Continue to Site

Laplace transform and fourier transform

Status
Not open for further replies.

xuexucheng

Full Member level 2
Hi, nice guys
I have a question about lapalase transform and fourier transform.
if the TF is 1/(1+s), then this system is stable, and we change s to jw and get 1/(1+jw)
we can get the a-f and phase-f characteristic.
However, if the is 1/(1-s), then this system is unstable, and we change s to jw and get 1/(1-jw)
we can also get the a-f and phase-f characteristic.
the a-f characteristic of there two system is same. and just the phase is lagged or ahead.
what i can not understand is that dose the 1/(1-jw) system stable?
(1/(1-s) is unstable, but I can not find 1/(1-jw) unstable, why???)
dose this system exist ?
Oh, hope I express clear.

Last edited:
xuexucheng

xuexucheng

Points: 2
Laplace transform is a more general form of fourier transform. Laplace transform can be applied to all signals, and all systems. Fourier transform of a system exists only if a system is stable, i.e, only if the region of convergence in the pole - zero plot of the laplace transform includes the jw- axis. If it includes, then it means the fourier transform exists.

If the transform is 1/(1-s), there exists a pole in the right half of the s-plane at s=1. This system has a time domain representation of exp(t)*u(t) or exp(-t)*u(-t). If you notice, these signals are not absolutely
integrable in their respective intervals. So, it doesn't satisfy dirichlet's condition. Hence, the fourier transform doesn't exist.

refer oppenheim's signals book for a better explanation.

Last edited:

Points: 2

Points: 2

rgamma

Points: 2
I have gain what you have said.
but How to explain the system of 1/(1-jw),
its amplitude frequency is same as 1/(1+jw),
its phase frequency is just ahead or lagged,
dose this system exitst?
how to implement it?

Laplace transform is a more general form of fourier transform. Laplace transform can be applied to all signals, and all systems. Fourier transform of a system exists only if a system is stable, i.e, only if the region of convergence in the pole - zero plot of the laplace transform includes the jw- axis. If it includes, then it means the fourier transform exists.

If the transform is 1/(1-s), there exists a pole in the right half of the s-plane at s=1. This system has a time domain representation of exp(t)*u(t) or exp(-t)*u(-t). If you notice, these signals are not absolutely
integrable in their respective intervals. So, it doesn't satisfy dirichlet's condition. Hence, the fourier transform doesn't exist.

refer oppenheim's signals book for a better explanation.

Yeah the system exists, but it can't be analysed in the fourier domain. By implementation, if you mean implementation using circuits, you can just use a diode, resistance, opamp circuit to implement it.

If at all, it exists, yeah both will have the same magnitude. The phase of the 1/(1+jw) system will be -tan inverse( w) and the phase of 1/(1-jw) will be -tan inverse(-w).

Points: 2

rgamma

Points: 2
Great!
you are a great and nice person!
use a diode, resistance, opamp circuit to implement it---------can you describe more detail?
it can't be analysed in the fourier domain----------- why?

Yeah the system exists, but it can't be analysed in the fourier domain. By implementation, if you mean implementation using circuits, you can just use a diode, resistance, opamp circuit to implement it.

If at all, it exists, yeah both will have the same magnitude. The phase of the 1/(1+jw) system will be -tan inverse( w) and the phase of 1/(1-jw) will be -tan inverse(-w).

rgamma

rgamma

Points: 2
Yeah, here's the circuit diag:

As i mentioned earlier, the system is not stable. Hence, according to the definition of fourier transform, it can't be analysed in that domain.

Points: 2

rgamma

Points: 2
Yeah, here's the circuit diag:

View attachment 57359

As i mentioned earlier, the system is not stable. Hence, according to the definition of fourier transform, it can't be analysed in that domain.

Points: 2

rgamma

Points: 2
but How to explain the system of 1/(1-jw),
its amplitude frequency is same as 1/(1+jw),
its phase frequency is just ahead or lagged,
dose this system exitst?
how to implement it?

A system that has a denumerator of the form (1-sT) results from an amplifier with positive feedback.
It cannot be used as a linear device since it has a pole in the right hand half of the s-plane.
Thus, it is unstable (saturation) which means that neither the magnitude nor the phase of such a system can be measured.

Points: 2

rgamma

Points: 2
thank you
I simulate the circuit.
but I still can not understand it.
the diode confusing me.
could you explain it?

Yeah, here's the circuit diag:

View attachment 57359

As i mentioned earlier, the system is not stable. Hence, according to the definition of fourier transform, it can't be analysed in that domain.

---------- Post added at 14:32 ---------- Previous post was at 14:22 ----------

No, No,
not 1/(1-s), but 1/(1-jw)

1/(1-jw) can not indicate unstable!?
A system that has a denumerator of the form (1-sT) results from an amplifier with positive feedback.
It cannot be used as a linear device since it has a pole in the right hand half of the s-plane.
Thus, it is unstable (saturation) which means that neither the magnitude nor the phase of such a system can be measured.

rgamma

rgamma

Points: 2
The input to the diode is voltage. In any diode, I=Is. exp(Vd/ n.Vt).

So, it produces an output current which is the exponential of the input voltage. You send this current through a resistor to measure the output voltage.

The point is, 1/(1-jw) can't exist. only 1/(1-sT) can exists. The fourier transform can not be applied in this case at all since the function is divergent in the time domain and hence can't be integrated.

Points: 2

rgamma

Points: 2
No, No,
not 1/(1-s), but 1/(1-jw)
1/(1-jw) can not indicate unstable!?

Do you know the meaning of the complex frequency variable s and the transfer to jw?
As I have mentioned, such a system is not stable and has no magnitude/phase response.
And - forget the circuit with the diode. It is a non-linear circuit and has nothing to do with your problem.

Points: 2

Points: 2

Devrra

Points: 2
The circuit is non linear, FT can't be applied.
first, thanks a lot.
I understand what you have said.
however, I still have a question.
ok, let's forget 1/(1-jw) first.
then, is there a system whose amplitude frequency is same as the RC low pass filter and phase frequency is ahead not lagged?
can this system be implemented by the RC low pass filter and then use a all pass filter to adjust the pahse?
if the above is right, then I think this system should be described as 1/(1-jw). oh, faint .

No, No,
not 1/(1-s), but 1/(1-jw)
1/(1-jw) can not indicate unstable!?

Do you know the meaning of the complex frequency variable s and the transfer to jw?
As I have mentioned, such a system is not stable and has no magnitude/phase response.
And - forget the circuit with the diode. It is a non-linear circuit and has nothing to do with your problem.

rgamma

rgamma

Points: 2
....................
then, is there a system whose amplitude frequency is same as the RC low pass filter and phase frequency is ahead not lagged?
can this system be implemented by the RC low pass filter and then use a all pass filter to adjust the pahse?
if the above is right, then I think this system should be described as 1/(1-jw). oh, faint .

An allpass exhibits one or more zeros with a positive real part, but all poles have a negative real part (located in the LHP). Thus, you never can expect a denominator like (1-s) resp. (1-jw).

I have understood what you have said.
you mean there is no system of 1/(1-jw)?
but I still want to find this system of 1/(1-jw).
haha, so stupid.

by the way, is there the system of 1/(1-s)?

An allpass exhibits one or more zeros with a positive real part, but all poles have a negative real part (located in the LHP). Thus, you never can expect a denominator like (1-s) resp. (1-jw).

Yeah there exists a system of 1/(1-s) but 1/(1-jw) can't exist. Laplace transform can be applied to any signal you want to apply it to, but fourier transform is for stable systems only.

but I still want to find this system of 1/(1-jw).
haha, so stupid.
..........
by the way, is there the system of 1/(1-s)?

Yesterday I have asked you if you are aware of the meaning of the complex frequency s.
Both functions (with s and with jw) belong to the same system.
As I told you, a denominator (1-sT) results from positive feedback only and is an unstable system.
But you can theoretically define it and perform some simulations - however, without any practical meaning.

Yeah there exists a system of 1/(1-s) but 1/(1-jw) can't exist.
Laplace transform can be applied to any signal you want to apply it to, but fourier transform is for stable systems only.
Wrong.

Any of 1/(1+s), 1/(1-s), 1/(1+j*w) and 1/(1-j*w) can exist actually.
Fourier Transformation and Laplace Transformation are same from point of view of complex function theory.
Consider "Analytic Continuation". Here w is complex number.
Difference is a trivial, we use complex variable "s" or "w".

(1) In studying stability of system, an important factor which we have to focus on is "Characteristics Polynomial" not "Input to Output Transfer Function"

(2) Fourier Transformation and Laplace Transformation are mathematically same.
Consider "Analytic Continuation" in complex function theory.
Especially they are completely same for causal function.

(3) Even if you charcaterize circuit by H(jw) instead of H(s), "eigen mode" is same.
So there is no difference between H(jw) and H(s) regarding "eigen mode".

(4) Even if gain is lesser than 0dB, circuit could be unstable.
Here you have to understand "eigen mode".

(5) In small signal AC analysis, "eigen mode" is never generated.
Consider time domain equation in stead of AC analysis.

(6) Consider simple RC-lowpass filter.
H(jw)=1/(1+j*w/wc), where wc=R*C.
If R is negative, wc is negative.
Consider time domain phenomena,.
If "eigen mode", exp(-t*wc) excited by some noise will grow infinitely since wc<0. This is due to negative resistance.

Again Fourier Transformation and Laplace Transformation are mathematically same.

Last edited:

V
Points: 2
Hi pancho_hideboo,

Any of 1/(1+s), 1/(1-s), 1/(1+j*w) and 1/(1-j*w) can exist actually.

What do you mean with "exist" ? On paper as a ficticious transfer function?
Or as a system that has this transfer function that can be measured? In this case, I disagree with you as far as the negative sign is concerned (as explained before: instability).

(1) In studying stability of system, an important factor which we have to focus on is "Characteristics Polynomial" not "Input to Output Transfer Function"

There is not much difference between both. The char. polynomial is identical to the denominator of H(s).

(2) Fourier Transformation and Laplace Transformation are mathematically same.
Consider "Analytic Continuation" in complex function theory.
Especially they are completely same for causal function.

I don't think so, but it is not the right place to explain.

(4) Even if gain is lesser than 0dB, circuit could be unstable.
Here you have to understand "eigen mode".

Do you refer to the (open) loop gain or to the closed-loop gain?

(6) Consider simple RC-lowpass filter.
H(jw)=1/(1+j*w/wc), where wc=R*C.
If R is negative, wc is negative.
Consider time domain phenomena,.
If "eigen mode", exp(-t*wc) excited by some noise will grow infinitely since wc<0. This is due to negative resistance.

Therefore, my question above: Does a system exist that has such a transfer function that can be measured?

Regards
LvW

Last edited:

Status
Not open for further replies.