Why do we need transforms (Fourier, Laplace, Z and wavelet etc.) for a signal to analyse?

Question

Why do we need transforms (Fourier, Laplace, Z and wavelet etc.) for a signal to analyse? Is it necessary for practical calculations and analysis?

Answer 1

No, transforms aren't "necessary", but they do make some types of calculations much simpler and more convenient.

It is possible to do all computation and analisys of a signal in either the time domain or the frequency domain. However, some operations are much simpler and more intuitive in one than the other.

This can be illustrated with something as simple as a R-C low pass filter. If it is on the input of a signal that you are measuring, then you may want to know how long of a delay it adds for the result to settle within some error of the input signal. That is best done in the time domain by writing the equation of the output signal in response to a unit step input. This could be done in frequency space, but would be quite convoluted, and you'd have to express some of the things you inherently know in the time domain in the frequency space.

On the other hand, if this is a audio application, you may want to know how the filter effects the amplitude of different frequencies. That is most easily and intuitively done by expressing the filter in frequency space. The response to any particular sine input could be computed in the time domain, but it would take much more computation and not be as intuitive.

In summary, both time domain and frequency space are whole and consistant ways of looking at a signal, but each gives different intuitive insights, and each makes different types of problems harder or easier.

Answer 2

All transforms are tools to make analysis easier. They are tools that engineers, scientists, and mathematicians have developed over the years to help make their jobs easier or to help them gain a greater understanding of the phenomenon they are looking at. The Laplace transform, for example, makes solving differential equations easier. The wavelet transform helps you analyze both frequency and time domains at the same time. I think the word you used - "practical" - is key. These transforms are used to take cumbersome problems and make them more practical.

Answer 3

There are some cases where frequency is directly important, such as radio communication and audio reproduction. But in general, the Laplace and Fourier transforms are nice because they convert certain difficult mathematical operations into easier ones:

Differentiation -> Multiply by s

Integration -> Divide by s

Convolution of two response functions -> Multiplication of two transfer functions

The last one in particular is very important for dealing with feedback control, so much so that Laplace transforms are used even when talking about time-domain phenomena like step responses. This is applicable to many areas of engineering, not just signal analysis.

Answer 4

Transformations are useful because it makes understanding the problem easier in one domain than in another. I'm sure you can do it in any domain, but it will be much more complex.

Think of a filter. What does a filter do ? Think about how difficult it would be to explain to someone or analyse the circuit in the time domain.

Sometimes, its just easier to work with one domain than the other. You can solve an RLC in the time domain but it will be a 2nd order differential equation equation. You can absolutely solve it using calculus and take the derivative of this and that. Or you can transform it into the S domain (Laplace transform), and solve the circuit with simple algebra and then convert your results from the S domain back into the time domain (inverse Laplace transform).

Some domains are just the digital equivalent, like the Z domain is to the S domain.

Answer 5

Transforms (Fourier, Laplace) are used in frequency automatic control domain to prove thhings like stability and commandability of the systems.

Answer 6

These transformations are mainly adopted to solve differentaial equations under different boundary conditions or you may call limits. For Laplace u can go for positive limits up to infinite but in case of Fourier limits may be from minus to plus infinity. Moreover the kernal also vary for both the functions as it contains iota in exponential of Fourier kernals whereas not in Laplace.