Author Topic: TT1 = Problem 5  (Read 9903 times)

Victor Ivrii

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TT1 = Problem 5
« on: October 16, 2012, 06:30:11 PM »
The functions $H$ and $Q$ are defined as follows:
\begin{equation*}
H(x) = \left\{\begin{aligned}
&1, \qquad&&x>0,\\
&0, \qquad&&x \leq 0,
\end{aligned}\right.
\end{equation*}   
\begin{equation*}
Q(x) = \left\{\begin{aligned}
&1, \qquad&&|x| \leq 1,\\
&0, \qquad&&|x| > 1.
\end{aligned}\right.
\end{equation*}
Consider the function
\begin{equation*}
C(x) = \int_{-\infty}^{+\infty} H(x-y) Q(y) dy.
\end{equation*}
  • (a) Graph the function $C(x)$.
  • (b) Identify the set of points $x$ where $C(x) >0.$

Ian Kivlichan

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Re: TT1 = Problem 5
« Reply #1 on: October 16, 2012, 06:50:29 PM »
Hopeful solution attached!

Aida Razi

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Re: TT1 = Problem 5
« Reply #2 on: October 16, 2012, 07:22:05 PM »
Solution is attached,

Victor Ivrii

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Re: TT1 = Problem 5
« Reply #3 on: October 16, 2012, 07:29:37 PM »
We have two rather different answers and the person who posted after should explain what's wrong with the previous posts. This is a general remark without endorsement of any particular solution.

Ian Kivlichan

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Re: TT1 = Problem 5
« Reply #4 on: October 16, 2012, 07:39:21 PM »
Solution is attached,

Aida, I think your solution is not correct - the integral from -1 to 0 will only be 0 if x < y in H(x-y), i.e. x < -1.

Also, consider extreme cases - for C(-100), H(-100-y) can never be greater than 0, so a solution with C(x) = 1 cannot be right (unless I've totally misunderstood something).

Djirar

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Re: TT1 = Problem 5
« Reply #5 on: October 16, 2012, 08:07:48 PM »
I solved the question in the same way as Aida Razi did. Of course for part b, C(x) > 0  for all x.
« Last Edit: October 16, 2012, 08:16:30 PM by Djirar »

Aida Razi

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Re: TT1 = Problem 5
« Reply #6 on: October 16, 2012, 08:10:14 PM »
I solve the question in the same way as Aida Razi did. Of course for part b, C(x) > 0  for all x.

Yes, as C(x) is a constant function and it is equal to one, then x is lR: it can be any real number.

Aida Razi

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Re: TT1 = Problem 5
« Reply #7 on: October 16, 2012, 08:16:11 PM »
Solution is attached,

Aida, I think your solution is not correct - the integral from -1 to 0 will only be 0 if x < y in H(x-y), i.e. x < -1.

Also, consider extreme cases - for C(-100), H(-100-y) can never be greater than 0, so a solution with C(x) = 1 cannot be right (unless I've totally misunderstood something).

I believe your solution is not correct; H(x) and H(x-y) has the same value for these two different domain because it is a constant function.

Qitan Cui

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Re: TT1 = Problem 5
« Reply #8 on: October 16, 2012, 08:23:50 PM »
I agree with Ian's solution. The value of H(x-y) depends on both x and y. So you have to break the scenarios regarding to x when doing the integral.

Ian Kivlichan

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Re: TT1 = Problem 5
« Reply #9 on: October 16, 2012, 08:25:25 PM »
Solution is attached,

Aida, I think your solution is not correct - the integral from -1 to 0 will only be 0 if x < y in H(x-y), i.e. x < -1.

Also, consider extreme cases - for C(-100), H(-100-y) can never be greater than 0, so a solution with C(x) = 1 cannot be right (unless I've totally misunderstood something).

I believe your solution is not correct; H(x) and H(x-y) has the same value for these two different domain because it is a constant function.

Shouldn't they be different though? For H(x), it's 1 for x>0 and 0 for x<=0, but H(x-y) is 1 for x-y>0, or x>y, and 0 for x<=y.

In any case, how can C(-100) be nonzero? x-y=-100-y>0 is impossible for -1<=y<=1, which is the only area where Q(y) is nonzero.

Jinchao Lin

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Re: TT1 = Problem 5
« Reply #10 on: October 16, 2012, 09:20:20 PM »
We are only asked to draw the graph but not the functional form, don't we?

Victor Ivrii

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Re: TT1 = Problem 5
« Reply #11 on: October 16, 2012, 10:04:23 PM »
Ian is correct. BTW,

Operation $f*g$, $(f*g)(x)=\int_{-\infty}^\infty f(x-y)g(y)\,dy$ is called a convolution (we implicitly met it and will discuss when study Fourier transform),

$H(x)$ is called Heaviside function and $(H*f)(x)=\int_{-\infty}^x f(y)\,dy$. Therefore the answer is such integral of $Q(x)$ and
\begin{equation*}
(Q*H)(x)=\left\{\begin{aligned}
& 0 \qquad&& x<-1,\\
&x+1 && -1<x<1,\\
&2 && x\ge 1.
\end{aligned}\right.
\end{equation*}

Aida Razi

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Re: TT1 = Problem 5
« Reply #12 on: October 16, 2012, 10:08:35 PM »
Solution is attached,

Aida, I think your solution is not correct - the integral from -1 to 0 will only be 0 if x < y in H(x-y), i.e. x < -1.

Also, consider extreme cases - for C(-100), H(-100-y) can never be greater than 0, so a solution with C(x) = 1 cannot be right (unless I've totally misunderstood something).

I believe your solution is not correct; H(x) and H(x-y) has the same value for these two different domain because it is a constant function.

Shouldn't they be different though? For H(x), it's 1 for x>0 and 0 for x<=0, but H(x-y) is 1 for x-y>0, or x>y, and 0 for x<=y.

In any case, how can C(-100) be nonzero? x-y=-100-y>0 is impossible for -1<=y<=1, which is the only area where Q(y) is nonzero.

Sorry Ian, You are right   ???