# Confusions regarding Laplace transform

We know that Laplace transform has two parts, $$\\sigma \$$ and $$\ j\omega\$$.

$$\\sigma\$$ is the real part while $$\ j\omega\$$ is the imaginary part.

Is it okay to say that the term $$\\sigma \$$ is a damping factor term, while $$\ j\omega\$$ is a pure sinusoidal (pure frequency) term?

I was reading another question on EE stack exchange at link below:

Relation and difference between Fourier, Laplace and Z transforms

I read a sentence and I highlighted that sentence in attached snap. The sentence is:

Fourier transform F(jω) which is essentially the frequency domain representation of f(t)

So what does that mean? That a Laplace transform is not a frequency domain representation of f(t)?

• When you use $s\rightarrow j\omega$ to get the frequency response, you are assuming that the transient, $e^{\sigma t}$, has decayed to zero (i.e it's the 'steady state' frequency response). $\sigma$ must be negative otherwise the system is unstable.
– Chu
Commented Aug 22, 2022 at 8:59

First, what is a transformation? A transform is a mapping of a function in one domain into another domain. To go from one domain to another, you need basis-functions (basis-vectors).

In Fourier transform, the basis-functions are complex exponentials; $$\e^{-j\omega t} = \cos(\omega t) - j\sin(\omega t)\$$.

In Laplace transform, the basis-functions are the product of a complex exponential and a decaying exponential; $$\e^{-\sigma t}e^{-j\omega t} = e^{-(\sigma+j\omega) t} = e^{-st}\$$.

Fourier transform: $$\F(j\omega) = \displaystyle\int_{-\infty}^\infty f(t)e^{-j\omega t} \: \text{d}t \tag1\$$

Laplace transform: $$\F(s) = \displaystyle\int_{-\infty}^\infty f(t)e^{-st} \: \text{d}t \tag2\$$

If you wanted to investigate how similar the vectors $$\\textbf{v}_1\$$ and $$\\textbf{v}_2\$$ were, you would compute their inner product, $$\\textbf{v}_1 \cdot \textbf{v}_2 \$$. For signals, it turns out that an operation equivalent to the inner product is

$$\int_{-\infty}^{\infty}v_1(t)v_2(t) \; \text{d}t \tag3$$

where $$\v_1(t)\$$ and $$\v_2(t)\$$ are functions of time. The result of the inner product is a number whose magnitude tells us the similarity between two signals (vectors).

In the Fourier transform, it is investigated how similar $$\e^{-j\omega t}\$$ is to $$\f(t)\$$. Since $$\e^{-j\omega t}\$$ is a function of both time and frequency, the result of the inner product yields a function $$\F(j\omega)\$$ that we can evaluate at any $$\\omega=\omega_0\$$ to compute the similarity between $$\f(t)\$$ and $$\e^{-j\omega t}\$$. If $$\F(j\omega_0)\$$ is large it means that the frequency $$\\omega_0\$$ is strongly present in $$\f(t)\$$. Plotting $$\F(j\omega)\$$ as a function of $$\\omega\$$ yields the frequency characteristic, and shows the frequency contents of $$\f(t)\$$. For this reason, the Fourier transform is an excellent tool for signal analysis.

The same thought process holds for the Laplace transform, however, instead of looking at how similar $$\f(t)\$$ is to sinusoids, now we look at how similar $$\f(t)\$$ is to exponentially decaying/growing sinusoids. So $$\F(s)\$$ is a function telling you how similar $$\f(t)\$$ is to $$\e^{-st}\$$ for any $$\s\$$ and doesn't have the same intuitive usefulness as the Fourier transform does for signal analysis.

However, the Laplace transform is very useful for analyzing systems since it allows you to use the initial conditions at $$\t=0^-\$$ for finding the response of a system. Furthermore, the Laplace transformation exists for systems that are asymptotically stable, marginally and unstable because the Laplace integral converges (for certain values of $$\\sigma\$$) for increasing signals - whereas the Fourier transform only exists for asymptotically stable systems (the Fourier integral only converges for decreasing signals).

Conclusion

$$\F(j\omega)\$$ tells you how similar a function $$\f(t)\$$ is to $$\e^{-j\omega t}\$$ and tells you the frequency contents of $$\f(t)\$$. The Fourier transform is excellent for signal analysis.

$$\F(s)\$$ tells you how similar a function $$\f(t)\$$ is to $$\e^{-st}\$$ and tells you the presence of exponentials in $$\f(t)\$$. The Laplace transform is excellent for system analysis.

• I think the limits on your Laplace integation are not right Commented Aug 22, 2022 at 19:37
• @ClaraDiazSanchez The equation for the Laplace I've written in my answer is known as the bilateral Laplace transform. However, there also exists the unilateral Laplace transform $F(s) = \displaystyle\int_{0^-}^\infty e^{-st}f(t) \: \text{d}t$ which is restricted to causal signals only. It turns out that every physical signal you can create irl is causal.
– Carl
Commented Aug 23, 2022 at 6:47

So what does that mean? That a Laplace transform is not a frequency domain representation of f(t)?

The Fourier transform is a subset of the Laplace transform. Fourier ignores σ. Laplace is the all-encompassing transform and yes, it encompasses the frequency domain representation of f(t).