If you understand Fourier transforms then Laplace transforms can usually be thought of in the same way. So I will try to show how they are related, using three examples that illustrate the three general cases.
Let's start with the Fourier transform of a function ##f(t)##
$$
\hat{F}(\omega) = \int_{-\infty}^\infty f(t) \, e^{-j\omega t} \, dt$$
In electrical engineering we require the variable ##\omega## to be real. If you think of the functions as vectors and the integral as a dot product, then the Fourier transform is projecting the function ##f(t)## onto the space of exponentials ##e^{j\omega t}##. The inverse transform then reconstructs the original function
$$
f(t) = \frac{1}{2\pi}\int_{-\infty}^\infty \hat{F}(\omega) \, e^{j\omega t} \, d\omega$$
So you are just adding up all of the frequency components with the correct amplitude and phase.
For the Laplace transform, I am assuming that you are using the unilateral Laplace transform, which is for functions that are zero for ##t<0##. In this case, for a given function ##f(t)## the Laplace transform is
$$ F(s) = \int_0^\infty f(t) \, e^{-st} dt $$
Here the variable ##s## is a complex variable, and the Laplace transform in general only converges for a right-half plane ##Re(s)>\sigma_0## (that is, the real part of ##s## larger than some real number ##\sigma_0##). This region over which the transform converges is called the Region of Convergence (ROC). The Laplace transform is projecting the function ##f(t)## onto the space of exponentials ##e^{st}##. The inverse Laplace transform reconstructs the original function, but is now a contour integral in the complex plane; the contour is a vertical line that is inside the ROC:
$$
f(t) = \frac{1}{j2\pi}\int_{\sigma_1-j\infty}^{\sigma_1+j\infty} F(s) \, e^{s t} \, ds$$
for any ##\sigma_1>\sigma_0##. So the weird thing is that when you reconstruct the original function, you only need to add up the components along ##s=\sigma_1 + j\omega## for a single value of ##\sigma_1##, but all values of ##\omega##.
Why use the Laplace transform at all, when we have the Fourier transform? Well, it allows us to analyze signals that grow exponentially, whereas Fourier does not. Let's look at a few examples. Below I will use the unit step function ##u(t)## which is ##1## for ##t>0## and 0 for ##t<0##.
1. A damped exponential, ##f(t) = u(t) e^{-t} e^{j\omega_0 t}##. In this case ##F(s) = 1/(s + 1 - j\omega_0)## with ROC ##Re(s)>-1##. The Fourier transform is ##\hat{F}(\omega) = 1/(j\omega - j\omega_0 + 1)##. Note that in this case ##\hat{F}(\omega) = F(j\omega)##; this will always be true when the imaginary axis ##Re(s)=0## is inside the ROC. The Fourier and Laplace transforms are equivalent for this case.
2. A growing exponential, ##f(t) = u(t) e^{t} e^{j\omega_0 t}##. In this case ##F(s) = 1/(s - 1 - j\omega_0)## with ROC ##Re(s)>1##. Note that the Fourier transform does not exist, which is always the case when the imaginary axis ##Re(s)=0## is outside of the ROC. So for this case the Laplace transform is useful, while the Fourier transform is not.
3. Your example, ##f(t) = u(t)##. In this case ##F(s) = 1/s## and the ROC is ##Re(s)>0##. This case brings up another difference between the way electrical engineers usually work with Fourier transforms, in that we allow the Fourier transforms to be generalized functions. For your example, the Fourier transform is ##\hat{F}(\omega) = 1/(j\omega) + \pi \delta(\omega)##, where ##\delta(w)## is the Dirac delta function. Note that
##\hat{F}(\omega) \neq F(j\omega)## for this case, which is usually true when the imaginary axis ##Re(s)=0## is the boundary of the ROC. The Fourier transform usually includes generalized functions such as the delta function (or its derivatives) for this case. Clearly the Fourier and Laplace transforms are not equivalent in this case; you should use whichever is more useful for your task at hand. For controls, Laplace is more useful. For some signal analysis problems Fourier may be more useful.
Now we can answer your question,
cnh1995 said:
So, if f(t)=1, F(s) is 1/s or 1/(σ+jω).
If σ= -1 and w= ±1, we get |F(s)|= 0.707 and phase angle Φ= -45°. This means the signal f(t)=1 contains a an exponentially decaying sinusoidal signal of time constant 1 whose peak magnitude is 0.707 and frequency of oscillations of this decaying signal is 1 Hz. Is this correct?
No. Since ##s=-1+j\omega## is outside the ROC, you cannot in general evaluate ##F## there and expect to get anything meaningful.
Hope that helped.
Jason