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iphone low pass filter

Asked 2010-12-22T01:11:28.257
13

I'm trying to implement a low-pass filter for an iphone app where I record a sound and then it gets played back slightly muffled; like the voice is coming from another room.

I've looked into the different options for audio recording and manipulation and have found it a bit confusing...digital signal processing isn't a strong point at all. I've mainly looked into OpenAL and inside the EFX library there is a filter that specifically does what I need, but EFX is not included on the iPhone. Is there a way of replicating that behaviour using OpenAL for the iPhone? Is there another option such as Audio Units that could provide a solution?

Thanks for your help

EDIT:

So after Tom's answer and links, I've come up with what I think is a correct implementation. However, I'm not getting a muffling effect at all, rather just a decrease in volume. Here's the (summarised) code I have currently:

File is recorded using AVAudioRecorder and the following settings:

[recordSetting setValue :[NSNumber numberWithInt:kAudioFormatLinearPCM] forKey:AVFormatIDKey];
[recordSetting setValue:[NSNumber numberWithFloat:44100] forKey:AVSampleRateKey];
[recordSetting setValue:[NSNumber numberWithInt: 1] forKey:AVNumberOfChannelsKey];

[recordSetting setValue :[NSNumber numberWithInt:16] forKey:AVLinearPCMBitDepthKey];
[recordSetting setValue :[NSNumber numberWithBool:NO] forKey:AVLinearPCMIsBigEndianKey];
[recordSetting setValue :[NSNumber numberWithBool:NO] forKey:AVLinearPCMIsFloatKey];

I then read in the file and transform it with the code below:

// Read in the file using AudioFileOpenURL
AudioFileID fileID = [self openAudioFile:filePath];

// find out how big the actual audio data is
UInt32 fileSize = [self audioFileSize:fileID];

// allocate the memory to hold the file
SInt16 * outData = (SInt16 *)malloc(fileSize); 

// Read in the file to outData
OSStatus result = noErr;
result = AudioFileReadBytes(fi
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1 Answer

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Tom's answer is the following recursive filter:

y[n] = (1 - a)*y[n-1] + a*x[n]

H(z) = Y(z)/X(z) = a / (1 - (1 - a)*1/z)

I'll plot this in Python/pylab for a=0.25, a=0.50, and a=0.75:

from pylab import *

def H(a, z):
    return a / (1 - (1 - a) / z)

w = r_[0:1000]*pi/1000
z = exp(1j*w)
H1 = H(0.25, z)
H2 = H(0.50, z)
H3 = H(0.75, z)
plot(w, abs(H1), 'r') # red
plot(w, abs(H2), 'g') # green
plot(w, abs(H3), 'b') # blue

alt text

Pi radians/sample is the Nyquist frequency, which is half the sampling frequency.

If this simple filter is inadequate, try a 2nd order Butterworth filter:

# 2nd order filter:
# y[n] = -a[1]*y[n-1] - a[2]*y[n-2] + b[0]*x[n] + b[1]*x[n-1] + b[2]*x[n-2]

import scipy.signal as signal

# 2nd order Butterworth filter coefficients b,a
# 3dB cutoff = 2000 Hz
fc = 2000.0/44100
b, a = signal.butter(2, 2*fc)

# b = [ 0.01681915,  0.0336383 ,  0.01681915]
# a = [ 1.        , -1.60109239,  0.66836899]

# approximately:
# y[n] = 1.60109*y[n-1] - 0.66837*y[n-2] + 
#        0.01682*x[n] + 0.03364*x[n-1] + 0.01682*x[n-2]

# transfer function
def H(b,a,z):
    num = b[0] + b[1]/z + b[2]/(z**2)
    den = a[0] + a[1]/z + a[2]/(z**2)
    return num/den

H4 = H(b, a, z)
plot(w, abs(H4))
# show the corner frequency
plot(2*pi*fc, sqrt(2)/2, 'ro')  
xlabel('radians')

alt text

Evaluate a test signal at the 3dB cutoff frequency fc=2000:

fc = 2000.0/44100
b, a = signal.butter(2, 2*fc)

# test signal at corner frequency (signed 16-bit)
N = int(5/fc)  # sample for 5 cycles
x = int16(32767 * cos(2*pi*fc*r_[0:N]))

# signed 16-bit output
yout = zeros(size(x), dtype=int16)

# temp floats
y  = 0.0    
y1 = 0.0
y2 = 0.0

# filter the input
for n in r_[0:N]:
    y = (-a[1] * y1 + 
         -a[2] * y2 + 
          b[0] *
answered 2011-01-16T01:06:39.270

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