INTRODUCTION
EXAMPLE 2
EXAMPLE 4
EXAMPLE 6
EXAMPLE 8
EXAMPLE 10
EXAMPLE 12
EXAMPLE 14
EXAMPLE 16
EXAMPLE 18
SUMMARY

EXAMPLE 1
EXAMPLE 3
EXAMPLE 5
EXAMPLE 7
EXAMPLE 9
EXAMPLE 11
EXAMPLE 13
EXAMPLE 15
EXAMPLE 17
EXAMPLE 19

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INTRODUCTION

In this page, numerous examples are presented. Normally a problem is solved using both the differential equations and the Laplace transforms.

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EXAMPLE 1

For the circuit in Fig. 47, it is given that R = 1 Ω, and L = 2 H. Obtain the following.

1. Zero-input response given that i(0)= 5 A.
2. Impulse Response.
3. Unit-step Response
4. Ramp Response
5. Response to sinusoidal input, 5.sin(t).u(t)
6. Response to complex sinusoid input, 10 ×e^{- t}·sin(t)·u(t)

Solution:

For the circuit in Fig. 47, the differential equation is obtained as shown below.

From the transformed circuit shown in Fig. 47, we get the following equations.

a. Zero-input Response

The zero-input response is obtained, as shown above.

b. Impulse Response

Using the Laplace transforms, the impulse response can be obtained, as shown above. Using the differential equation approach, we get the impulse response, as shown below.

c. Unit-step Response

The unit-step response can be obtained, as shown below. Since the input is a unit-step, the particular solution is unity and the unit-step response is then the sum of the particular solution and the complementary solution, as shown by equation (5.13). The constant K, can be evaluated from the zero initial condition, as illustrated by equation (5.14). The unit-step response can be obtained using the Laplace transforms, as shown below.

4. Ramp Response

The ramp response is obtained, as shown below.

Equation (5.18) displays the differential equation with ramp input. The ramp response, obtained using the differential equation, is displayed below.

5. Response to sinusoidal input, 5 ∙sin(t)∙u(t)

Equation (5.24) displays the differential equation with sinusoidal input. The steady-state response, obtained using the differential equation, is displayed below.

It is possible to combine the two terms in equation (5.25) and express it, as √5 ∙sin(t - tan^{- 1}(2)∙u(t).

The particular solution can be obtained using the jω notation.

The total solution is the sum of the complementary and the particular integral.

Using Laplace Transforms, we get that

Obtaining the solution using the Laplace transforms is not that easy, due to the effort needed to expand the function into partial fractions.

6. Response due to complex sinusoid input, 10 ×e^{- t}·sin(t)·u(t)

The solution obtained using the differential equation approach is displayed below.

The total solution is the sum of the complementary and the particular integral.

Using Laplace Transforms, we get that

This problem has illustrated how the response of a first-order system can obtained for different inputs.

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EXAMPLE 2

For the circuit shown in Fig. 48, assume that v_C(0) = 0. Find v_C(t) for the two inputs specified below.

1. i_S(t) = cos(t).u(t)
2. i_S(t) = sin(t).u(t)

Solution:

First we form a differential equation for the circuit. For the circuit in Fig. 48, we get that

Equation (5.39) specifies the capacitor voltage as a Laplace transform function. Next given that i_S(t) = cos(t).u(t), an expression for the capacitor voltage is obtained, as shown below.

It is possible to obtain particular solution or the steady-state response using phasors.

The response to the sinusoidal input is then obtained, as shown below.

The solution using Laplace transforms can be obtained as shown below.

Next given that i_S(t) = sin(t).u(t), an expression for the capacitor voltage is obtained as shown below.

It is possible to obtain particular solution or the steady-state response using phasors.

The response to the sinusoidal input is then obtained, as shown below.

The solution using Laplace transforms can be obtained as shown below.

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EXAMPLE 3

For the circuit in Fig. 49, determine i_L(t).

Solution:

Equation (5.58) is obtained by equating the voltage across the resistor 10 Ω with the voltage across the series combination of the resistor and the inductor. Equation (5.59 is the differential equation of this circuit. The value of current through the inductor is 1 A at t = 0. The initial current is due to 2.u(-t). On solving equation (5.59), equation (5.60) is obtained.

From the transformed circuit displayed in Fig. 50, the solution using Laplace transforms can be obtained.

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EXAMPLE 4

For the circuit in Fig. 51, find the instant t, at which v_R(t) = v_R(t).

Solution:

An expression for the inductor current is obtained, as shown above.

By equating the inductor voltage with the resistor voltage, as shown by equation (5.66), we can obtain the instant t, at which the inductor voltage equals the resistor voltage. The result is expressed by equation (5.68).

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EXAMPLE 5

For the circuit in Fig. 52, a pulse voltage is applied. Find the capacitor voltage by using:

• Differential equations approach
• Laplace Transforms, and
• Convolution Integral.

Solution:

Differential Equations Approach

Equation (5.69) presents the differential equation of the circuit. It can be split into two equations, as shown by equation (5.70). The solution is presented below.

Using Laplace Transforms

Equation (5.73) presents the Laplace transform function of the capacitor voltage. Note how the pulse voltage is represented, using the transform for shifted time function. Equation (5.75) expands the capacitor voltage into partial fractions. The inverse transform is shown by equation (5.76).

Equation (5.75) can be presented as two equations, as shown by equations (5.76) and (5.77).

Using Convolution Integral

To use convolution integral, we need to know the impulse response. The impulse response is the derivative of the unit step response, presented by equation (5.71).

The convolution integral is obtained using two equations, one for 0 ≤ t ≤ 1 s, and the other for t > 1 s. When t < 1 s, the lower limit for integral is zero, as the impulse response is zero for t < 0 s. When t > 1 s, the lower limit for integral is (t - 1).

Equations (5.78) and (5.79) present the expression for the capacitor voltage.

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EXAMPLE 6

For the circuit shown in Fig. 54, find the zero-state response i(t). The shape of the input pulse is shown on the right side.

Solution:

Equation (5.80) descries the input pulse. Equation (5.81) presents the differential equation of the circuit. The zero-state response is presented by equations (5.82) and (5.83).

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EXAMPLE 7

For the circuit shown in Fig. 55, find the zero-state response i(t). The shape of the input pulse is shown on the right side.

Solution:

Equation (5.84) descries the input pulse. Equation (5.85) presents the differential equation of the circuit. The zero-state response is presented by equations (5.86) and (5.87). Equation (5.86) presents the sum of three responses. The term, (1 - e^-t ) is the forced response due to the unit step. The term, (t - 1) is the forced response due to the ramp input. The third term, an exponential function, is the complementary solution, solved for zero-initial condition. The response for period t > 1 second is obtained by treating the inductor current at t = 1 second as the initial condition. Only natural response remains after t > 1 second.

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EXAMPLE 8

For each of the input signals listed below, obtain the output voltage of the circuit shown in Fig. 56.

Solution:

a. Response to v_S(t) = 5 × cos(t)

Equations (5.88) and (5.89) present the differential equation of the circuit. In an ideal op-amp, the voltage the non-inverting terminal is the same as the voltage at the inverting terminal. For the circuit shown in Fig. 56, the inverting input terminal acts as virtual ground. Equation (5.90) presents the output voltage as a Laplace transform function.

Since an everlasting sinusoidal signal is presented as the input, the jω; notation can be used, as shown by equation (5.91). The corresponding time-domain expression is presented by equation (5.92).

b. Response to v_S(t) = 5 × cos(t)∙u(t).

The input signal is present only for t > 0 second. Hence the solution is the sum of the particular solution and the complementary solution. The particular solution is presented by equation (5.92). Since the initial capacitor voltage is zero, the constant associated with the complementary solution can be obtained, and the zero-state response to sinusoidal is presented by equation (5.94).

c. Response to v_S(t) = 10∙e^-2t × cos(3t)∙u(t).

Obtaining the zero-input response to complex sinusoid input signal is no more complicated than obtaining the response to sinusoidal input. Equation (5.95) presents the differential equation of the circuit. The particular solution is obtained as shown by equation (5.96). Equation (5.97) presents the zero-state response to the complex sinusoid input signal.

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EXAMPLE 9

The capacitor has zero charge for t < 0. The switch is thrown open a t = 0. Find the value of t at which the energy stored in the capacitor equals 25% of the maximum energy stored in it.

Solution:

The value of current i² at t = 0^- is zero. Hence i²(t) is u(t). Note that v_S(t) = 20 V. The Laplace transform function for the capacitor voltage can be obtained, as shown by equation (5.99) and equation (5.100) presents an expression for the capacitor voltage in domain.

The maximum capacitor voltage is 40 V. The maximum energy stored in the capacitor is expressed by equation (5.101). At t= t₁, the energy stored in the capacitor equals 25% of the maximum energy. Hence the capacitor voltage at t= t₁ is 20 V. The value of t₁ is expressed by equation (5.103).

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EXAMPLE 10

Find the zero-state response i₂(t) for the circuit in Fig. 58, if i_S(t) = 2∙ r(t).

Solution:

Equation (5.105) shows the first step in getting the differential equation. The source current is the sum of the currents through the inductors. The voltage across the inductor in the middle equals the voltage across the series combination of the resistor and the inductor, as shown by equation (5.106). Equation (5.107) is obtained from equation (5.106).

Equation (5.108) presents the response of i₂(t). The value of constant A can be obtained,and equation (5.109) presents the zero-state response of i₂(t). We can get he same response, using the Laplace transforms, as shown below.

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EXAMPLE 11

Find the zero-state response v_O(t) of the output voltage of the circuit in Fig. 59. It is given that v_S(t) = u(t).

Solution:

Equation (5.112) is obtained by applying Kirchoff's current law, at the node where the capacitor is incident. On re-arranging, we get equations (5.113) and (5.114).

Equation (5.115) shows the expression for the capacitor voltage. The value of A can be obtained from the initial voltage across the capacitor. The zero-state response of capacitor voltage is expressed by equation (5.116). Equation (5.117) presents the output voltage.

The solution can be obtained using the Laplace transforms. The transformed circuit is shown in Fig. 60.

Equation (5.118) is obtained by using the Kirchoff's current law. This equation can be re-arranged and we can obtain equations (5.119) and (5.120). The inverse transform of equation (5.120) is presented by equation (5.116).

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EXAMPLE 12

Find the zero-state response v_O(t) for the circuit in Fig. 61 for each of the two input signals, specified below.

1. e^-t ∙ u(t)
2. e^-2t ∙ u(t)

Solution:

Response to input, e^-t ∙ u(t)

Note the differential equation should be formed for the capacitor voltage and not for the output voltage.

Once an expression for the capacitor voltage is obtained, the output voltage can be obtained as the sum of the source voltage and the capacitor voltage, as shown by equation (5.127).

Response to input, e^-2t ∙ u(t)

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EXAMPLE 13

For the circuit in Fig. 62, v_S(0)= 0. It is given that v_S(t) = t ∙u(t).

a. Find the zero-state response v_C(t).

b. Construct a Norton’s equivalent for the circuit by replacing all the elements to the left of the capacitor by a single resistor and a single independent current source.

Solution:

The circuits in Fig. 63 show how Thevenin's voltage and Norton's current can be obtained. From the circuit drawn for obtaining Thevenin's voltage, we get that

From the circuit drawn for obtaining Norton's current, we get that

Thevenin's resistance is obtained as the ratio of Thevenin's voltage to Norton's current. Now the circuit in Fig. 62 can be presented using Norton's equivalent circuit.

An expression for the capacitor voltage can be obtained as shown below.

We can obtain the same solution, using the Laplace transforms.

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EXAMPLE 14

i. A resistance of 1 Ω and a capacitance of 1 F are in series. An ac voltage of v(t)=2∙sin (t + Φ) is applied suddenly. Find an expression for the voltage across the capacitor .
b. When does the transient response become zero ?

Solution:

The particular solution is obtained, as shown below.

The complementary response is obtained as shown below.

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EXAMPLE 15

A relay circuit is shown in Fig. 64. It is energized at t = 0 when the coil is at rest and the relay contacts close at t = 10 ms when the current through the coil is 16 mA. The resistance of coil is 500 Ω. Find its inductance.

Solution:

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EXAMPLE 16

Obtain i(t) for t > 0 for the circuit in Fig. 66.

Solution:

The problem is solved in two steps.

• Step 1: Obtain v_C(t)
• Step 2: Obtain i(t)

Represent the ramp source and the 2 Ω resistor by the Norton equivalent, and obtain the circuit shown in Fig. 67. From Fig. 67,

Equation (5.150) is the differential equation for the circuit in Fig. 67. The complementary solution is expressed by equation (5.151). The particular solution is expressed by equation (5.152).

The zero-state response of the capacitor voltage is expressed by equations (5.153) and (5.154). Now we can obtain an expression for i(t) from Fig. 66.

The transformed circuit is presented below to illustrate the use of Laplace transforms.

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EXAMPLE 17

For the circuit in Fig. 69, the source voltage is √(200)∙cos(10t) for all t . The switch is open for t < 0 and is closed at t = 0 s. Get the value of i₁(0) and i₂(0) . Find expressions for i₁(t) and i₂(t) for t > 0 s.

Solution:

The value of i₂(0) can be found out as follows.

The source voltage equals the sum of voltage drops across the 3 Ω resistor and the 5 Ω resistor. The only unknown, i₂(0) , can be determined.

An expression for i₂(t) can be obtained, as shown by equation (5.163). A loop equation can be formed and this loop includes the source, the 3 Ω resistor, the 2 Ω resistor and the inductor. Equation (5.165) presents the differential equation of the circuit.

Equation (5.166) presents the time domain expression for i₁(t) . The value of constant K can be obtained from the value of i₁(0). Equation (5.167) presents the total response of i₁(t).

An expression for i₂(t) can be obtained from equations (5.167) and (5.168). Equation (5.169) presents the total response of i₂(t). It is verified that the value of i₂(0) evaluated from equation (5.169) agrees with the previously estimated value of i₂(0).

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EXAMPLE 18

A network can have a periodic non-sinusoidal signal as the excitation signal. Such circuits are encountered frequently in electronics. The response of the RC network to a step input with non-zero initial condition is presented first and then it is explained how the periodic response is obtained for repetitive non-sinusoidal signals.

For the network in Fig. 70, it is given that R = 1 Ω, C = 1 F. The waveform in Fig. 70 is defined by these parameters: T = 1 s, and E = 10 V. Obtain

1. the transient response for the first two cycles and
2. the periodic response.
   

The initial voltage across the capacitor is zero.

Solution:

Equations (5.170), (5.171) and (5.172) describe how the step response can be obtained. The capacitor voltage at t = 0.5 s can be obtained, and this value is the initial condition for the next half-cycle.

Equation (5.173) is the differential equation of the circuit for 0.5 < t < 1.0 s. The response is the sum of the response due to the step input and the response due to the initial condition, as shown by equation (5.174). We can repeat the process for the next cycle.

After some cycles, the response becomes periodic. The time taken for the response to become periodic is about 5 times the time constant of the circuit. The time constant of this circuit is 1 s and it takes 5 cycles for the response to become periodic since the cycle period is 1 s. It is possible to obtain the periodic response directly as follows.

Periodic Response

Let the capacitor voltage at the beginning of the positive half-cycle be v_C(0). Then

When the response is periodic, the capacitor voltage at the ned of the first half-cycle has the same magnitude as that of v_C(0), but its polarity is opposite to that of v_C(0). This is so, because the excitation signal is a symmetric square wave signal. Equation (5.183) is not valid for any periodic input signal, but it is valid for this input signal, since it has half-wave symmetry. Let a signal f(t) be periodic and symmetric. When f(t) is symmetric, f(t) = - f(t ∓ 0.5T), where T is the period of a cycle.

The value of v_C(0) is obtained, as shown by equations (5.183) and (5.184), since the values of E and T are known. Then

A waveform is periodic if , f(t) = f(t ∓T). Let the response be periodic and let it be called y(t). Then, when f(t) is periodic,y(t) =y(t ∓T) and y(0) = y(T). It is seen that v_C(0) = v_C(1), the period being equal to 1 second.

In this case, obtaining the response to the periodic square wave input signal is not difficult. If the periodic signal is a a more complex signal and if the order of the system is higher than 1, obtaining the response to the periodic input signal can be quite difficult. But the complexity numerical routine needed to solve the problem does not increase drastically. The Matlab script to simulate the performance is presented below.

% Obtaining response to square-wave input
% Response over 5 cycles
% 100 Calculations performed for every cycle
% cycle period 1 s
% Time constant 1 s
incr=0;   % Initializing
Vini=0;   % Initializing
for n=1:500;
   x(n)=n/100;     % array to hold time in 10s of ms
   % define input at each instant
   if (n<=500) 	vsource(n)= -10.0; end;
   if (n<=450) 	vsource(n)= 10.0; end;
   if (n<=400) 	vsource(n)= -10.0; end;
   if (n<=350) 	vsource(n)= 10.0; end;
   if (n<=300) 	vsource(n)= -10.0; end;
   if (n<=250) 	vsource(n)= 10.0; end;
   if (n<=200) 	vsource(n)= -10.0; end;
   if (n<=150) 	vsource(n)= 10.0; end;
   if (n<=100) 	vsource(n)=-10.0; end;
   if (n<=50) 	vsource(n)=10.0; end;
   if (n==1) vCap(n) =Vini; 
      else vCap(n)=vCap(n-1)+incr; 
      end;
   incr=(vsource(n)-vCap(n))/100; % compute increment
 end;
 axis([0 5 -5 5])
 plot(x,vCap)
 grid 

The response is computed over five cycles and the input is defined for each step. The increment to the capacitor voltage is calculated and it is added to the previous capacitor voltage after each step. The results are stored in an array and the response is plotted.

It can be seen that response is almost periodic after five cycles.

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EXAMPLE 19

For the network in Fig. 72, given that i(0)= 0 obtain

• a. the transient current response for 0 ≤ t ≤ 2 s, and
• b. the periodic response.

Solution:

The response for the first cycle can be obtained, as shown above. Get the value of i(1), as shown.

The same solution is obtained as the particular integral. For 1 ≤ t < 2 s, as shown by equation (5.193). The value of K(1) is obtained from the particular solution, the complementary solution and the value of i(1). Get the value of i(2), as shown. We can calculate the response for third cycle in the same manner.

Periodic Response

It can be seen that obtaining the periodic is not that difficult. Since the input signal is not symmetric, we have to equate the values of i(t) at the start and the end of a cycle.

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SUMMARY

This page has presented a number of worked examples. In the next page, two applications are presented.

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