Given a differential equation with an excitation signal, the steps involved in obtaining the solution are the same. The first step is to obtain the particular solution. In the case of an exponential input signal, it can be determined directly. The second step is to get the assumed complementary solution. The third step is to write the total solution as the sum of the particular solution and the complementary solution. The fourth step is to determine the unknown constants in the complementary solution from the initial conditions.
An exponential signal is a continuous signal and the derivatives of an exponential signal are also exponential, with the same time constant. Equation (7.1) defines an exponential signal, with an amplitude of A at time t = 0. The time constant of this function is reciprocal of m, where m is known as the neper frequency. The differential equation of a first order system with exponential input is expressed by equation (7.2). The pole of the first order system is at - k, and the pole belonging to the excitation function is at - m. Given that m and k are not equal to each other, we can obtain the particular solution as follows.
We know that the excitation function dictates what the response should be. Since the excitation function is an exponential function, we can assume that the particular solution is also an exponential function, with its neper frequency being the same as that of the excitation function. Equation (7.3) presents the assumed particular solution. Since the assumed solution satisfies the given differential equation, substitute the particular solution and its derivative into equation (7.2), in place of y and its derivative. We get then equation (7.4). We can the value of C based on the initial condition. Given that the initial condition is of zero value, the particular solution is expressed by (7.5). The procedure outlined is straight forward, but still an alternative procedure is advocated here, since that technique can be used for other excitation signals, such as the sinusoidal function, and complex sinusoid function.
The differential equation of a first order system with exponential input is expressed by equation (7.2). We can use letter D to represent differentiation. In this case, we call it as the D operator. Using this operator, the differential equation can be represented, as shown by equation (7.6). The polynomial in D can be called as function Q(D), and we get equation (7.7). What has been achieved here? We can develop a technique, which can be used to solve differential equations with inputs, such as the sinusoidal input and the complex sinusoid. In this context, we have defined a function which is a first order polynomial in D. In fact, we can use this notation to represent a polynomial in D of any order.
Now let us assume the particular solution, as stated by equation( 7.8). We can replace y and its derivative by the particular solution and its derivative. Then we get equation (7.9). It turns out that the expression within brackets can be called as a function Q(- m). Given a polynomial in D, we can replace D, by - m. Equation (7.10) shows how we can equate the particular solution and the excitation signal. Then the particular solution is obtained as shown by equation (7.11).
The zero-state response or the response to an exponential input can then be obtained as the sum of the particular solution and the complementary solution, as expressed by equation (7.12). The value of B can be determined from the initial value of zero-state response. Given zero initial value, we get the value of B, as shown by equation (7.13). Equation (7.14) expresses the zero-state response.
We can apply this technique for getting the particular solution to second order systems also. The only catch is that the polynomial in - m, should not be equal to zero. Then the question can be asked how we can obtain the particular solution when function Q( - m) is zero. This question is answered next.
Equation (7.15) specifies function Q( D) . We can replace D by - m, and get equation (7.16). This function is not equal to zero, when m is not equal to k. When m is equal to k, the function is of zero value, as shown by equation (7.17). In this case, the differential equation with the exponential input is expressed by equation (7.18). It can be seen that the first order system has a pole at - k. The excitation function also has a pole at - k.
Hence the particular solution that we get is due to two poles at the same location. We have seen in the case of homogeneous equation of a critically-damped second order system, that the particular solution is the product of an exponential function, corresponding to the pole, and a first order polynomial in time t, due to multiplicity of pole being 2. The situation here is the same and we can guess the particular solution to be of the same form.
The first order differential equation with exponential input is shown by equation (7.19). Both the system and the excitation function have a pole at the same location. The assumed particular solution is expressed by equation (7.20). Using the particular solution and its derivative, we get equation (7.21) from the given differential equation. The value of B can be evaluated based on the initial value of yP(t), and equation (7.22) defines the particular solution. This procedure is simple, but it is preferable to use the D operator, since that technique can be extended to higher order systems.
The first order differential equation with exponential input is shown by equation (7.23). The assumed particular solution is expressed by equation (7.24). Using this particular solution, we get equation (7.25). Equation (7.25) can be simplified, and we get equation (7.26). After canceling the exponential function from both sides, we get equation( 7.27). We know that D-operator represents differentiation. Therefore 1/D represents integration. The constant of integration can be ignored, because of the nature of the complementary solution. Now two numerical examples are presented.
An example is presented now. Equation (7.29) presents a first order differential equation, with an exponential input. Equation defines function Q( D). Replace D by - 3. We get equation (7.30), and it is seen that function Q(-3) = -1. Hence the particular solution is obtained as shown by equation (7.31). The response to the exponential input signal is the sum of the particular solution and the complementary solution, as shown by equation (7.32). The value of constant A can be obtained from the initial value of the zero-state response. Since the initial value of the zero-state response is zero, we get that A equals 10. The zero-state response is expressed by equation (7.33). It can be seen that obtaining the total response, when the initial condition is not zero, is quite easy. As before, the value of constant A can be obtained from the initial value of the total response.
Another example is presented now. Equation (7.34) presents a first order differential equation, with an exponential input. Equation defines function Q( D). Replace D by - 2. We get equation (7.35), and it is seen that function Q(- 2) = 0. Hence the particular solution is obtained as shown by equation (7.36). Remember that 1/D represents integration. Integral of 1 is time t. The response to the exponential input signal is the sum of the particular solution and the complementary solution, as shown by equation (7.37). The value of constant A can be obtained from the initial value of the zero-input response. Since the initial value of the zero-input response is zero, we get that A is equal to zero. The zero-input response is expressed by equation (7.38).
The differential equation of a second order system with exponential input is shown by equation (7.39). This equation can be over-damped, critically-damped, under damped, or undamped, based on the values of b and c. The particular solution is obtained in the same fashion, as outlined for the first order system. If the pole due to the excitation function is distinct from the poles of the second order system, then the solution is straight forward. If the pole of the excitation function is at the same location as the pole of the system, then the technique used is different. We have had an example of this technique, with the first order system where the pole of the system and the pole of the excitation function are at the same location.
Since the excitation function is an exponential function with its pole at - k, the assumed particular solution is also an exponential function with its pole at the same location, as shown by equation (7.41). Since the assumed particular solution satisfies the given differential equation, we get equation (7.42). When function Q(- k), is not zero, the solution obtained is displayed by equation (7.43).
As stated by equation (7.44), let function Q(- k), is be equal to zero. Given that the system is over-damped, we can express function Q(D) as the product of two terms, as shown by equation (7.46). Given that Q1(- k) = 0, we get equation (7.47). Since Q1(- k) = 0, the particular solution is obtained as shown by equations (7.48) and (7.49). Since there are two poles at - k, the particular solution contains the product of an exponential function and time t, as shown by equation (7.49).
Now we take up an example. Here the differential equation of an over-damped system with an exponential input function is shown by equation (7.50). The system has its poles at - 2, and - 3, whereas the pole of the excitation function is at - 4. We can evaluate function Q(- 4), as shown by equation (7.51). Equation (7.52) expresses the particular solution.
Since the initial conditions are not of zero value, the total response is obtained. The assumed complementary solution is presented by equation (7.53). The total solution is the sum of the particular solution and the complementary solution, as shown by equation (7.54). The values of constants, A and B, are obtained from the initial conditions, as shown by equations (7.55) and (7.56). The total response is expressed by equation (7.57).
Another example is presented now. Here the differential equation of an over-damped system with an exponential input function is shown by equation (7.58). The system has its poles at - 2, and - 4, whereas the pole of the excitation function is at - 2. We can express Q(D) as the product of two functions, Q1(D) and Q2(D), as shown by equations (7.59) and (7.60). .We can evaluate function Q( - 2) and it is equal to zero. In such a case, we get equation (7.61) wherein is Q2(- 2) is evaluated. Then we get the particular solution by replaced D by (D - 2) in Q1(D), as shown by equation (7.62). One over D represents integration, and the integral of 1 is time t.
Since the initial conditions are not of zero value, the total response is obtained. The total solution is the sum of the particular solution and the complementary solution, as shown by (7.63). The values of constants, A and B, are obtained from the initial conditions, as shown by equation (7.64). The total response is expressed by equation (7.65).
The differential equation of a second order critically-damped system with exponential input is shown by equation (7.66). Equation (7.67) presents the same differential equation using Q(D). Given that Q(- m ) is not equal to zero, the particular solution is obtained, as shown by equation (7.68).
It is possible that the pole of the exponential function is at the same location as one or more poles of the system. For the differential equation defined by equation (7.69), both the poles of the system and the pole due to the excitation function are all at the same location. Hence Q(- k ) is zero, and the particular solution is assumed to be of the form stated by equation (7.70) .
From equation (7.70), we get the first derivative and the second derivative of the assumed solution, as shown by equations (7.71), (7.72), (7.73) and (7.74). We get equation (7.75) by differentiating by parts.
When function Q(- k) = 0, equation (7.76) is used. We can cancel the exponential function on both sides, and then we can evaluate function v(t), as shown by equation (7.77). If there are two poles at the same location, 1/Q(D - k) will be equal to 1/D, as shown in the case of an over-damped system. In the case of critically-damped system, Q(D - k) will be equal to D2 , and hence we need double integration, as shown by equation (7.78).
One over D represents integration, and the integral of 1 is time t. We can ignore the constant of integration, since the complementary solution will have the corresponding term. If there are three poles at the same location, 1/Q(D - k) will be equal to 1/(D2). Double integration of 1 yields (t2)/2. Again we ignore the constants of integration. This case occurs when the exponential input function to the critically-damped system has its pole at the same location as that of the system.
It is not essential that you understand the proof. But remember the result. Proof is presented here, since this formula is used in the subsequent sections. You need to understand how the technique used for the critically-damped system is different from the technique used for the over-damped system.
Here the differential equation of a critically-damped system with an exponential input function is shown by equation (7.79). The system has both its poles at - 3, whereas the pole of the excitation function is at - 2. We can evaluate function Q( - 2) , as shown by equation (7.80). Equation (7.81) expresses the particular solution.
Since the initial conditions are not of zero value, the total response is obtained. The total solution is the sum of the particular solution and the complementary solution, as shown by equation (7.82). The values of constants, A and B, are obtained from the initial conditions, as shown by equation (7.83). The total response is expressed by equation (7.84).
Here the differential equation of a critically damped system with an exponential input function is shown by equation (7.85). The system itself has two roots at -3, and the pole of the excitation function is also at - 3. Hence the multiplicity of poles is equal to three. In such a case, the particular solution has to contain t2, as one of its parts.
The value of function Q( - 3), is zero, as shown by equation (7.86). Then D in function Q(D), is replaced by ( D - 3), as shown by equation (7.87), and the function Q(D - 3) equals D2. The particular solution is obtained, as shown by equation (7.88). The double integral of 1 is obtained, as shown by equation (7.89).
Since the initial conditions are not of zero value, the total response is obtained. The total solution is the sum of the particular solution and the complementary solution, as shown by equation (7.90). The values of constants, A and B, are obtained from the initial conditions, as shown by equation (7.91). The total response is expressed by equation (7.92).
Here the differential equation of an under-damped system with an exponential input function is shown by equation (7.93). The system has a pair of complex conjugate poles, whereas the pole of the excitation function is at - 2. We can evaluate function Q( - 2), as shown by equation (7.94). Equation (7.95) expresses the particular solution. Equation (7.96) expresses the assumed complementary solution can be obtained.
Since the initial conditions are not of zero value, the total response is obtained. The total solution is the sum of the particular solution, and the complementary solution, as shown by equation (7.97). The values of constants, A and B, are obtained from the initial conditions, as shown by equations (7.98) and (7.99). The total response is expressed by equation (7.100).
Here the differential equation of an undamped system with an exponential input function is shown by equation (7.101). The system has a pair of conjugate poles on the imaginary axis, whereas the pole of the excitation function is at - 2. We can evaluate function Q(- 2), as shown by equation (7.102). Equation (7.103) expresses the particular solution.
Since the initial conditions are not of zero value, the total response is obtained. The total solution is the sum of the particular solution and the complementary solution, as shown by equation (7.104). The values of constants, A and B, are obtained from the initial conditions, as shown by equations (7.105) and (7.106). The total response is expressed by equation (7.107).
This page has described how the response of a first and a second order system can be obtained, when the system has an exponential input signal. The next page describes how the response of a first and a second order system can be obtained, if the input is a sinusoidal signal.