Differential-Equation Models (Solution of Differential ...

Differential-Equation Models (Solution of Differential Equations - Natural and Forced

response)

An important class of continuous time LTI systems are those modeled by ordinary linear differential equations with constant coefficients.

Given an input x(t) and an output y(t),

is a linear first-order (the order of the derivatives is one) differential equation with constant coefficients (as long as a and b are constants). A general nth order linear differential equation with constant coefficients is:

which we can write as:

Solution of Differential Equations

A classical method for the solution of our differential equation is called the method of undetermined coefficients. We express the output y(t) as the sum of complementary or natural (yc(t)) and particular or forced (yp(t)) solutions:

y(t) = yc(t) + yp(t)

Natural response The natural response yc(t) is the solution to the homogeneous equation:

Assume that the solution of the homogeneous equation is:yc(t)= Cest

Substituting in the homogeneous equation yields:

and we get:

This is the characteristic equation and it may be factored as:

The solution is of the form:

assuming there are no repeated roots (which is all we will cover here).

Ex. Given a first-order differential equation

find its homogeneous solution. Your answer should be in terms of a constant C.

Forced response The forced response yp(t) solves the equation

The form of the solution is determined by the input x(t). For an exponential input x(t)= A eat, the solution would be yp(t) = P eat where A, a, and P are constants.

Ex. For the previous example, assume an input x(t) = 6 e3 t. Find the particular solution yp(t).

Ex. Now, assuming that the system is initially at rest (the initial conditions are 0), i.e., y(0) = 0 , solve for the constant C in your overall solution y(t) = yc(t) + yp(t):

Ex. Given y '(t) + 2y(t) = x(t)where x(t) = 4 e2t, find y(t). Assume y(0) = 0.

Homogeneous solution Particular solution

Total solution

Recall that the natural solution to our differential equation was:

where si is the root of the characteristic equation

The root of the characteristic equation si can be either real or complex. If it is complex, it must occur in conjugate pairs because the coefficients of the characteristic equation are real.

• If si is complex, then is exponential in form. • If si is complex, then let and and the conjugate pair of these terms

can be expressed as:

where

StabilityThe roots si will determine if the overall system is BIBO stable.Assume we have a causal LTI system. The solution to our differential equation is of the form

where y(t) = 0 for t < t0 (t0 is the start time). Since yp(t) is always of the same form as the input x(t), if the input x(t) is bounded, then yp(t) will also be bounded. (Alternatively, since we are only considering BIBO stability, our input is guaranteed to be stable.)

Thus, since the stability of the system will only depend on the system itself and not on the input, let's examine the solutions to the homogeneous equation:

Clearly, as long as the real part of all roots (also called poles) of this equation, σi < 0 (since we've assumed that the system is causal), then each term in yc(t) will be bounded. Thus, we require for BIBO stability that:

Thus, a necessary and sufficient condition for stability of a causal LTI system is that all roots of the system characteristic equation lie in the left half plane of the s-plane.

Ex. Given a causal LTI system described by the differential equation

determine if the system is BIBO stable.

The system is not BIBO stable.

Given an input x(t) = X est to a BIBO stable LTI system modeled by an nth order linear differential equation with constant coefficients, we will examine the steady-state system response. Assume that X and s are complex.

The forced (particular or steady-state) response of the system to this input is of the same form of the input, i.e.

From the differential equation describing the system,

or,

Plugging in x(t)= X est and yss(t)= Y est , we get

which we can write as Y = H(s) X

where is a transfer function.

So given an input to an LTI system, the steady-state response is

Similar to the differential equation and the impulse response, the transfer function H(s) completely characterizes the LTI system (we can derive the differential equation from H(s) and vice versa).

In general, by superposition, given an input , the output of the system is

Eigenfunctions of CT LTI systems are complex exponentials:

Check: What is est*h(t)?

Where is the eigenvalue. This motivates the Laplace

transform and the Fourier Transform:

H(s) is known as the bilateral Laplace transform of h(t).

If s = jω, then we get H(jω), the Fourier Transform of h(t).

Ex. Given an input x(t) = est, andh(t) = e3t u(t)

find its steady-state outputyss(t) = H(s) est

YOU FINISH: