Regarding a Paper

3 Design task : Part 1

The code and output of the question is under: Code:

x=logspace(3.0910,7.7187,5) w1=x(1)/(2*pi) w2=x(2)/(2*pi) w3=x(3)/(2*pi) w4=x(4)/(2*pi) w5=x(5)/(2*pi)

Result :

Part 2

We have made the code for this question which is provide in file form. After that go step wise to get the output.

1. Create the five filter in matlab using the bpw function.

We have already create it in file it is provide in code form.

2. Create five gain factor and multiply each with impulse response.

Make a gain factor as:

{2,3,5,1,7,} and multiply them with impulse response as {2h,3h1,5h2,h3,7h4}

3. Sum them all together to get total impulse response.

Now add all the impulse response as a new variable. Y=2h+3h1+5h2+h3+7h4.

Part 3:

After completing the part 2 we have to read the audio file as provide us using the command

“audioread ” as:

[a,Fs]=audioread(‘undone_clip.wav’). it will read the audio and will show it as output . Take a single row form thr output and convolve it with the total response of part 2. X=a(:,1)

For convolution q=conv(x,y)

use the sound command and listen the sound.

Question 1

Filter !l (radians/second) !h (radians/second)

h1[n] 0 1000 h2[n] 0 18000 h3[n] 0 254000 h4[n] 0 3646000 h5[n] 0 52324000

Filter !l (radians/sample) !l (radians/sample)

h1[n] 0 196.2547 h2[n] 0 2.8167e+03 h3[n] 0 4.0427e+04 h4[n] 0 5.8022e+05 h5[n] 0 8.3276e+06

Question 3:

Impulse response, h3[n], using both the rectangular windows.

(part a)Impulse response, h3[n], using both the (Part b)Impulse response, h3[n], using both

( Question 2 )

Rectangular windows. The Blackman windows. While comparing both of the graph we can see that the result is same.

Part (1)a

Frequency response jH3(ej!)j for the middle band pass lter using the rectangular.

Part (1) b

Frequency response jH3(ej!)j for the middle band pass lter using the Blackman windows

( Question 4 )

Part (2)

N = 4096 points in the fft.

Figure 2

Question 5

Filter Gain (V/V) Gain (dB)

h[n] 20 59.9146

h1[n] 30 68.0239 h2[n] 40 73.7776 h3[n] 50 78.2405

h4[n] 60 81.8869

Question 6

the magnitude frequency response for each individual lter, i.e., jHk (e j!)j for k = 1; 2; : : : ; 5 (again use N = 4096 points in the fft). The graph show below show the different for all the five plot.

(Part 1) (Part 2)

(Part 3) (Part 4)

(Part 5) Figure 3

Question 7:

Now for the total system magnitude frequency response jH(ej!)j as under:

Question 8

Figure 4

The plot for the input signal of sound The plot for the output signal of sound

Question 9

frequency spectrum of the input audio signals

frequency spectrum of the output audio signals

Figure 6