What is the relationship between a step input and an integrator?
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2
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While trying to understand control engineering from first principles I came across the following which I cannot yet explain intuitively or mathematically.
What is the relationship, between a step input and an integrator?
Why are they identical to each other?
I kept on seeing $1/s$ being used to both represent a step input and an integrator.
- The Laplace transform of the unit-step function is $1/s$.
- An integrator symbol is also $1/s$.
Step Function:
Integrator Block:
Multiplication by s in Frequency (Laplace) domain is differentiation in time.
Dividing by s in Frequency (Laplace) domain is equivalent to integration in time.
Is a step input equivalent to integrating in the time domain, or is it purely coincidental that they both have a spectrum that falls as frequency increases?
Why the Laplace transform of the integral is 1/s?
$int$ in Time Domain = $ 1/s$ in Freq Domain
AND
$mathscr{L} {1/s} = 1$
EDIT:
If I am understanding the answers correctly, there is not relationship between a step INPUT and an integrator, but there is a relationship between a step FUNCTION and integrator, as explained below.
control control-system
asked Nov 24 at 18:36
Rrz0
956226
add a comment |
up vote
2
down vote
favorite
While trying to understand control engineering from first principles I came across the following which I cannot yet explain intuitively or mathematically.
What is the relationship, between a step input and an integrator?
Why are they identical to each other?
I kept on seeing $1/s$ being used to both represent a step input and an integrator.
- The Laplace transform of the unit-step function is $1/s$.
- An integrator symbol is also $1/s$.
Step Function:
Integrator Block:
Multiplication by s in Frequency (Laplace) domain is differentiation in time.
Dividing by s in Frequency (Laplace) domain is equivalent to integration in time.
Is a step input equivalent to integrating in the time domain, or is it purely coincidental that they both have a spectrum that falls as frequency increases?
Why the Laplace transform of the integral is 1/s?
$int$ in Time Domain = $ 1/s$ in Freq Domain
AND
$mathscr{L} {1/s} = 1$
EDIT:
If I am understanding the answers correctly, there is not relationship between a step INPUT and an integrator, but there is a relationship between a step FUNCTION and integrator, as explained below.
control control-system
asked Nov 24 at 18:36
Rrz0
956226
2
The step response is the integral of the impulse response.
– jonk
Nov 24 at 19:33
To answer the question you just added to the bottom: yes.
– Hearth
Nov 24 at 20:27
And the step input signal is the differential of a ramp input signal LOL.
– Andy aka
Nov 24 at 21:52
Just to cap it off.... The impulse response is $hleft(tright)=mathscr{L}^{-1}{Hleft(sright)}$, which is the derivative of the step response, $h left( t right)= frac{ text{d} }{ text{d} t }y_{ gamma } left( t right)$. So: $y_{ gamma } left(tright)=int h left( t right) : text{d}t$.
– jonk
Nov 24 at 21:58
Just let's remind everyone what this question is: What is the relationship if any, between a step input and an integrator?. The key word here is INPUT and not output.
– Andy aka
Nov 25 at 9:39
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
While trying to understand control engineering from first principles I came across the following which I cannot yet explain intuitively or mathematically.
What is the relationship, between a step input and an integrator?
Why are they identical to each other?
I kept on seeing $1/s$ being used to both represent a step input and an integrator.
- The Laplace transform of the unit-step function is $1/s$.
- An integrator symbol is also $1/s$.
Step Function:
Integrator Block:
Multiplication by s in Frequency (Laplace) domain is differentiation in time.
Dividing by s in Frequency (Laplace) domain is equivalent to integration in time.
Is a step input equivalent to integrating in the time domain, or is it purely coincidental that they both have a spectrum that falls as frequency increases?
Why the Laplace transform of the integral is 1/s?
$int$ in Time Domain = $ 1/s$ in Freq Domain
AND
$mathscr{L} {1/s} = 1$
EDIT:
If I am understanding the answers correctly, there is not relationship between a step INPUT and an integrator, but there is a relationship between a step FUNCTION and integrator, as explained below.
control control-system
asked Nov 24 at 18:36
Rrz0
956226
While trying to understand control engineering from first principles I came across the following which I cannot yet explain intuitively or mathematically.
What is the relationship, between a step input and an integrator?
Why are they identical to each other?
I kept on seeing $1/s$ being used to both represent a step input and an integrator.
- The Laplace transform of the unit-step function is $1/s$.
- An integrator symbol is also $1/s$.
Step Function:
Integrator Block:
Multiplication by s in Frequency (Laplace) domain is differentiation in time.
Dividing by s in Frequency (Laplace) domain is equivalent to integration in time.
Is a step input equivalent to integrating in the time domain, or is it purely coincidental that they both have a spectrum that falls as frequency increases?
Why the Laplace transform of the integral is 1/s?
$int$ in Time Domain = $ 1/s$ in Freq Domain
AND
$mathscr{L} {1/s} = 1$
EDIT:
If I am understanding the answers correctly, there is not relationship between a step INPUT and an integrator, but there is a relationship between a step FUNCTION and integrator, as explained below.
control control-system
control control-system
asked Nov 24 at 18:36
Rrz0
956226
asked Nov 24 at 18:36
Rrz0
956226
edited Nov 26 at 9:15
asked Nov 24 at 18:36
Rrz0
956226
asked Nov 24 at 18:36
Rrz0
956226
asked Nov 24 at 18:36
Rrz0
956226
956226
2
The step response is the integral of the impulse response.
– jonk
Nov 24 at 19:33
To answer the question you just added to the bottom: yes.
– Hearth
Nov 24 at 20:27
And the step input signal is the differential of a ramp input signal LOL.
– Andy aka
Nov 24 at 21:52
Just to cap it off.... The impulse response is $hleft(tright)=mathscr{L}^{-1}{Hleft(sright)}$, which is the derivative of the step response, $h left( t right)= frac{ text{d} }{ text{d} t }y_{ gamma } left( t right)$. So: $y_{ gamma } left(tright)=int h left( t right) : text{d}t$.
– jonk
Nov 24 at 21:58
Just let's remind everyone what this question is: What is the relationship if any, between a step input and an integrator?. The key word here is INPUT and not output.
– Andy aka
Nov 25 at 9:39
add a comment |
2
The step response is the integral of the impulse response.
– jonk
Nov 24 at 19:33
To answer the question you just added to the bottom: yes.
– Hearth
Nov 24 at 20:27
And the step input signal is the differential of a ramp input signal LOL.
– Andy aka
Nov 24 at 21:52
Just to cap it off.... The impulse response is $hleft(tright)=mathscr{L}^{-1}{Hleft(sright)}$, which is the derivative of the step response, $h left( t right)= frac{ text{d} }{ text{d} t }y_{ gamma } left( t right)$. So: $y_{ gamma } left(tright)=int h left( t right) : text{d}t$.
– jonk
Nov 24 at 21:58
Just let's remind everyone what this question is: What is the relationship if any, between a step input and an integrator?. The key word here is INPUT and not output.
– Andy aka
Nov 25 at 9:39
2
2
The step response is the integral of the impulse response.
– jonk
Nov 24 at 19:33
The step response is the integral of the impulse response.
– jonk
Nov 24 at 19:33
To answer the question you just added to the bottom: yes.
– Hearth
Nov 24 at 20:27
To answer the question you just added to the bottom: yes.
– Hearth
Nov 24 at 20:27
And the step input signal is the differential of a ramp input signal LOL.
– Andy aka
Nov 24 at 21:52
And the step input signal is the differential of a ramp input signal LOL.
– Andy aka
Nov 24 at 21:52
Just to cap it off.... The impulse response is $hleft(tright)=mathscr{L}^{-1}{Hleft(sright)}$, which is the derivative of the step response, $h left( t right)= frac{ text{d} }{ text{d} t }y_{ gamma } left( t right)$. So: $y_{ gamma } left(tright)=int h left( t right) : text{d}t$.
– jonk
Nov 24 at 21:58
Just to cap it off.... The impulse response is $hleft(tright)=mathscr{L}^{-1}{Hleft(sright)}$, which is the derivative of the step response, $h left( t right)= frac{ text{d} }{ text{d} t }y_{ gamma } left( t right)$. So: $y_{ gamma } left(tright)=int h left( t right) : text{d}t$.
– jonk
Nov 24 at 21:58
Just let's remind everyone what this question is: What is the relationship if any, between a step input and an integrator?. The key word here is INPUT and not output.
– Andy aka
Nov 25 at 9:39
Just let's remind everyone what this question is: What is the relationship if any, between a step input and an integrator?. The key word here is INPUT and not output.
– Andy aka
Nov 25 at 9:39
add a comment |
2 Answers
2
active
oldest
votes
up vote
4
down vote
Consider what happens when an integrator gets a unit impulse as its input. What is the output waveform? What's the Laplace transform of a unit impulse?
answered Nov 24 at 19:06
Neil_UK
72.5k274159
When an integrator gets a unit impulse, according to my simulink the output rises from 0 to 1 over a 1 second duration and then remains constant at 1. The Laplace transform of a unit impulse is 1, but I don't intuitively see any connection.
– Rrz0
Nov 24 at 19:21
2
A discrete impulse is different from a really-o truly-o Dirac delta functional, which is an "impulse" in continuous-time signal processing. $delta(t)$ has no width, is zero everywhere except for $t=0$, and $int_{0^-}^{0^+} delta(t) dt = 1$. If you haven't seen it before and it's boggling your mind -- relax, you have company.
– TimWescott
Nov 24 at 20:44
2
If you do the math to answer the question "what is the impulse response of an integrator?" then you will be the width of a Dirac impulse away from understanding the answer to your question.
– TimWescott
Nov 24 at 20:50
1
The Dirac impulse is indeed the relationship. This contradicts the answer of Andy aka does it not?
– Rrz0
Nov 24 at 21:38
1
@Rrz0 I never let contradicting Andy stand in the way of my posting the correct answer.
– Neil_UK
Nov 25 at 5:38
|
show 6 more comments
up vote
2
down vote
What is the relationship if any, between a step INPUT and an
integrator?
For the purpose of reminding folk what this question is about I have edited the quote above (from the OP) to highlight the word INPUT. Folk seem to be reading response (or output) instead and that is somewhat baffling.
Consider this:
- White noise has a spectrum of "x" at all frequencies i.e. it is spectrally flat
- A perfect amplifier with a gain of "x" has a transfer function of "x" at all frequencies.
Does anyone get in a muddle about this? Do they have a relationship?
So, a unit step has a spectrum that falls as frequency increases and an integrator also has a transfer function that happens to do the same. Should this be a big deal?
And a final reminder about the question asked: -
answered Nov 24 at 18:43
Andy aka
236k10173401
4
While it's true that there doesn't have to be a relationship between two things, it can often be enlightening to consider why two things look similar. I don't disagree with your answer, but I think it might be too dismissive; this is, after all, the type of question that leads to great insight in mathematics. Honestly, after reading your answer, now I find myself thinking about whether there is some mathematical relationship between a perfect amplifier and white noise.
– Hearth
Nov 24 at 18:48
@Felthry I insist you make this an answer!!
– Andy aka
Nov 24 at 18:49
I would, but it doesn't answer the question! And I'm not currently able to write it up into a form that does so.
– Hearth
Nov 24 at 18:51
2
Thanks for the answer. It feels as though I am still missing something fundamental, which I can't yet explain. It could be because I don't fully understand your last sentence. @Felthry I think that would actually be very helpful.
– Rrz0
Nov 24 at 18:57
@Andyaka if there really is no relationship then is the thought process presented above incorrect?Am I going off on a tangent?
– Rrz0
Nov 24 at 21:16
|
show 1 more comment
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
Consider what happens when an integrator gets a unit impulse as its input. What is the output waveform? What's the Laplace transform of a unit impulse?
answered Nov 24 at 19:06
Neil_UK
72.5k274159
When an integrator gets a unit impulse, according to my simulink the output rises from 0 to 1 over a 1 second duration and then remains constant at 1. The Laplace transform of a unit impulse is 1, but I don't intuitively see any connection.
– Rrz0
Nov 24 at 19:21
2
A discrete impulse is different from a really-o truly-o Dirac delta functional, which is an "impulse" in continuous-time signal processing. $delta(t)$ has no width, is zero everywhere except for $t=0$, and $int_{0^-}^{0^+} delta(t) dt = 1$. If you haven't seen it before and it's boggling your mind -- relax, you have company.
– TimWescott
Nov 24 at 20:44
2
If you do the math to answer the question "what is the impulse response of an integrator?" then you will be the width of a Dirac impulse away from understanding the answer to your question.
– TimWescott
Nov 24 at 20:50
1
The Dirac impulse is indeed the relationship. This contradicts the answer of Andy aka does it not?
– Rrz0
Nov 24 at 21:38
1
@Rrz0 I never let contradicting Andy stand in the way of my posting the correct answer.
– Neil_UK
Nov 25 at 5:38
|
show 6 more comments
up vote
4
down vote
Consider what happens when an integrator gets a unit impulse as its input. What is the output waveform? What's the Laplace transform of a unit impulse?
answered Nov 24 at 19:06
Neil_UK
72.5k274159
When an integrator gets a unit impulse, according to my simulink the output rises from 0 to 1 over a 1 second duration and then remains constant at 1. The Laplace transform of a unit impulse is 1, but I don't intuitively see any connection.
– Rrz0
Nov 24 at 19:21
2
A discrete impulse is different from a really-o truly-o Dirac delta functional, which is an "impulse" in continuous-time signal processing. $delta(t)$ has no width, is zero everywhere except for $t=0$, and $int_{0^-}^{0^+} delta(t) dt = 1$. If you haven't seen it before and it's boggling your mind -- relax, you have company.
– TimWescott
Nov 24 at 20:44
2
If you do the math to answer the question "what is the impulse response of an integrator?" then you will be the width of a Dirac impulse away from understanding the answer to your question.
– TimWescott
Nov 24 at 20:50
1
The Dirac impulse is indeed the relationship. This contradicts the answer of Andy aka does it not?
– Rrz0
Nov 24 at 21:38
1
@Rrz0 I never let contradicting Andy stand in the way of my posting the correct answer.
– Neil_UK
Nov 25 at 5:38
|
show 6 more comments
up vote
4
down vote
up vote
4
down vote
Consider what happens when an integrator gets a unit impulse as its input. What is the output waveform? What's the Laplace transform of a unit impulse?
answered Nov 24 at 19:06
Neil_UK
72.5k274159
Consider what happens when an integrator gets a unit impulse as its input. What is the output waveform? What's the Laplace transform of a unit impulse?
answered Nov 24 at 19:06
Neil_UK
72.5k274159
answered Nov 24 at 19:06
Neil_UK
72.5k274159
answered Nov 24 at 19:06
Neil_UK
72.5k274159
answered Nov 24 at 19:06
Neil_UK
72.5k274159
72.5k274159
When an integrator gets a unit impulse, according to my simulink the output rises from 0 to 1 over a 1 second duration and then remains constant at 1. The Laplace transform of a unit impulse is 1, but I don't intuitively see any connection.
– Rrz0
Nov 24 at 19:21
2
A discrete impulse is different from a really-o truly-o Dirac delta functional, which is an "impulse" in continuous-time signal processing. $delta(t)$ has no width, is zero everywhere except for $t=0$, and $int_{0^-}^{0^+} delta(t) dt = 1$. If you haven't seen it before and it's boggling your mind -- relax, you have company.
– TimWescott
Nov 24 at 20:44
2
If you do the math to answer the question "what is the impulse response of an integrator?" then you will be the width of a Dirac impulse away from understanding the answer to your question.
– TimWescott
Nov 24 at 20:50
1
The Dirac impulse is indeed the relationship. This contradicts the answer of Andy aka does it not?
– Rrz0
Nov 24 at 21:38
1
@Rrz0 I never let contradicting Andy stand in the way of my posting the correct answer.
– Neil_UK
Nov 25 at 5:38
|
show 6 more comments
When an integrator gets a unit impulse, according to my simulink the output rises from 0 to 1 over a 1 second duration and then remains constant at 1. The Laplace transform of a unit impulse is 1, but I don't intuitively see any connection.
– Rrz0
Nov 24 at 19:21
2
A discrete impulse is different from a really-o truly-o Dirac delta functional, which is an "impulse" in continuous-time signal processing. $delta(t)$ has no width, is zero everywhere except for $t=0$, and $int_{0^-}^{0^+} delta(t) dt = 1$. If you haven't seen it before and it's boggling your mind -- relax, you have company.
– TimWescott
Nov 24 at 20:44
2
If you do the math to answer the question "what is the impulse response of an integrator?" then you will be the width of a Dirac impulse away from understanding the answer to your question.
– TimWescott
Nov 24 at 20:50
1
The Dirac impulse is indeed the relationship. This contradicts the answer of Andy aka does it not?
– Rrz0
Nov 24 at 21:38
1
@Rrz0 I never let contradicting Andy stand in the way of my posting the correct answer.
– Neil_UK
Nov 25 at 5:38
When an integrator gets a unit impulse, according to my simulink the output rises from 0 to 1 over a 1 second duration and then remains constant at 1. The Laplace transform of a unit impulse is 1, but I don't intuitively see any connection.
– Rrz0
Nov 24 at 19:21
When an integrator gets a unit impulse, according to my simulink the output rises from 0 to 1 over a 1 second duration and then remains constant at 1. The Laplace transform of a unit impulse is 1, but I don't intuitively see any connection.
– Rrz0
Nov 24 at 19:21
2
2
A discrete impulse is different from a really-o truly-o Dirac delta functional, which is an "impulse" in continuous-time signal processing. $delta(t)$ has no width, is zero everywhere except for $t=0$, and $int_{0^-}^{0^+} delta(t) dt = 1$. If you haven't seen it before and it's boggling your mind -- relax, you have company.
– TimWescott
Nov 24 at 20:44
A discrete impulse is different from a really-o truly-o Dirac delta functional, which is an "impulse" in continuous-time signal processing. $delta(t)$ has no width, is zero everywhere except for $t=0$, and $int_{0^-}^{0^+} delta(t) dt = 1$. If you haven't seen it before and it's boggling your mind -- relax, you have company.
– TimWescott
Nov 24 at 20:44
2
2
If you do the math to answer the question "what is the impulse response of an integrator?" then you will be the width of a Dirac impulse away from understanding the answer to your question.
– TimWescott
Nov 24 at 20:50
If you do the math to answer the question "what is the impulse response of an integrator?" then you will be the width of a Dirac impulse away from understanding the answer to your question.
– TimWescott
Nov 24 at 20:50
1
1
The Dirac impulse is indeed the relationship. This contradicts the answer of Andy aka does it not?
– Rrz0
Nov 24 at 21:38
The Dirac impulse is indeed the relationship. This contradicts the answer of Andy aka does it not?
– Rrz0
Nov 24 at 21:38
1
1
@Rrz0 I never let contradicting Andy stand in the way of my posting the correct answer.
– Neil_UK
Nov 25 at 5:38
@Rrz0 I never let contradicting Andy stand in the way of my posting the correct answer.
– Neil_UK
Nov 25 at 5:38
|
show 6 more comments
up vote
2
down vote
What is the relationship if any, between a step INPUT and an
integrator?
For the purpose of reminding folk what this question is about I have edited the quote above (from the OP) to highlight the word INPUT. Folk seem to be reading response (or output) instead and that is somewhat baffling.
Consider this:
- White noise has a spectrum of "x" at all frequencies i.e. it is spectrally flat
- A perfect amplifier with a gain of "x" has a transfer function of "x" at all frequencies.
Does anyone get in a muddle about this? Do they have a relationship?
So, a unit step has a spectrum that falls as frequency increases and an integrator also has a transfer function that happens to do the same. Should this be a big deal?
And a final reminder about the question asked: -
answered Nov 24 at 18:43
Andy aka
236k10173401
4
While it's true that there doesn't have to be a relationship between two things, it can often be enlightening to consider why two things look similar. I don't disagree with your answer, but I think it might be too dismissive; this is, after all, the type of question that leads to great insight in mathematics. Honestly, after reading your answer, now I find myself thinking about whether there is some mathematical relationship between a perfect amplifier and white noise.
– Hearth
Nov 24 at 18:48
@Felthry I insist you make this an answer!!
– Andy aka
Nov 24 at 18:49
I would, but it doesn't answer the question! And I'm not currently able to write it up into a form that does so.
– Hearth
Nov 24 at 18:51
2
Thanks for the answer. It feels as though I am still missing something fundamental, which I can't yet explain. It could be because I don't fully understand your last sentence. @Felthry I think that would actually be very helpful.
– Rrz0
Nov 24 at 18:57
@Andyaka if there really is no relationship then is the thought process presented above incorrect?Am I going off on a tangent?
– Rrz0
Nov 24 at 21:16
|
show 1 more comment
up vote
2
down vote
What is the relationship if any, between a step INPUT and an
integrator?
For the purpose of reminding folk what this question is about I have edited the quote above (from the OP) to highlight the word INPUT. Folk seem to be reading response (or output) instead and that is somewhat baffling.
Consider this:
- White noise has a spectrum of "x" at all frequencies i.e. it is spectrally flat
- A perfect amplifier with a gain of "x" has a transfer function of "x" at all frequencies.
Does anyone get in a muddle about this? Do they have a relationship?
So, a unit step has a spectrum that falls as frequency increases and an integrator also has a transfer function that happens to do the same. Should this be a big deal?
And a final reminder about the question asked: -
answered Nov 24 at 18:43
Andy aka
236k10173401
4
While it's true that there doesn't have to be a relationship between two things, it can often be enlightening to consider why two things look similar. I don't disagree with your answer, but I think it might be too dismissive; this is, after all, the type of question that leads to great insight in mathematics. Honestly, after reading your answer, now I find myself thinking about whether there is some mathematical relationship between a perfect amplifier and white noise.
– Hearth
Nov 24 at 18:48
@Felthry I insist you make this an answer!!
– Andy aka
Nov 24 at 18:49
I would, but it doesn't answer the question! And I'm not currently able to write it up into a form that does so.
– Hearth
Nov 24 at 18:51
2
Thanks for the answer. It feels as though I am still missing something fundamental, which I can't yet explain. It could be because I don't fully understand your last sentence. @Felthry I think that would actually be very helpful.
– Rrz0
Nov 24 at 18:57
@Andyaka if there really is no relationship then is the thought process presented above incorrect?Am I going off on a tangent?
– Rrz0
Nov 24 at 21:16
|
show 1 more comment
up vote
2
down vote
up vote
2
down vote
What is the relationship if any, between a step INPUT and an
integrator?
For the purpose of reminding folk what this question is about I have edited the quote above (from the OP) to highlight the word INPUT. Folk seem to be reading response (or output) instead and that is somewhat baffling.
Consider this:
- White noise has a spectrum of "x" at all frequencies i.e. it is spectrally flat
- A perfect amplifier with a gain of "x" has a transfer function of "x" at all frequencies.
Does anyone get in a muddle about this? Do they have a relationship?
So, a unit step has a spectrum that falls as frequency increases and an integrator also has a transfer function that happens to do the same. Should this be a big deal?
And a final reminder about the question asked: -
answered Nov 24 at 18:43
Andy aka
236k10173401
What is the relationship if any, between a step INPUT and an
integrator?
For the purpose of reminding folk what this question is about I have edited the quote above (from the OP) to highlight the word INPUT. Folk seem to be reading response (or output) instead and that is somewhat baffling.
Consider this:
- White noise has a spectrum of "x" at all frequencies i.e. it is spectrally flat
- A perfect amplifier with a gain of "x" has a transfer function of "x" at all frequencies.
Does anyone get in a muddle about this? Do they have a relationship?
So, a unit step has a spectrum that falls as frequency increases and an integrator also has a transfer function that happens to do the same. Should this be a big deal?
And a final reminder about the question asked: -
answered Nov 24 at 18:43
Andy aka
236k10173401
edited Nov 25 at 9:46
answered Nov 24 at 18:43
Andy aka
236k10173401
answered Nov 24 at 18:43
Andy aka
236k10173401
answered Nov 24 at 18:43
Andy aka
236k10173401
236k10173401
4
While it's true that there doesn't have to be a relationship between two things, it can often be enlightening to consider why two things look similar. I don't disagree with your answer, but I think it might be too dismissive; this is, after all, the type of question that leads to great insight in mathematics. Honestly, after reading your answer, now I find myself thinking about whether there is some mathematical relationship between a perfect amplifier and white noise.
– Hearth
Nov 24 at 18:48
@Felthry I insist you make this an answer!!
– Andy aka
Nov 24 at 18:49
I would, but it doesn't answer the question! And I'm not currently able to write it up into a form that does so.
– Hearth
Nov 24 at 18:51
2
Thanks for the answer. It feels as though I am still missing something fundamental, which I can't yet explain. It could be because I don't fully understand your last sentence. @Felthry I think that would actually be very helpful.
– Rrz0
Nov 24 at 18:57
@Andyaka if there really is no relationship then is the thought process presented above incorrect?Am I going off on a tangent?
– Rrz0
Nov 24 at 21:16
|
show 1 more comment
4
While it's true that there doesn't have to be a relationship between two things, it can often be enlightening to consider why two things look similar. I don't disagree with your answer, but I think it might be too dismissive; this is, after all, the type of question that leads to great insight in mathematics. Honestly, after reading your answer, now I find myself thinking about whether there is some mathematical relationship between a perfect amplifier and white noise.
– Hearth
Nov 24 at 18:48
@Felthry I insist you make this an answer!!
– Andy aka
Nov 24 at 18:49
I would, but it doesn't answer the question! And I'm not currently able to write it up into a form that does so.
– Hearth
Nov 24 at 18:51
2
Thanks for the answer. It feels as though I am still missing something fundamental, which I can't yet explain. It could be because I don't fully understand your last sentence. @Felthry I think that would actually be very helpful.
– Rrz0
Nov 24 at 18:57
@Andyaka if there really is no relationship then is the thought process presented above incorrect?Am I going off on a tangent?
– Rrz0
Nov 24 at 21:16
4
4
While it's true that there doesn't have to be a relationship between two things, it can often be enlightening to consider why two things look similar. I don't disagree with your answer, but I think it might be too dismissive; this is, after all, the type of question that leads to great insight in mathematics. Honestly, after reading your answer, now I find myself thinking about whether there is some mathematical relationship between a perfect amplifier and white noise.
– Hearth
Nov 24 at 18:48
While it's true that there doesn't have to be a relationship between two things, it can often be enlightening to consider why two things look similar. I don't disagree with your answer, but I think it might be too dismissive; this is, after all, the type of question that leads to great insight in mathematics. Honestly, after reading your answer, now I find myself thinking about whether there is some mathematical relationship between a perfect amplifier and white noise.
– Hearth
Nov 24 at 18:48
@Felthry I insist you make this an answer!!
– Andy aka
Nov 24 at 18:49
@Felthry I insist you make this an answer!!
– Andy aka
Nov 24 at 18:49
I would, but it doesn't answer the question! And I'm not currently able to write it up into a form that does so.
– Hearth
Nov 24 at 18:51
I would, but it doesn't answer the question! And I'm not currently able to write it up into a form that does so.
– Hearth
Nov 24 at 18:51
2
2
Thanks for the answer. It feels as though I am still missing something fundamental, which I can't yet explain. It could be because I don't fully understand your last sentence. @Felthry I think that would actually be very helpful.
– Rrz0
Nov 24 at 18:57
Thanks for the answer. It feels as though I am still missing something fundamental, which I can't yet explain. It could be because I don't fully understand your last sentence. @Felthry I think that would actually be very helpful.
– Rrz0
Nov 24 at 18:57
@Andyaka if there really is no relationship then is the thought process presented above incorrect?Am I going off on a tangent?
– Rrz0
Nov 24 at 21:16
@Andyaka if there really is no relationship then is the thought process presented above incorrect?Am I going off on a tangent?
– Rrz0
Nov 24 at 21:16
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The step response is the integral of the impulse response.
– jonk
Nov 24 at 19:33
To answer the question you just added to the bottom: yes.
– Hearth
Nov 24 at 20:27
And the step input signal is the differential of a ramp input signal LOL.
– Andy aka
Nov 24 at 21:52
Just to cap it off.... The impulse response is $hleft(tright)=mathscr{L}^{-1}{Hleft(sright)}$, which is the derivative of the step response, $h left( t right)= frac{ text{d} }{ text{d} t }y_{ gamma } left( t right)$. So: $y_{ gamma } left(tright)=int h left( t right) : text{d}t$.
– jonk
Nov 24 at 21:58
Just let's remind everyone what this question is: What is the relationship if any, between a step input and an integrator?. The key word here is INPUT and not output.
– Andy aka
Nov 25 at 9:39