$1 н (e^{\omega} ) 1 < 0.01 \space \space 0.5 \pi. < I \omega ! < π$

Design a linear phase FIR filter to meet tinese specifications using the window design method

Mumbai University > EXTC > Sem 6 > Discrete Time Signal Processing

Marks : 08

Year : MAY 2015

Consider the following specification for a low pass filter

$0.99 ≤ |H(e^{jw} )| ≤ 1.01 \hspace{1.5cm} for 0 ≤ w \\ |H(e^{jw} )| ≤0.01 \hspace {1.5cm} for 0.35π ≤ w ≤ π$

Design a linear phase FIR filter to meet these specifications using the window design method.

Solution):

$(i) A_p=20 \log⁡(1.01)-20 \log⁡(0.99)=0.1737dB \\ (ii) A_s=20 \log⁡(1.01)-20 \log⁡(0.01)=40.086dB \\ (iii) W_p=0.3π \\ (iv) w_s=0.35π \\ (v) F_s=1Hz \\ (vi) LPF$

(1) select window function $As=53dB \gt 40.086dB$ (required value)

Select hamming window function

(2)Calculate N

$N ≥ \dfrac {C}{f2-f1}$

1) For hamming function $c=0.347$

2) $f2-f1=fs-fp$

$$w_p=0.3π =2πf_p \space\space\space f_p=0.15 $$

$w_s=0.35 π = 2 π f_p \space\space\space f_s=0.175 \\ N ≥ \dfrac {3.47}{0.175-0.15}\\ N ≥ 125.6 \\ Let\space\space\space N = 127(odd)$

(3) calculate $w_c \\ w_c=\dfrac {w_p+w_s}2=0.325π rad$

Step1 : Find Hd(w)

Let $Hd(w)=|Hd(w)| e^{jΦ} $

Where (i)Magnitude response:

$ H_d(w)= \{ 1 \text { for} –w_c ≤ |w| ≤ w_c \\ 0 \space\space\space\text{ otherwise} \}$

(ii)Phase response: $e^{jΦ}$

For Linear Phase LPF with symmetric )=h[n]

$Φ=-(\dfrac {N-1}2)w=-αw \\ e^{jΦ}=e^{-jαw} $

By substituting

$$H_d (w)=\{(e^{-jαw} \space\space\space for –w_c ≤ |w| ≤ w_c\\ 0 \space\space\space \text {otherwise}\} \\ \text { where} α=63 \text {and} w_c=0.325π$$ Step2: Find hd[n]

By inverse $DTFT,h_d [n]=\dfrac 1{2π} ∫\limits_{-π}^π H(w) e^{jπw} dw$

Derivation of $ h_d [n] $ for LPF

$h_d [n]= \dfrac {w_c}π \dfrac { \sin(n-α) w_c}{(n-α) w_c} \text { where} α=63 \text {and } w_c=0.325π $

Step 3: Find h[n]

Linear phase FIR filter with impulse response h[n] is given by

$h[n]=h_d [n] w[n]\\ h[n]=[0.325(\dfrac {\sin⁡(n-63)0.325π}{(n-83)0.325π})] [0.54-0.46 \cos\dfrac {2πn}{126} ] for 0≤n≤126 $