Question

Design a 5-tap FIR bandpass filter with a lower cutoff frequency of
1,600 Hz, an upper cutoff frequency of 1,800 Hz, and a sampling rate
of 8,000 Hz using

a. rectangular window function

b. Hamming window function.

Determine the transfer function and difference equation of the designed
FIR system, and compute and plot the magnitude frequency response
for Ω= 0, π/4, π/2, 3π/4, and π radians.

PLEASE SHOW STEPS CLEARLY

Answer 1

* a) Denýn of FIR filters using Retangular window :- The desired frequency response of a fitter is given by , Haleiw) {! for We < W< we for wealwisho - step 1 - Inverse discrete time frequency time transform for Ad (sw) ✓ Ho (etw) ellu ↑ Haleiw) և انا- we We ha(n)z I ar jon do -Wc - 2 we son le j dw. an -we jo e 1 jun z I Coso tj lino so se e dw QA اليأس Sino- e -e i un 25 [com n WL jwen -j weh 27 m jwen diten juch е. AR 23 swch -j wch - е. [ . an 21

ha(n)= sin (1) Cochem.-0 in wp-?! Kunrating to N sampler hin) = 4. min () [N = 5] Given Tin Rectangle window, -240 12 for imi 2 [wc = x/2 a thing on ] for n=0; hlo) Sit sino =1] 2010 o L-Hospital rule . FIR is symmetrical because B (w) = {Haleiw) = 0. [w) = 6 (N-|-n) {from Ø Mz») h (1) - h(-1) 0.3183. M= 2 =) h (2)=h(-2) stup -3: Transfer function H(2), H(7) = hlo) + sin 1/2 0. (N-1)/2 nal "+2") ne! hlm) (2" + 2) [N=5 =h(0) + 2/2 him) (2" + 2 + (0.3163)(*"++) +0 +6013183)(7+7") 2 TH(2) et

- (N-1)/2 Relizable filter, multiply by Z N-5 - 2 > Z M -2 -1 H(Z) = 0.52² » Ž (5-1)/2 #la). *[+ +(0.3181)(2+2"] +0.3183 zi 7.2² +0,3183 z ² 20.57 +0,31837' +0,318373. 14+)7 0.91013 0.2187 2'70.58 +0-31812-17 Altiplayering Incense 2 - Juer forms, h(1) = h(3) = 0.3183 n (2) - Frequency response ; From eq ④; Put z = Hiew) = 0.5 + 0.3183 Leute tew] 0.5 m + z 0.5 * 0:3183 x2 (اع پائی) 0.5 + 0.6366 cosao. Euler's formula -50 Iloso enote = 0.5 + 016366 Cosy Hacedo)

* Plot the Response: يا : -2 نما indey (0) * = 15° | R2 = 40" an/ 1013E ۔ (مائع) + 4 N=1&2 = iss" | 0،049 |-0.34 0.9 گ,0 in dB ال ۰۱۱ يك 6.4 9. 2- ي و6.02 - (0) (پہلے) 4 2 ۱ + + Wlindeg) 90 13° (8 -2 - ۔ 1

*b). Design of FIR filters wings Hemming Window :- The desired frequency response of a low pass filter is given by, Hd (ew) - Halw). ; 3774 ; #3: ha(n) = sin / (3 (n-3)] * (n-3) of n=3. ha(3)= j(3-3) 27 stly 1 ہے - جمال -انے stly lidw at dw 2 ves A(n-3) na 3. here the +993 - 1914): iha (n) = sin ( (n-3)) ; n&3. -D 314 The impulse response of h En) = ha(n). Wham (n) : Osn<4-③ (N=5]. Witam (n) = 0,54 -0.46 come around my ). Osnsm-l. FIR filter is (0) och SN- Osney W sam (n) = 0.54 -0.46 cos ( amin Hence, the impulse resporue of the FI R. sin {z (n-3))*(0.59-0046 cos (2 h(n) x foisy-- Do yo los ( anam); n = 3 filter for Osnshis 25h ht T(n-3) coefficiento M haln) 0.075 Whamin's hin) = hain). Wham (n) 0:08 0.006 0 11-0:159/0.54 Quyla -0.048586 2 0.225 0.08 0.018 مر 0.405 0.75 0.54 4 0.225 0.08 0.018 n=0; halo) rin 13 (x-1)) min (321-3)) *(n-3) Waamino) 0.54 – 0,46 Cos ( en ). (-3) = 0.075 z 0.54-0,46 Cos(o) = 0.0%, h(n)z halu). Witam (n) 0.006(1) 6x10; NE!), holt) - sink (1-3)) -0.159 *(1-3) 0154 - 0.46 Cos ()= 0.54. 1=2, hold) = sin (17 (2-3)) z o.aas +(2-3) 0.54 – 0.46600 (RL)) 2 0.080 Wham (1) Wham (2) 1 .n=3; ha (3) 3/4 0.75 Ww cm (3) 7 0.54 – 9.46 cos [ 20 (3) 0,54, z h (B) = hd (3). Wham (3) -0.75X0.54 201405 n=u; hd (u) = sin ( 2 (4-3)) 0.225. T(4-3) Wham (4), 0154-0146 Cos Can(u)) 0.084 h14) holu) x Wilano (4) 2 = 0.018 0.225 x 0.08 + Zahin). since , N= 5 = Odd, frequency response of the lentire symmetur PIM( (N=2) Confo{nad 03) Na5; Hlw) h=0 Hw). esaw [H(a) + 2H16) Con aw +2661.0 w] estaw [Co.018) +2 (0.006) 6 w +2.6-0.0858) a -j2u Cosw] 0.018 +0.012 Cos2w of 0.1716 losw] H(WE e w usº 180° Hley -0.02149 dB Ido H10) 1354 -0.0214 2159621621152x103 3.77x104 1-71.7108 -53-343 -68,450B. [0.018 +0.012 – 0.1916) (2009(Hle))-d8] -0.1416 0.018 +0 -011213 13] H() 218)[ z 2 2:02149 H() e 0.018 + (-0.012) - 0] 2.596x264 +(3^), 2 () 10.018 to +0.1213] - 3112 {0, 1395] = 2.152x10 H(A) (0.018 .018 +0.012 +0:1716] 3.7767 x 10 3 -27 e -4 2 08- of- 09- as- 180' 135' १० Sh (M (or) + AP 0 0 Add a comment

; #3: ha(n) = sin / (3 (n-3)] * (n-3) of n=3. ha(3)= j(3-3) 27 stly 1 ہے - جمال -انے stly lidw at dw 2 ves A(n-3) na 3. here the +993 - 1914): iha (n) = sin ( (n-3)) ; n&3. -D 314 The impulse response of h En) = ha(n). Wham (n) : Osn<4-③ (N=5]. Witam (n) = 0,54 -0.46 come around my ). Osnsm-l. FIR filter is (0) och SN- Osney W sam (n) = 0.54 -0.46 cos ( amin Hence, the impulse resporue of the FI R. sin {z (n-3))*(0.59-0046 cos (2 h(n) x foisy-- Do yo los ( anam); n = 3 filter for Osnshis 25h ht T(n-3)

coefficiento M haln) 0.075 Whamin's hin) = hain). Wham (n) 0:08 0.006 0 11-0:159/0.54 Quyla -0.048586 2 0.225 0.08 0.018 مر 0.405 0.75 0.54 4 0.225 0.08 0.018 n=0; halo) rin 13 (x-1)) min (321-3)) *(n-3) Waamino) 0.54 – 0,46 Cos ( en ). (-3) = 0.075 z 0.54-0,46 Cos(o) = 0.0%, h(n)z halu). Witam (n) 0.006(1) 6x10; NE!), holt) - sink (1-3)) -0.159 *(1-3) 0154 - 0.46 Cos ()= 0.54. 1=2, hold) = sin (17 (2-3)) z o.aas +(2-3) 0.54 – 0.46600 (RL)) 2 0.080 Wham (1) Wham (2) 1

.n=3; ha (3) 3/4 0.75 Ww cm (3) 7 0.54 – 9.46 cos [ 20 (3) 0,54, z h (B) = hd (3). Wham (3) -0.75X0.54 201405 n=u; hd (u) = sin ( 2 (4-3)) 0.225. T(4-3) Wham (4), 0154-0146 Cos Can(u)) 0.084 h14) holu) x Wilano (4) 2 = 0.018 0.225 x 0.08 + Zahin). since , N= 5 = Odd, frequency response of the lentire symmetur PIM( (N=2) Confo{nad 03) Na5; Hlw) h=0 Hw). esaw [H(a) + 2H16) Con aw +2661.0 w] estaw [Co.018) +2 (0.006) 6 w +2.6-0.0858) a -j2u Cosw] 0.018 +0.012 Cos2w of 0.1716 losw] H(WE e w usº 180° Hley -0.02149 dB Ido H10) 1354 -0.0214 2159621621152x103 3.77x104 1-71.7108 -53-343 -68,450B. [0.018 +0.012 – 0.1916) (2009(Hle))-d8] -0.1416 0.018 +0 -011213 13] H() 218)[ z 2 2:02149 H() e 0.018 + (-0.012) - 0] 2.596x264 +(3^), 2 () 10.018 to +0.1213] - 3112 {0, 1395] = 2.152x10 H(A) (0.018 .018 +0.012 +0:1716] 3.7767 x 10 3 -27 e -4 2 08- of- 09- as- 180' 135' १० Sh (M (or) + AP

esaw [H(a) + 2H16) Con aw +2661.0 w] estaw [Co.018) +2 (0.006) 6 w +2.6-0.0858) a -j2u Cosw] 0.018 +0.012 Cos2w of 0.1716 losw] H(WE e w usº 180° Hley -0.02149 dB Ido H10) 1354 -0.0214 2159621621152x103 3.77x104 1-71.7108 -53-343 -68,450B. [0.018 +0.012 – 0.1916) (2009(Hle))-d8] -0.1416 0.018 +0 -011213 13] H() 218)[ z 2 2:02149 H() e 0.018 + (-0.012) - 0] 2.596x264 +(3^), 2 () 10.018 to +0.1213] - 3112 {0, 1395] = 2.152x10 H(A) (0.018 .018 +0.012 +0:1716] 3.7767 x 10 3 -27 e -4 2

08- of- 09- as- 180' 135' १० Sh (M (or) + AP
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