Name of Window |
transition Width (Hz) Normalized |
Passband Ripple |
Stopband Attenuation (dB) |
Window Function |
---|---|---|---|---|

Rectangular or Boxcar |
.9/N | .7416 | 21 | 1 |

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)) |

Kaizer | 2.93/N (beta=4.54) 4.32/N (beta=6.76) 5.71/N (beat=8.96) |
.0274 .00275 .000275 |
50 70 90 |
I(beta*(1-[(2*n)/(N-1)]^2)^.5) / (I*beta) |

**Advantages / Disadvantages of FIR filtering**

**Advantages**- Linear Phase Response
- Stability

**Disadvantages**- More memory requirements (large number of coefficients for sharp cutoffs)
- processing speed concerns (due to large number of coefficients)

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: rwb@scs.howard.edu.