System in Differential Equation

Previously we introduced the properties of signals and systems in the time domain. Here we focus on a common case, where a system is defined by a differential equation and how can we solve them.

We use $D$ as the differential operator for convenience.

TLI system is always defined as a linear differential equation, that is,

$$y(t) = \mathcal{S}(x(t))$$ $$P(D)y(t) = x(t)$$

Where $P(D)$ is a polynomial of operator $D$. For convenience, we force the constant term to exists. Or else, we can always integrate the both side to get an equivalent equation with our desired $P(D)$.

We can usually solve this easily with the following method for common inputs,

Eigen Decomposition

We decompose the result $y(t)$ we are looking for into two parts,

$$y(t) = \overline{y}(t) + y^*(t)$$

Where,

$$P(D)\overline{y}(t) =0$$

Called the eigen solution.

And,

$$P(D)y^*(t) = x(t)$$

Called the special solution.

Eigen Solution

Let's first look for,

$$P(D)\overline{y}(t) = 0$$

If we take,

$$L(D) = P(D) - \mu$$

Where $\mu$ is the coefficient of the zeroth order term in $P$, then,

$$L(D)\overline{y}(t) = \mu \overline{y}(t)$$

That is equal to say we should look for the eigen function of $L(D)$.

info

The deduction done below is no where rigorous.

Let's use,

$$P(D)\overline{y}(t)=0$$

Because it is easier to solve.

We can decompose $P(D)$ into,

$$P(D) = \prod_{i=0}^{i=m} (D-a_i)^{p_i}$$

So long that any of the factor results in zero after applying to a function, we can have the eigen function.

For,

$$(D-a_i)^{p_i} y_i(t) = 0$$

If,

$$p_i = 1$$

Then obviously,

$$y_i(t) = A_i \exp(a_i t)$$

For,

$$p_i \geq 2$$

It is,

$$y_i(t) = \sum_{j=0}^{j=p_i-1} A_{ij} x^j \exp(a_i t)$$

So when can have,

$$\overline{y}(t) = \sum_{i=0}^{i=m} y_i(t) = \sum_{i=0}^{i=m} \sum_{j=0}^{j=p_i-1} A_{ij} x^j \exp(a_i t)$$

Special Solution

Now we want,

$$P(D)y^*(t) = x(t)$$

We need to look for,

$$P(D)^{-1}x(t)$$

Because we can always use Taylor expansion to make,

$$P(D)^{-1} = P'(D)$$

So formally,

$$y(t) = P'(D)x(t)$$

But $P'(D)$ has infinite terms, so this is actually hard to solve.

But if $x(t)$ is only a polynomial of $t$ with the highest order of $n$, we can truncate $P'(D)$ to order $n$, so that it is easy to solve.

However, if there exists $e^{at}$ in $x(t)$, truncate can't be done. But we can use the following of what we call the shifting theorem,

Because,

$$D(e^{at} x(t)) = (D+a)e^{at}x(t)$$

So we can conclude,

$$P(D) e^{at}x(t) = e^{at}P(D+a)x(t)$$

This is the shifting theorem.

Example

Let's solve,

$$(D + 1)^2(D-1) y(t) = e^{-at}t$$

We can have,

$$\overline{y}(t) = Ae^{t} + (B + Ct) e^{-t}$$

Then we look after,

$$y^*(t) = \frac{1}{(D+1)^2(D-1)}e^{-at}t$$

We apply the shifting theorem,

$$y^*(t) = e^{-at} \frac{1}{(D-a+1)^2(D-a-1)}t$$

We can do some simplification,

$$y^*(t) = e^{-at} \frac{1}{(D-a+1)^2(D-a-1)}t \\ = e^{-at} \frac{(D-a+1)-(D-a-1)}{2(D-a+1)^2(D-a-1)}t \\ = e^{-at} (\frac{1}{2(D-a+1)(D-a-1)} - \frac{1}{2(D-a+1)^2}) t \\ = \frac{1}{2} e^{-at} (\frac{1}{D^2-2aD+a^2-1} - \frac{1}{D^2 - 2(a-1)D + (a-1)^2}) t$$

Consider,

$$\frac{1}{D^2-2aD+a^2-1}$$

We can use the polynomial devision, that is,

$$\frac{1}{D^2-2aD+a^2-1} \\ = \frac{\frac{1}{a^2-1}(D^2-2aD+a^2-1) - \frac{D^2-2aD}{a^2-1}}{D^2-2aD+a^2-1} \\ = \frac{1}{a^2 -1} -D\frac{-2a + D}{(D^2-2aD+a^2-1)(a^2 -1)} \\ = \frac{1}{a^2 -1} - D\frac{\frac{-2a}{a^2-1}(D^2-2aD+a^2-1) - (1 - \frac{4a^2}{a^2-1})D + \frac{2a}{a^2-1}D^2}{(D^2-2aD+a^2-1)(a^2 -1)} \\ = \frac{1}{a^2 - 1} + \frac{2a}{(a^2-1)^2}D + D^2\frac{\frac{2a}{a^2-1}D + \frac{4a^2}{a^2 - 1}-1}{(D^2-2aD+a^2-1)(a^2 -1)}$$

Because as we expand further, the order will be higher than $D^2$, but $D^2 t = 0$, so we can truncate the result.

Similarly,

$$\frac{1}{D^2 - 2(a-1)D + (a-1)^2} = \frac{1}{(a-1)^2} + \frac{2}{(a-1)^3}D + D^2 R(D)$$

Let's go back to,

$$y^*(t) = \frac{1}{2} e^{-at} (\frac{1}{D^2-2aD+a^2-1} - \frac{1}{D^2 - 2(a-1)D + (a-1)^2}) t \\ = \frac{e^{-at}}{2}(\frac{1}{a^2 - 1}t + \frac{2a}{(a^2-1)^2} + \frac{1}{(a-1)^2}t + \frac{2}{(a-1)^3})$$

So we have the solution,

$$y(t) = y^*(t)+\overline{y}(t)$$

Differential Equation with Pulse signals

Integration Example

The previous method is very nice. However, when come to equations like,

$$P(D)y(t) = \sum_{i=0}^{+\infty} A_i D^{i} \delta(t) + u(t) s(t)$$

Of course, we can solve this by parts with the previous method. That is, we can solve,

$$P(D)y(t) = s(t) \quad t > 0 \\ P(D)y(t) = 0 \quad t < 0$$

Instead.

This seems to have solved the issue. But we will have multiple sets of solutions and undefined coefficients- but according to the theory, we should only need one initial state.

We illustrate this by,

$$Dy + y = \delta(t) + u(t)$$

For $t > 0$,

$$y = Ae^{-t} + 1$$

For $t < 0$,

$$y = (B-A)e^{-t}$$

However, since this equation is one order, the following initial state,

$$y(-0) = 4$$

Should give us a complete solution- but we have two coefficients here.

This is because, $A$ and $B$ are not independent of each other.

We can rewrite the above solution into one with step function,

$$y(t) = (A + Bu(t))e^{-t} + u(t)$$

Let's consider have the differential equation behaves around $0$.

$$Dy + y = \delta(t) + u(t)$$

We integrate on the both sides from $-\epsilon$ to $\epsilon$,

$$\int_{-\epsilon}^{\epsilon} (Dy+y) \mathrm{d}t = \int_{-\epsilon}^{\epsilon} (\delta(t) + u(t)) \mathrm{d}t$$ $$y(\epsilon) - y(-\epsilon) + \int_{-\epsilon}^{\epsilon} y(t) \mathrm{d}t = \int_{-\epsilon}^{\epsilon} (\delta(t) + u(t)) \mathrm{d}t$$

We make $\epsilon \to 0$, so,

$$y(+0) - y(-0) = 1$$

That is to say,

$$y(-0) = 4 \quad y(+0) = 5$$

Now we can have,

$$A = 4 \quad A+B+1 = 5$$

Now we can solve $A$ and $B$.

This is how we can solve differential equation with pulse signals.

Similarly, for equations of higher order, there will be $y^{(n)}(+0) - y^{(n)}(-0)$ as well.

You can always analyze by integration, the method above is too grinding- we have a better trick called pulse balancing.

Pulse Balancing

Let's consider the differential equation,

$$P(D)y(t) = \sum_{i=0}^{+\infty} A_i D^{i} \delta(t) + Cu(t) + s(t)$$

What results in the jump value? It is the discontinuous part of the right side, that is, all $\delta(t)$ related function and $Cu(t)$.

tip

Why discontinuous part on the right side causes jump value? We can integrate on the both sides, just like before.

We can always consider moving the right side to into the $y(t)$, that is,

$$P(D)(y(t) - \sum_{i=0}^{+\infty} A'_i D^{i} \delta(t) - C'u(t)) = s(t)$$

If $s(t)$ is continuous, then $y(t) - \sum_{i=0}^{+\infty} A'_i D^{i} \delta(t) - C'u(t)$ is continuous. Because the $\sum_{i=0}^{+\infty} A'_i D^{i} \delta(t) + C'u(t)$ part is discontinuous, and so is $y(t)$, their jump value must be the same so that they can be deduced. So we can just integrate $\sum_{i=0}^{+\infty} A'_i D^{i} \delta(t) + C'u(t)$ for the jump value- and better off, the integration is easy. For the zeroth order, it is obvious $C'$, for the $i$ th order, it is, $(i)!A'_{i-1}$.

This is kind of abstract- let's demonstrate with an example,

Example

Let's solve,

$$D^2y+4Dy+3y = \delta'(t) + \delta(t) + u(t)$$

And,

$$y(-0) = 2 \quad y'(-0) = 2$$

We first calculate the solution by parts.

$$y(t) = (A+Bu(t))e^{-3t} + (C + Du(t))e^{-t} + u(t)$$

Now let's calculate the jump values. We want to find,

$$P(D)(y(t) - \sum_{i=0}^{+\infty} A'_i D^{i} \delta(t) - C'u(t)) = s(t)$$

Obviously, the highest derivate from $\delta'(t)$ should be a result from the highest term on the left side, that is, we can conclude that,

$$D^2(y-u(t))+4D(y-u(t))+3(y-u(t)) = -3\delta(t) - 2u(t)$$

Then repeat for $-3\delta(t)$,

$$D^2(y-u(t)+3tu(t))+4D(y-u(t)+3tu(t))+3(y-u(t)+3tu(t)) = 14u(t) + 9tu(t)$$

Finally, the $u(t)$ terms,

$$D^2(y-u(t)+3tu(t)-14\frac{1}{2}t^2u(t))+\\4D(y-u(t)+3tu(t)-14\frac{1}{2}t^2u(t))+\\3(y-u(t)+3tu(t)-14\frac{1}{2}t^2u(t)) = 9tu(t) - 28tu(t) - 21t^2u(t)$$

Now that the right side is continuous, we can conclude,

$$y_c(t) = y-u(t)+3tu(t)-14\frac{1}{2}t^2u(t)$$

is continuous.

So we have,

$$y(+0)-y(-0) = 1$$

Then we calculate the derivate,

$$y_c'(t) = y'(t) - \delta(t)+3u(t)+14tu(t)$$

So,

$$y'(+0)-y'(-0) = -3$$

We can add these equations for the final result.

Please note that the solution does not actually preserve how the $y$ behaves at $0$. That is, it is only correct for $t < 0$ and $t > 0$.

info

If you want what happens at $t = 0$, we can always transform the equation into the form,

$$P(D)y_c(t) = S(t)u(t)$$

Where $S(0) = 0$.

If you want to preserve what had happened at $t = 0$, just solve this equation for $t > 0$ and $t \leq 0$ separately. And finally ensemble them into one solution with step function.