Lowpass Frequency Response.
Highpass Frequency Response.
Bandpass Frequency Response.
Bandstop Frequency Response.
|Name of Window||transition Width (Hz) Normalized||Passband Ripple||Stopband Attenuation (dB)||Window Function|
|Hanning||3.1/N||.0546||44||.5 + .5 cos((2*pi*n)/N)|
|Hamming||3.3/N||.0194||53||.54 + .46 cos((2*pi*n)/N)|
|Blackman||5.5/N||.0017||74||.42 + .5 cos((2*pi*n)/(N-1)) + .08cos((4*pi*n)/(N-1))|
|I(beta*(1-[(2*n)/(N-1)]^2)^.5) / (I*beta)|
Advantages / Disadvantages of FIR filtering
Take this second order filter
H(z) = .345(1 + 2z^-1 +z^-2) / (1 + .5z^-1 + 2z^-2)
changing H(z) = Y(z)/X(z) and multiplying the denominator by Y(z) and numerator by X(z) we get:
Y(z)(1 + .5z^-1 + 2z^-2) = X(z)(.345 + .69z^-1 + .345z^-2)
(Y(z) + .5Y(z)z^-1 + 2Y(z)z^-2) = (.345X(z) + .69X(z)z^-1 + .345X(z)z^-2)
taking the inverse Z transform, we get:
Y(n) + .5Y(n-1) + 2Y(n-2) = .345X(n) + .69X(n-1) + .345X(n-2)
solving for Y(n) we get out difference equation:
Y(n) = .345X(n) + .69X(n-1) + .345X(n-2) - .5Y(n-1) - 2Y(n-2)
Now, this difference equation can be implemented. Also note that X(n-1) and X(n-2) are previous samples which mean that at the start of the filter they must be defined. These are called initial conditions. Usually you can set them to 0 since in most signals there isn't any information prior to time 0 seconds.
Assignment: Plot the Z plane, frequency, and phase response of the above second order filter and email me back and let me know if its stable and tell me its cutoff frequency. Matlab is a great tool for this. I would recommend using some of the scripts I have on this site. Assume a sampling frequency of 48 Khz and the initial conditions are 0.
Also, if it is unstable, try using this: H(z) = .32(1 + 2z^-1 + 1) / (1 + .5z^-1 -2z^-2) and tell me what the cutoff frequency is. Hint: If you are using my scripts the frequency response is normalized so you must denormalize the x-axis to half the sampling frequency to obtain the cutoff frequency. My email is: email@example.com.