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Fourier series Method
1. Frequency response of a discrete-time filter is a periodi function with period Ωs (sampling freq).
2. From the F.S analysis we know that any periodic function can be expressed as a linear combination of complex exponentials.
Therefore desired freqency response of a discrete time filter can be represented by F.S as
T = sampling period
The F.S co-efficient or impulse response samples of filter can be obtained using
clearly if we wish to realize this filter with impulse response h(n), then it must have finite no. of co-efficient, which is equivalent to truncating the infinite expansion of H (e jΩT ) ,which leads to approximation of (e jΩT ) which is denoted by
We choose M=N-1/2, in order to keep ‘N’ no of samples in h(n).
However, this filter can’t be physically realizable due to the presence of +ve powers of Z, means that the filter must produce an output that is advanced in time with respect to the i/p. this difficulty can be overcome by introducing a delay M=N-1/2, samples.
Therefore H(z) = Z-M H1(z) = Z-M
H(z) = h(-M)Z0 + h(-M+1) Z-1 +…. +h(M) Z-2M
Let bi = h(i-M) i=0 to 2M
H(z) = be the transfer function of discrete filter that is physically realizable.
1. N=2M+1, impulse response co-eff, bi = 0 to 2M.
2. h(n) is symmetric about bM
3. The duration of impulse response is Ti = 2MT
4. Its magnitude and time delay function can be found in the following way
This implies that magnitude response of the filter we have desired approximates the desire magnitude response. The time delay of H(ejw) is a constant M. thus sinusoids of different frequencies are delayed by the same amount as they are processed by the filter, we have designed. Consequently, this is a linear phase filter, which means that it does not introduce phase distortion.
Design a LPF (FIR) filter with frequency response
Design LPF that approximate following freq response.
H(F) = 1 0≤ F ≤ 1000Hz
= 0 else where 1000 ≤ F≤ Fs/2
When the sampling frequency is 8000 SPS. The impulse response duration is to be
limited to 2.5ms
h(0) = 0.25 h(6) = -0.05305
h(1) = 0.22508 h(7) = -0.03215
h(2) = 0.15915 h(8) = 0
h(3) = 0.07503 h(9) = 0.02501
h(4) = 0 h(10) = 0.03183
h(5) = -0.04502
bi = h(i-10)
Desing a BPF for H(f) = 1 160≤ F ≤ 200Hz
= 0 else where
Fs = 800SPS
Ti = 20 ms
N = 17
h(0) = 0.1 h(4) = 0.07568
h(1) = 0.01558 h(5) = 0.06366
h(2) = -0.09355 h(6) = -0.05046
h(3) = -0.04374 h(7) = -0.07220 h(8) = 0.02338
bi = h(i-8) h(-n) = h(n)