Impulse function is basically the derivative of the step function u(t), which is 0 for t<0, and 1 for t>0.

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 profess goes to infinity.

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

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 $$\int^{\infty}_{0}\delta(t)e^{-st}dt=e^{-s\times 0}=1$$

Impulse function is basically the derivative of the step function u(t), which is 0 for t<0, and 1 for t>0.

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 profess goes to infinity.

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

Impulse function is basically the derivative of the step function u(t), which is 0 for t<0, and 1 for t>0.

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 profess goes to infinity.

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