written 6.5 years ago by | • modified 5.5 years ago |
H(ejω)=ej3ω;0≤|ω|≤π2 H(ejω)=0;otherwise
written 6.5 years ago by | • modified 5.5 years ago |
H(ejω)=ej3ω;0≤|ω|≤π2 H(ejω)=0;otherwise
written 6.5 years ago by | • modified 6.5 years ago |
We know that
H(k)=H(ω)(ω=2πkN) ………(1)
H(ejω)=e−j3ω
Here ∝=3
And∝=(N−1)2
∴N=7
Now, By using equation (1)
H(k)=e(−j6πk)7;0≤2πk7≤π2
=0;π2≤2πk7≤π
∴H(k)=e^{\frac{(-j6πk)}{7}} ;0≤k≤\frac{7π}{4π}
=0 ; \frac{7π}{4π}≤k≤\frac{7π}{2π}
∴H(k)=e^{\frac{(-j6πk)}{7}} ;0≤k≤2
=0 ;2≤k≤4
Now , Here K is vary from 0 to 2
When k=0
∴H(0)=1
When k=1
∴H(1)=e^{\frac{(-j6π)}{7}}
When K=2
∴H(2)=e^{\frac{(-j12π)}{7}}
According to transfer function equation of Frequency Sampling Realization:
H(z)=[\frac{(1-z^{-7})}{7}][\frac{H(0)}{(1-z^{-1}}+\frac{H(1)}{(1-e^{\frac{j2π}{7}} z^{-1}} +(\frac{H(2))}{(1-e^{\frac{j6π}{7}} z^{-1} )}]