• We’re currently investigating an issue related to the forum theme and styling that is impacting page layout and visual formatting. The problem has been identified, and we are actively working on a resolution. There is no impact to user data or functionality, this is strictly a front-end display issue. We’ll post an update once the fix has been deployed. Thanks for your patience while we get this sorted.

EE's: Convolution Problem

J

jmcoreymv

Diamond Member
I just had this problem on a quiz (a week ago actually). I had to convolute the ramp function ( t u(t) ) with a step function ( u(t) - u(t-4) ) and evaluate the answer at t = 2. I get different answers when I do it graphically versus analytically. Any ideas?
 
BTW, the teachers answer was 2, he did it graphically and it makes sense to me. But I should be able to get the same answer analytically.
 
Originally posted by: jaydee
What class is this? I'm a junior and have no idea what you're talking about.
Click to expand...

Woah. I would think you'd know what convolution is. You first learn it in basic DSP classes and network analysis classes.
 
Laplace to the rescue!

While there's no way I'm going to do convolution just to telp you 🙂evil: ), my prognosis is that you're doing something wrong. If you would stop screwing up on either the analytical solution, the grpahical one, or both, you'd surely find they evaluate to the same answer.

😉
 
Originally posted by: NutBucket
Originally posted by: jaydee
What class is this? I'm a junior and have no idea what you're talking about.
Click to expand...

Woah. I would think you'd know what convolution is. You first learn it in basic DSP classes and network analysis classes.
Click to expand...

Ahh, must be Sig & Sys then (taking it in the winter). The last EE core class I've yet to take.
 
Originally posted by: jaydee
Originally posted by: NutBucket
Originally posted by: jaydee
What class is this? I'm a junior and have no idea what you're talking about.
Click to expand...

Woah. I would think you'd know what convolution is. You first learn it in basic DSP classes and network analysis classes.
Click to expand...

Ahh, must be Sig & Sys then (taking it in the winter). The last EE core class I've yet to take.
Click to expand...

Haven't you done RLC filters and what not? How about Fourier? You should have learned convolution along with those things.
 
Well... at least write down the equations you used for both methods so we can point out what you did wrong. I'm not in the mood to do the convolution myself.
 
Nope, haven't touched those yet. Taken Circuits, Digitals, Micros, Electronics, Fields and all the math/science. Actually, haven't taken Circuits II either, that will be winter as well.
 
Nope, haven't touched those yet. Taken Circuits, Digitals, Micros, Electronics, Fields and all the math/science. Actually, haven't taken Circuits II either, that will be winter as well.
 
Originally posted by: jaydee
Nope, haven't touched those yet. Taken Circuits, Digitals, Micros, Electronics, Fields and all the math/science. Actually, haven't taken Circuits II either, that will be winter as well.
Click to expand...

Jeez. You guys move slow eh? I think I did convolution 2nd year.
 
Originally posted by: jmcoreymv
I just had this problem on a quiz (a week ago actually). I had to convolute the ramp function ( t u(t) ) with a step function ( u(t) - u(t-4) ) and evaluate the answer at t = 2. I get different answers when I do it graphically versus analytically. Any ideas?
Click to expand...

Do you get either the graphing or analytical correctly?
 
Originally posted by: NutBucket
Originally posted by: jaydee
Nope, haven't touched those yet. Taken Circuits, Digitals, Micros, Electronics, Fields and all the math/science. Actually, haven't taken Circuits II either, that will be winter as well.
Click to expand...

Jeez. You guys move slow eh? I think I did convolution 2nd year.
Click to expand...

Well Sig & Sys has Diff Eq's for a pre-req, which has Calc III as a pre-req here, so unless you come in with Calc, you can't take it until your 3rd year.
 
Originally posted by: Skiguy411
Originally posted by: TuxDave
Originally posted by: Skiguy411
Doh, is this stuff hard? Im planning on getting a degree in EE and NCSU.
Click to expand...

🙂 No more sleep for you!
Click to expand...


Sigh, so its pretty difficult eh? What about computer engineering?
Click to expand...

Hahaha... so what are you looking for a major. Are you looking for an easy major to graduate with? Or do you want something challenging that'll make you valuable.
 
Originally posted by: JetBlack69
Originally posted by: jmcoreymv
I just had this problem on a quiz (a week ago actually). I had to convolute the ramp function ( t u(t) ) with a step function ( u(t) - u(t-4) ) and evaluate the answer at t = 2. I get different answers when I do it graphically versus analytically. Any ideas?
Click to expand...

Do you get either the graphing or analytical correctly?
Click to expand...

Well the teacher did it graphically and his answer was 2.

When I do it analytically I did it like this (i think)

x(t-L) = u(t-L) - u(t-L-4)
h(L) = L u(L)

so the integral is
integral( L u(L) u(t-L) - L u(L) u(t-L-4) dL)

Then I separated that integral and chose my limits.

integral from 0 to t( L dL) - integral from 0 to t-4( L dL)

so thats .5t^2 - .5(t-4)^2

Evaluate that at t=2 and I get 0.
 
funny, i just had a test on convolution / fourier series last friday... i was well prepared but the directions really got me confused... finally figured out what i was supposed to do with less than a minute left.... grrr trying to forget about it and move on...
 
Originally posted by: jmcoreymv
Originally posted by: JetBlack69
Originally posted by: jmcoreymv
I just had this problem on a quiz (a week ago actually). I had to convolute the ramp function ( t u(t) ) with a step function ( u(t) - u(t-4) ) and evaluate the answer at t = 2. I get different answers when I do it graphically versus analytically. Any ideas?
Click to expand...

Do you get either the graphing or analytical correctly?
Click to expand...

Well the teacher did it graphically and his answer was 2.

When I do it analytically I did it like this (i think)

x(t-L) = u(t-L) - u(t-L-4)
h(L) = L u(L)

so the integral is
integral( L u(L) u(t-L) - L u(L) u(t-L-4) dL)

Then I separated that integral and chose my limits.

integral from 0 to t( L dL) - integral from 0 to t-4( L dL)

so thats .5t^2 - .5(t-4)^2

Evaluate that at t=2 and I get 0.
Click to expand...


Well... the only flaw that I see is that if you integrate from 0 to -2, that second function is always zero.
 
You must log in or register to reply here.

TRENDING THREADS

Back
Top