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Why do we need Laplace and other transform for Signal processing

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aruns11152

Newbie level 5
I have Signal and System course in the past semester. But I was not able to resolve why we need to transform the signal to other form and what problems a z-transform solve which cant a laplace transform cant do??

deepmak

Member level 2
Good day Aruns
if i remember well , Z transform help us to solve problem in discrete time systems, while Laplace transform is for continuous time system , thus working with sampled signals / digital signal is done using Z transform.
i hope this was helpful .. the last time i had to deal with both of transforms was 13 /14 years ago
Regards

aruns11152

Newbie level 5
Thanks for the information Deepmak and it really helped also
But can u please answer why we cant deal all the signal in time domain only why we have to go for frequency domain ??

deepmak

Member level 2
Greetings and glad it helped
as far as i remember , working in time domain can get very complex (the equations) , however transforming to frequency domain is meant to make solving the problem easier by obtaining simpler equations ( functions of f), after doing the required math we usually revert results to time domain
if i remember anything from the past i will post
regards

aruns11152

aruns11152

Points: 2

Full Member level 4
the main reason of using these techniques is that if u try to solve a continuous eq that is a differential eq with order 3 or above and u try solving this differentail eq ull know why this laplace transform was invented where u simply convert a heck of a job differential eq into a simple algerbric problem.

aruns11152

aruns11152

Points: 2

deepmak

Member level 2
i found this in some note,it may helps :
"Why study systems in the Frequency Domain?
1) Helps us develop intuition about how linear time-invariant systems “process” their inputs. The key picture to have in your mind is that systems behave like filters. In particular:
a) a sinusoidal input produces a sinusoidal output that is scaled and shifted, but at the same frequency as the input. Scaling and shifting is described by the systems “frequency response”.
b) any signal can viewed as a sum of sinusoids
c) Thus, we can think of a linear system as “breaking down” its input signal into a sum of sinusoids; processing each sinusoid according to its “frequency response”; then reassembling the processed individual frequencies into the output signal. In this framework, we can view systems as filters such as “low-pass”, “high-pass”, “band-pass”, and “notch” filters.
2) Makes solving for the system response easier. "
probably if you can read some about time domain and frequency domain you would find it more meaningful , try the links if you wish :

http://mysite.du.edu/~etuttle/electron/elect6.htm

Regards

aruns11152

aruns11152

Points: 2

Eshal

Advanced Member level 1
Hello,
I had Signals and System in last semester. My teacher told me that we use Laplace because it can reduce differential equations in algebraic form. Differential equations are not so easy to solve directly and they take much steps to solve. Even we need to find complementary and particular solutions both in order to find the proper solution of any differential equation. While if we use Laplace Transforms then we can easily solve any equation. There is no need to find complementary and particular solution separately. Both are integrated in one solution when using Laplace transform. Its my personal experience. I used it personally.

And I agree with deepmak. He is right. z-transform is used for digital signals. I am not 100% sure but I am 90% sure about this that z-transform is used for sampled signals.

Regards,
Princess.

aruns11152

aruns11152

Points: 2

deepmak

Member level 2
also in response to why we have to deal with frequency domain :
suppose we want to send an analog signal g(t) which maximum frequency is f1 (speech signal , sensor signal ) into a digital transmission channel ;as you know it cannot sent as it is and it needs to be conditioned ( or prepared ) for the next process , the first step would be sampling the signal , (then will be Quantization and Digitizing) but first how will i know i will be able to recover the same signal (receive it ) at the end of the transmission channel ? as you know sampling is like multiplying the analog signal with Dirac pulse at a rate or frequency of f (sampling frequency ) (f=2f1)
if g(t) was the analog signal then the sampled one will be gs(t)=g(t)*dT(t) :dT(T) is Dirac pulse.
if i move to frequency domain ( take Fourier transform of gs(t) ):Gs(W) = f*Sigma (n=-infinity to n=+infinity) of G(w-nw1)
this would mean that we can get original signal using a filter of width 2W1 (after scaling by 1/f1) .. if you look at it in graphical representation , you would see that sampled signal spectrum in frequency domain is a sequence of original signal specturm
i dunno if i was lucky choosing this example , but this is what comes to my mind
best luck
regards
Deep

aruns11152

aruns11152

Points: 2

aruns11152

Newbie level 5
So this means that it have no relation conceptually except solving the differential equations only??

Eshal

Advanced Member level 1
Let me ask you. Can you tell me what type of equation can you make from the circuit? Only differential equation. Inductor gives you differential equations for voltage and current, capacitor gives you differential equation for voltage and current. Even these are basic elements.
So what do you want to do more if those differential equations are get simplified by any technique. Those techniques are Laplace or z-transform.
Even Laplace is much better than that you solve differential equations directly and find separate solutions.

aruns11152

aruns11152

Points: 2

deepmak

Member level 2
as a matter of fact , the Laplace transform ( and its special case , fourier transform) are related to the signal function f(t) , you can see from Laplace definition that it has the f(t) in the integral , you can always see Laplace transform as an image of the signal function in another domain , and it represent the signal in that domain otherwise solving problems or equations in the frequency domain cannot be applied to original signal , like if you find values of an active filter components in Laplace domain , will those values work in time domain if the Laplace transform of the filter function wasn't related to the time domain function of the filter ?
i hope i was clear in some way , smiles
sorry posting from work , so if miss something let me know

aruns11152

aruns11152

Points: 2

albert22

Full Member level 6
As have being said the laplace transform gives us the tool for converting a differential equation into an algebraic equation. Much simpler to solve. Integrals are converted to f(t)/s and derivatives to f(t)*s An important application in engineering is to get H(s), the transfer function of a LTI (linear time invariant) system or circuit. H(s) is also the impulse response of the system. The study of the poles and zeroes of H(s) permits to characterize the behavior of the system and study its stability. Evaluating H(s) on the jw axis gives us the frequency response.
For a circuit that includes R,C and Ls you dont even have to write the differential equation but only use 1/sC and sL in the kirchoff equations, in the same way that you apply the impedance abstraction Xc=1/jwC and Xl=jwL.
Other important property is that the convolution in the time domain (t) maps to the product in the laplace domain (s). That makes easier to study cascade systems for example.
The fourier transform is a special case of the laplace transform (or the laplace transform is a generalization of the fourier transform) where s=jw.
The z-transform is the equivalent tool for discrete time LTI which are characterized by LCDE ( difference equations ) instead of differential equations. Properties are similar to the fourier transform. The reason for applying the z-transform here and not the Fourier transform is that the later may not converge for certain sequences (discrete signals)

goldsmith

Advanced Member level 6
Other important property is that the convolution in the time domain (t) maps to the product in the laplace domain (s). That makes easier to study cascade systems for example.
The fourier transform is a special case of the laplace transform (or the laplace transform is a generalization of the fourier transform) where s=jw.
Hi to all
In addition , when we are talking about sine wave S=J*omega in fact S=sigma+j*omega and for sine wave sigma is zero .
In a short answer , Laplace is some kind of math , which can help to simplify the analysis and help us to don't go through the complicated equations .

Best Wishes
Goldsmith

Points: 2

Points: 2