CHAPTER 5. ANALYSIS AND DESIGN
5-8
is the unit-sample response of a bandpass filter with center frequency u0 . From
property (iii) in Table 3.1,
HBP (u) = HLP (u − u0 ).
(5.14)
Note that in general, hBP [x] as defined above will be complex, except for special
frequencies on the boundary of a unit cell of Λ∗ , where 2u0 ∈ Λ∗ (and so, −u0 and
u0 represent the same discrete-domain frequency on the lattice Λ).
A real bandpass filter will have a frequency response that is symmetric about
zero, with center frequencies ±u0 . Such a filter can be obtained by
hBP [x] = 2hLP [x] cos(2πu0 · x)
= hLP [x](exp(j2πu0 · x) + exp(−j2πu0 · x))
HBP (u) = HLP (u − u0 ) + HLP (u + u0 ).
(5.15)
(5.16)
Note that in both cases discussed above, the frequency response retains its periodicity with respect to the reciprocal lattice.
A bandstop filter can be obtained from a bandpass filter by simply subtracting the unit-sample response of the bandpass filter from that of an all-pass filter
hAP [x] = δ[x]. Thus
hBS [x] = δ[x] − hBP [x]
HBS (u) = 1 − HBP (u).
(5.17)
(5.18)
Example 5.2. Design a real Gaussian bandstop filter with center frequencies ±u0
and 3 dB bandwidth uc = 15 c/ph for a 512 × 512 image on a square sampling
lattice. Use it to remove the specific frequency u0 = (150, −86) c/ph from the image
‘barbara’.
Solution. From the specification, r = 0.1325/uc = 0.00883 ph. Using the approximation c = d(λ)/2πr2 = 0.0078, the desired unit-sample response is
(x2 + y 2 )
hBS [x] = δ[x] − 2c cos(2π(u0 x + v0 y)) exp −
2r2
−(x2 + y 2 )
= δ[x] − .0156 cos(2π(150x − 86y)) exp
0.0001378