testing header
Math Goodies is a free math help portal for students, teachers, and parents.
Free Math
Newsletter
 
 
Interactive Math Goodies Software

Buy Math Goodies Software
testing left nav
Math Forums @ Math Goodies
Math Forums @ Math Goodies
Home | Profile | Register | Active Topics | Members | Search | FAQ
 All Forums
 Homework Help Forums
 Pre-Calculus and Calculus
 CONVOLUTION! need help desperately will pay money!

Note: You must be registered in order to post a reply.
To register, click here. Registration is FREE!

Screensize:
UserName:
Password:
Format Mode:
Format: BoldItalicizedUnderlineStrikethrough Align LeftCenteredAlign Right Horizontal Rule Insert HyperlinkInsert EmailInsert Image Insert CodeInsert QuoteInsert List Insert Special Characters Insert Smilie
   
Message:
* HTML is OFF
* Forum Code is ON

Math Symbols
Squared [squared] Cubed [cubed] Square Root [sqrt] Cube Root [cbrt] Pi [pi]
Alpha [alpha] Beta [beta] Gamma [gamma] Theta [theta] Angle [angle]
Degrees [degrees] Times [times] Divide [divide] Less Than or Equal To [less-than] Greater Than or Equal To [greater-than]
Plus Minus [plus-minus] Integral [integral] Sum [sum] Sub 1 [sub-1] Sub 2 [sub-2]
Element Of [element-of] Union [union] Intersect [intersect] Subset [subset] Empty Set [empty-set]

  Check here to subscribe to this topic.
 
   

T O P I C    R E V I E W
nirvanaguy Posted - 01/29/2011 : 22:06:04
Hi guys,
I really need to understand this as soon as possible so I don't get screwed and left behind in the lectures..

I have2 problems:

z(t)=[e^(-t)*u(t)] * [e^(-t)*u(t)]

I got the answer 1 but for some reason I don't think that's right.. I know that when you analyze it graphically z=0 for t<0 and for t>0 i had my bounds go from 0 to infinity of e^-(t)dt


2nd) [t*e^(-t)*u(t)]*[2*u(t+.5)-2*u(t-.5)]

For this one for my boundaries for when 0<t<1 i used 0 to t
and for t>1 I used t-1 to t as my boundaries since thats the width of the unit pulse. however, i'm know thats not right because i graphed them and the area comes out all funky. I think it's wrong because it has somtehing to do with the .5 shifts on both sides instead of just the shift of 1 to the right..

If someone could help out that would be amazing and I will honestly pay you if you'd like
1   L A T E S T    R E P L I E S    (Newest First)
Ultraglide Posted - 02/11/2011 : 17:27:22
I am sure you will find all kinds of material on the net. I did.

Math Forums @ Math Goodies © 2000-2004 Snitz Communications Go To Top Of Page
This page was generated in 0 seconds. Snitz Forums 2000
testing footer
About Us | Contact Us | Advertise with Us | Facebook | Blog | Recommend This Page




Copyright © 1998-2014 Mrs. Glosser's Math Goodies. All Rights Reserved.

A Hotchalk/Glam Partner Site - Last Modified 21 May 2014