What is the relationship between a step input and an integrator?
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ margin-bottom:0;
}
$begingroup$
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 '18 at 18:36
Rrz0Rrz0
983429
$endgroup$
add a comment |
$begingroup$
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 '18 at 18:36
Rrz0Rrz0
983429
$endgroup$
2
$begingroup$
The step response is the integral of the impulse response.
$endgroup$
– jonk
Nov 24 '18 at 19:33
$begingroup$
To answer the question you just added to the bottom: yes.
$endgroup$
– Hearth
Nov 24 '18 at 20:27
$begingroup$
And the step input signal is the differential of a ramp input signal LOL.
$endgroup$
– Andy aka
Nov 24 '18 at 21:52
$begingroup$
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: $y_{ gamma } left(tright)=int h left( t right) : text{d}t$.
$endgroup$
– jonk
Nov 24 '18 at 21:58
$begingroup$
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.
$endgroup$
– Andy aka
Nov 25 '18 at 9:39
add a comment |
$begingroup$
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 '18 at 18:36
Rrz0Rrz0
983429
$endgroup$
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 '18 at 18:36
Rrz0Rrz0
983429
asked Nov 24 '18 at 18:36
Rrz0Rrz0
983429
edited Nov 26 '18 at 9:15
Rrz0
asked Nov 24 '18 at 18:36
Rrz0Rrz0
983429
asked Nov 24 '18 at 18:36
Rrz0Rrz0
983429
asked Nov 24 '18 at 18:36
Rrz0Rrz0
983429
983429
2
$begingroup$
The step response is the integral of the impulse response.
$endgroup$
– jonk
Nov 24 '18 at 19:33
$begingroup$
To answer the question you just added to the bottom: yes.
$endgroup$
– Hearth
Nov 24 '18 at 20:27
$begingroup$
And the step input signal is the differential of a ramp input signal LOL.
$endgroup$
– Andy aka
Nov 24 '18 at 21:52
$begingroup$
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: $y_{ gamma } left(tright)=int h left( t right) : text{d}t$.
$endgroup$
– jonk
Nov 24 '18 at 21:58
$begingroup$
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.
$endgroup$
– Andy aka
Nov 25 '18 at 9:39
add a comment |
2
$begingroup$
The step response is the integral of the impulse response.
$endgroup$
– jonk
Nov 24 '18 at 19:33
$begingroup$
To answer the question you just added to the bottom: yes.
$endgroup$
– Hearth
Nov 24 '18 at 20:27
$begingroup$
And the step input signal is the differential of a ramp input signal LOL.
$endgroup$
– Andy aka
Nov 24 '18 at 21:52
$begingroup$
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: $y_{ gamma } left(tright)=int h left( t right) : text{d}t$.
$endgroup$
– jonk
Nov 24 '18 at 21:58
$begingroup$
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.
$endgroup$
– Andy aka
Nov 25 '18 at 9:39
2
2
$begingroup$
The step response is the integral of the impulse response.
$endgroup$
– jonk
Nov 24 '18 at 19:33
$begingroup$
The step response is the integral of the impulse response.
$endgroup$
– jonk
Nov 24 '18 at 19:33
$begingroup$
To answer the question you just added to the bottom: yes.
$endgroup$
– Hearth
Nov 24 '18 at 20:27
$begingroup$
To answer the question you just added to the bottom: yes.
$endgroup$
– Hearth
Nov 24 '18 at 20:27
$begingroup$
And the step input signal is the differential of a ramp input signal LOL.
$endgroup$
– Andy aka
Nov 24 '18 at 21:52
$begingroup$
And the step input signal is the differential of a ramp input signal LOL.
$endgroup$
– Andy aka
Nov 24 '18 at 21:52
$begingroup$
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: $y_{ gamma } left(tright)=int h left( t right) : text{d}t$.
$endgroup$
– jonk
Nov 24 '18 at 21:58
$begingroup$
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: $y_{ gamma } left(tright)=int h left( t right) : text{d}t$.
$endgroup$
– jonk
Nov 24 '18 at 21:58
$begingroup$
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.
$endgroup$
– Andy aka
Nov 25 '18 at 9:39
$begingroup$
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.
$endgroup$
– Andy aka
Nov 25 '18 at 9:39
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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 '18 at 19:06
Neil_UKNeil_UK
79.3k285182
$endgroup$
$begingroup$
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.
$endgroup$
– Rrz0
Nov 24 '18 at 19:21
2
$begingroup$
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.
$endgroup$
– TimWescott
Nov 24 '18 at 20:44
2
$begingroup$
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.
$endgroup$
– TimWescott
Nov 24 '18 at 20:50
1
$begingroup$
The Dirac impulse is indeed the relationship. This contradicts the answer of Andy aka does it not?
$endgroup$
– Rrz0
Nov 24 '18 at 21:38
1
$begingroup$
@Rrz0 I never let contradicting Andy stand in the way of my posting the correct answer.
$endgroup$
– Neil_UK
Nov 25 '18 at 5:38
|
show 6 more comments
$begingroup$
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 '18 at 18:43
Andy akaAndy aka
245k11186425
$endgroup$
4
$begingroup$
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.
$endgroup$
– Hearth
Nov 24 '18 at 18:48
$begingroup$
@Felthry I insist you make this an answer!!
$endgroup$
– Andy aka
Nov 24 '18 at 18:49
$begingroup$
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.
$endgroup$
– Hearth
Nov 24 '18 at 18:51
2
$begingroup$
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.
$endgroup$
– Rrz0
Nov 24 '18 at 18:57
$begingroup$
@Andyaka if there really is no relationship then is the thought process presented above incorrect?Am I going off on a tangent?
$endgroup$
– Rrz0
Nov 24 '18 at 21:16
|
show 1 more comment
Your Answer
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',
autoActivateHeartbeat: false,
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
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
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
$begingroup$
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 '18 at 19:06
Neil_UKNeil_UK
79.3k285182
$endgroup$
$begingroup$
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.
$endgroup$
– Rrz0
Nov 24 '18 at 19:21
2
$begingroup$
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.
$endgroup$
– TimWescott
Nov 24 '18 at 20:44
2
$begingroup$
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.
$endgroup$
– TimWescott
Nov 24 '18 at 20:50
1
$begingroup$
The Dirac impulse is indeed the relationship. This contradicts the answer of Andy aka does it not?
$endgroup$
– Rrz0
Nov 24 '18 at 21:38
1
$begingroup$
@Rrz0 I never let contradicting Andy stand in the way of my posting the correct answer.
$endgroup$
– Neil_UK
Nov 25 '18 at 5:38
|
show 6 more comments
$begingroup$
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 '18 at 19:06
Neil_UKNeil_UK
79.3k285182
$endgroup$
$begingroup$
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.
$endgroup$
– Rrz0
Nov 24 '18 at 19:21
2
$begingroup$
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.
$endgroup$
– TimWescott
Nov 24 '18 at 20:44
2
$begingroup$
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.
$endgroup$
– TimWescott
Nov 24 '18 at 20:50
1
$begingroup$
The Dirac impulse is indeed the relationship. This contradicts the answer of Andy aka does it not?
$endgroup$
– Rrz0
Nov 24 '18 at 21:38
1
$begingroup$
@Rrz0 I never let contradicting Andy stand in the way of my posting the correct answer.
$endgroup$
– Neil_UK
Nov 25 '18 at 5:38
|
show 6 more comments
$begingroup$
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 '18 at 19:06
Neil_UKNeil_UK
79.3k285182
$endgroup$
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 '18 at 19:06
Neil_UKNeil_UK
79.3k285182
answered Nov 24 '18 at 19:06
Neil_UKNeil_UK
79.3k285182
answered Nov 24 '18 at 19:06
Neil_UKNeil_UK
79.3k285182
answered Nov 24 '18 at 19:06
Neil_UKNeil_UK
79.3k285182
79.3k285182
$begingroup$
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.
$endgroup$
– Rrz0
Nov 24 '18 at 19:21
2
$begingroup$
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.
$endgroup$
– TimWescott
Nov 24 '18 at 20:44
2
$begingroup$
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.
$endgroup$
– TimWescott
Nov 24 '18 at 20:50
1
$begingroup$
The Dirac impulse is indeed the relationship. This contradicts the answer of Andy aka does it not?
$endgroup$
– Rrz0
Nov 24 '18 at 21:38
1
$begingroup$
@Rrz0 I never let contradicting Andy stand in the way of my posting the correct answer.
$endgroup$
– Neil_UK
Nov 25 '18 at 5:38
|
show 6 more comments
$begingroup$
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.
$endgroup$
– Rrz0
Nov 24 '18 at 19:21
2
$begingroup$
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.
$endgroup$
– TimWescott
Nov 24 '18 at 20:44
2
$begingroup$
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.
$endgroup$
– TimWescott
Nov 24 '18 at 20:50
1
$begingroup$
The Dirac impulse is indeed the relationship. This contradicts the answer of Andy aka does it not?
$endgroup$
– Rrz0
Nov 24 '18 at 21:38
1
$begingroup$
@Rrz0 I never let contradicting Andy stand in the way of my posting the correct answer.
$endgroup$
– Neil_UK
Nov 25 '18 at 5:38
$begingroup$
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.
$endgroup$
– Rrz0
Nov 24 '18 at 19:21
$begingroup$
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.
$endgroup$
– Rrz0
Nov 24 '18 at 19:21
2
2
$begingroup$
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.
$endgroup$
– TimWescott
Nov 24 '18 at 20:44
$begingroup$
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.
$endgroup$
– TimWescott
Nov 24 '18 at 20:44
2
2
$begingroup$
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.
$endgroup$
– TimWescott
Nov 24 '18 at 20:50
$begingroup$
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.
$endgroup$
– TimWescott
Nov 24 '18 at 20:50
1
1
$begingroup$
The Dirac impulse is indeed the relationship. This contradicts the answer of Andy aka does it not?
$endgroup$
– Rrz0
Nov 24 '18 at 21:38
$begingroup$
The Dirac impulse is indeed the relationship. This contradicts the answer of Andy aka does it not?
$endgroup$
– Rrz0
Nov 24 '18 at 21:38
1
1
$begingroup$
@Rrz0 I never let contradicting Andy stand in the way of my posting the correct answer.
$endgroup$
– Neil_UK
Nov 25 '18 at 5:38
$begingroup$
@Rrz0 I never let contradicting Andy stand in the way of my posting the correct answer.
$endgroup$
– Neil_UK
Nov 25 '18 at 5:38
|
show 6 more comments
$begingroup$
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 '18 at 18:43
Andy akaAndy aka
245k11186425
$endgroup$
4
$begingroup$
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.
$endgroup$
– Hearth
Nov 24 '18 at 18:48
$begingroup$
@Felthry I insist you make this an answer!!
$endgroup$
– Andy aka
Nov 24 '18 at 18:49
$begingroup$
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.
$endgroup$
– Hearth
Nov 24 '18 at 18:51
2
$begingroup$
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.
$endgroup$
– Rrz0
Nov 24 '18 at 18:57
$begingroup$
@Andyaka if there really is no relationship then is the thought process presented above incorrect?Am I going off on a tangent?
$endgroup$
– Rrz0
Nov 24 '18 at 21:16
|
show 1 more comment
$begingroup$
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 '18 at 18:43
Andy akaAndy aka
245k11186425
$endgroup$
4
$begingroup$
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.
$endgroup$
– Hearth
Nov 24 '18 at 18:48
$begingroup$
@Felthry I insist you make this an answer!!
$endgroup$
– Andy aka
Nov 24 '18 at 18:49
$begingroup$
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.
$endgroup$
– Hearth
Nov 24 '18 at 18:51
2
$begingroup$
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.
$endgroup$
– Rrz0
Nov 24 '18 at 18:57
$begingroup$
@Andyaka if there really is no relationship then is the thought process presented above incorrect?Am I going off on a tangent?
$endgroup$
– Rrz0
Nov 24 '18 at 21:16
|
show 1 more comment
$begingroup$
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 '18 at 18:43
Andy akaAndy aka
245k11186425
$endgroup$
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 '18 at 18:43
Andy akaAndy aka
245k11186425
edited Nov 25 '18 at 9:46
answered Nov 24 '18 at 18:43
Andy akaAndy aka
245k11186425
answered Nov 24 '18 at 18:43
Andy akaAndy aka
245k11186425
answered Nov 24 '18 at 18:43
Andy akaAndy aka
245k11186425
245k11186425
4
$begingroup$
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.
$endgroup$
– Hearth
Nov 24 '18 at 18:48
$begingroup$
@Felthry I insist you make this an answer!!
$endgroup$
– Andy aka
Nov 24 '18 at 18:49
$begingroup$
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.
$endgroup$
– Hearth
Nov 24 '18 at 18:51
2
$begingroup$
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.
$endgroup$
– Rrz0
Nov 24 '18 at 18:57
$begingroup$
@Andyaka if there really is no relationship then is the thought process presented above incorrect?Am I going off on a tangent?
$endgroup$
– Rrz0
Nov 24 '18 at 21:16
|
show 1 more comment
4
$begingroup$
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.
$endgroup$
– Hearth
Nov 24 '18 at 18:48
$begingroup$
@Felthry I insist you make this an answer!!
$endgroup$
– Andy aka
Nov 24 '18 at 18:49
$begingroup$
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.
$endgroup$
– Hearth
Nov 24 '18 at 18:51
2
$begingroup$
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.
$endgroup$
– Rrz0
Nov 24 '18 at 18:57
$begingroup$
@Andyaka if there really is no relationship then is the thought process presented above incorrect?Am I going off on a tangent?
$endgroup$
– Rrz0
Nov 24 '18 at 21:16
4
4
$begingroup$
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.
$endgroup$
– Hearth
Nov 24 '18 at 18:48
$begingroup$
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.
$endgroup$
– Hearth
Nov 24 '18 at 18:48
$begingroup$
@Felthry I insist you make this an answer!!
$endgroup$
– Andy aka
Nov 24 '18 at 18:49
$begingroup$
@Felthry I insist you make this an answer!!
$endgroup$
– Andy aka
Nov 24 '18 at 18:49
$begingroup$
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.
$endgroup$
– Hearth
Nov 24 '18 at 18:51
$begingroup$
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.
$endgroup$
– Hearth
Nov 24 '18 at 18:51
2
2
$begingroup$
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.
$endgroup$
– Rrz0
Nov 24 '18 at 18:57
$begingroup$
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.
$endgroup$
– Rrz0
Nov 24 '18 at 18:57
$begingroup$
@Andyaka if there really is no relationship then is the thought process presented above incorrect?Am I going off on a tangent?
$endgroup$
– Rrz0
Nov 24 '18 at 21:16
$begingroup$
@Andyaka if there really is no relationship then is the thought process presented above incorrect?Am I going off on a tangent?
$endgroup$
– Rrz0
Nov 24 '18 at 21:16
|
show 1 more comment
Thanks for contributing an answer to Electrical Engineering Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
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
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
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
2
$begingroup$
The step response is the integral of the impulse response.
$endgroup$
– jonk
Nov 24 '18 at 19:33
$begingroup$
To answer the question you just added to the bottom: yes.
$endgroup$
– Hearth
Nov 24 '18 at 20:27
$begingroup$
And the step input signal is the differential of a ramp input signal LOL.
$endgroup$
– Andy aka
Nov 24 '18 at 21:52
$begingroup$
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: $y_{ gamma } left(tright)=int h left( t right) : text{d}t$.
$endgroup$
– jonk
Nov 24 '18 at 21:58
$begingroup$
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.
$endgroup$
– Andy aka
Nov 25 '18 at 9:39