Signals and systems

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Vikram50517

Joined Jan 4, 2020
33
Hello all!
My doubt is a vague one.Actually how is it possible that an operator in the time domain (say, d/dt) is a variable in Laplace domain ? (I am aware of the mathematical definitions) what is the "intuition" behind it? Coming to a simple control system an input x(t) is modelled as X(S) and output is modelled as Y(S) and the system i.e the transfer function is also modelled as say T(S) "which is a variable in s".As I previously mentioned a system is an function mapper but how is it a variable ? It is very perplexing
 

BobaMosfet

Joined Jul 1, 2009
1,398
There is nothing complex about the Laplace transform- In the time domain you're looking at a single instance, where as in the Laplace domain you're looking at many instances at different times unionized together via function. A complex variable.
 

bogosort

Joined Sep 24, 2011
596
Hello all!
My doubt is a vague one.Actually how is it possible that an operator in the time domain (say, d/dt) is a variable in Laplace domain ? (I am aware of the mathematical definitions) what is the "intuition" behind it? Coming to a simple control system an input x(t) is modelled as X(S) and output is modelled as Y(S) and the system i.e the transfer function is also modelled as say T(S) "which is a variable in s".As I previously mentioned a system is an function mapper but how is it a variable ? It is very perplexing
You ask good questions; I can tell that you're actively developing your intuition.

The short answer to your question is that the Laplace transform projects functions of time onto complex exponentials, and complex exponentials are eigenfunctions of differential operators.

Let's look at the mechanisms behind these ideas. A differential operator D acts on exponentials as so:
\[ D\{ e^{\lambda t} \} = \lambda e^{\lambda t} \]
In other words, the action of differentiation on an exponential is to return a scaled version of the exponential.

Fundamentally, the Laplace transform is an inner product between some function f and a complex exponential:
\[ \mathcal{L}\{ f(t) \} = \int_0^{\infty} f(t) \, e^{-st} \, dt \]
In other words, the Laplace integral finds the projection of the function f along the direction of exp(-st), for some (constant) parameter s. For instance, with s = 2, the integral gives us the projection of f in the direction of exp(-2t). Parametrizing over all s gives us a proper function F(s), and we say that the Laplace transform is a linear transformation from functions in t to functions in s.

Consider what happens inside the integral when we take the Laplace transform of a differential operator:
\[ \mathcal{L}\{ Df \} = \int_0^\infty f'(t) \, e^{-st} \, dt \]
One way to solve this is to integrate by parts, using the identity
\[ \int u \, dv = uv - \int v \, du \]
Setting u = exp(-st), we have du = -s exp(-st) dt; likewise, setting dv = f'(t) dt, we have v = f(t). Thus,
\[ \mathcal{L}\{ Df \} = e^{-st} f(t) - \int_0^\infty f(t) \underbrace{[-s e^{-st}]}_{D\{ e^{-st} \}} \, dt \]
In other words, the Laplace transform of a differential operator (which acts on a function f) is essentially the projection of f on the differential action on a complex exponential parametrized by s. And since complex exponentials are eigenfunctions of differential operators, the projection is scaled by s:
\[ e^{-st} f(t) - \int_0^\infty f(t) [-s e^{-st}] \, dt = e^{-st} f(t) + \overbrace{s}^\text{scale} \underbrace{ \int_0^\infty f(t) e^{-st} \, dt }_{\mathcal{L}\{f(t)\} } \]
Evaluating over the limits of integration, we find that exp(-st) f(t) = -f(0), giving us the expected result:
\[ \mathcal{L}\{ Df \} = s \mathcal{L}\{ f \} - f(0) \]
Taken over all values of s, we see that s is not a variable, but a scaling operator. The Laplace transform takes differential operators to scaling operators; equivalently, its transforms differential equations to polynomial equations.

Hope that helps you develop your intuition.
 
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