Electronic – Does an impulse function at t=0 go to infinity, or 1

laplace transform

Impulse function is basically the derivative of the step function u(t):

$$
u(t)=
\begin{cases}
0 & \text{if $t<0$} \\
1 & \text{if $t\geq0$}
\end{cases}
$$

I was taught that in laplace domain, if the transfer function $$H(s)=\frac{N(s)}{D(s)}$$ has a higer degree in the numerator than in the denominator, then the circuit is unstable because that would mean $$H(s)=1+\frac{R(s)}{D(s)}$$ and the inverse laplace of 1 is the impulse function, which my professor says goes to infinity.

But what I don't get is why the laplace transform of the impulse function is equal to $$\int^{\infty}_{0}\delta(t)e^{-st}dt=e^{-s\times 0}=1$$

Best Answer

The \$\delta(t)\$ is defined as positive infinite amplitude and infinitesimal width with an area of 1 at \$\delta(0)\$, and 0 otherwise.

As Wikipedia states:

\$\int^\infty_{-\infty}\delta(t)~dt = 1\$

Since it only has one point that has a non-zero value, which is at \$\delta(0)\$, then that is the only point we need to consider. And the area at that point is 1 per definition.


And \$x^0=1\$ which is why \$e^{-s×0}=1\$

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