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Before the example it has been stated that in a system described by $$ Q(D)y(t) = P(D)x(t), \quad (1.) \iff \\ (D^N + a_1D^{N-1} + ... + a_{N-1}D + a_N)y(t) = (b_{N-M}D^M + b_{N-M+1}D^{M-1} + ...+b_{N-1}D + b_N)x(t) $$

if $N = M$, setting $y(t) = h(t)$ = impulse response, we will get
$$ h(t) = b_0 \delta(t) + \text{characteristic modes} $$
If $N > M$, $b_0 = 0$ so we get $$ h(t) = \text{characteristic modes}, \quad (2.)$$ Where the characteristic modes has been defined as the exponential terms/functions in the solution to the differential equation $Q(\lambda) = 0$. Then comes an example (Example 2.3) of a method, "impulse matching", to find the unkown coefficients of the $N$ characteristic modes in $h(t)$ in equation (2.), which I do not understand.

Find the impulse response $h(t)$ for a system specified by $$(D^2 + 5D + 6)y(t) = (D+1)x(t), \quad (2.20)$$ In this case, $b_0$ = 0. Hence, $h(t)$ consists of only the characteristic modes. The characteristic polynomial is $\lambda^2 + 5\lambda + 6 = (\lambda + 2)(\lambda + 3)$. The roots are $-2$ and $-3$. Hence, the impulse response $h(t)$ is $$ h(t) = (c_1e^{-2t} + c_2e^{-3t})u(t) , \quad (2.21)$$ letting $x(t)$ = $\delta(t)$ and $y(t) = h(t)$ in Eq. (2.20), we obtain $$ \ddot h(t) +5\dot h(t) + 6h(t) = \dot \delta(t) + \delta(t), \quad (2.22)$$ Recall that initial conditions $h(0^-)$ and $\dot h(0^-)$ are bot zero. But the application of an impulse at $t=0$ creates new initial conditions at $t=0^+$. Let $h(0^+) = K_1$ and $\dot h(0^+) = K_2$. These jump discontinuities in $h(t)$ and $\dot h(t)$ at $t=0$ result in impulse terms $\dot h(t) = K_1 \delta(t)$ and $\ddot h(0) = K_1 \dot \delta(t) + K_2 \delta(t)$ on the left-hand side. Matching the coefficients of impulse terms on both sides of Eq. (2.22) yields $$ 5K_1 + K_2 = 1, \quad K_1 = 1 \implies K_1=1, K_2=-4$$ We now use these values $h(0^+) = K_1 = 1$ and $\dot h(0^+) = K_2 = -4$ in Eq. (2.21) to find $c_1$ and $c_2$. Setting $t=0^+$ in Eq. (2.21), we obtain $c_1 + c_2 = 1$. Also setting $t=0^+$ in $h(t)$, we obtain $-2c_1 - 3c_1 = -4$. These two simultaneous equations yield $c_1 = -1$ and $c_2 = 2$. Therefore $$ H(t) = (-e^{-2t} + 2e^{-3t})u(t) $$

I do not understand the part in boldface. Why does the jump discontinuities result in impulse terms and what is the logic behind the calculation of the impulse terms from the jump discontinuities?

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From Eq. $(2.21)$ you can see that the impulse response $h(t)$ jumps from zero to some non-zero value at $t=0$. The derivative of a jump is a Dirac delta impulse, so there's a Dirac delta impulse in the derivative of the impulse response. Also the derivative of $h(t)$ jumps from zero to some possibly non-zero value at $t=0$, so there's also a Dirac delta impulse in the second derivative of $h(t)$.

A slightly different way to see this is the following. Write $h(t)$ as

$$h(t)=f(t)u(t)\tag{1}$$

with

$$f(t)=c_1e^{-2t}+c_2e^{-3t}\tag{2}$$

The derivative of $h(t)$ is then

$$h'(t)=f'(t)u(t)+f(0)\delta(t)\tag{3}$$

because $u'(t)=\delta(t)$ and $f(t)\delta(t)=f(0)\delta(t)$. The second derivative is

$$h''(t)=f''(t)u(t)+f'(0)\delta(t)+f(0)\delta'(t)\tag{4}$$

From the given difference equation we can match the coefficients of $\delta(t)$ and $\delta'(t)$ on both sides of the equation:

$$\begin{align}f'(0)+5f(0)&=1\\ f(0)&=1\end{align}\tag{5}$$

Using $(2)$ this results in

$$\begin{align}3c_1+2c_2&=1\\c_1+c_2&=1\end{align}\tag{6}$$

with the solution

$$\begin{align}c_1&=-1\\c_2&=2\end{align}\tag{7}$$

From $(1)$ and $(2)$ we obtain

$$h(t)=\left(-e^{-2t}+2e^{-3t}\right)u(t)\tag{8}$$

Also take a look at this answer to a related question.

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