Infinite sum of discrete unit-step signals
$begingroup$
Trying to sketch the following signal:
$$sum_{k=-infty}^infty (u[k]-u[k-3])(u[n-k]-u[n-k-3])$$
Where $u[n]$ is the unit step signal (the Heaviside function, $1$ when $nge 0$ and $0$ otherwise).
My issue is that once $n$ gets involved I have no idea what to do. I believe that the first portion is just a DC signal of $1$ for all $n$ (when summed from $-infty$ to $infty$) but when multiplied by the second half I'm not sure what to do.
summation graphing-functions signal-processing
asked Feb 20 '16 at 21:30
JaccJacc
815
$endgroup$
add a comment |
$begingroup$
Trying to sketch the following signal:
$$sum_{k=-infty}^infty (u[k]-u[k-3])(u[n-k]-u[n-k-3])$$
Where $u[n]$ is the unit step signal (the Heaviside function, $1$ when $nge 0$ and $0$ otherwise).
My issue is that once $n$ gets involved I have no idea what to do. I believe that the first portion is just a DC signal of $1$ for all $n$ (when summed from $-infty$ to $infty$) but when multiplied by the second half I'm not sure what to do.
summation graphing-functions signal-processing
asked Feb 20 '16 at 21:30
JaccJacc
815
$endgroup$
$begingroup$
How are you defining $u[0]$?
$endgroup$
– John Barber
Feb 20 '16 at 21:38
$begingroup$
Well u[n] is 0 for all n<0 and 1 for all n>=0. Is that what you mean?
$endgroup$
– Jacc
Feb 20 '16 at 21:39
$begingroup$
This is just the convolution of the first expression $u[k]-u[k-3]$ with itself.
$endgroup$
– Matt L.
Feb 21 '16 at 10:36
add a comment |
$begingroup$
Trying to sketch the following signal:
$$sum_{k=-infty}^infty (u[k]-u[k-3])(u[n-k]-u[n-k-3])$$
Where $u[n]$ is the unit step signal (the Heaviside function, $1$ when $nge 0$ and $0$ otherwise).
My issue is that once $n$ gets involved I have no idea what to do. I believe that the first portion is just a DC signal of $1$ for all $n$ (when summed from $-infty$ to $infty$) but when multiplied by the second half I'm not sure what to do.
summation graphing-functions signal-processing
asked Feb 20 '16 at 21:30
JaccJacc
815
$endgroup$
Trying to sketch the following signal:
$$sum_{k=-infty}^infty (u[k]-u[k-3])(u[n-k]-u[n-k-3])$$
Where $u[n]$ is the unit step signal (the Heaviside function, $1$ when $nge 0$ and $0$ otherwise).
My issue is that once $n$ gets involved I have no idea what to do. I believe that the first portion is just a DC signal of $1$ for all $n$ (when summed from $-infty$ to $infty$) but when multiplied by the second half I'm not sure what to do.
summation graphing-functions signal-processing
summation graphing-functions signal-processing
asked Feb 20 '16 at 21:30
JaccJacc
815
asked Feb 20 '16 at 21:30
JaccJacc
815
edited Feb 20 '16 at 23:30
user147263
asked Feb 20 '16 at 21:30
JaccJacc
815
asked Feb 20 '16 at 21:30
JaccJacc
815
asked Feb 20 '16 at 21:30
JaccJacc
815
815
$begingroup$
How are you defining $u[0]$?
$endgroup$
– John Barber
Feb 20 '16 at 21:38
$begingroup$
Well u[n] is 0 for all n<0 and 1 for all n>=0. Is that what you mean?
$endgroup$
– Jacc
Feb 20 '16 at 21:39
$begingroup$
This is just the convolution of the first expression $u[k]-u[k-3]$ with itself.
$endgroup$
– Matt L.
Feb 21 '16 at 10:36
add a comment |
$begingroup$
How are you defining $u[0]$?
$endgroup$
– John Barber
Feb 20 '16 at 21:38
$begingroup$
Well u[n] is 0 for all n<0 and 1 for all n>=0. Is that what you mean?
$endgroup$
– Jacc
Feb 20 '16 at 21:39
$begingroup$
This is just the convolution of the first expression $u[k]-u[k-3]$ with itself.
$endgroup$
– Matt L.
Feb 21 '16 at 10:36
$begingroup$
How are you defining $u[0]$?
$endgroup$
– John Barber
Feb 20 '16 at 21:38
$begingroup$
How are you defining $u[0]$?
$endgroup$
– John Barber
Feb 20 '16 at 21:38
$begingroup$
Well u[n] is 0 for all n<0 and 1 for all n>=0. Is that what you mean?
$endgroup$
– Jacc
Feb 20 '16 at 21:39
$begingroup$
Well u[n] is 0 for all n<0 and 1 for all n>=0. Is that what you mean?
$endgroup$
– Jacc
Feb 20 '16 at 21:39
$begingroup$
This is just the convolution of the first expression $u[k]-u[k-3]$ with itself.
$endgroup$
– Matt L.
Feb 21 '16 at 10:36
$begingroup$
This is just the convolution of the first expression $u[k]-u[k-3]$ with itself.
$endgroup$
– Matt L.
Feb 21 '16 at 10:36
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
To begin with, consider each factor separately.
$(u[k]-u[k-3])$ is equal to zero unless $k$ is one of the numbers $0,1,2$. For these numbers it is equal to $1$.
$(u[n-k]-u[n-k-3])$ is equal to zero unless $n-k$ is one of the numbers $0,1,2$.
So, you are summing over $k=0,1,2$. For each $k$, you have a function of $n$ which is nonzero only when $n-kin{0,1,2}$, which means $nin {k,k+1,k+2}$.
So, you have:
- a bump at $0,1,2$ when $k=0$
- a bump at $1,2,3$ when $k=1$
- a bump at $2,3,4$ when $k=2$
and are adding these up.
$endgroup$
add a comment |
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.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
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: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
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
},
noCode: 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%2fmath.stackexchange.com%2fquestions%2f1664701%2finfinite-sum-of-discrete-unit-step-signals%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
To begin with, consider each factor separately.
$(u[k]-u[k-3])$ is equal to zero unless $k$ is one of the numbers $0,1,2$. For these numbers it is equal to $1$.
$(u[n-k]-u[n-k-3])$ is equal to zero unless $n-k$ is one of the numbers $0,1,2$.
So, you are summing over $k=0,1,2$. For each $k$, you have a function of $n$ which is nonzero only when $n-kin{0,1,2}$, which means $nin {k,k+1,k+2}$.
So, you have:
- a bump at $0,1,2$ when $k=0$
- a bump at $1,2,3$ when $k=1$
- a bump at $2,3,4$ when $k=2$
and are adding these up.
$endgroup$
add a comment |
$begingroup$
To begin with, consider each factor separately.
$(u[k]-u[k-3])$ is equal to zero unless $k$ is one of the numbers $0,1,2$. For these numbers it is equal to $1$.
$(u[n-k]-u[n-k-3])$ is equal to zero unless $n-k$ is one of the numbers $0,1,2$.
So, you are summing over $k=0,1,2$. For each $k$, you have a function of $n$ which is nonzero only when $n-kin{0,1,2}$, which means $nin {k,k+1,k+2}$.
So, you have:
- a bump at $0,1,2$ when $k=0$
- a bump at $1,2,3$ when $k=1$
- a bump at $2,3,4$ when $k=2$
and are adding these up.
$endgroup$
add a comment |
$begingroup$
To begin with, consider each factor separately.
$(u[k]-u[k-3])$ is equal to zero unless $k$ is one of the numbers $0,1,2$. For these numbers it is equal to $1$.
$(u[n-k]-u[n-k-3])$ is equal to zero unless $n-k$ is one of the numbers $0,1,2$.
So, you are summing over $k=0,1,2$. For each $k$, you have a function of $n$ which is nonzero only when $n-kin{0,1,2}$, which means $nin {k,k+1,k+2}$.
So, you have:
- a bump at $0,1,2$ when $k=0$
- a bump at $1,2,3$ when $k=1$
- a bump at $2,3,4$ when $k=2$
and are adding these up.
$endgroup$
To begin with, consider each factor separately.
$(u[k]-u[k-3])$ is equal to zero unless $k$ is one of the numbers $0,1,2$. For these numbers it is equal to $1$.
$(u[n-k]-u[n-k-3])$ is equal to zero unless $n-k$ is one of the numbers $0,1,2$.
So, you are summing over $k=0,1,2$. For each $k$, you have a function of $n$ which is nonzero only when $n-kin{0,1,2}$, which means $nin {k,k+1,k+2}$.
So, you have:
- a bump at $0,1,2$ when $k=0$
- a bump at $1,2,3$ when $k=1$
- a bump at $2,3,4$ when $k=2$
and are adding these up.
answered Feb 20 '16 at 23:30
user147263
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics 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%2fmath.stackexchange.com%2fquestions%2f1664701%2finfinite-sum-of-discrete-unit-step-signals%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
$begingroup$
How are you defining $u[0]$?
$endgroup$
– John Barber
Feb 20 '16 at 21:38
$begingroup$
Well u[n] is 0 for all n<0 and 1 for all n>=0. Is that what you mean?
$endgroup$
– Jacc
Feb 20 '16 at 21:39
$begingroup$
This is just the convolution of the first expression $u[k]-u[k-3]$ with itself.
$endgroup$
– Matt L.
Feb 21 '16 at 10:36