Question Solved1 Answer Task 1 1. An analog signal contains frequencies upto 10 kHz. This signal is sampled at 50 kHz. Design an FIR filter using windowing technique for a transition band of 5 kHz. The filter should provide minimum –21 dB attenuation at the end of transition band. Task 1 1. An analog signal contains frequencies upto 10 kHz. This signal is sampled at 50 kHz. Design an FIR filter using windowing technique for a transition band of 5 kHz. The filter should provide minimum –21 dB attenuation at the end of transition band.

ABHBUG The Asker · Advanced Mathematics

Transcribed Image Text: Task 1 1. An analog signal contains frequencies upto 10 kHz. This signal is sampled at 50 kHz. Design an FIR filter using windowing technique for a transition band of 5 kHz. The filter should provide minimum –21 dB attenuation at the end of transition band.
More
Transcribed Image Text: Task 1 1. An analog signal contains frequencies upto 10 kHz. This signal is sampled at 50 kHz. Design an FIR filter using windowing technique for a transition band of 5 kHz. The filter should provide minimum –21 dB attenuation at the end of transition band.
See Answer
Add Answer +20 Points
Community Answer
ERJ7O4 The First Answerer
See all the answers with 1 Unlock
Get 4 Free Unlocks by registration

Answer:- 3dB cut-off at 30 pirad//secOmega_(p)=2pi xx10 xx10^(3)rad//sec Omega_(s)=2pi xx(10+5)xx10^(3)rad//secSampling frequency F_(SF)=100HzStopband attenuation of 50dB at 45 pirad//secA_(s)=50dB for omega_(s)=45 pirad//secomega=(Omega)/(F_(sf))omega_(p)=(Omega_(p))/(F_(sf))=(2pi xx10 xx10^(3))/(50 xx10^(3))=0.4 piomega_(s)=(Omega_(s))/(F_(sf))=(2pi xx(10+5)xx10^(3))/(50 xx10^(3))=0.6 piFigure 26: Frequency response of LPFomega_(p)=0.4 pirad//sampleomega_(s)=0.6 pirad//sampleType of window isThe stopband attenuation of 50dBquadH_(d)(omega)={[e^(-j omega tau)" for "-omega_(c) <= omega <= omega_(c)],[0quad" ... See the full answer