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Convolutional integral of square signal and triangle signal.

ttokoro
21-Topaz I

Convolutional integral of square signal and triangle signal.

Convolutional integral of square signal and triangle signal is well knowen probrem.
How to find the maximum value of the output signal? Prime's symbolic engine sometimes fail for convolution integral.

7/6 is the answer I want to get. Prime 11 sheet attached.

image.png

ACCEPTED SOLUTION

Accepted Solutions
Werner_E
25-Diamond I
(To:Werner_E)

Luc's approach sure has advantages. Not only the much faster numerical evaluation but also significantly better precision as this picture shows:

Werner_E_0-1749161793025.png

Decreasing the value of TOL significantly makes it better, but still ...

Werner_E_1-1749161902975.png

 

 

View solution in original post

5 REPLIES 5
Werner_E
25-Diamond I
(To:ttokoro)

Why would you need a symbolical solution ??  We know that Mathcad's and Prime's symbolic engines are not top notch and state of the art. Wouldn't a numerical result suffice for practical needs?

Werner_E_1-1749095580431.png

 

 

LucMeekes
23-Emerald IV
(To:ttokoro)

Mathcad 11 (not Prime 11) is lazy:

LucMeekes_0-1749145766148.png

Try the laplace route:

LucMeekes_2-1749145890366.png

Yep, at t=5/2=2.5 that's where the summed length of x(t) and h(t) ends, so y(t) is 0 and y'(t) is 0.

Success!
Luc

 

Werner_E
25-Diamond I
(To:LucMeekes)

Here is your approach done with MC15.

While MC11 seems to find 5/2 twice (exact and numerically, a bit strange IMHO), MC15 finds the solution 0.

None of the two finds the solution 7/6 😞

Werner_E_0-1749150902558.png

 

Werner_E
25-Diamond I
(To:Werner_E)

Luc's approach sure has advantages. Not only the much faster numerical evaluation but also significantly better precision as this picture shows:

Werner_E_0-1749161793025.png

Decreasing the value of TOL significantly makes it better, but still ...

Werner_E_1-1749161902975.png

 

 

ttokoro
21-Topaz I
(To:ttokoro)

image.png

 

With using range tau=0.0.01..3 plot showes no error.

image.png

But without using range tau=0.0.01..3 plot shows the errors for symbolic engine's results. (Red and dotted lines are errors.)

So all errors comes from how to treat the unit step functions, Φ(t). If we use derivative, it shows impulse function Δ(t).

And derivative of convolution integral shows bug.

But we want to use Φ(t) such as if function to use the function for only the separated period.

This convolution error or bug is not fixed for long time.

 

image.png

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