You're very welcome, thanks for watching. Make sure to check out my website adampanagos.org for additional content (600+ videos) you might find helpful. Thanks, Adam.
Glad I could help, thanks for watching. Make sure to check out my website adampanagos.org for additional content (600+ videos) you might find helpful. Thanks, Adam
Glad I could help, thanks for watching. Make sure to check out my website adampanagos.org for additional content (395+ videos) you might find helpful. Thanks, Adam
Hi Adam! Thanks for the great content! A quick question, when we write down the phase response in terms of tan^-1, does the range limit of tan^-1 also applies (i.e. tan^-1 returns a value between -pi/2 to pi/2)? If that applies, phase response 0 and pi will be the same (since tan(0)=tan(pi) periodic with pi), which contradicts the intuition that they are Not the same. Could you elaborate on that a bit? Great thanks!!!!
Yes, that's correct. If you're trying to go from the Z-domain into the frequency domain, you just replace z with e^(jOmega). So, if you have z^(-2) we have (e^(jOmega))^-2 = e^-j2Omega like you said. hope that helps. Adam
I cannot thank you enough for your awesome and step by step explanation
You're very welcome, thanks for watching. Make sure to check out my website adampanagos.org for additional content (600+ videos) you might find helpful. Thanks, Adam.
Wow. I was surprised to learn that the magnitude of a complex fraction is the fraction of the magnitudes. That gem slipped by me long ago. Thanks!
Glad I could help, thanks for watching. Make sure to check out my website adampanagos.org for additional content (600+ videos) you might find helpful. Thanks, Adam
Thank you Adam. You are better than my professor :)
Glad I could help, thanks for watching!
@@AdamPanagos SO CLUTCH MAN. TIP OF THE HAT
Thank you sir! Very clear explanation and well constructed!
Glad I could help, thanks for watching. Make sure to check out my website adampanagos.org for additional content you might find helpful. Thanks, Adam
Thank you very much, very clear and HQ video!
Glad I could help, thanks for watching. Make sure to check out my website adampanagos.org for additional content (395+ videos) you might find helpful. Thanks, Adam
Hi Adam! Thanks for the great content! A quick question, when we write down the phase response in terms of tan^-1, does the range limit of tan^-1 also applies (i.e. tan^-1 returns a value between -pi/2 to pi/2)? If that applies, phase response 0 and pi will be the same (since tan(0)=tan(pi) periodic with pi), which contradicts the intuition that they are Not the same. Could you elaborate on that a bit? Great thanks!!!!
I have exam tomorrow and i blv my teacher copied your lecture, thanks sir
I hope your exam went well! Where are you studying?
@@AdamPanagos it was... Not bad. Wish I found the lecture earlier. I am studying in BD
what would happen if you have a z^-2 or higher?
if im right it's just e^-j2omega
Yes, that's correct. If you're trying to go from the Z-domain into the frequency domain, you just replace z with e^(jOmega). So, if you have z^(-2) we have (e^(jOmega))^-2 = e^-j2Omega like you said. hope that helps.
Adam
@@AdamPanagos thank you