System in S Domain

We will say how Laplace transform solves the issue of jump value. But before that, we need some trivial work done.

System Response Function in S Domain

Because fourier transform and laplace transform follows the same rule of convolution, and LTI system can always be described by the system impulse response, that is,

$$y(t) = x(t) \ast h(t)$$

Just like how we did with fourier transformation, we can use,

$$Y(s) = X(s)H(s)$$

$H(s)$ is the system response function in s domain.

Solving System in Differential Equation by Laplace Transform

To describe this part in a more general way is hard. So let's only look at a system described by a differential equation of the second order. What we are introducing is valid for higher orders, it's just more complex.

$$(aD^2+bD+c)y = x$$

Now let's apply laplace transform. Please note that, for laplace transform, the derivative will result in an extra term.

Since,

$$Df \xrightarrow{\mathcal{L}} s L(s) - f(-0)$$

Thus,

$$D^2f = DDf \\ \xrightarrow{\mathcal{L}} s(sL(s) - f(-0)) - f'(-0) \\ = s^2 L(s) - sf(-0) - f'(-0)$$

Put this into the original equation,

$$as^2 Y(s) - asf(-0) - af'(-0) + b s Y(s) - bf(-0) + c Y(s) = X(s)$$

Let's group them,

$$(as^2 + bs + c) Y(s) = X(s) + asf(-0) + af'(-0) + bf(-0)$$

We can tell that the right hand side is made up of two parts,

$X(s)$ is the external input. Where as $asf(-0) + a'f(-0) + bf(-0)$ is caused by internal state.

We treat both the external input and the internal state as the input of the system, so we can write,

$$H(s) = \frac{1}{as^2 +bs+c}$$

Due to the linearity, the solution is also in two parts,

$$Y_{0s}(s) = \frac{X(s)}{as^2 +bs+c} \\ Y_{0i}(s) = \frac{asf(-0) + af'(-0) + bf(-0)}{as^2 +bs+c}$$

The $Y_{0s}(s)$ is the zero state response. Because it presumes $f^{(n)}(-0) = 0$.

The $Y_{0i}(s)$ is the zero input response. Because it presumes $Y(s) = 0$.

The final $Y(s) = Y_{0s}(s) + Y_{0i}(s)$. To get the time domain result, we simply perform a inverse laplace transform.

Please note that this solution is only dependent on $f^{(n)}(-0)$. And thus there is no need to calculate the jump value.

In addition, please note the fact that,

$$Y(s) = Y_{0s}(s) + Y_{0i}(s)$$

Is only valid for $t > -0$. We can't have any information for $t < -0$ because we use single side laplace transform. If we want the convergence of the single side, we lose half of the solution.

info

You can also do this separation in the time domain. It is just that it's not obvious.

For a differential equation,

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

With initial values, $y^{(n)}(-0)$ given.

If we presume $x(t) = 0$, the solution is the zero input response in the time domain, which is in pair with the zero input response in s domain. And vice versa for zero state response solved by letting $y^{(n)}(-0)=0$ instead of the given values.

In the end, the solution is the sum of the two solutions.

$$y(t) = y_{0s}(t) + y_{0i}(t)$$

However, please note that, because we used single side laplace transformation, we can only solve the response from $-0$ to $+\infty$. So we have only $y(t)$ on the positive side.

Example

If we choose,

$$a = 1 \quad b = 4 \quad c = 3 \quad f(-0) = 1 \quad f'(-0) =1 \quad x(t) = \delta(t) + u(t)$$

We will have,

$$X(s) = 1 + \frac{1}{s}$$

Thus, for the zero state response,

$$Y_{0s}(s) = \frac{1+\frac{1}{s}}{s^2 +4s+3} \\ = \frac{s+1}{(s+3)(s+1)s} \\ = \frac{1}{(s+3)(s+1)} \\ = \frac{1}{2}(\frac{1}{s+1} - \frac{1}{s+3})$$

Because,

$$e^{-at} \xrightarrow{\mathcal{L}} \frac{1}{s+a}$$

Thus, for the positive time,

$$y_{0s}(t) = \frac{1}{2}(e^{-t} - e^{-3t})$$

And for the zero input response,

$$Y_{0i} = \frac{s+5}{s^2 + 4s + 3} \\ = \frac{s+5}{(s+3)(s+1)} \\ = \frac{2}{s+1} - \frac{1}{s+3}$$

So,

$$y_{0i} = 2 e^{-t} - e^{-3t}$$

This is why laplace transform doesn't need a jump value- we only need the information at $-0$.