I Getting zeros and poles for Laplace transform

bnich
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Getting zero's and poles for Laplace transform of damped cosine
I'm following the intuition behind getting the zero's and poles of a damped cosine function with this video


At around 11:50, he shows some graphics pertaining to multiplying the probing function with the impulse response, but the graphics don't seem correct.

For example, in the B+B' graphic, the probing function exactly equals the impulse response, but should the probing function be the inverse of the impulse response in order for the sum to be "just barely infinite" as depicted in the screenshot
laplace.png

Image source: http://www.dspguide.com/CH32.PDF

And for the D graphic, the complex part should be zero, but shouldn't the real part of the exponential be the same as where the poles are? Image below describes what I think it should be
Screenshot 2024-02-02 at 3.55.40 PM.png

image source: https://www.dummies.com/article/tec...s-understanding-poles-and-zeros-of-fs-166275/

Thanks for your time. I look forward to hearing some feedback!
 
I have the equation ##F^x=m\frac {d}{dt}(\gamma v^x)##, where ##\gamma## is the Lorentz factor, and ##x## is a superscript, not an exponent. In my textbook the solution is given as ##\frac {F^x}{m}t=\frac {v^x}{\sqrt {1-v^{x^2}/c^2}}##. What bothers me is, when I separate the variables I get ##\frac {F^x}{m}dt=d(\gamma v^x)##. Can I simply consider ##d(\gamma v^x)## the variable of integration without any further considerations? Can I simply make the substitution ##\gamma v^x = u## and then...

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