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[Electroenthusiast]

Control Systems>

> What is the use of transformations like z-transforms, laplace transforms?
> How does it make a problem easy to solve?
> How exactly does Frequency Domain Signal look like (representation) and how does it help in CS?
> Any links that make me understand all these?

Why cant i post this thread here ? Electronic Theory and other forum links?

 

[Nigel Goodwin]

Why cant i post this thread here ? Electronic Theory and other forum links?

Perhaps you should try reading what it says at the top of that page?

Note: New threads now go in the Articles section (click Home tab at top of page).

 

[Electroenthusiast]

Perhaps you should try reading what it says at the top of that page?

OK;

So anyone with answers to the above questions?

 

[Electroenthusiast]

OK;

So anyone with answers to the above questions?

Anyone with answer???
I badly require to understand those concepts...
At least book referral would do...

 

[MrAl]

Control Systems>

> What is the use of transformations like z-transforms, laplace transforms?
> How does it make a problem easy to solve?
> How exactly does Frequency Domain Signal look like (representation) and how does it help in CS?
> Any links that make me understand all these?

Why cant i post this thread here ? Electronic Theory and other forum links?

Hello there,


I will try to answer your questions with a brief overview, and also include an example at the end.



? What is the use of transformations like z-transforms, laplace transforms ?
Laplace Transforms are used in continuous time systems to get a different view of the system and hopefully simplify some aspect of the original analysis problem.
Z Transforms are used in a similar manner but are used in discrete time (sampled) systems.

There are also transforms that can transform a Laplace Transform into a Z Transform and thus allow us, for one thing, to transform familiar filter circuits into microcontroller code that can perform the filtering action completely in software...a very useful thing to have when working with DSP for example.


? How does it make a problem easy to solve ?
It makes the problem easier in many cases because of a variety of reasons. One of the most important reason is the Laplace Transform can transform a differential equation into an algebraic equation and thus makes the solution easier to obtain. We end up picking up an extra dimension when we do this, but it's almost always worth the extra effort because it's still simpler in the long run. Another reason is the different perspective allows us to inspect important qualities about a system with a sort of birds eye view.

? How exactly does Frequency Domain Signal look like (representation) and how does it help in CS ?
Probably the most often used frequency domain representation is the Bode Plot. It is one way that helps to understand the stability of a system which is very important. A second technique that helps is a Root Locus plot, of which i have included an example below.

? Any links that make me understand all these ?
A quick search in Google will probably turn up hundreds of pages on Laplace and related, as well as Bode Plots too.

As a quick example i've included a root locus plot of a feedback system where we want to know if the system breaks into oscillation, and if so, what conditions cause this behavior. The plot is for a simple third order system and the red lines indicate where the system oscillates. Normally the x and y axis would be marked, but here each point represents one new value of K, a system parameter, where that one lone dot toward the middle represents K=0, and each successive dot moving to the right represents K=1, K=2, K=3, etc. We can see that when K=8 the circuit becomes a perfect oscillator because it crosses the imaginary axis, and because of the unique relationship between this frequency representation and the time domain representation we can also tell that when K is greater than 8 the circuit oscillates with more and more amplitude until in real life it would saturate the system and render the whole system uncontrollable.
The transformation of the circuit from the time domain to the frequency domain makes this kind of relatively simple analysis possible which gives us a very quick view of the system stability.

If you would like to see an example of a circuit analysis simplification due to the Laplace Transform and it's properties i intend to show a quick little example in another thread where someone was asking about how to do this. I am sure there must be examples on the web too, but the one i intend to show will be super simple so the beauty of this kind of transform will immediately become apparent.

 

[Electroenthusiast]

Thanks Mr AI, Many questions were answered here; but not in the class room!
Why are systems represented in Differential Equations?


After a Laplace Transform we'll get something 'S' and 'Z' in case of Z transform...
What does that Signify? What does that mean?WHY Z', WHY S'?
Does this subject has Wide Application in Electronics? How do you rate it?

 

[MrAl]

Hi again,


Well, any system that deals with rate of change will probably use DE's, and in electrical circuits the capacitor and inductor have characters that involve such rates of change, so it makes sense that DE's would be used with electrical circuits. Not all systems need be represented this way, but if they are, they are

I dont know who first used the lower case 's' for the Laplace variable, but in any case it is physically the complex frequency:
s=a+j*w
with a being the real part and w being the imaginary part. The real and imaginary parts form a complex plane and this is where we can plot the roots.

The lower case 'z' might have come from the name of the place where the first guy to use these transforms came from, but in any case it represents a delay.
For example, z^-1 represents a delay of one time period, while z^-2 represents a delay of two time periods, etc.
We end up with a number of z's in the equation for a system. I guess an example would be in order here but it's going to have to wait a bit.

Laplace has a rather wide application for sure, and Z Transforms as well. I rate them both pretty highly, but we also have to keep in mind that many systems are non linear and with the advent of digital computers calculations on somewhat high order systems can happen pretty quick these days, so it makes sense to include numerical methods in with Laplace and Z. Numerical solutions will help when other methods are difficult to apply, but pure Laplace will help more with theory.

So we would end up with these three or four as being important:
Laplace Transforms
Z Transforms
State Vector Differential Equations
Numerical Solutions to State Vector Differential Equations



A quick example of using Laplace Transforms to solve a simple RC circuit driven by a voltage source E, where we want the solution to be the voltage across the capacitor at time t, given initial conditions are all zero...

First, a resistor circuit with two resistors, R1 and R2, forming a voltage divider, R1 on top and R2 on bottom, driven by source E.
First, the total impedance is Z=R1+R2, and the current is I=E/Z, and the voltage across R2 is vR2=I*R2 which worked out is also vR2=E*R2/Z
which is Vo=E*R2/(R1+R2). Compare this last expression to the one obtained below for the RC network.

  +--R1--+--oVo
  |      |
  E      R2    Simple voltage divider
  |      |
  +------+--oGND

Next, a resistor R1 in series with a cap C1 driven by a source E that is a dc source, and at time t=0 we turn the circuit on. The initial cap voltage is zero volts.

  +--R1--+--oVc
  |      |
  E      C1    Simple RC network with step voltage input
  |      |
  +------+--oGND

First, the complex impedance of the capacitor (which we will call 'sC1' for short) is:
sC1=1/(s*C1)

and the total impedance is again the sum of the two in series, which is Z=R1+sC1, and so the current I=(1/s)*E/Z as before except we multiply E times 1/s because it 'steps' up to E from 0 at time t=0, so the voltage across the cap is simply Vc=I*sC1, which when multiplied out comes out to:
Vc=(1/s)*E*sC1/(R1+sC1)

(notice this is the same expression as before with the resistors except sC1 replaces R2 and we added (1/s) to account for the step change in E)

and now substituting the actual complex impedance for both sC1's we get:
Vc=(1/s)*E*(1/(s*C1))/(R1+1/(s*C1))

and after simplifying we get:
Vc=E/(s*(s*R1*C1+1))

which is really:
Vc(s)=E/(s*(s*R1*C1+1))

Now we look in a table of Laplace Transforms, and we find
1/(s+a)=>e^(-a*t)

so we divide top and bottom of Vc(s) by R1*C1 and we get:
Vc(s)=E/(R1*C1)/(s*(s+1/(R1*C1)))

which if we let a=1/(R1*C1) we get:
Vc(s)=a*E/(s*(s+a))

and notice that
(1/E)*s*Vc(s)/a=1/(s+a)

which is in the form we need, so:
(1/E)*s*Vc(s)/a => e^(-a*t)

and now divide both sides by s and we get:
(1/E)*Vc(s)/a => (1/s)*e^(-a*t)

which means we can use a property of the Laplace Transform to get the solution, namely, a multiplication by 1/s in the frequency domain is equivalent to an integration in the time domain:
(1/E)*Vc(s)/a => -e^(-a*t)/a+K

and now multiply to again isolate Vc(s) we get:
Vc(s) => E*(K-e^(-a*t))

or
Vc(t)=E*(K-e^(-a*t))

and we can use the initial conditions Vc(0)=0 to obtain the value of K which comes out to equal 1, so we end up with:
Vc(t)=E*(1-e^(-a*t))

where again a=1/(R1*C1).

We ended up having to integrate in this problem because we usually wont find this exact transform we needed in a table, but a simpler way would be to use partial fractions on Vc(s) and find the solution that way. Partial fractions are often used to solve the inverse transform problem.

So we started with
Vc(s)=E/(R1*C1)/(s*(s+1/(R1*C1)))

and used Inverse Laplace Transform methods to get the required time domain solution:
Vc(t)=E*(1-e^(-a*t))

Probably most noteworthy here though is how easy we got Vc(s) above, in exactly the same way we solve for the voltage across R2 in the two resistor circuit. The only difference here is instead of starting with R2, we started with 1/(s*C1), after which we took exactly the same algebraic steps to solve for Vc(s) except for multiplying E by 1/s to account for the step change in voltage at t=0.

In fact, we could have started with the general topology circuit with all resistors:
Vo=E*R2/(R1+R2)

and simply changed the R2 resistor to 1/(s*C1) to get the same result (after also multiplying E by 1/s).
This shows the significance of the topology over the actual type of components used and how much the Laplace Transform can simplify the procedure for finding a solution.

Some component transformations:
Capacitor C1 => 1/(s*C1)
Inductor L1 => s*L1

The capacitor is defined:
dv/dt=i/C

which is a differential equation, while
1/(s*C1)

is algebraic.