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!
Can i use fourier series to find the H?
You make some great videos. Thank you.
Thank you sir.It was really helpfull.
perfect and precise.
Thank you
Thanks man, it is really useful!!
thank you
Thank You !
Thank you very much...
Really useful thank you sir ! :)