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Need help to understand the integral rules used solving the convolution of two functions

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4

1

$begingroup$

I am teaching myself how convolution works, there's a question which looks like this - find the convolution of the following two functions $f$ and $g$.

I understand the problem intuitively that the resulting function should be essentially the product of $f$ sweeping over $g$, and since the functions are quite simple, I can find key points like when $x = 0, 1, 2, 3$ and interpolate the graph of the resulting function easily.

While reading through the "solution" of this problem in my textbook, for the intersection of $0 le x lt 1$, the author wrote this: for $0leq x<1$,
$$int_{-infty}^{infty} g(t)cdot f(x-t)dt=int_{0}^x 2tcdot dt=frac{x^2}{2}cdot 2.$$
which I'm having trouble to understand. How exactly did he replace the $infty$ and $-infty$ with $0$ and $x$, and how exactly did he turn the whole $g(t) cdot f(x-t)$ into $2t$?

integration convolution

share|cite|improve this question

edited 8 hours ago

Robert Z

112k1010 gold badges7979 silver badges153153 bronze badges

asked 8 hours ago

maranicmaranic

3744 bronze badges

$endgroup$

add a comment | 

4

1

$begingroup$

I am teaching myself how convolution works, there's a question which looks like this - find the convolution of the following two functions $f$ and $g$.

I understand the problem intuitively that the resulting function should be essentially the product of $f$ sweeping over $g$, and since the functions are quite simple, I can find key points like when $x = 0, 1, 2, 3$ and interpolate the graph of the resulting function easily.

While reading through the "solution" of this problem in my textbook, for the intersection of $0 le x lt 1$, the author wrote this: for $0leq x<1$,
$$int_{-infty}^{infty} g(t)cdot f(x-t)dt=int_{0}^x 2tcdot dt=frac{x^2}{2}cdot 2.$$
which I'm having trouble to understand. How exactly did he replace the $infty$ and $-infty$ with $0$ and $x$, and how exactly did he turn the whole $g(t) cdot f(x-t)$ into $2t$?

integration convolution

share|cite|improve this question

edited 8 hours ago

Robert Z

112k1010 gold badges7979 silver badges153153 bronze badges

asked 8 hours ago

maranicmaranic

3744 bronze badges

$endgroup$

add a comment | 

$begingroup$

I am teaching myself how convolution works, there's a question which looks like this - find the convolution of the following two functions $f$ and $g$.

I understand the problem intuitively that the resulting function should be essentially the product of $f$ sweeping over $g$, and since the functions are quite simple, I can find key points like when $x = 0, 1, 2, 3$ and interpolate the graph of the resulting function easily.

While reading through the "solution" of this problem in my textbook, for the intersection of $0 le x lt 1$, the author wrote this: for $0leq x<1$,
$$int_{-infty}^{infty} g(t)cdot f(x-t)dt=int_{0}^x 2tcdot dt=frac{x^2}{2}cdot 2.$$
which I'm having trouble to understand. How exactly did he replace the $infty$ and $-infty$ with $0$ and $x$, and how exactly did he turn the whole $g(t) cdot f(x-t)$ into $2t$?

integration convolution

share|cite|improve this question

edited 8 hours ago

Robert Z

112k1010 gold badges7979 silver badges153153 bronze badges

asked 8 hours ago

maranicmaranic

3744 bronze badges

$endgroup$

share|cite|improve this question

edited 8 hours ago

Robert Z

112k1010 gold badges7979 silver badges153153 bronze badges

asked 8 hours ago

maranicmaranic

3744 bronze badges

share|cite|improve this question

edited 8 hours ago

Robert Z

112k1010 gold badges7979 silver badges153153 bronze badges

asked 8 hours ago

maranicmaranic

3744 bronze badges

edited 8 hours ago

Robert Z

112k1010 gold badges7979 silver badges153153 bronze badges

edited 8 hours ago

Robert Z

112k1010 gold badges7979 silver badges153153 bronze badges

asked 8 hours ago

maranicmaranic

3744 bronze badges

asked 8 hours ago

maranicmaranic

3744 bronze badges

3 Answers
3

active

oldest

votes

4

$begingroup$

Hint. Note that $f(x-t)=2$ when $0leq x-tleq 1$, i.e. $x-1leq tleq x$, otherwise it is zero. Hence
$$int_{-infty}^{infty} g(t)f(x-t)dt=int_{x-1}^x g(t)2 dt.$$
Now if $xin[0,1)$ then what is $g(t)$ for $tin [x-1,0]$? And for $tin [0,x]$?

share|cite|improve this answer

edited 8 hours ago

answered 8 hours ago

Robert ZRobert Z

112k1010 gold badges7979 silver badges153153 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So let me think out loud - if $x in [0,1)$ then $g(t) = t$ for the positive part $t in [0,x)$, but $0$ for the negative part $t in [x-1, 0)$, and we can write that in two parts because $0$ must be inside $[x-1, x]$. So it makes sense to break $int_{x-1}^{x} g(t) 2dt$ into two parts - $int_{x-1}^{0} g(t) 2dt$ and $int_{0}^{x} g(t) 2dt$, where the first part is simply $0$, no matter what $x$ we choose. Hence our original integral is now $int_{0}^{x} g(t) 2dt$ which is actually $int_{0}^{x} tcdot 2dt$ due to $g(t) = t$ for the positive part. Is my reasoning correct? :D
  $endgroup$
  – maranic
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @maranic It's perfect!
  $endgroup$
  – Robert Z
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It was quite tough to wrap my head around this, because we are dealing with 2 variables here. :D
  $endgroup$
  – maranic
  7 hours ago
  

  
  
  
  

add a comment | 

1

$begingroup$

Observe that integrand $g(t)f(x-t)$ (where $x$ is fixed and $t$ is ranging over $mathbb R$) takes value $0$ for every $tnotin[0,x]$.

This justifies to replace $int_{-infty}^{infty}cdots$ by $int_0^xcdots$.

Further for any fixed $xin[0,1)$ it is true that $g(t)f(x-t)=t2$ on interval $[0,x]$.

share|cite|improve this answer

answered 8 hours ago

drhabdrhab

113k55 gold badges4949 silver badges141141 bronze badges

$endgroup$

add a comment | 

1

$begingroup$

I'm assuming

$$
f(x)=begin{cases} 2 & xin [0,1] \ 0 & else end{cases}
$$

And

$$
g(x)=begin{cases} x & xin [0,1] \ 2-x & xin [1,2)\ 1 & xin [2,3] \0 & else end{cases}
$$

Adding this together, we see
$$
int_{-infty}^{infty} g(t)f(x-t)textrm{d}t=2int_{1-x}^x g(t)textrm{d}t,
$$
since for these values of $t,$ $f(x-t)=2$ and for all other values of $t$, $f(x-t)$ is $0$.

Now, $g(t)=0$ for $tleq 0$ so
$$
2int_{1-x}^x g(t)textrm{d}t=2int_0^x g(t)textrm{d}t=2int_0^x ttextrm{d}t,
$$
simply by plugging into the definition of $g$.

share|cite|improve this answer

edited 8 hours ago

answered 8 hours ago

WoolierThanThouWoolierThanThou

99611 silver badge77 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You're correct.
  $endgroup$
  – WoolierThanThou
  8 hours ago
  

  
  
  
  

add a comment | 

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4

$begingroup$

Hint. Note that $f(x-t)=2$ when $0leq x-tleq 1$, i.e. $x-1leq tleq x$, otherwise it is zero. Hence
$$int_{-infty}^{infty} g(t)f(x-t)dt=int_{x-1}^x g(t)2 dt.$$
Now if $xin[0,1)$ then what is $g(t)$ for $tin [x-1,0]$? And for $tin [0,x]$?

share|cite|improve this answer

edited 8 hours ago

answered 8 hours ago

Robert ZRobert Z

112k1010 gold badges7979 silver badges153153 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So let me think out loud - if $x in [0,1)$ then $g(t) = t$ for the positive part $t in [0,x)$, but $0$ for the negative part $t in [x-1, 0)$, and we can write that in two parts because $0$ must be inside $[x-1, x]$. So it makes sense to break $int_{x-1}^{x} g(t) 2dt$ into two parts - $int_{x-1}^{0} g(t) 2dt$ and $int_{0}^{x} g(t) 2dt$, where the first part is simply $0$, no matter what $x$ we choose. Hence our original integral is now $int_{0}^{x} g(t) 2dt$ which is actually $int_{0}^{x} tcdot 2dt$ due to $g(t) = t$ for the positive part. Is my reasoning correct? :D
  $endgroup$
  – maranic
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @maranic It's perfect!
  $endgroup$
  – Robert Z
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It was quite tough to wrap my head around this, because we are dealing with 2 variables here. :D
  $endgroup$
  – maranic
  7 hours ago
  

  
  
  
  

add a comment | 

4

$begingroup$

Hint. Note that $f(x-t)=2$ when $0leq x-tleq 1$, i.e. $x-1leq tleq x$, otherwise it is zero. Hence
$$int_{-infty}^{infty} g(t)f(x-t)dt=int_{x-1}^x g(t)2 dt.$$
Now if $xin[0,1)$ then what is $g(t)$ for $tin [x-1,0]$? And for $tin [0,x]$?

share|cite|improve this answer

edited 8 hours ago

answered 8 hours ago

Robert ZRobert Z

112k1010 gold badges7979 silver badges153153 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So let me think out loud - if $x in [0,1)$ then $g(t) = t$ for the positive part $t in [0,x)$, but $0$ for the negative part $t in [x-1, 0)$, and we can write that in two parts because $0$ must be inside $[x-1, x]$. So it makes sense to break $int_{x-1}^{x} g(t) 2dt$ into two parts - $int_{x-1}^{0} g(t) 2dt$ and $int_{0}^{x} g(t) 2dt$, where the first part is simply $0$, no matter what $x$ we choose. Hence our original integral is now $int_{0}^{x} g(t) 2dt$ which is actually $int_{0}^{x} tcdot 2dt$ due to $g(t) = t$ for the positive part. Is my reasoning correct? :D
  $endgroup$
  – maranic
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @maranic It's perfect!
  $endgroup$
  – Robert Z
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It was quite tough to wrap my head around this, because we are dealing with 2 variables here. :D
  $endgroup$
  – maranic
  7 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

Hint. Note that $f(x-t)=2$ when $0leq x-tleq 1$, i.e. $x-1leq tleq x$, otherwise it is zero. Hence
$$int_{-infty}^{infty} g(t)f(x-t)dt=int_{x-1}^x g(t)2 dt.$$
Now if $xin[0,1)$ then what is $g(t)$ for $tin [x-1,0]$? And for $tin [0,x]$?

share|cite|improve this answer

edited 8 hours ago

answered 8 hours ago

Robert ZRobert Z

112k1010 gold badges7979 silver badges153153 bronze badges

$endgroup$

share|cite|improve this answer

edited 8 hours ago

answered 8 hours ago

Robert ZRobert Z

112k1010 gold badges7979 silver badges153153 bronze badges

answered 8 hours ago

Robert ZRobert Z

112k1010 gold badges7979 silver badges153153 bronze badges

answered 8 hours ago

Robert ZRobert Z

112k1010 gold badges7979 silver badges153153 bronze badges

1

$begingroup$

Observe that integrand $g(t)f(x-t)$ (where $x$ is fixed and $t$ is ranging over $mathbb R$) takes value $0$ for every $tnotin[0,x]$.

This justifies to replace $int_{-infty}^{infty}cdots$ by $int_0^xcdots$.

Further for any fixed $xin[0,1)$ it is true that $g(t)f(x-t)=t2$ on interval $[0,x]$.

share|cite|improve this answer

answered 8 hours ago

drhabdrhab

113k55 gold badges4949 silver badges141141 bronze badges

$endgroup$

add a comment | 

1

$begingroup$

Observe that integrand $g(t)f(x-t)$ (where $x$ is fixed and $t$ is ranging over $mathbb R$) takes value $0$ for every $tnotin[0,x]$.

This justifies to replace $int_{-infty}^{infty}cdots$ by $int_0^xcdots$.

Further for any fixed $xin[0,1)$ it is true that $g(t)f(x-t)=t2$ on interval $[0,x]$.

share|cite|improve this answer

answered 8 hours ago

drhabdrhab

113k55 gold badges4949 silver badges141141 bronze badges

$endgroup$

add a comment | 

$begingroup$

Observe that integrand $g(t)f(x-t)$ (where $x$ is fixed and $t$ is ranging over $mathbb R$) takes value $0$ for every $tnotin[0,x]$.

This justifies to replace $int_{-infty}^{infty}cdots$ by $int_0^xcdots$.

Further for any fixed $xin[0,1)$ it is true that $g(t)f(x-t)=t2$ on interval $[0,x]$.

share|cite|improve this answer

answered 8 hours ago

drhabdrhab

113k55 gold badges4949 silver badges141141 bronze badges

$endgroup$

share|cite|improve this answer

answered 8 hours ago

drhabdrhab

113k55 gold badges4949 silver badges141141 bronze badges

answered 8 hours ago

drhabdrhab

113k55 gold badges4949 silver badges141141 bronze badges

answered 8 hours ago

drhabdrhab

113k55 gold badges4949 silver badges141141 bronze badges

1

$begingroup$

I'm assuming

$$
f(x)=begin{cases} 2 & xin [0,1] \ 0 & else end{cases}
$$

And

$$
g(x)=begin{cases} x & xin [0,1] \ 2-x & xin [1,2)\ 1 & xin [2,3] \0 & else end{cases}
$$

Adding this together, we see
$$
int_{-infty}^{infty} g(t)f(x-t)textrm{d}t=2int_{1-x}^x g(t)textrm{d}t,
$$
since for these values of $t,$ $f(x-t)=2$ and for all other values of $t$, $f(x-t)$ is $0$.

Now, $g(t)=0$ for $tleq 0$ so
$$
2int_{1-x}^x g(t)textrm{d}t=2int_0^x g(t)textrm{d}t=2int_0^x ttextrm{d}t,
$$
simply by plugging into the definition of $g$.

share|cite|improve this answer

edited 8 hours ago

answered 8 hours ago

WoolierThanThouWoolierThanThou

99611 silver badge77 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You're correct.
  $endgroup$
  – WoolierThanThou
  8 hours ago
  

  
  
  
  

add a comment | 

1

$begingroup$

I'm assuming

$$
f(x)=begin{cases} 2 & xin [0,1] \ 0 & else end{cases}
$$

And

$$
g(x)=begin{cases} x & xin [0,1] \ 2-x & xin [1,2)\ 1 & xin [2,3] \0 & else end{cases}
$$

Adding this together, we see
$$
int_{-infty}^{infty} g(t)f(x-t)textrm{d}t=2int_{1-x}^x g(t)textrm{d}t,
$$
since for these values of $t,$ $f(x-t)=2$ and for all other values of $t$, $f(x-t)$ is $0$.

Now, $g(t)=0$ for $tleq 0$ so
$$
2int_{1-x}^x g(t)textrm{d}t=2int_0^x g(t)textrm{d}t=2int_0^x ttextrm{d}t,
$$
simply by plugging into the definition of $g$.

share|cite|improve this answer

edited 8 hours ago

answered 8 hours ago

WoolierThanThouWoolierThanThou

99611 silver badge77 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You're correct.
  $endgroup$
  – WoolierThanThou
  8 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

I'm assuming

$$
f(x)=begin{cases} 2 & xin [0,1] \ 0 & else end{cases}
$$

And

$$
g(x)=begin{cases} x & xin [0,1] \ 2-x & xin [1,2)\ 1 & xin [2,3] \0 & else end{cases}
$$

Adding this together, we see
$$
int_{-infty}^{infty} g(t)f(x-t)textrm{d}t=2int_{1-x}^x g(t)textrm{d}t,
$$
since for these values of $t,$ $f(x-t)=2$ and for all other values of $t$, $f(x-t)$ is $0$.

Now, $g(t)=0$ for $tleq 0$ so
$$
2int_{1-x}^x g(t)textrm{d}t=2int_0^x g(t)textrm{d}t=2int_0^x ttextrm{d}t,
$$
simply by plugging into the definition of $g$.

share|cite|improve this answer

edited 8 hours ago

answered 8 hours ago

WoolierThanThouWoolierThanThou

99611 silver badge77 bronze badges

$endgroup$

share|cite|improve this answer

edited 8 hours ago

answered 8 hours ago

WoolierThanThouWoolierThanThou

99611 silver badge77 bronze badges

answered 8 hours ago

WoolierThanThouWoolierThanThou

99611 silver badge77 bronze badges

answered 8 hours ago

WoolierThanThouWoolierThanThou

99611 silver badge77 bronze badges