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
Answer question

Know the answer?

Add Answer to:

Design a 5-tap FIR bandpass filter

Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Log In

Sign Up

Post as anonymous

Post Your Answer

Not the answer you're looking for? Ask your own homework help question. Our experts will answer your question WITHIN MINUTES for Free.

Similar Homework Help Questions

• 7.29. Design a 41-tap bandpass FIR filter with lower and upper cutoff frequencies of 2,500 Hz...

  7.29. Design a 41-tap bandpass FIR filter with lower and upper cutoff frequencies of 2,500 Hz and 3,000 Hz, respectively, using the following window functions. Assume a sampling frequency of 8,000 Hz. a. Hanning window function b. Blackman window function. List the FIR filter coefficients and plot the frequency responses for each design. 7.30 Design a 41-tap band reject FIR filter with cutoff frequencies of 2,500 Hz and 3,000 Hz, respectively, using the Hamming window function. Assume a sampling frequency...

• 7.3. Design a 5-tap FIR lowpass filter with a cutoff frequency of 100 Hz and a...

  7.3. Design a 5-tap FIR lowpass filter with a cutoff frequency of 100 Hz and a sampling rate of 1,000 Hz using a 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 ?--0, ?/4, ?/2, 3r/4, and ? radians.

• Styles Paragraph 6. Given the difference equation y(n)-x(n-1)-0.75y(n-1)-0.125(n-2) a. Use MATLAB function filterl) and filticl) to...

  Styles Paragraph 6. Given the difference equation y(n)-x(n-1)-0.75y(n-1)-0.125(n-2) a. Use MATLAB function filterl) and filticl) to calculate the system response y(n)for n 0, 1, 2, 3, 4 with the input of x(n (0.5) u(n)and initial conditions x(-1)--1, y(-2) -2, and y(-1)-1 b. Use MATLAB function filter!) to calculate the system response y(n) for n-0, 1, 2, 3,4 with the input of x(n) (0.5)"u(n)and zero initial conditions x(-1)-0, (-2)-0, and y(-1)-0 Design a 5-tap FIR low pass filter with a cutoff...

• In this problem, you are asked to design a length-16 FIR low-pass filter with cutoff frequency...

  In this problem, you are asked to design a length-16 FIR low-pass filter with cutoff frequency ωc = π 2 radians, using the window design method. 2. [FIR Filter Design) In this problem, you are asked to design a length-16 FIR low-pass filter with cutoff frequency We = radians, using the window design method. (a) Find an expression for the coefficients {hn}n using a truncation (rectangular) window. (b) Find an expression for the coefficients {n}=l using a Hamming window. (c)...

• 3. Design a bandpass FIR filter using Kaiser's formula for filter order, using Hamming window with...

  3. Design a bandpass FIR filter using Kaiser's formula for filter order, using Hamming window with the following specifications: the lower passband and stopband edge frequencies are fpi- 700 Hz, fs1 - 300 Hz, the upper passband and stopband edge frequencies fp2 - 2 kHz fs2 - 2400 Hz, the sampling frequency fs-10 kHz, and 6p-0.03, ando0.004.

• Design a 31-tap highpass FIR filter whose cutoff frequency is 2,500 Hz using the following window...

  Design a 31-tap highpass FIR filter whose cutoff frequency is 2,500 Hz using the following window functions. Assume that the sampling frequency is 8,000Hz. a. Hanning window function c. Blackman window function

• 1. Design a 10th-order lowpass FIR filter using the window method (fir1) to cut frequencies above...

  1. Design a 10th-order lowpass FIR filter using the window method (fir1) to cut frequencies above 30Hz in an application where the sampling frequency is 125 Hz. 2. Plot the filter coefficients that define the filter (stem). 3. Plot the frequency response of the FIR filter designed (freqz) 4. Design a 100th-order lowpass FIR filter using the window method (fir1) to cut frequencies above 30Hz in an application where the sampling frequency is 125 Hz. Plot the filter coefficients that...

• matlab code as well please. 7. (100) Design a bandpass FIR filter with the following Spec:...

  matlab code as well please. 7. (100) Design a bandpass FIR filter with the following Spec: (a) Lower cut off frequency: 1250Hz, (b) lower transition width: 1500Hz, (c) upper cutoff frequency: 2850 Hz, (d) upper transition width: 1300 Hz, (e) stop band attenuation: 60dB, (f) passband ripple 0.02 dB, and (g) sampling frequency: 8000Hz. Your answer needs to include (i) normalized frequencies, (ii) Window type, (iii) order of the filter and their numerical values computed by matlab command firwd(), and...

• please need correct answer. I will upvote. Design a second-order digital bandpass Butterworth filter with a...

  please need correct answer. I will upvote. Design a second-order digital bandpass Butterworth filter with a lower cutoff frequency of 1.9 kHz, an upper cutoff frequency 2.1 kHz, and a passband ripple of 3dB at a sampling frequency of 8,000 Hz. a. Determine the transfer function and difference equation. b. Use MATLAB to plot the magnitude and phase frequency respon

• b) When designing a FIR filters, the impulse response of the ideal low-pass filter is usually modified by multiplying i...

  b) When designing a FIR filters, the impulse response of the ideal low-pass filter is usually modified by multiplying it by a windowing function such as the Hamming window which is defined, for an odd number N of samples, by: (2n)-(N-I)-ns(N-1) N-12 wlnl 0.54 + 0.46 cos i What are the advantages of windowing with this function compared 2 with a standard rectangular window? ii) Design a 10th Order Hamming windowed FIR low-pass filter with cut- off frequency at 1000...