This page outlines how the particular solution is to be obtained. Both the zero-state response and the total response can be obtained as the sum of the particular solution and the complementary solution. The total response is the sum of the zero-state response and the zero-input response. When obtaining the zero-state response, the initial conditions are set to be of zero value. On the other hand, the zero-input response depends on the non-zero initial conditions, with the external input set to zero. It is not necessary to obtain the total response only as the sum of the zero-state response and the zero-input response. It is possible to get the total response, as the sum of the particular solution and the complementary solution, wherein the non-zero initial conditions are taken into account. If you do not understand what is being stated here, read further. After going through a couple of worked examples, you will understand what is being talked about.
This page mainly outlines the technique that is used for getting the particular solution. It has no worked example. The subsequent pages contain a sufficient number of worked examples, to clarify and make you understand the technique. Since the technique is common, it is present here.
Equation (1) presents an n-th order linear differential equation. The D-operator is defined by equation (2). Using the D-operator , we get equations (2), (3), (4), (5) and (6).
Now let us consider a second-order linear differential equation, since we confine ourselves to second-order or less. In equation (124), the order of functions on both sides is 2. The right-hand side can be expressed as shown by equation (25). If the order of both sides is the same, then the solution due to the highest order is proportionate to the excitation function. Hence it is sufficient to consider the type of equation described by equation (126). Note that b1 and b0 in equations (125) and (126) are not the same. The purpose of presenting these equations is to show that the order of function on the right-hand side need not be higher than one.
Let us start off with equation (126). You need to remember that the particular solution is the result of applying a forcing function. Hence the particular solution reflects the forcing function. If the forcing function is sinusoidal, then the particular solution is also sinusoidal, at the same frequency. If the forcing function is an exponential function, then the particular solution also contains an exponential function, with the same neper frequency or the same time constant. If the forcing function is a complex sinusoid, then the particular solution is also a function, containing the same complex sinusoid. If the forcing function is a unit-step function, then the particular solution also contains a step function, but its magnitude may or may not be unity. If the forcing function is a ramp function, then the particular solution also contains a ramp function and a step function, but the slope of the ramp may or may not be unity. The only input or the excitation function that does leave its imprint on the particular solution is the impulse function, and therein lies the importance of the impulse function. The impulse response reflects the nature of the system.
Assume that the excitation function or the forcing function f(t) is an exponential function, as described by equation (127). In this case, we assume that the particular solution yP(t) is also an exponential function, as shown by equation (128). Then the first derivative and the second derivative of the assumed particular solution are obtained, as shown by equation (129). Equation (130) defines Q(D) and Q(s). Equation (131) is obtained from equations (126), (127), (128), (129) and (130). When Q(s) is not zero, we get the particular solution, as expressed by equation (132). It is relatively simple to get the particular solution in this case.
Now let us consider the case when Q(s) is zero, as stated by equation (133). In this case, the assumed solution is stated by equation (134). Get the first and the second derivative of the assume solution, as shown by equations (135) and (136). Hence we get equation (137). If you do not understand the mechanism, it is not critical. You need to know what to do, which will become clear after some examples.
Let us extend what we have learned so far to a second-order system. Given an over-damped system, equation (138) can be represented, as shown by equation (139). Since b is not equal to a, we get equation (140). It is a valid operation.
To evaluate the particular solution, assume it to be of the form shown by equation (141). Since the root is at - a and the neper frequency of the assumed solution is also at the same place, we make use of equation (137) and equation (142) is obtained from equation (140). The presence of D in the denominator for the right-hand expression of equation (143) represents integration. We get t by integrating one with respect to time. Ignore the constant of integration. If you are observant, you will find that we apply very much the same technique when we use the Laplace transforms. But in the case of the Laplace transforms, you have to contend with partial fractions and remembering the appropriate inverse transform. What has been outlined is really the simplest technique you can think of.
We can obtain the particular solution due to the singularity functions, as explained now. The unit step function can be viewed as an exponential function, with s = 0, as shown by equation (144). We can apply the same technique and the particular solution for the unit-step input is obtained as shown by equation (145). It is a constant, as expressed by equation (146). Since the impulse function is the derivative of the unit-step function, the particular solution is the derivative of the expression in equation (146). As shown by equation (146), the particular solution due to the impulse input is zero. Since the ramp function is the integral of the unit-step function, the particular solution due to the ramp function is the integral of the expression in equation (146). Hence, as shown by equation (148), the particular solution due to the ramp function contains a part that increase linearly with time and a constant reflecting the constant of integration.
The following pages present examples to illustrate how the zero-state response and the total response can be obtained for different excitation signals.