuit structure and parameters.

The solutions of the equation are discussed below in three cases.

(1) When , that is , , the transition process is a non-periodic case, also known as the over-damped case. In this case, the characteristic equation has two unequal negative real roots. The general form of the general solution is:

(Formula 8-7-5)

Current:

(Formula 8-7-6)

The integral constant A1 is determined by the initial conditions, and the time value of (Formula 8-7-5) (Formula 8-7-6) is taken as :

From the initial value:

Solving the above two equations together, we get:

(Formula 8-7-7)

Substitute A1 into (Formula 8-7-5) (Formula 8-7-6) to obtain:

Capacitor voltage:

(Formula 8-7-8)

Current:

Inductor voltage:

And because , therefore:

(Formula 8-7-9)

(Formula 8-7-10)

Figure 8-7-2

The curves of , , and time variation are shown in Figure 8-7-2. In (Formula 8-7-8), there are two components, S1 and S2 are both negative values, and , so decays faster than . These two monotonically decreasing exponential functions determine that the discharge process of the capacitor voltage is non-periodic.

The initial value of the inductor voltage at is . At , due to the continuous negative increase of the current , it is negative; after , the current decreases negatively and becomes positive, and finally decays to zero.

If , or , the analysis process is the same as above.

(2) When , that is, , the transition process is critically damped, and the characteristic equation has two equal negative real roots.

(Formula 8-7-11)

The general form of capacitor voltage is:

(Formula 8-7-12)

Current:

(Formula 8-7-13)

Determine the integration constants, , from the initial conditions :

The solution is:

therefore:

(Formula 8-7-14), (Formula 8-7-15) (Formula 8-7-16)

The curves that change with time are similar to those shown in Figure 8-7-2 , and the response is still non-periodic and non-oscillatory.

(3) When , that is , , the transition process is underdamped and is a periodic oscillation. At this time, the characteristic equation has two conjugate complex roots with negative real parts. Let , called the attenuation coefficient, be the resonant angular frequency, and be called the oscillation angular frequency, then the characteristic root is:

(Formula 8-7-17)

The general form of capacitor voltage is:

(Formula 8-7-18)

Current:

(Formula 8-7-19)

Determine the integral constants A and from the initial values , and take the values of (Formula 8-7-18) and (Formula 8-7-19) at time to obtain:

Solving the equation together:

, (Formula 8-7-20)

then:

(Formula 8-7-21)

(Formula 8-7-22)

(Formula 8-7-23)

Figure 8-7-3

The waveforms of , , and are shown in Figure 8-7-3. They are all sinusoidal waves with exponentially decaying amplitudes. The dotted line in the figure is the envelope. When reaches a maximum value, it is zero; when reaches a maximum value, I is zero. This kind of oscillation with gradually decreasing amplitude is called damped oscillation or attenuated oscillation. The larger the attenuation coefficient b, the faster the amplitude decays; the smaller b, the slower the amplitude decays. The damped oscillation angular frequency is determined by the parameters of the circuit itself. When the resistance decreases, the attenuation coefficient decreases, and the attenuation slows down. In the extreme case of , the attenuation coefficient , the response becomes an equal-amplitude oscillation, also known as an undamped oscillation. The undamped oscillation angular frequency is equal to the resonant angular frequency . At this time, (Formula 8-7-21) (Formula 8-7-22) (Formula 8-7-23) becomes:

(Formula 8-7-24)

(Formula 8-7-25)

(Formula 8-7-26)

The above undamped oscillation is not formed by the forced action of the excitation source, and is a zero input response, so it is called free oscillation. The following is an analysis of the long-damped periodic oscillation process of the circuit from the perspective of energy conversion.

Example 8-7-1 For the circuit shown in Figure 8-7-4, when , switch S is closed. Given , , , . Try to calculate , and , respectively .