What is the relationship between a step input and an integrator?











up vote
2
down vote

favorite
3












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 if any, 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:



enter image description here



Integrator Block:



enter image description here




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



$mathscr{L} {1/s} = 1$






control control-system







share|improve this question




















  • 1




    The step response is the integral of the impulse response.
    – jonk
    5 hours ago










  • To answer the question you just added to the bottom: yes.
    – Felthry
    4 hours ago












  • And the step input signal is the differential of a ramp input signal LOL.
    – Andy aka
    3 hours ago










  • 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
    3 hours ago

















up vote
2
down vote

favorite
3












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 if any, 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:



enter image description here



Integrator Block:



enter image description here




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



$mathscr{L} {1/s} = 1$






control control-system







share|improve this question




















  • 1




    The step response is the integral of the impulse response.
    – jonk
    5 hours ago










  • To answer the question you just added to the bottom: yes.
    – Felthry
    4 hours ago












  • And the step input signal is the differential of a ramp input signal LOL.
    – Andy aka
    3 hours ago










  • 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
    3 hours ago















up vote
2
down vote

favorite
3









up vote
2
down vote

favorite
3






3





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 if any, 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:



enter image description here



Integrator Block:



enter image description here




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



$mathscr{L} {1/s} = 1$






control control-system







share|improve this question















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 if any, 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:



enter image description here



Integrator Block:



enter image description here




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



$mathscr{L} {1/s} = 1$






control control-system




control control-system






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited 1 hour ago

























asked 6 hours ago









Rrz0

939226




939226








  • 1




    The step response is the integral of the impulse response.
    – jonk
    5 hours ago










  • To answer the question you just added to the bottom: yes.
    – Felthry
    4 hours ago












  • And the step input signal is the differential of a ramp input signal LOL.
    – Andy aka
    3 hours ago










  • 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
    3 hours ago
















  • 1




    The step response is the integral of the impulse response.
    – jonk
    5 hours ago










  • To answer the question you just added to the bottom: yes.
    – Felthry
    4 hours ago












  • And the step input signal is the differential of a ramp input signal LOL.
    – Andy aka
    3 hours ago










  • 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
    3 hours ago










1




1




The step response is the integral of the impulse response.
– jonk
5 hours ago




The step response is the integral of the impulse response.
– jonk
5 hours ago












To answer the question you just added to the bottom: yes.
– Felthry
4 hours ago






To answer the question you just added to the bottom: yes.
– Felthry
4 hours ago














And the step input signal is the differential of a ramp input signal LOL.
– Andy aka
3 hours ago




And the step input signal is the differential of a ramp input signal LOL.
– Andy aka
3 hours ago












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
3 hours ago






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
3 hours ago












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?






share|improve this answer





















  • 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
    5 hours ago












  • @Rrz0r. Are you sure that isn't a one sample duration?
    – Scott Seidman
    5 hours ago










  • No I am not sure about this, I am using a discrete impulse block on Simulink.
    – Rrz0
    5 hours ago






  • 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
    4 hours ago






  • 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
    4 hours ago




















up vote
2
down vote














What is the relationship if any, between a step input and an
integrator?




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?






share|improve this answer

















  • 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.
    – Felthry
    6 hours ago












  • @Felthry I insist you make this an answer!!
    – Andy aka
    6 hours ago










  • 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.
    – Felthry
    6 hours ago






  • 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
    6 hours ago












  • @Andyaka if there really is no relationship then is the thought process presented above incorrect?Am I going off on a tangent?
    – Rrz0
    4 hours ago











Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["\$", "\$"]]);
});
});
}, "mathjax-editing");

StackExchange.ifUsing("editor", function () {
return StackExchange.using("schematics", function () {
StackExchange.schematics.init();
});
}, "cicuitlab");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "135"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














 

draft saved


draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2felectronics.stackexchange.com%2fquestions%2f408636%2fwhat-is-the-relationship-between-a-step-input-and-an-integrator%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























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?






share|improve this answer





















  • 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
    5 hours ago












  • @Rrz0r. Are you sure that isn't a one sample duration?
    – Scott Seidman
    5 hours ago










  • No I am not sure about this, I am using a discrete impulse block on Simulink.
    – Rrz0
    5 hours ago






  • 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
    4 hours ago






  • 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
    4 hours ago

















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?






share|improve this answer





















  • 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
    5 hours ago












  • @Rrz0r. Are you sure that isn't a one sample duration?
    – Scott Seidman
    5 hours ago










  • No I am not sure about this, I am using a discrete impulse block on Simulink.
    – Rrz0
    5 hours ago






  • 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
    4 hours ago






  • 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
    4 hours ago















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?






share|improve this answer












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?







share|improve this answer












share|improve this answer



share|improve this answer










answered 6 hours ago









Neil_UK

72.3k274159




72.3k274159












  • 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
    5 hours ago












  • @Rrz0r. Are you sure that isn't a one sample duration?
    – Scott Seidman
    5 hours ago










  • No I am not sure about this, I am using a discrete impulse block on Simulink.
    – Rrz0
    5 hours ago






  • 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
    4 hours ago






  • 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
    4 hours ago




















  • 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
    5 hours ago












  • @Rrz0r. Are you sure that isn't a one sample duration?
    – Scott Seidman
    5 hours ago










  • No I am not sure about this, I am using a discrete impulse block on Simulink.
    – Rrz0
    5 hours ago






  • 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
    4 hours ago






  • 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
    4 hours ago


















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
5 hours ago






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
5 hours ago














@Rrz0r. Are you sure that isn't a one sample duration?
– Scott Seidman
5 hours ago




@Rrz0r. Are you sure that isn't a one sample duration?
– Scott Seidman
5 hours ago












No I am not sure about this, I am using a discrete impulse block on Simulink.
– Rrz0
5 hours ago




No I am not sure about this, I am using a discrete impulse block on Simulink.
– Rrz0
5 hours ago




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
4 hours ago




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
4 hours ago




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
4 hours ago






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
4 hours ago














up vote
2
down vote














What is the relationship if any, between a step input and an
integrator?




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?






share|improve this answer

















  • 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.
    – Felthry
    6 hours ago












  • @Felthry I insist you make this an answer!!
    – Andy aka
    6 hours ago










  • 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.
    – Felthry
    6 hours ago






  • 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
    6 hours ago












  • @Andyaka if there really is no relationship then is the thought process presented above incorrect?Am I going off on a tangent?
    – Rrz0
    4 hours ago















up vote
2
down vote














What is the relationship if any, between a step input and an
integrator?




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?






share|improve this answer

















  • 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.
    – Felthry
    6 hours ago












  • @Felthry I insist you make this an answer!!
    – Andy aka
    6 hours ago










  • 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.
    – Felthry
    6 hours ago






  • 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
    6 hours ago












  • @Andyaka if there really is no relationship then is the thought process presented above incorrect?Am I going off on a tangent?
    – Rrz0
    4 hours ago













up vote
2
down vote










up vote
2
down vote










What is the relationship if any, between a step input and an
integrator?




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?






share|improve this answer













What is the relationship if any, between a step input and an
integrator?




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?







share|improve this answer












share|improve this answer



share|improve this answer










answered 6 hours ago









Andy aka

235k10173400




235k10173400








  • 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.
    – Felthry
    6 hours ago












  • @Felthry I insist you make this an answer!!
    – Andy aka
    6 hours ago










  • 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.
    – Felthry
    6 hours ago






  • 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
    6 hours ago












  • @Andyaka if there really is no relationship then is the thought process presented above incorrect?Am I going off on a tangent?
    – Rrz0
    4 hours ago














  • 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.
    – Felthry
    6 hours ago












  • @Felthry I insist you make this an answer!!
    – Andy aka
    6 hours ago










  • 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.
    – Felthry
    6 hours ago






  • 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
    6 hours ago












  • @Andyaka if there really is no relationship then is the thought process presented above incorrect?Am I going off on a tangent?
    – Rrz0
    4 hours ago








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.
– Felthry
6 hours ago






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.
– Felthry
6 hours ago














@Felthry I insist you make this an answer!!
– Andy aka
6 hours ago




@Felthry I insist you make this an answer!!
– Andy aka
6 hours ago












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.
– Felthry
6 hours ago




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.
– Felthry
6 hours ago




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
6 hours ago






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
6 hours ago














@Andyaka if there really is no relationship then is the thought process presented above incorrect?Am I going off on a tangent?
– Rrz0
4 hours ago




@Andyaka if there really is no relationship then is the thought process presented above incorrect?Am I going off on a tangent?
– Rrz0
4 hours ago


















 

draft saved


draft discarded



















































 


draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2felectronics.stackexchange.com%2fquestions%2f408636%2fwhat-is-the-relationship-between-a-step-input-and-an-integrator%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







-