The window method of low-pass FIR filter design can be used as the first step in designing a bandpass FIR filter. Let's say we want a 31-tap FIR filter with the frequency response shown in Figure 5-22(a), but instead of being centered about zero Hz, we want the filter's passband to be centered about fs/4 Hz. If we define a low-pass FIR filter's coefficients as hlp(k), our problem is to find the hbp(k) coefficients of a bandpass FIR filter. As shown in Figure 5-28, we can shift Hlp(m)'s frequency response by multiplying the filter's hlp(k) low-pass coefficients by a sinusoid of fs/4 Hz. That sinusoid is represented by the sshift(k) sequence in Figure 5-28(a), whose values are a sinewave sampled at a rate of four samples per cycle. Our final 31-tap hbp(k) FIR bandpass filter coefficients are


Figure 5-28. Bandpass filter with frequency response centered at fs/4: (a) generating 31-tap filter coefficients hbp(k); (b) frequency magnitude response |Hbp(m)|.

whose frequency magnitude response |Hbp(m)| is shown as the solid curves in Figure 5-28(b). The actual magnitude of |Hbp(m)| is half that of the original |Hlp(m)| because half the values in hbp(k) are zero when sshift(k) corresponds exactly to fs/4. This effect has an important practical implication. It means that, when we design an N-tap bandpass FIR filter centered at a frequency of fs/4 Hz, we only need to perform approximately N/2 multiplications for each filter output sample. (There's no reason to multiply an input sample value, x(n–k), by zero before we sum all the products from Eq. (5-6) and Figure 5-13, right? We just don't bother to perform the unnecessary multiplications at all.) Of course, when the bandpass FIR filter's center frequency is other than fs/4, we're forced to perform the full number of N multiplications for each FIR filter output sample.

Notice, here, that the hlp(k) low-pass coefficients in Figure 5-28(a) have not been multiplied by any window function. In practice, we'd use an hlp(k) that has been windowed prior to implementing Eq. (5-20) to reduce the passband ripple. If we wanted to center the bandpass filter's response at some frequency other than fs/4, we merely need to modify sshift(k) to represent sampled values of a sinusoid whose frequency is equal to the desired bandpass center frequency. That new sshift(k) sequence would then be used in Eq. (5-20) to get the new hbp(k).


Prev don't be afraid of buying books Next

Chapter One. Discrete Sequences and Systems

Chapter Two. Periodic Sampling

Chapter Three. The Discrete Fourier Transform

Chapter Four. The Fast Fourier Transform

Chapter Five. Finite Impulse Response Filters

Chapter Six. Infinite Impulse Response Filters

Chapter Seven. Specialized Lowpass FIR Filters

Chapter Eight. Quadrature Signals

Chapter Nine. The Discrete Hilbert Transform

Chapter Ten. Sample Rate Conversion

Chapter Eleven. Signal Averaging

Chapter Twelve. Digital Data Formats and Their Effects

Chapter Thirteen. Digital Signal Processing Tricks

Appendix A. The Arithmetic of Complex Numbers

Appendix B. Closed Form of a Geometric Series

Appendix C. Time Reversal and the DFT

Appendix D. Mean, Variance, and Standard Deviation

Appendix E. Decibels (dB and dBm)

Appendix F. Digital Filter Terminology

Appendix G. Frequency Sampling Filter Derivations

Appendix H. Frequency Sampling Filter Design Tables

Understanding Digital Signal Processing
Understanding Digital Signal Processing (2nd Edition)
ISBN: 0131089897
EAN: 2147483647
Year: 2004
Pages: 183 © 2008-2020.
If you may any questions please contact us: