What is the result of convolving a triangle signal with a delta function signal?

Click For Summary
Convolving a triangle signal with a delta function results in a constant value of 1 due to the properties of convolution and the nature of delta functions. The convolution operation in one domain corresponds to multiplication in the frequency domain, leading to a transformation involving sinc functions. The triangle signal, when convolved with an impulse train, primarily contributes at the origin, resulting in a single impulse. This process effectively integrates the triangle signal across the delta functions, yielding a constant output. Understanding this relationship clarifies why the convolution yields a value of 1.
stanigator
Messages
3
Reaction score
0
For the question stated in the pictures attached to this message, for part d of the question, I just can't picture why the convolution of the two signals is 1. Can someone please explain how this would be the case as the triangular signal is sweeping through all the delta functions? Thanks.
 

Attachments

  • CIMG0218.jpg
    CIMG0218.jpg
    28.4 KB · Views: 843
  • CIMG0219.jpg
    CIMG0219.jpg
    25.7 KB · Views: 1,220
Engineering news on Phys.org
stanigator said:
For the question stated in the pictures attached to this message, for part d of the question, I just can't picture why the convolution of the two signals is 1. Can someone please explain how this would be the case as the triangular signal is sweeping through all the delta functions? Thanks.

Without looking at the problem, I will take a wild guess. Convolution is one domain is multiplication in the other. The transform of a delta-train is another delta-train with different spacing. I don't know what the convolution of a triangle is, but I'm guessing it's something close to a sinc function, because a triangle can be thought of as the integral of two box functions which each transform to sinc functions. So now you multiply an impulse-train with something sinc-esque. I'm guessing the zeros of the sinc-esque function will hit the impulse-train everywhere except at the origin, so the result will be a single impulse at 0. Then transform that back and you get a constant.
 
I am trying to understand how transferring electric from the powerplant to my house is more effective using high voltage. The suggested explanation that the current is equal to the power supply divided by the voltage, and hence higher voltage leads to lower current and as a result to a lower power loss on the conductives is very confusing me. I know that the current is determined by the voltage and the resistance, and not by a power capability - which defines a limit to the allowable...

Similar threads

Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
4K
Replies
8
Views
2K
Replies
26
Views
5K
  • · Replies 15 ·
Replies
15
Views
4K
  • · Replies 1 ·
Replies
1
Views
1K
Replies
1
Views
2K
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K