Laplace Initial Value Theorem
We use the Initial Value Theorem:
\[ f(0^+) = \lim_{s\to\infty} sF(s) \]
Instead of doing long calculations, we can use a quick shortcut by comparing the highest powers of \(s\).
Shortcut Rule
After multiplying by s:
| Power of Numerator | Power of Denominator | Initial Value |
|---|---|---|
| Numerator < Denominator | 0 | |
| Numerator = Denominator | Ratio of coefficients | |
| Numerator > Denominator | ∞ (Impulse present) |
Apply to the Problem
\[ F(s)=\frac{2s^2+2s}{s^2+4s+7} \]
Step 1: Multiply by \(s\)
\[ sF(s)=\frac{2s^3+2s^2}{s^2+4s+7} \]
Step 2: Compare Highest Powers
Numerator power = 3
Denominator power = 2
Initial Value = ∞
Super Fast Mental Trick (5 seconds)
Look at the original function:
\[ F(s)=\frac{2s^2+2s}{s^2+4s+7} \]
Multiplying by \(s\) makes the numerator highest power \(s^3\).
Since numerator degree becomes larger than denominator degree, the limit becomes infinite.
Answer: Infinite initial value (Impulse at \(t=0\))
Common Laplace initial values:
| Laplace Function | Initial Value |
|---|---|
| \(\frac{\omega}{s^2+\omega^2}\) | 0 |
| \(\frac{s}{s^2+\omega^2}\) | 1 |
| \(\frac{1}{s+a}\) | 1 |