Transfroms Why are there So many of them
Why in the world are we having so many transforms to map a signal in time domain into frequency domain
We have Fourier series, the Laplace transform, The fourier transform for the continuous signals alone and the Z transform and DFT for the Discrete signals
And not to forget series expansions in many names as well
WHy are we having so many of them any advantage of one over the other
We have Fourier series, the Laplace transform, The fourier transform for the continuous signals alone and the Z transform and DFT for the Discrete signals
And not to forget series expansions in many names as well
WHy are we having so many of them any advantage of one over the other
Replies

Kaustubh KatdareðŸ˜¨

Jeffrey Arulraj
Well strange expression sir this is not chillax Why is so many transform there in signal processing ?Kaustubh KatdareðŸ˜¨ 
Anil JainConquerorWell strange expression sir this is not chillax Why is so many transform there in signal processing ?
Probably he is as confuse as I am.
Just a suggestion,
After typing your post, you should read it once just to make sure that others will be able to understand what do you want to convey, right?
CB 
Ahsanul haqueWe are Ideal Follower of them...

Anand TamariyaBeauty of these Mathematical expressions is that they convert complex problems into simpler ones which can be solved easily. e.g. Fourier transform converts a time domain problem to frequency domain problem.
Now as much as we can understand the term continuous, for any practical application we need to break it down to discrete values which can be handled by computing machines. e.g. integral is a sum of series of discrete values with number of terms tending to infinity. Similarly, DFT/FFT is for Fourier transform and Ztransform is for Laplace transform. 
Halil Alper ONAYI agree with Anand. As an example, if you want to multiply two signal in time domain,it is too difficult. However, if you do that in laplace domain it is much more easier. At the end, transforms exist for easiness.

Jeffrey ArulrajLaplace transform converted the signal to s domain mate and not laplace domain A really nice way to emphasis though
Multiplication in time domain will result in convolution in frequency domain right
Halil Alper ONAYI agree with Anand. As an example, if you want to multiply two signal in time domain,it is too difficult. However, if you do that in laplace domain it is much more easier. At the end, transforms exist for easiness.
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