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SOME SIMPLE CIRCUITS

SUMMARY

This page illustrates how the zero-state response and the total response of first-order circuits can be obtained. It is shown how we can obtain a solution using either Laplace transforms or differential equation. There is more stress on how to form a differential equation than on its solution, since the chapter on differential equations has illustrated how to obtain the solution. In the case of Laplace transforms, the first step is draw the transformed circuit. Expanding the Laplace transform function into partial fractions and then obtaining the inverse time domain functions should be considered as routine operations.

We start off with an *RC* circuit to illustrate how to obtain the zero-state and the total response.

An *RC* circuit is presented in Fig. 38a. The switch remains open for **t****<** 0 s and it is closed at **t**** =** 0 s. In this section, we find an expression for the capacitor voltage for different types of input. In some cases, its is assumed that *v*_{C}(0) = 0 and in some cases, it assumed that *v*_{C}(0) ≠ 0.

To form the differential equation, we use the circuit in Fig. 38b. The transformed circuit is shown in Fig. 38c and we derive an expression of the capacitor voltage as a Laplace transform function using the circuit in Fig. 38c.

The differential equation that describes the behaviour of the circuit in Fig. 38b is obtained first.

Equation (4.1) equates the applied voltage to the sum of the resistor voltage and capacitor voltage. This equation is based on Kirchoff's Voltage Law. The resistor voltage is the product of resistance and the current through the resistor. Since the resistor and the capacitor are in series, the resistor current is the same as the capacitor current. The capacitor current is equal to the product of capacitance and the rate of change of its voltage. Hence we get equation (4.2). Equation (4.3) is obtained from equations (4.1) and (4.2). On re-arranging equation (4.3), we get equation (4.4). Either equation (4.3) or equation (4.4) can be called as the differential equation for the circuit in Fig. 38b, but equation (4.4) is in the preferred format. It is appropriate to use the capacitor voltage as the variable of the differential equation, since it reflects energy stored in the system. In such a case, the solution is continuous, unless there is impulse current flowing into the circuit.

To express the capacitor voltage as a Laplace transform function, the circuit in Fig. 38c is used. It is easier to obtain an expression for the capacitor voltage using the principle of superposition.

The circuits in shown how the principle of superposition can be applied. We can consider one source at a time and we can add the contribution due to each of the sources acting alone.

Equation (4.5) presents an expression for the capacitor voltage. This expression can be re-arranged and we get equation (4.6). Divide both the numerator and the denominator of the first term in equation (4.5) by *sR* and then we get equation (4.6).

We can obtain equation (4.6) in a different way also, and this technique is shown below.

We can express equation (4.4) using Laplace transforms, as shown by equation (4.7). On r-arranging equation (4.7), we get equation (4.8), presented as equation (4.6) earlier.

Now let us the response of the circuit when the input is a unit-step, with the initial capacitor voltage being assumed to be non-zero.

Equation (4.9) state that the excitation function is a unit-step and that *v*_{C}(0) ≠ 0. Equation (4.10) is the differential equation of the circuit in Fig. 38b, when the excitation function is a unit-step. The solution is the sum of complementary solution and the particular solution. The complementary solution is obtained from the homogeneous equation, represented by equation (4.12). The complementary solution is called *v*_{CC}(*t*) and since the complementary solution satisfies the homogeneous equation, we get equation (4.12).

The complementary solution is presented by equation (4.13). If you have difficulty with this equation, you need to go through the chapter on differential equations. The particular solution satisfies the differential equation, and hence we get equation (4.14). The input dictates what the response should be. Since the unit-step has a value of 1 for ** t** > 0, the response also is a constant and the rate of change of response has to be zero, as stated by equation (4.15) and the value of particular solution is obtained to be unity.

We can express the capacitor voltage as the sum of complementary solution and the particular solution, as shown by equation (4.16). The particular solution is known and the initial voltage of capacitor is known. Hence we obtain equation (4.17) and the unknown coefficient present in the complementary solution can be determined, as shown by equation (4.17). The total response is presented by equation (4.18). Now the response is obtained using Laplace transforms.

The Laplace transform of unit-step function is presented by equation (4.19). From equations (4.6) and (4.19), we get equation (4.20). Equation (4.20) can be expanded into partial fractions. If you do not know how to get equation (4.21) from equation (4.20), go through the chapter on Laplace Transforms. The inverse transform, representing the capacitor voltage in time domain, is presented by equation (4.22).

Equation (4.22) can be presented in a quite a few ways, as shown below.

When *v*_{C}(0) = 0, the total response is the same as the zero-state response or the unit-step response, as expressed by equation (4.23). The response due to the non-zero initial condition is the zero-input response, as expressed by equation (4.24). From equation (4.18), it can be seen that the total response is the sum of the zero-state response and the zero-input response.

The particular solution represents the forced response, as expressed by equation (4.25). The part of the response due to the pole of the system is the natural response, as expressed by equation (4.26). From equation (4.18), it can be seen that the total response is the sum of the forced response and the natural response.

When the input is a step input or a sinusoidal input, the particular solution represents the steady-state response, as expressed by equation (4.27). The terms that have an exponentially decaying component are grouped together and the sum of these responses represents the transient response, as expressed by equation (4.28). From equation (4.18), it can be seen that the total response is the sum of the steady-state response and the transient response.

It is incomplete to leave out the solution that can be obtained using the convolution integral. The convolution integral is presented formally later. However, the equation that represents the convolution integral can be used without a formal introduction. Equation (4.29) presents the convolution integral. According to the convolution integral, the response *y*(*t*) of a system to any input, say *x*(*t*), can be obtained if its response to impulse input is known. Normally impulse response is called as *h*(*t*). For this circuit, the input signal is a unit-step function. Even through the impulse response has not obtained yet, the expression for the impulse response that is obtained in the next sub-section can be presented now. The impulse response and the input function now being considered are presented by equation (4.30). The convolution integral can be evaluated, as shown by equations (4.31).

The convolution integral can be represented in terms of Laplace transforms also.

The Laplace transform of the convolution integral is expressed by equation (4.32). The corresponding terms for the circuit in Fig38c are expressed by equation (4.33).

Hence the Laplace transform of the capacitor voltage is obtained, as shown by equation (4.34). The capacitor voltage in time domain is expressed by equation (4.36).

A Matlab program can be used to evaluate the response. Let *R* be 10 Ohms and *C* = 0.1 F. The program can be stored as a *.m file and then run under Matlab.

%.. Unit Step Response

%.. Enter the resistor value in Ohms

R=10;

%.. Enter the capacitor Value in Farad

C=0.1;

%.. Enter the numerator polynomial

num=[1/(R*C)];

%.. Enter the denominator polynomial

den=[1 1/(R*C)];

%.. Enter the command for computing step response

step(num,den)

%.. Enter grid and title of the plot

grid

title('Unit-step Response of given G(s)')

The above Matlab script makes use of the built-in* step*(num,den) function, where num is the numerator polynomial row vector and den is the denominator polynomial row vector. For

example, if the polynomial in

If the differential equation approach is to be used, it is preferable to obtain the response to unit step input first. Let this be response be the unit step response. We know that the impulse signal is the derivative of the step signal and when the impulse input is applied to a linear system, the impulse response is the derivative of the unit step response. This approach is illustrated first. Subsequently it is shown how the impulse response can be obtained from the basics.

Equation (4.36) expresses the unit-step response obtained in the previous sub-section. By differentiating the unit-step response with respect to time, we get the impulse response, shown in equation (4.37). This is the preferred technique.

Next it is shown how we can obtain the impulse response using the traditional approach.

Equation (4.37a) presents the differential equation, with impulse input and the capacitor voltage prior to application of impulse is zero. We can multiply both sides by the integrating factor and obtain equation (4.37b). We obtain equation (4.38) from equation (4.37b), as illustrated.

Integrate equation (4.38) and we get equation (4.39). Equation (4.40) is obtained from equation (4.39) using the property of an impulse function. The capacitor voltage is then obtained, as shown by equation (4.41).

This technique can be shown to be the same as the * Heaviside operator* technique and it can be extended to higher order systems. But getting the solution turns out to be somewhat difficult. Hence this technique is not used any further.

The importance of impulse response lies in that the response contains only the characteristic modes of the system. The impulse response reveals the nature of the system. Hence we can assume that the impulse response is the same as the homogeneous response. In the case of homogeneous response, the unknown constants are obtained from the initial values of the system. In the case of impulse response, we make use of the fact that the impulse response satisfies the differential equation and the unknown constants are determined.

Let the impulse response be called *h*(*t*) and let *k* be equal to the reciprocal of time constant, as shown by equation (4.42). Equation (4.43) presents the differential equation. Obtain the derivative of *h*(*t*) from *h*(*t*), as shown by equation (4..44). Substitute *h*(*t*) and its derivative into equation (4.43), and we get equation (4.45). The value of *A* can then be determined, as shown by equation (4.47). The trick is in getting the derivative of *h*(*t*).

Equation (4.47) presents *h*(*t*). Equations (4.48), (4.49) and (4.50) show how to obtain the derivative of *h*(*t*). This technique gets complicated for a second-order system and it is not used in the subsequent part of this text.

It can be seen that there is a jump in the capacitor voltage. Since the impulse is required to be obtained, it is assumed that *v*_{C}(0^{-}) = 0. Because the source is **d**(t) , it can inject an impulse current of magnitude **d**(*t*)/*R* and hence *v*_{C}(0^{+}) can be greater than 0. Recall that unless an impulse current flows through a capacitor at* t* = 0, *v*_{C}(0^{-}) = *v*_{C}(0^{+}). That is, if an impulse current flows through a capacitor at *t* = 0, *v*_{C}(0^{-}) = *v*_{C}(0^{+}); otherwise, *v*_{C}(0^{-}) ≠ *v*_{C}(0^{+}).

Equations (4.51), (4.52) and (4.53) illustrate how we make use of the impulse current to obtain *v*_{C}(0^{+}) and then obtain *v*_{C}(*t*). This technique is useful and it can be extended to other circuits also.

We can make use of Laplace transforms and obtain *v*_{C}(*t*), as shown by equations (4.54) and (4.55). The Matlab script for obtaining the impulse response is presented below.

%.. Unit Impulse Response

%.. To obtain impulse response, multiply

% G(s) by s and then use step response

%.. Enter the resistor value in Ohms

R=10;

%.. Enter the capacitor Value in Farad

C=0.1;

%.. Enter the numerator polynomial

num=[1/(R*C)];

%.. Enter the denominator polynomial

den=[1 1/(R*C)];

%.. Enter the command for computing step response

impulse(num,den)

%.. Enter grid and title of the plot

grid

title('Unit-Impulse Response of given G(s)')

The plot of impulse response is shown above. Next we take up the ramp response of the circuit.

In this section, the response of the *RC* circuit to other inputs is presented in a cursory manner. If you have difficulty in understanding the material, go through the chapters on differential equations and Laplace transforms.

**Ramp Response**

The ramp response can be obtained either by integrating the unit-step response or by using the classical approach. The solution using Laplace transforms can be obtained as shown below. In the case of ramp response, the particular solution is the forced response and the complementary solution is the natural response. Equation (4.58) presents the zero-state response. When the input is not a step input or a sinusoidal input, there is no relevance to steady-state response or transient response. A first-order system tracks the ramp input with the steady-state error being equal to its time constant.

**Response to Exponential Input****:**** Case ****1:*** k * ≠ (1/

**Response to Exponential Input: Case ****2:*** k * = (1/

**Response to Sinusoidal Input**

The solution can be obtained using the phasor notation also, as illustrated by Fig. 40.

The source voltage is assigned to have an amplitude of one and zero phase. Equation (4.71) shows how the capacitor voltage can be obtained. It can be converted to an expression in time domain.

Equation (4.72) appears to be different from equation (4.70). They are both equal, as shown now. Expand equation (4.72), as shown by the equation below equation (4.72). The impedance diagram in Fig. 40 defines the value of phase angle, **a**.

Equations (4.73), (4.74) and ( 4,75) illustrate the further steps. It can be seen that equations (4.75) and (4.70) are the same. In the case of sinusoidal input, the particular solution represents the forced response and the steady-state response. In this case, the zero-state response has been determined and the complementary solution represents both the natural response and the transient response.

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

**Response to Sinusoidal Input****:**** Numerical Routine **

From the current values of the source voltage and the capacitor voltage, the rate of change of capacitor voltage is found out. Based on the rate of change of capacitor voltage, the next value of capacitor voltage is found out. Again, the calculations are repeated.

A flow chart is shown in Fig. 41 to illustrate the numerical routine. Initially the capacitor voltage is set to zero. Then the rate of capacitor voltage is obtained as shown below.

Equation (4.79) presents the rate of change of capacitor voltage with respect to time. It is preferable to express the above equation as shown by equation (4.80).

Equation (4.81) expresses the increment in capacitor voltage can be determined and this increment is added to the previous capacitor voltage to obtain the next capacitor voltage. The angle is also incremented, as shown by equation (4.82).

The Matlab script is presented next.The amplitude of sinusoidal input is assigned to be 100 Volts and the angular time constant is set to be 1 radian.

% Obtaining response to sinusoidal input

% Transient response : 3 cycles

% Calculations performed for every degree

rad = pi/180.0; % step size in radians

wRC=1.0; % The time constant in radians

Vini=0.0; % Initializing the voltage

incr=0; % variable to hold the increment

for n=1:1080; % 1080 deg equals 3 cycles

x(n)=n; % Array representing angle

angle=n;

% Preferable to restrict angle to be

% between 0 and 360 degrees.

% Specify input voltage for each step

if (angle>360) angle =angle-360; end;

vsource(n)=100*sin(angle*rad); % input voltage

if (n==1) vCap(n) =Vini; % Capacitor voltage zero

else vCap(n)=vCap(n-1)+incr; end; % Compute increment

incr=(vsource(n)-vCap(n))/wRC*rad; % add increment

end;

plot(x,vCap)

axis([0 1080 -80 80])

grid

It can be seen that the steady-state response is reached in third cycle, since a time period greater than 5 times the time constant has already elapsed at the start of the third cycle.

In this section, it is shown how the differential equation can be obtained for some simple circuits.

**R & C in Parallel**

A circuit with a resistor and a capacitor in parallel is shown in Fig. 43. The variable for the differential equation should be the capacitor voltage. The differential equation that describes the behaviour of this circuit is obtained as follows.

The source current equals the sum of the resistor current and the capacitor current, as expressed by equation (4.83). Equation (4.84) presents expression for the resistor current and the capacitor current. We get equation (4.85) from equations (4.83) and (4.84). Equation (4.86) presents the differential equation in the desired format.

From the transformed circuit in Fig. 43, equations (4.87) and (4.88) can be derived. The impulse response of the circuit in Fig. 43 can be obtained as shown below.

**R & L in Series **

A circuit with a resistor and an inductor in series is shown in Fig. 44. The variable for the differential equation should be the inductor current. The differential equation that describes the behaviour of this circuit is obtained as follows.

The Laplace transform of the inductor current can be obtained from the transformed circuit in Fig. 44.

The impulse response of this circuit is obtained as follows.

**R & L in Parallel**

A circuit with a resistor and an inductor in parallel is shown in Fig. 45. The variable for the differential equation should be the inductor current. The differential equation that describes the behaviour of this circuit is obtained as follows.

The Laplace transform of the inductor current can be obtained from the transformed circuit in Fig. 45.

Getting an expression for the inductor current due to the initial inductor current is not that easy. When the contribution to the inductor current due to the initial inductor current is to be determined, the source current is replaced by an open circuit and the inductor current is the same as the current through the resistor. The term with the initial inductor current in equation (4.104) reflects the current through the resistor. It is preferable not to use this model for the inductor. Replace the inductor by its reactance and impulse voltage source, as shown below in Fig. 46.

From the circuit in Fig. 46, we get that

The impulse response of the circuit in Fig. 45 is obtained as follows.

Some examples have been presented in this page. The main aim is to show how a differential equation can be formed that describes the behaviour of the circuit. It is also shown how the transformed circuit can be obtained. based on the transformed circuit, a Laplace transform function for the output variable can be obtained. Once either the differential equation or the Laplace transform function is obtained, further steps are as explained in the chapters on differential equations and Laplace transforms. The next page presents some worked examples. After that, a few applications are presented.