ECE 3620Lecture 5 - Zero-input solution of differential equations

To see how to solve unforced LTI differential equations. To meet the characteristic equation. To handle multiple and complex roots of the characteristic equation.

Also, to see how to determine the constants that arise in the solution of the differential equations. When given the initial conditions, this amounts to solving a set of equations. For circuit problems, where the initial conditions are given in terms of circuit values, the initial conditions of the function must first be found, then the constants must be found. pp. 105-114

We have learned that differential equations can be used to represent a circuit response. We are interested in finding solutions of the differential equations, because the solution represents what the system is doing. We will begin by looking at the differential equation in the traditional way, from the ``time domain''. Later we will cover other very elegant methods of solving the equations using Laplace transforms.

Differential equations.

Mathematicians usually treat the solution of differential equations by finding the homogeneous solution and the forced solution. We often break the solution into the ``zero-input'' response and the ``zero-state'' response.

Zero-input response

The response of the system when there is no input forcing function. The output resulting is due to initial conditions of the system. (For example, charge on a capacitor.)

Zero-state response

When a system is in zero-state, it is assumed that the initial conditions are all zero (the ``zero state''). Thus, the response of the system is due to some kind of external input.

The output is the sum of the zero-input response and the zero-state response, because we are dealing with linear systems:

$$\text{ {\bf Total response = zero-input response $+$ zero-state response.}}$$

In general, a LTIC (linear, time-invariant, continuous time) can be expressed in terms of differential equation

$$\frac{d^n y}{dt^n} + a_{n-1}\frac{d^{n-1}y}{dt^{n-1}} + \cdot... ...c{d^{m-1} f}{dt^{m-1}} + \cdots + b_1 \frac{df}{dt} + b_0 f(t)$$

In operator notation

$$(D^n + a_{n-1}D^{n-1} + \cdots + a_1D + a_0)y(t) = (b_mD^m + b_{m-1}D^{m-1} + \cdots + b_1 D + b_0) f(t)$$

Let

$$Q(D) = D^n + a_{n-1}D^{n-1} + \cdots + a_1D + a_0$$

$$P(D) = b_mD^m + b_{m-1}D^{m-1} + \cdots + b_1 D + b_0$$

Then

$$Q(D)y(t) = P(D)f(t)$$

Usually we will have $m \leq n$.

Let $y_{\rm0}(t)$ be the zero-input response, i.e., the response when $f(t) = 0$. Let $y_{\rm z-s}(t)$ be the zero-state response, when $f(t) \not \equiv 0$ but the initial conditions are zero.

Let us start with $y_{\rm z-i}(t)$. That is, the input $f(t)$ is assumed to be zero, so we have

$$Q(D)y(t) = 0$$

or

$$Q(D)y_0(t) = 0$$

($y_0(t)$ is what the book uses as the zero-input response.) Take a simple example: (RL circuit)

$$(D+a_0)y_0(t) = 0$$

$$Dy_0 = -a_0y_0$$

This is the key equation: the derivative of the solution has the same form as the solution.

$$\frac{dy_0}{y_0} = -a_0 dt$$

Integrate:

$$\ln y_0(t) = -a_0 t + k$$

$$y_0(t) = e^k e^{-a_0 t} = C e^{-a_0 t}$$

Observe that in general if

$$y_0(t) = C e^{\lambda t}$$

that

$$\frac{dy_0}{dt} = C\lambda e^{\lambda t}$$

So all we have to do is to match coefficients.

What is $C$? We must have some extra information to tell us -- some specific value of the solution at some time. This is often the initial condition. We must know, for example, the initial current on the inductor. (Plot some examples.) If $y_0(0)$ is known, then we have

$$y_0(0) = C$$

More generally, take $Q(D) y_0(t) = 0$, or

$$(D^n + a_{n-1}D^{n-1} + \cdots + a_1D + a_0)y_0(t) = 0$$

Building on our experience and intuition, assume a solution of the form

$$y_0(t) = c e^{\lambda t}$$

for some $c$ and some $\lambda$, to be determined. Substituting (expand the derivatives) we get

$$c(\lambda^n + a_{n-1}\lambda^{n-1} + \cdots + a_1\lambda + a_0)e^{\lambda t} = 0$$

which could be written as

$$cQ(\lambda)e^{\lambda t} = 0$$

There are two possible ways this could happen: either $c=0$, would be the case if there is no zero-input response (all the initial conditions are zero); or

$$(\lambda^n + a_{n-1}\lambda^{n-1} + \cdots + a_1\lambda + a_0) = 0 = Q(\lambda)$$

Note that this is a polynomial in $\lambda$ of degree $n$. It has $n$ solutions, which we will write as $\lambda_1, \lambda_2, \ldots, \lambda_n$. That is,

$$Q(\lambda_1) = 0 \quad Q(\lambda_2) = 0\quad \ldots \quad Q(\lambda_n) = 0$$

Aside: how do we find the roots? If it is quadratic equation, you can do it yourself, lickety split. Anything higher, plan to use a computer (that's why they were invented!). Use MATLAB, for example.

So let $y_0(t) = c_1 e^{\lambda_1 t}$. Does this work? Of course: $\lambda_1$ is one of the roots. How about $y_0(t) = c_2 e^{\lambda_2 t}$. Does this work? Yes. How in the world are we to handle this wealth of riches? We now have many solutions; what to do?

Take them all -- this is, after all, a linear system. The sum of any solutions is another solution. So we will take

$$y_0(t) = c_1 e^{\lambda_1 t} + c_2 e^{\lambda_2 t} + \cdots c_n e^{\lambda_n t}$$

This provides the total solution, provided that we can find all of the constants $c_1, c_2,\ldots,c_n$.

Recap: Assume solution, substitute in the DE, then find roots to equation and combine together. Don't lose sight of the fact that there is a physical system behind all this.

The equation

$$Q(\lambda) = 0$$

is called the characteristic equation. Recall from the example we did on linear algebra that we have the same denominator in all solutions. It turns out that this denominator is also the characteristic equation.

The overall system behavior of the system is determined by the characteristic equation. The values of $\lambda$ that are the roots are known as the modes or natural frequencies or eigenmodes or eigenvalues or characteristic values of the system.

This example with complex roots is true in general. If we have a 2nd order system so that

$$y_0(t) = c_1 e^{(\alpha + j\beta)t} + c_2 e^{(\alpha - j\beta)t}$$

that is real, then $c_1$ and $c_2$ must be conjugates; let

$$c_1 = \frac{c}{2}e^{j\theta}\quad\quad c_2 = \frac{c}{2}e^{-j\theta}$$

Then

$$y_0(t) = c e^{\alpha t} \cos(\beta t + \theta)$$

Discuss. Plot some pole locations and discuss. Stability...

In the general case, if there is a repeated root

$$(D-\lambda)^2 y_0(t) = 0$$

then

$$y_0(t) = (c_1 + c_2 t)e^{\lambda t}$$

satisfies the DE. If there is a repeated root

$$(D-\lambda)^r y_0(t) = 0$$

then

$$y_0(t) = (c_1 + c_2 t + c_3 t^2 + \cdots + c_rt^{r-1})e^{\lambda t}$$

satisfies the DE. The corresponding characteristic modes are $e^{\lambda t}, te^{\lambda t}, \ldots, t^{r-1}e^{\lambda t}$.