Consider the H(z) equation at 3:37 Now we can write the denominator as a0+sum from k=1 to N ak z^{-k}. Now, divide numerator and denominator by a0, we would get H(z)=[sum k=0 to M (bk/a0) z^{-k}]/[1+sum k=1 to N ak z^{-k}]. Now put bk/a0=b'k, so, we have the new modified expression as H(z)=[sum k=0 to N b'k z^{-k}]/[1+sum k=1 to N ak z^{-k}], which is the same format that you are expecting. Hence, both the ways of representing IIR filters is fine. The one discussed in the video is a more general form.
At 16:15 you have told that duration will be from -inf to +inf but u(n) range will be from 0to inf and u(n-1) range will be from 1to inf hence h(n) will be only from 0to inf right ????
Thank you mam....
Really you helped a lot....
What an effort.. Thanks a lot
Very informative 👍🏻
Nice mam
Plz upload full lectures
Will be uploaded soon...
In the IIR filter h(z) denominator term includes +1 right.? why you didn't add that.? and k is also starts from 1 to N
Consider the H(z) equation at 3:37
Now we can write the denominator as a0+sum from k=1 to N ak z^{-k}. Now, divide numerator and denominator by a0, we would get H(z)=[sum k=0 to M (bk/a0) z^{-k}]/[1+sum k=1 to N ak z^{-k}]. Now put bk/a0=b'k, so, we have the new modified expression as H(z)=[sum k=0 to N b'k z^{-k}]/[1+sum k=1 to N ak z^{-k}], which is the same format that you are expecting. Hence, both the ways of representing IIR filters is fine. The one discussed in the video is a more general form.
@@shivanidhok3831 ohoo okeoke....... thank you for your response and the explanation
At 16:15 you have told that duration will be from -inf to +inf but u(n) range will be from 0to inf and u(n-1) range will be from 1to inf hence h(n) will be only from 0to inf right ????
Yes...In general, the h(n) is defined for all values of n, I.e. from -inf to inf, but due to u(n) it will have non-zero values from 0 to inf.
Can you tell me the editor which is used by you to teach in this online mode
OpenBoard for screen recording.
Mam plz