The window method of lowpass FIR filter design can be used as the first step in designing a bandpass FIR filter. Let's say we want a 31tap FIR filter with the frequency response shown in Figure 522(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 lowpass 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 528, we can shift Hlp(m)'s frequency response by multiplying the filter's hlp(k) lowpass coefficients by a sinusoid of fs/4 Hz. That sinusoid is represented by the sshift(k) sequence in Figure 528(a), whose values are a sinewave sampled at a rate of four samples per cycle. Our final 31tap hbp(k) FIR bandpass filter coefficients are
Figure 528. Bandpass filter with frequency response centered at fs/4: (a) generating 31tap filter coefficients hbp(k); (b) frequency magnitude response Hbp(m).
whose frequency magnitude response Hbp(m) is shown as the solid curves in Figure 528(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 Ntap 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. (56) and Figure 513, 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) lowpass coefficients in Figure 528(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. (520) 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. (520) to get the new hbp(k).
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