What about, for the second example, we have the bounded input Sin(t)? The input is bounded from -1 to 1, and the output (the integral of the function from 0 to t) is bounded from -2 to 2 so we could choose a = 3 for example. Or, are we making an assumption that the signal is being rectified?
Very good vids. Congrats and tons of gratitude here, I think that the system at 6:34 computes the algebraic area under x(τ), not the actual area. The parts below the y-axis are being subtracted from the positive ones.
Excellent lecture, concept-wise. However, you seem to be having problems with audio quality = wavering and crackly. Either that or you have the world's biggest frog in your throat, in which case get well soon.
Thanks for the explanation!
What about, for the second example, we have the bounded input Sin(t)? The input is bounded from -1 to 1, and the output (the integral of the function from 0 to t) is bounded from -2 to 2 so we could choose a = 3 for example. Or, are we making an assumption that the signal is being rectified?
Very good vids. Congrats and tons of gratitude
here, I think that the system at 6:34 computes the algebraic area under x(τ), not the actual area. The parts below the y-axis are being subtracted from the positive ones.
Excellent lecture, concept-wise. However, you seem to be having problems with audio quality = wavering and crackly. Either that or you have the world's biggest frog in your throat, in which case get well soon.
How do you plug in u(t) in the integral ?
why is exp[at] unbounded signal? It IS bounded whenever t is bounded, right?
It is BIBO stable only when both the input and output are bounded for ALL t. So you can't consider when t is bounded.
is x(tau) = x(t)? meaning x(tau) the input?
could u teach me cross and auto correlation??? post something on it
that is more like digital signal processing, not control systems.
So nice sir
Thank you sir!
Thank you, very clear
Not clear :(
maşaallah,
Allah sana hayırlar versin
Good work matey
feeling sleepy....