can someone help me with this exercise? The simplest LTI processor which approximates a digital differentiator has the difference equation: y[n]=x[n]-x[n-1] A less-widely used alternative is to estimate the central difference, using the equation: y[n] : 0.5[x[n] - x[[n-2]] Sketch the magnitude responses of the two aPProximations on the same diagram, over tñe range 0 < omgea < pi. Contrast their performance with that of an ideãl differentiator. By how many dB is each resPonse lower than that of the ideal differentiator at the frequency omega = 0.2pi?
hey sir, at around 5:30-5:50. how did you get the ω/2 on the frequency response? is it because there's only 2 terms? 3 terms would be ω/3? etc..? thanks
factor out exp(-j*omega/2) from both terms; this gives: RHS = exp(-j*omega/2) {1/2*exp(j*omega/2) + 1/2*exp(-j*omega/2)} we can rearrange Euler's formula to find that: cos(theta) = 1/2 * (exp(j*theta)+exp(-j*theta)) and so the right hand side is: RHS = exp(-j*omega/2) {cos(omega/2)} as shown in the video :)
this was really helpful for my signals midterm, thank you
can someone help me with this exercise?
The simplest LTI processor which approximates a digital differentiator has the difference equation:
y[n]=x[n]-x[n-1]
A less-widely used alternative is to estimate the central difference, using the equation:
y[n] : 0.5[x[n] - x[[n-2]]
Sketch the magnitude responses of the two aPProximations on the same diagram, over tñe range 0 < omgea < pi. Contrast their performance with that of an ideãl differentiator. By how many dB is each resPonse lower than that of the ideal differentiator at the frequency omega = 0.2pi?
please question :
what is it Frequency Response x(t) = e^−|t| ???
hey sir, at around 5:30-5:50. how did you get the ω/2 on the frequency response? is it because there's only 2 terms? 3 terms would be ω/3? etc..? thanks
factor out exp(-j*omega/2) from both terms; this gives:
RHS = exp(-j*omega/2) {1/2*exp(j*omega/2) + 1/2*exp(-j*omega/2)}
we can rearrange Euler's formula to find that:
cos(theta) = 1/2 * (exp(j*theta)+exp(-j*theta))
and so the right hand side is:
RHS = exp(-j*omega/2) {cos(omega/2)}
as shown in the video :)
Really useful and clearly explained, thanks!
You make some great videos. Thank you.
Thank you sir.It was really helpfull.
Can i use fourier series to find the H?
perfect and precise.
Thank you
Thanks man, it is really useful!!
thank you
Thank you very much...
Thank You !
Really useful thank you sir ! :)