laplace transform

[PG1995]

Hi

Could you please help me with these queries? Thank you.

Regards
PG

 

[MrAl]

Hello there,

's' is the complex frequency, not just the frequency, but it's often just called the frequency domain.

If we tried to ignore the discontinuity at t=0 we would loose the entire solution in the case of the impulse forcing function for example.

The sinusoidal part doesnt matter as much as the exponential part because the exponential part acts as the envelope of the sinusoidal part, and the sinusoidal part always goes up and down forever, but the exponential part tames it so that it does go to zero at infinity, and that's only because of the exponential part that we get a zero eventually...the sinusoidal part has nothing to do with that and cant do that anyway because it's cycling is non stop.

 

[steveB]

PG,

In Q3 it is not true that |exp(-σt)|=1, as you stated.

Steve

 

[PG1995]

Thank you, MrAl, Steve.

How do I visualize this, s, complex angular frequency? Kindly help me.

I'm not still clear about the Q2. Could you please help me with it?

If |exp(-σt)|=1 is incorrect, then what should it be? Please let me know. Thanks.

Regards
PG

 

[steveB]

How do I visualize this, s, complex angular frequency? Kindly help me.

With various transforms, you can imagine that you are applying a particular "test" function to a system. When we use Fourier Transforms, we can imagine that we are applying sine and cosine waves (or complex exponentials) to the system, and we look at how a system responds to that input signal. Laplace Transforms are one step more general than Fourier transforms because you get the Fourier Transform when σ=0.

So, imagine that you are applying signals of the type exp(-σt) sin(ωt) u(t) or exp(-σt) exp(jtω) u(t). In other words, imagine a decaying sine wave for starters. Now, the class of test signals is more general than this because you can have pure sine waves, pure decaying exponentials, exploding exponentials and increasing sine waves etc.

In essence, the σ is a decay (or expand) rate (units of 1/sec are sometimes called rates), and ω is a frequency (units of 1/sec are also sometimes called frequencies). The two are combined and called a complex frequency traditionally, but don't let the word frequency throw you off what the concept is.

I'm not still clear about the Q2. Could you please help me with it?

I think MrAl hit the nail on the head when he mentioned the impulse function. The Laplace transform is an integral and impulses are objects that generate non-zero numbers when integrated. If you start your integration at 0+, you miss the contribution of the impulse function.

Keep in mind that some systems depend, not only on the input signal, but also on derivatives of the input signal. Hence, even responses to step functions, ramps etc. have information in them from 0- to 0+ because derivatives generate impulses.

If it is still not clear, just be aware of the general importance of the distinction and then as you do more and more examples, it will become obvious.

If |exp(-σt)|=1 is incorrect, then what should it be? Please let me know. Thanks.

I guess it should be |exp(-σt)| is equal to 1 if σ equals zero and |exp(-σt)| not equal to 1 if σ is not equal to zero. The σ is a parameter that can take on any real value.

 

[MrAl]

Thank you, MrAl, Steve.

How do I visualize this, s, complex angular frequency? Kindly help me.

I'm not still clear about the Q2. Could you please help me with it?

If |exp(-σt)|=1 is incorrect, then what should it be? Please let me know. Thanks.

Regards
PG

Hello again,

For Q2 i mentioned that the impulse forcing function occurs at t=0 and only at t=0, so if we ignore that we loose the whole signal.

You can visualize complex frequency by looking at the complex frequency plane as shown in the attachment.

Each waveform shown represents what the time domain looks like at that point in the complex frequency plane (at the approximate center of the waveform). Ignore the phases of the waveforms, pay attention to the shape (ramping up or down or steady) and the frequency (higher or lower or none at all just a curved line or straight line).
As we move up or down in the plane we increase in frequency, and as we move from 0 to the right we become more unstable (ramping up faster), and as we move from 0 to the left we become more stable (ramping down faster eventually reaching zero amplitude). What isnt clear from this diagram is that larger abs(sigma) means we ramp faster (need to make a bigger drawing maybe).
Note along the jw (j omega) axis we see only sinusoidal waves, and along the sigma axis we see only ramping exponentials with no sinusoidals.
Also ignore the thicker or thinner lines as this was a rough hand drawing. All the waveforms lines have the same thickness if drawn a little better.

 

[PG1995]

Re: Q2

Thank you, Steve, MrAl.

As MrAl mentioned impulse function occurs at t=0, so if Laplace transform starts from "0" instead of "0-" then how would the Laplace miss the contribution of the impulse? Kindly help me with it if possible. Thank you.

Regards
PG

 

[steveB]

Re: Q2

Thank you, Steve, MrAl.

As MrAl mentioned impulse function occurs at t=0, so if Laplace transform starts from "0" instead of "0-" then how would the Laplace miss the contribution of the impulse? Kindly help me with it if possible. Thank you.

Regards
PG

Well, we can get into precise mathematical definitions involving limits, but dammit - we're engineers, so let's not do that.

I think it suffices to say, the intent of the 0- notation is to be an instruction to capture the impulse during integration. The notation 0+ is an instruction to not include the impulse during integration. The notation 0 without the + or - qualifier would be ambiguous in cases where there is an impulse.

When we deal with the one-sided transform, we basically want to define a function that is zero for t<0- and then capture all information about the signal after that (t>0-). The impulse function, despite the fact that it is a weird and almost nonsensical function, starts at zero and then grows to infinity with finite area between 0- and 0+. Hence, we need to start integrating at 0- to capture all information about the function.

It's a simple idea on an intuitive level, but still it requires crossing treacherous mathematical ground to deal with this rigorously. Fortunately for us, practical problems do not require us to be very sophisticated.

 

[PG1995]

Thanks a lot.

The impulse function, despite the fact that it is a weird and almost nonsensical function, starts at zero and then grows to infinity with finite area between 0- and 0+. Hence, we need to start integrating at 0- to capture all information about the function.

In my humble opinion if it starts at zero then there is no need to include 0- which is infinitesimal degree less than zero. So, I would say the impulse function starts at 0- and then grows to infinity with finite area between 0- and 0+. Kindly guide me. Thanks.

Regards
PG

 

[steveB]

Thanks a lot.



In my humble opinion if it starts at zero then there is no need to include 0- which is infinitesimal degree less than zero. So, I would say the impulse function starts at 0- and then grows to infinity with finite area between 0- and 0+. Kindly guide me. Thanks.

Regards
PG

If you start at 0, you are in some sense starting in the middle of the impulse function, at best. More properly, you are just being ambiguous about what you mean, because an impulse function is a very weird function. It is a function with no real shape, or a function with an infinite number of shapes, depending on your point of view. It really only has meaning when integrated. In physics, an impulse function is something that happens so fast that the details of the shape don't matter and only the area (or energy) of the function matter.

You need to go back to the way we defined impulse functions as limits of ordinary functions. Remember when we had a pulse of amplitude A that started at t=-1/(2A) and ended at +1/(2A). The area of this pulse is one, of course. Then, we took the limit as A goes to infinity. So, think of t=0- as t=-1/(2A) in the limit as A goes to infinity and think of t=0+ as t=+1/(2A) in the limit as A goes to infinity.

I'll stress again that 0- and 0+ should be thought of as "instructions" more so than actual numbers. They both represent the number zero, but added to that is the instruction to include the information about the weird behavior at t=0 (i.e. 0-) or to exclude the information about the weird behavior at t=0 (i.e. t=0+).

The term "instruction" is appropriate if you remember how we do integration with impulses. We don't actually manually integrate impulses as we might ordinary functions. What we do is remember the property (or really the definition) of the impulse when integrated. If you integrate the impulse alone its area is one, but if you integrate it with another function, it extracts the value of the function at t=0 (i.e. the sifting property).

 

[MrAl]

Hello again,


I think that's a pretty good explanation. I'll add a little to that.


The 0+ symbol is meant to show the place just to the right of zero. 0- is meant to show the place just to the left of zero. It means an infinitesimally close distance away from zero, but not zero.

Note also that these symbols 0+ and 0- seem to indicate something relative to zero, but not zero itself. This suggests (in a way) movement, and so these can be interpreted as:
0+ is the value as we approach zero from the right, and
0- is the value as we approach zero from the left.

For a quick example, say we have the function:
y(t)=x(t), {x>0}

which is just a ramp that starts from 0+ and goes to infinity. Note we cant really state that it goes from 0 to infinity, because 0 itself is excluded in the definition of the function, but because for every tiny increment away from zero to the right we do have x(t), so we want to be able to state that without directly stating that it is zero.

This really shows up when we want to find the value of the derivative at zero. What is the derivative of y(t) at x=0?
Well, x=0 is not defined, but since x=0+ is defined we can use that to help us. Since the derivative is 1 at x(0+) we have our initial value of the derivative at t=0+. Knowing what the derivative is just to the right of zero helps us solve many many problems even though the function is not defined at t=0. We just can not say it is so at t=0 because y(t) is not defined there.

So you'll often see the initial value shown for t=0+ rather than t=0 because sometimes the function is not defined at t=0, or something else just doesnt work at t=0 but it always works at t=0+.

Ask yourself how else could we show this. Since we cant actually state that something is so at t=0 for a function that isnt defined there but everything works out wonderfully if we assume we are CLOSE to that point, we find another way to show this by using the symbols t=0+ and t=0-, and that tells us what is happening on either side.

 

[PG1995]

Hi

I'm sorry to dig up this old thread but I need to ask something. Kindly be patient.

With various transforms, you can imagine that you are applying a particular "test" function to a system. When we use Fourier Transforms, we can imagine that we are applying sine and cosine waves (or complex exponentials) to the system, and we look at how a system responds to that input signal. Laplace Transforms are one step more general than Fourier transforms because you get the Fourier Transform when σ=0.

So, imagine that you are applying signals of the type exp(-σt) sin(ωt) u(t) or exp(-σt) exp(jtω) u(t). In other words, imagine a decaying sine wave for starters. Now, the class of test signals is more general than this because you can have pure sine waves, pure decaying exponentials, exploding exponentials and increasing sine waves etc.

In essence, the σ is a decay (or expand) rate (units of 1/sec are sometimes called rates), and ω is a frequency (units of 1/sec are also sometimes called frequencies). The two are combined and called a complex frequency traditionally, but don't let the word frequency throw you off what the concept is.

Here is my query. Please guide me. Thank you.

Regards
PG

 

[steveB]

No, I'm not trying to mathematically equate sine or cosine to complex exponentials. I'm just talking in generality about the types of functions the Laplace transform deals with as "test functions". For the most part, engineers are concerned with real signals, so the sine or cosine form is the best to think about. If we use complex exponentials we usually either use Re(exp(...)) or add two complex conjugate signals to make a real signal (i.e. Aexp(-j(...))+Aexp(+j(...)).

However, one should not exclude the use of complex signals even though they don't exist in the real world. More than once I've been able to analyze and design a system using a trick with complex domain signals. But, this is an advance way of thinking. I bring it up for you now only because, when I was a student, this point was made to me by my professors. I didn't understand the significance of it then, but the idea stayed in my mind until 10 years later when I needed it. So, don't worry about this if it doesn't make sense now, but keep it in the back of your mind in case it becomes useful in the future.