音频去噪 Enhancing Speech by removing noise

L ABORATORY 5 Enhancing Speech by removing noisePurposeThe main purpose of this lab is to design different IIR and FIR filters in the context of enhancing a peech corrupted by an additive noise. By doing so, this will help us to better understand the properties of different types of filters, the meanings of the design parameters, and some of the trade‐offs between filter classes.In order to see the behavior of these different types of filters, we were given a noisy speech signal and were asked to design a low-pass filter in order to remove the high-pass noise. After doing this, we had to pass the audio file through each of these filters in order to see the if the speech was filtered correctly and what is the effect of each filter on the filtered signal. The noisy signal was sampled at a frequency Fs = 44100 Hz and the filter specifications are:Part I : Matlab RipplePart II: Butterworth filterIn this part, we had to design a Butterworth filter that satisfies the given specifications.We used Matlab fdatool to design this filter. In order to use this Matlab built-in tool, we had to convert the above given specifications into IIR filter specifications. Thus, we took:Apass = Gpbmax – Gpbmin = 40 – 37 = 3dBAstop = Gpbmax – Gsbmax = 40 -(- 55) = 95dBAs for Fpass, Fstop and Fs, we put 2500 Hz, 4000Hz and 44100 Hz respectively.After designing this filter, we found by using the Matlab ‘’order’’ function that the the order of this filter is 23.The next step was to determine the number of multiplication operations per input sample required to implement the filter. In order to do this, we used the Ma tlab ‘’cost’’ function and found that the number of multiplication operations per input sample required to implement the filter is 46.We also had to plot the magnitude response of this filter in dB and in linear scale along with the group delay. Figure 1 shows the corresponding results.Figure 1 – Magnitude and group delay of the Butterworth filterAfter plotting the magnitude and group delay of this filter, we had to plot the pole-zero diagram. Figure 2 shows the corresponding result.Figure 2 – Pole-zero plot of the Butterworth filterAnother important aspect of the filter that we have to consider is the impulse response. Using the Matlab ‘’filter’’ and ‘’stem’’ functions, we plotted the impulse response if this filter. Figure 3 shows the corresponding result.Figure 3 – Impulse response of the Butterworth filterFinally, we filtered the noisy signal with this filter and noticed that the noise was almost completely removed and that we can listen clearly to the speech without any noise. However, the filtered signal was not strong enough and needed to be amplified en order for the sound to be loud enough. So we amplified the filtered signal by a factor of 100 and noticed that the speech was loud enough and very clear.Part III: Chebyshev Type I filterIn this part, we had to design a Chebyshev Type I filter that satisfies the given specifications. We used Matlab fdatool to design this filter. In order to use this Matlab built-in tool, we had to convert the above given specifications into IIR filter specifications. Thus, we took the same parameters as we used with the Butterworth filter:Apass = Gpbmax – Gpbmin = 40 – 37 = 3dBAstop = Gpbmax – Gsbmax = 40 -(- 55) = 95dBAs for Fpass, Fstop and Fs, we put 2500 Hz, 4000Hz and 44100 Hz respectively.After designing this filter, we found by using the Matlab ‘’order’’ function that the the order of this filter is 11.The next step was to determine the number of multiplication operations per input sample required to implement the filter. In order to do this, we used the Matlab ‘’cost’’ function and found that the number of multiplication operations per input sample required to implement the filter is 22.We also had to plot the magnitude response of this filter in dB and in linear scale along with the group delay. Figure 4 shows the corresponding results.Figure 4 – Magnitude and group delay of the Chebyshev Type I filterAfter plotting the magnitude and group delay of this filter, we had to plot the pole-zero diagram. Figure 5 shows the corresponding result.Figure 5 – Pole-zero plot of the Chebyshev Type I filterAnother important aspect of the filter that we have to consider is the impulse response. Using the Matlab ‘’filter’’ and ‘’stem’’ functions, we plotted the impulse response if this filter. Figure 6 shows the corresponding result.Figure 6 – Impulse response of the Chebyshev Type I filterFinally, we filtered the noisy signal with this filter and noticed that the noise was almost completely removed and that we can listen clearly to the speech without any noise. However, the filtered signal was not strong enough and needed to be amplified en order for the sound to be loud enough. So we amplified the filtered signal by a factor of 100 and noticed that the speech was loud enough and very clear.Part IV: Chebyshev Type II filterIn this part, we had to design a Chebyshev Type II filter that satisfies the given specifications. We used Matlab fdatool to design this filter. In order to use this Matlab built-in tool, we had to convert the above given specifications into IIR filter specifications. Thus, we took the same parameters as we used with the Butterworth filter:Apass = Gpbmax – Gpbmin = 40 – 37 = 3dBAstop = Gpbmax – Gsbmax = 40 -(- 55) = 95dBAs for Fpass, Fstop and Fs, we put 2500 Hz, 4000Hz and 44100 Hz respectively.After designing this filter, we found by using the Matlab ‘’order’’ function that the the order of this filter is 11.The next step was to determine the number of multiplication operations per input sample required to implement the filter. In order to do this, we used the Matlab ‘’cost’’ function and found that the number of multiplication operations per input sample required to implement the filter is 22.We also had to plot the magnitude response of this filter in dB and in linear scale along with the group delay. Figure 7 shows the corresponding results.Figure 7– Magnitude and group delay of the Chebyshev Type II filterAfter plotting the magnitude and group delay of this filter, we had to plot the pole-zero diagram. Figure 8 shows the corresponding result.Figure 8 – Pole-zero plot of the Chebyshev Type II filterAnother important aspect of the filter that we have to consider is the impulse response. Using the Matlab ‘’filter’’ and ‘’stem’’ functions, we plotted the impulse response if this filter. Figure 9 shows the corresponding result.Figure 9 – Impulse response of the Chebyshev Type II filterFinally, we filtered the noisy signal with this filter and noticed that the noise was almost completely removed and that we can listen clearly to the speech without any noise. However, the filtered signal was not strong enough and needed to be amplified en order for the sound to be loud enough. So we amplified the filtered signal by a factor of 100 and noticed that the speech was loud enough and very clear.Part V: Elliptic filterIn this part, we had to design an Elliptic filter that satisfies the given specifications. We used Matlab fdatool to design this filter. In order to use this Matlab built-in tool, we had to convert the above given specifications into IIR filter specifications. Thus, we took the same parameters as we used with the Butterworth filter:Apass = Gpbmax – Gpbmin = 40 – 37 = 3dBAstop = Gpbmax – Gsbmax = 40 -(- 55) = 95dBAs for Fpass, Fstop and Fs, we put 2500 Hz, 4000Hz and 44100 Hz respectively.After designing this filter, we found by using the Matlab ‘’order’’ function that the the order of this filter is 8.The next step was to determine the number of multiplication operations per input sample required to implement the filter. In order to do this, we used the Matlab ‘’cost’’ function and found that the number of multiplication operations per input sample required to implement the filter is 16.We also had to plot the magnitude response of this filter in dB and in linear scale along with the group delay. Figure 10 shows the corresponding results.Figure 10– Magnitude and group delay of the Elliptic filterAfter plotting the magnitude and group delay of this filter, we had to plot the pole-zero diagram. Figure 11 shows the corresponding result.Figure 11 – Pole-zero plot of the Elliptic filterAnother important aspect of the filter that we have to consider is the impulse response. Using the Matlab ‘’filter’’ and ‘’stem’’ functions, we plotted the impulse response if this filter. Figure 12 shows the corresponding result.Figure 12 – Impulse response of the Elliptic filterFinally, we filtered the noisy signal with this filter and noticed that the noise was almost completely removed and that we can listen clearly to the speech without any noise. However, the filtered signal was not strong enough and needed to be amplified en order for the sound to be loud enough. So we amplified the filtered signal by a factor of 100 and noticed that the speech was loud enough and very clear.ConclusionThe purpose of this experiment was to verify a conjecture that states that the ear is insensitive to phase. In order to do this, we had to implement all-pass filters that would keep the same magnitude of the frequency response of the audio signal but will introduce a phase shift.By doing this, we had the opportunity to better understand some concepts relating to all pass filters. We also saw the behaviour of these filters when we have an impulse input. Also, we plotted the magnitude, the group delay and the pole-zero plot of the all-pass filters in order tostudy the magnitude of the frequency response and the phase introduced by an all pass filter. We also had the opportunity to see the relation that exists between the poles and zeros in an all pass filters and noticed that the poles and zeros are at conjugate reciprocal locations. Finally, we concluded that the ear is insensitive to phase shift. However, for very big phase shifts, the ear can detect some distortions in the sound.AppndiceMain MATLAB codeload projIA.mat%Part a)%Impulse responsefigure(1)impz(b,a,100)%Magnitudefigure(2)[H,W]=freqz(b,a)plot(W,20*log10(abs(H)))xlabel('Frequency in rad/s');ylabel('Magnitude in DB');title('Magnitude Response of the all-pass filter with N=1')%Group Delayfigure(3)grpdelay(b,a)title('Group delay of the all-pass filter with N=1')%Part b)%Pole-zero plotfigure(4)zplane(b,a)title('Pole-zero plot of the all-pass filter with N=1')%Part c)%Listening to soundy = filter(b,a,speech)sound(y,fs)%Part d)%Implementing the filter with N=50n = [0:5000];h2 = [n == 0]% First, we define h2 as an impulse input signalfor (n=1:50)h2 = filter(b,a,h2); %We filter the impulse input with the all pass filter with N=1 50 timesend%Impulse responsefigure (5)stem(h2)title('Impulse response');xlabel('n samples');ylabel('Amplitude')%Magnitudefigure(6)[H,W]=freqz(b,a)plot(W,(20*log10((abs(H)).^50)))xlabel('Frequency in rad/s');ylabel('Magnitude in DB');title('Magnitude Response of the all-pass filter with N=50')%Group Delayfigure(7)plot(W/pi,50.*grpdelay(b,a))xlabel('Normalized frequency (x pi rad/sample)');ylabel('Group delay (samples)');title('Group Delay of the all-pass filter with N=50')grid%Listening to soundy2 = filter(b,a,speech)for(n=1:49)y2=filter(b,a,y2);endsound(y2,fs)。