What is the relationship between a step input and an integrator?
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 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:
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
$mathscr{L} {1/s} = 1$
control control-system
asked 6 hours ago
Rrz0
939226
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 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:
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
$mathscr{L} {1/s} = 1$
control control-system
asked 6 hours ago
Rrz0
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
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 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:
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
$mathscr{L} {1/s} = 1$
control control-system
asked 6 hours ago
Rrz0
939226
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:
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
$mathscr{L} {1/s} = 1$
control control-system
control control-system
asked 6 hours ago
Rrz0
939226
asked 6 hours ago
Rrz0
939226
edited 1 hour ago
asked 6 hours ago
Rrz0
939226
asked 6 hours ago
Rrz0
939226
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
add a comment |
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
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 6 hours ago
Neil_UK
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
|
show 3 more comments
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?
answered 6 hours ago
Andy aka
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
|
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 6 hours ago
Neil_UK
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
|
show 3 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 6 hours ago
Neil_UK
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
|
show 3 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 6 hours ago
Neil_UK
72.3k274159
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 6 hours ago
Neil_UK
72.3k274159
answered 6 hours ago
Neil_UK
72.3k274159
answered 6 hours ago
Neil_UK
72.3k274159
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
|
show 3 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
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
|
show 3 more comments
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?
answered 6 hours ago
Andy aka
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
|
show 1 more comment
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?
answered 6 hours ago
Andy aka
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
|
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?
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?
answered 6 hours ago
Andy aka
235k10173400
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?
answered 6 hours ago
Andy aka
235k10173400
answered 6 hours ago
Andy aka
235k10173400
answered 6 hours ago
Andy aka
235k10173400
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
|
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.
– 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
|
show 1 more comment
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
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