In this page, several examples are presented to illustrate how the time response of second-order circuits can be obtained. For most of the examples, the solution is obtained by using more than a single technique.
Find zero-state unit-step response vO(t) for the circuit in Fig. 33.
SOLUTION:
Solution using Laplace transforms
KCL equation at node with voltage V1(s):
KCL equation at the non-inverting input:
Solution using Differential Equations
The first equation is the KCL equation at the non-inverting input. The second equation is the KCL equation at the node with voltage v1(t). From the two equations, obtain the differential equations and its solution. The initial conditions are of zero value.
Solution using the State Variable Approach
We can obtain the state equation from the KCL equations at the non-inverting node and the capacitor voltage v2(t). Once we know the A and B matrices, the solution can be obtained as shown below.
The Laplace transform function obtained for the output voltage is the same as that obtained earlier.
For the circuit in Fig. 35,
Find zero-state unit-step response vO(t) for the circuit in Fig. 35.
SOLUTION:
Solution using Laplace transforms
Form two node equations, one for v2 and the other at the inverting input of op amp. Eliminate v2 and get the transfer function.
Solution using Differential Equations
Form KCL at the inverting input node and at the with voltage v2.
Solution using the State Variable Approach
For the circuit in Fig. 36, it is given that
Find expressions for v1(t), and v2(t) for t ≥ 0 s. Assume that the capacitors are initially uncharged.
SOLUTION:
Solution using Laplace transforms
We can eliminate V1(s) from the above equations.
Solution using Differential Equations
Form the KCL at the two nodes where the capacitors are incident, as shown below.
Differentiate equation (3.9) and use the expression in (3.10) for the derivative of v1.
Obtain the particular solution and the complementary solution. Obtain the response as the sum of these two solutions.
Solution using the State Variable Approach
The state-equation can be expressed as follows, when the initial conditions are of zero value.
For the circuit in Fig. 37, find expressions for vO(t) for t ≥ 0 s.
SOLUTION:
Solution using Laplace transforms
Get the initial conditions due to the 5 Volts source.
The transformed circuit is shown in Fig. 38.
The KCL equation at the output node can be formed.
Solution using Differential Equations
We need to form the differential equation first.
The solution is obtained in two steps. We determine the response due to the initial conditions first. We get the impulse response next. We add the two responses to get the response of the circuit in Fig. 37.
Let the unit-step response be α(t). Let the impulse response be h(t).
The total response of the circuit in Fig. 37 is obtained as the sum of responses, expressed by equations (4.3) and (4.11).
Solution using the State Variable Approach
Hence we get the following equations.
Equations (4.15) and (4.2) are the same. The solution is the same as that obtained earlier.
For the circuit in Fig. 40, it is given that
Obtain an expression for the capacitor voltage for t > 0.
SOLUTION:
Solution using Differential Equations
The first equation in (5.2) expresses the source current as the sum of the inductor current and the capacitor current. Since the capacitor current is the product of capacitance and the derivative of its voltage, we get the remaining equations in (5.2). We can differentiate equation (5.2) and then we get equation (5.3). Equation (5.4) equates the voltage drop across the two branches, since they are connected in parallel.
Replace the inductor current and its derivative in terms of the capacitor current and its derivatives. Then we get equation (5.5) and (5.6). The source current is expressed by equation (5.1) and hence the right-hand-side expression becomes zero. The solution is obtained as shown below.
The solution is easy, except that you need to figure out what the initial conditions. You need to understand equation (5.8). The capacitor voltage is expressed by equation (5.9).
Solution using Laplace transforms
From the transformed circuit in Fig. 41, we get that
The solution obtained using the Laplace transforms is shown above. It is relatively easy.
Solution using the State Variable Approach
From the circuit in Fig. 40, we can form the following equations.
From the above equations, the state equation can be obtained.
Express the state equation in terms of Laplace transforms, as shown below.
Then obtain an expression for the capacitor voltage. We get the same expression, as that obtained earlier.
For the circuit in Fig 42, it is given that
vS(t)= 10 × cos(t)∙ u(t), iL(0) = 1 A, vC(0) = 5 V.
Find expressions for iL(t) and vC(t).
SOLUTION:
Solution using Differential Equations
Solution using the State Variable Approach
Solution using Laplace transforms
From the transformed circuit, we can form the following equations.
Equate the voltage across the 2 Ω resistor with the voltage across the capacitor.
Then form an equation as shown below.
Equation (6.26) is the same equation, derived earlier for the inductor current.
Equation (6.29) is the same equation, derived earlier for the capacitor voltage.
The switch shown in Fig. 44 is open before t = 0 and the circuit is at rest for t < 0. If the switch is closed at t = 0, find i2(t) fort ≥ 0.
SOLUTION:
Solution using Laplace transforms
Prior to t = 0, the switch is open. In steady-state, the rate of change of inductor current is zero if the excitation is a dc voltage. Then, the initial inductor current at t = 0 is 0.5 A. With this initial current for 2 H inductor, the transform network can be drawn as shown in Fig. 45. From Fig. 45a, the KVL equation for the two loops can be formed.
From the two equations, I1(s) can be eliminated. From the KVL equation for the loop 2, we obtain that
With this value of I1(s) used in the KVL equation for loop 1, we get that
On simplifying and re-arranging,
Then
Alternatively, this problem can be solved with the help of the Thevenin’s theorem. With the switch in Fig. 44 open, the open-circuit voltage can be obtained. The open-circuit voltage is 1 Volt. Hence
The value of ZTh(s) is the impedance measured across the load terminals with the source short-circuited. In terms of Laplace transforms,
The equivalent circuit obtained using the Thevenin’s theorem is shown in Fig. 45b. From this circuit,
We get the same expression for I2(s) obtained earlier. This problem can be also solved using the differential equations. For t ≥0, the differential equations for the circuit in Fig. 44 are expressed below
From equation (ii) ,
Differentiate equation (iii). Then
For the above equation, the auxiliary equation is:
The complementary integral is then obtained to be:
Since the excitation signal is a dc signal, the particular solution is 0.5. The total solution for i2(t) is then
When the switch becomes ON at t = 0, the current through the 3H inductor does not rise suddenly and hence i2(0) = 0. As the current through the 2H inductor is 0.5 A at t = 0, the voltage across the 2 Ω resistor located before the switch is 1 V. Hence
The values of A and B can be now obtained.
Thus
Solution using the State Variable Approach
From the circuit in Fig. 44, we get that
The above two equations can be presented as follows
Since i1(0) = 0.5 Amp and i2(0) = 0 Amp, we get that
Using Cramer’s rule, we get that
The inverse transform of I2(s) yields i2(t).
The switch shown in Fig. 46 is in position ‘a’ before t = 0 and the circuit is at rest for t < 0. If the position of the switch is thrown from position ‘a’ to position ‘b’ at t = 0, find i1(0+), i2(0+), d i1(0+)/dt, and di2(0+)/dt for t ≥0. Find also expressions for i1(t) for t ≥ 0.
Solution:
Prior to t = 0, the circuit is at rest. From the circuit in Fig. 46, i1(0-) = 1 A and v2(0-) = 1 V, since in a circuit at rest with dc excitation, the inductor voltages are and the capacitor currents zero.
When the switch is thrown to position ‘b’, the circuit state changes. For t > 0, two loop equations can be formed as shown below. Note that i1(t) is not the loop current.
As there is no impulse source, i1(0+) = i1(0-) = 1 A and v2(0+) = v2(0-) = 1 V. That is,
Hence
On solving the above equations, we get that i2(0+) = -1 A, and
On subtracting equation (ii) from equation (i), we get that
The derivative of the above equation is:
For t > 0
Differentiate both equations. We get that
Eliminate i2 and its derivative from equations (a), ( c) and (d). Then we get
Then
From the initial conditions, we get that
This example shows how Matlab can be used to solve a problem that is too difficult to be solved using analytical approach.
A quasi-square wave voltage signal is applied to an RLC circuit shown in Fig. 47. Since the excitation signal is periodic, the responses present in the circuit are also periodic, after the transients have died down. The periodic current response and the periodic response of the voltage across the capacitor are to be obtained. The quasi-square wave signal is defined over a period as follows:
Since vin(t) is periodic,
Obtain the following periodic responses given that R =1 Ω, L = 1 H and C = 1 F:
i. Inductor current,
ii Capacitor voltage.
Solution Using Matlab
Before the program is presented, it is explained how the problem can be solved using a program. The input waveform is periodic, with a period of 1.2 s. A step size of 0.01 s is used in the program. Hence 120 iterations are required to compute the response of one cycle. A count variable, called n, is defined. It varies from 1 to 120. Hence the input voltage can be defined as shown.
Because the input is periodic, the inductor current becomes periodic. Let the inductor current at the start of a cycle be startAmp and let its value at the end of a cycle be finishAmp. When the inductor current is periodic, startAmp = finishAmp. The program keeps calculating the response till these values are close to each other. Calculations are repeated, using a while loop. If both startAmp and finishAmp are initially set to zero, the loop will not be executed. To start the execution of the loop, finishAmp is initially set to zero, whereas startAmp is arbitrarily set to 0.2 A. Once the program enters the loop, the first instruction is to make startAmp equal to finishAmp. Then a for loop is executed within the while loop. Here the count variable is n. For every step, the increments to capacitor voltage and the inductor current are computed. It is seen from the circuit that
At any step, the present values of vin, vo and iL are known. Hence the rates of change of capacitor voltage and the inductor current can be computed. The increment is obtained by multiplying the rate of change by the step size. The increments are added to the present values and the next values of vo and iL are obtained. The count n is then incremented automatically by the program each time the program executes one for loop. After n has got incremented to 120, the program comes out of the loop and the finishAmp is set equal to the inductor current. Then the while loop checks whether the absolute difference between startAmp and finishAmp is less than a prescribed small value. In this program, the prescribed value is 0.002 A. If the absolute difference between startAmp and finishAmp is less than the prescribed small value, the program exits the while loop; otherwise it repeats the while loop.
The program is executed for one more cycle after exiting the while loop. This time the values of capacitor voltage, the input voltage and the inductor current are stored in arrays and the plot command is used to obtain the plots.
The Matlab program used to solve the problem is presented below
% The Response of a series RLC circuit to a quasi-square wave input % The components selected to yield an underdamped response % This program yields periodic response
% Component values specified
R=1.0; L=1.0; C=1.0;
% step-size defined hh=0.01;
startamp=0.2; finishamp=0.0; vcap=0.0; amp=0.0;
while (abs(finishamp-startamp)>0.002),
startamp=finishamp;
for n=1:120;
if (n<11) vin=0.0; end;
if ((n>10) & (n<51)) vin=1.0; end;
if ((n>50) & (n<71)) vin=0.0; end;
if ((n>70) & (n<111)) vin=-1.0; end;
if (n>110) vin=0.0; end;
incrI=(vin-vcap)/L*hh;
incrV=(amp - vcap/R)*hh/C;
amp=amp+incrI;
vcap=vcap+incrV;
end;
finishamp=amp;
end;
zeit=0.0;
for n=1:120;
if (n<11) vin=0.0; end;
if ((n>10) & (n<51)) vin=1.0; end;
if ((n>50) & (n<71)) vin=0.0; end;
if ((n>70) & (n<111)) vin=-1.0; end;
if (n>110) vin=0.0; end;
incrI=(vin-vcap)/L*hh;
incrV=(amp - vcap/R)*hh/C;
amp=amp+incrI;
vcap=vcap+incrV;
zeit=zeit+hh;
InVolt(n)=vin;
Cur(n)=amp;
CapV(n)=vcap;
x(n)=zeit;
end;
subplot(311)
axis([0 1.2 -1.5 1.5])
plot(x,InVolt)
ylabel('Input Volt')
grid
subplot(312)
axis([0 1.2 -0.05 0.05])
plot(x,CapV)
ylabel('Cap. Volt')
grid
subplot(313)
axis([0 1.2 -0.25 0.25])
plot(x,Cur)
xlabel('Time in sec')
ylabel('Ind. Current')
grid
The responses obtained are shown below.
Responses to the quasi-square wave input
For the circuit in Fig. 48, obtain the expressions for
i. the inductor current and
ii. the capacitor voltage.
Assume that the inductor current is zero and the capacitor is in a discharged state at t = 0.
SOLUTION:
Let the source current be iS(t). Using Laplace transforms, we get that
After substituting values for the source, R, L and C, we get
The differential equation needed for solution can be formed as follows.
Differentiate both equations.
Substitute for the derivatives of capacitor voltage. The second derivative of capacitor voltage in equation (iv) can be replaced by the right-hand side value of equation (iii). The first derivative of capacitor voltage in equation (iv) can be replaced by the right-hand side value of equation (i). We get then
At t = 0, the inductor current stays at zero value. The source current of 5 A flows through the resistor and capacitor. Since the capacitor voltage at t = 0 is zero volts, the voltage across the inductor at t = 0 is the voltage across the resistor, which is 25 Volts.
The expression for the capacitor voltage is obtained as follows.
The capacitor voltage can be obtained from its differential equation.
In this page, some examples have been presented to illustrate how the response of second-order circuits can be obtained. The topic of Frequency Response is explained in the next chapter.