Respuesta :
Complete Question:
Suppose the impulse response of an FIR filter of order m =5 is as follows where the X terms are to be determined.
h = [2,4,3,X,X,X]
a) Assuming H(z) is a linear- phase filter, find the complete impulse response. If there are multiple solutions, find each of them
b) For each solution in part (a), indicate the linear-phase FIR filter type
c) For each solution in part (a), find the phase offset, α, and the group delay, D(f)
Answer:
a1) For symmetric impulse response, h(n) = [2,4,3,3,4,2]
a2)For asymmetric impulse response h(n) = [2,4,3,-3,-4,-2]
b1) For symmetric impulse response,the FIR filter is a low pass filter
b2) For symmetric impulse response,, the FIR filter is a differentiator
c1) Phase offset = -2.5w
c2) D(f) = π/2 - 2.5w
Step-by-step explanation:
a) The impulse response, h can either be symmetric or anti-symmetric about α = m/2
i) When the impulse response, h is symmetric about α = m/2, h becomes:
h(n) = [2,4,3,3,4,2]
ii) When the impulse response, h is anti-symmetric about α = m/2, h becomes:
h(n) = [2,4,3,-3,-4,-2]
b) Relationship for converting h(n) into the frequency domain using Fourier transform: [tex]H(e^{jw} ) = \sum h(n) e^{-jwn}[/tex], n= 1,2,3,4,5
i) For h(n) = [2,4,3,3,4,2]
[tex]H(e^{jw} ) = h(0) + h(1) e^{-jwn} + h(2) e^{-2jw} + h(3) e^{-3jw} + h(4) e^{-4jw} + h(5) e^{-5jw}[/tex]
[tex]H(e^{jw} ) = 2 + 4 e^{-jw} + 3 e^{-2jw} + 3 e^{-3jw} + 4 e^{-4jw} + 2 e^{-5jw}[/tex]
When [tex]w=0[/tex], [tex]H(e^{j0}) = 18[/tex]
When w = π, [tex]H(e^{j\pi }) = 0[/tex]
This is the property of a low pass filter, hence the FIR filter is a low pass filter
ii) For h(n) = [2,4,3,-3,-4,-2]
[tex]H(e^{jw} ) = h(0) + h(1) e^{-jwn} + h(2) e^{-2jw} + h(3) e^{-3jw} + h(4) e^{-4jw} + h(5) e^{-5jw}[/tex]
[tex]H(e^{jw} ) = 2 + 4 e^{-jw} + 3 e^{-2jw} - 3 e^{-3jw} - 4 e^{-4jw} - 2 e^{-5jw}[/tex]
When [tex]w=0, H(e^{jo}) = 0[/tex]
When [tex]w=\pi , H(e^{j \pi}) = 0[/tex]
This is the property of a differentiator, hence the FIR filter is a differentiator.
c1) Phase offset = -wα
α = 5/2 = 2.5
Phase offset = -2.5w
c2) Group delay, D(f) = π/2 - αw
D(f) = π/2 - 2.5w
Answer:
Step-by-step explanation:
x n = e j0n – n
Consider an FIR filter when the input is a complex sinusoid of the form
x n = Ae^((j)) e^((jn)) – n
Where it could be that was obtained by sampling the complex sinusoid
xt Ae^(j) e^(j0t) and 0=0Ts
From the difference equation for an tap FIR filter,
y_n=∑_(k=0)^5▒〖bxn-k〗
H(e^((j)))= ∑_(k=0)^5▒〖b_k e^((jk)) 〗
