Discussion :: Signals and Systems - Section 1 (Q.No.4)
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|Nabil said: (Nov 15, 2014)|
|I didn't understand the logic. Kindly provide me the detailed solution with proper reasoning.|
|Susee said: (Feb 10, 2015)|
|I cannot understood. Could you please explain me once?|
|Brc said: (Mar 6, 2015)|
|Can I take z=s and s=jw?|
|Neha said: (Jan 4, 2016)|
|If you consider x(z) = (z-1)(z+1)/z2.
Then if z = ejw = cosw + jsinw.
Then x(ejw) = (e2jw - 1)/e2jw.
Now magnitude is X(w) = ((Cos 2w-1)2 + (sin 2w)2) 1/2.
As magnitude = (Real 2+image 2) 1/2.
So the plot is given for its magnitude similarly when you change the poles there won't be change in magnitude, hence filter remains the same.
Hope you will understand this.
|Bibo said: (Apr 21, 2016)|
|For digital domain, in low frequency w = 0 hence z = exp(j0) = 1.
High-frequency w = pi, z = -1 now put z = 1 and -1 and find response, you will get zero every time.
Hence, both are Band Pass filter.
|Ghanshyam said: (Sep 17, 2016)|
|How does it take to pole at the origin?|
|Kishore said: (Feb 5, 2018)|
|In denominator how did z^2 came, poles should come in denominator and zeroes should come in the numerator. Explain me clearly.|
|K.Ramesh said: (Mar 29, 2018)|
|What is the LPF?|
|Rachana said: (Feb 12, 2019)|
|Thank you all for providing best explanation.|
|Lavanya said: (Mar 15, 2019)|
|I am not getting this, can anyone explain me in clear?|
|Geetha said: (Jun 16, 2019)|
|I didn't understand, Please anyone, explain me.|
|Fatima Noor said: (Jan 22, 2020)|
|Zeros are same but poles are changed. Then How we get the same response?|
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