Chapter 7: DFT Filter Bank Solutions
à Problem 7.1
Problem
An FIR Filter has Transfer function H z 2 z1 z2 0.5z3 and frequency response H . a) Determine the Impulse Response of the filter F z with frequency response F H 0.2 . Is the impulse response going to be real? b) Determine the impulse response of the filter G z with frequency response H 0.2 H 0.2 . How do you relate the two impulse responses f n and g n ? Solution a) The transfer function F z is determined as F z H zej0.2 . In fact you can verify that j j0.2 F F z zej H e e H 0.2 . Substituing for the z-Transform we obtain F z 2 ej0.2z1 ej0.4z2 0.5ej0.6z3 This yields an impulse response f n 2 n ej0.2 n 1 ej0.4 n 2 0.5ej0.6 n 3 Notice that it is computed as f n h n ej0.2n , with h n the impulse response of H z . b) By the same argument the transfer function can be determined as G z H zej0.2 H zej0.2 . Therefore the transfer function becomes G z 4 2cos 0.2 z1 2cos 0.4 z2 cos 0.6 z3 4 1.61803z1 0.6180z2 0.3090z3 and the impulse response
g n 4 n 1.61803 n 1 0.6180 n 2 0.3090 n 3 It is computed as g n ej0.2n ej0.2n h n 2cos 0.2n h n . Therefore g n 2Real f n . 2 Solutions_Chapter_7[1].nb
à Problem 7.2
Problem
You want to determine the low frequency and high frequency components of a signal x n . Design an efficient filter bank where the prototype filter has at least 50dB attenuation in the stopband and a transition region of 0.1. Use the window method. Solution
The prototype filter is a low pass filter with bandwidth 2, with impulse response sin 2 n h0 n n w n
with w n the non causal window sequence. Since we need 50dB attenuation in the stopband we use a Blackman window, which has a transition region 12 N. Therefore we determine the filter length N from the transition band as 12 0.1 N This yields N 121. The Low Pass Filter H0 z has the polyphase decomposition 2 1 2 H0 z E0 z z E1 z where n E0 z h0 2n z w 0 1 n
since h0 2n 0 for n 0. Similarly n sin n2 n E1 n h0 2n 1 z w 2n 1 z n n 2n1
Problem 7.3 à
Solution
The prototype filter has frequency response H with bandwith M 8 . Therefore the non causal impulse response of the prototype filter is given by h n sin 8 n w n n
with w n being a window of length N 1 21. For example let w n be a hamming window, which has the expression w n N 0.54 0.46 cos 2 n for 0 n N 2 N
Solutions_Chapter_7[1].nb 3
where we listed the causal expression generally found in most tables. From this expression it is easy to see that
2 w n 0.54 0.46cos 20 n for 10 n 10 Finally the expression of the impulse response h n becomes sin 8 n h n n 0.54 0.46cos 10 n for 10 n 10
and zero otherwise. The transfer function of the prototype filter is then given by H z 0.0018z10 0.0014z9 0.0000z8 0.0047z7 0.0149z6 0.0318z5 0.0543z4 0.0794z3 0.1027z2 0.1191z 0.1250 0.1191 z1 0.1027z2 0.0794z3 0.0543z4 0.0318z5 0.0149z6 0.0047z7 0.0000z8 0.0014z9 0.0018z10 The eight polyphase components of the prototype filter then become as follows: 8 8n Ek z h 8n k z , for k 0, 1, ..., 7 n which yields
8 8 0 8 E0 z h 8 z h 0 z h 8 z 0.1250 8 8 0 8 8 8 E1 z h 9 z h 1 z h 7 z 0.0014z 0.1191 0.0047z 8 8 0 8 8 8 E2 z h 10 z h 2 z h 6 z 0.0018z 0.1027 0.0149z 8 0 8 8 E3 z h 3 z h 5 z 0.0794 0.0318z 8 0 8 8 E4z h4z h4 z 0.0543 0.0543z 8 0 8 8 E5z h5 z h 3 z 0.0318 0.0794z 8 0 8 16 8 16 E6z h6z h2z h 10 z 0.0149 0.1027z 0.0018z 8 0 8 16 8 16 E7z h7z h1z h 9 z 0.0047 0.1191z 0.0014z The implementation is shown below. 4 Solutions_Chapter_7[1].nb
x[n] v [n] 8 0 E0 (z ) z
8 E1(z ) v1[n] z DFT
z
8 E7 (z ) v7[n]
à Problem 7.4
Solution
a) , b), c) H z is M Band, since h 4n 0 for n 0; d) H z is not M Band since H k is not a constant for all , as shown below. k 2
H k 2 k
4 2
e) H z is M Band since H k is a constant as shown below. k 2
Solutions_Chapter_7[1].nb 5
H k 2 k
2
à Problem 7.5
Solution
Let 1 5 and 2 5 for any 0 5 . Then H z is an M Band filter with 1 A 5 as shown below. H k 2 5 k
2 5 5
à Problem 7.6
Solution
With M 16 the prototype filters for both Analysis and Synthesis have a bandwidth c 16. From what we have seen about maximally decimated DFT Filter banks, if we want to use FIR filters, the only possibility for perfect reconstruction is that both filters h n and g n in the analysis and synthesis network have length M = 16. In this way we would have
6 Solutions_Chapter_7[1].nb
H z = h 0 + h -1 z + ... + h -15 z15 G z = g 0 + g 1 z-1 + ... + g 15 z-15 with the Perfect Reconstruction condition 1 h n g n = Å1ÅÅ6ÅÅÅÅ What makes this problem a bit different from the standard FIR window based design problem is the fact the filter order is odd, ie the total filter length is 16, which is even. In Chapter 4 we have consid- ered only the case where the total filter length is odd as N = 2 L + 1. Although most of the time this is not a major restriction, in this case we have to design a filter with the precise length, and none of the filter coefficients can be zero. In other words we cant use (say) a filter with length 14, and assume h -15 = 0, since this would require g 15 =¶, clearly not feasable.
In order to design a filter with even length, we can call Hd w the frequency response of an ideal Low Pass Filter with bandwidth wc , and compute +wc w w - j ÅÅÅÅÅ2 1 - j ÅÅÅÅÅ2 jwn hd n = IDTFT H w e = ÅÅÅÅÅÅÅÅ2 p e e dw -wc This leads to the impulse response
1 sin wc n- ÅÅÅÅ2 hd n = ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ1 p n- ÅÅÅÅ2 Now the goal is to find a linear phase approximation with a finite number of coefficients. In particular let L ` -jwn HL w =S hd n e n=-L+1 which can be written as L-1 L ` jwn -jwn HL w =S hd -n e +S hd n e n=0 n=1
It is easy to see that hd -n = hd n + 1 and therefore we can write L L ` -jw jwn -jwn HL w = e S hd n e +S hd n e n=1 n=1 j ÅwÅÅÅÅ ` This shows that e 2 H L w is real, since L L j ÅÅÅÅÅw ` - j ÅÅÅÅÅw jwn - j ÅÅÅÅÅw -jwn e 2 HL w = e 2 S hd n e + e 2 S hd n e n=1 n=1 ` and therefore H L w has linear phase. As a consequence a causal translation has linear phase too, which leads to the linear phase FIR filter with frequency response 2 L-1 ` -jw L-1 -jwn HL w e =S hd n - L + 1 e n=0
Solutions_Chapter_7[1].nb 7
p In our case, the bandwidth is wc = ÅÅÅ16ÅÅÅÅ , the filter order is 15 = 2 L - 1, which yields L = 8, and the FIR filter bacomes
p 1 sin ÅÅÅÅÅÅÅ16 n-7- ÅÅÅÅ2 hd n = ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ1 ÅÅÅÅÅ , for n = 0, ..., 15 p n-7- ÅÅÅÅ2 without including the window. Finally the filters h n and g n of the analysis and synthesis networks become p 1 sin ÅÅÅÅÅÅÅ16 n-7- ÅÅÅÅ2 2 p h -n = hd n w n = ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ1 ÅÅÅÅÅ 0.54 - 0.46 cos ÅÅÅÅÅÅÅÅ15 n for 0 n N p n-7- ÅÅÅÅ2 and 16 g n = ÅhÅÅÅÅÅÅ-ÅÅnÅÅÅÅ , for 0 n N In terms of the polyphase decomposition, every term is a constant, as E-k z = h -k Fk z = g k for k = 0, ..., 15. It is just a matter of computing the coefficients to determine the final result shown in the figure below for the analysis network.
0.0034 v0[n] 0.0056 0.0119 0.0218 0.0339 0.0462 x[n] 0.0562 S / P 0.0618 DFT 0.0618 0.0562 0.0462 0.0339 0.0218 0.0119 0.0056 0.0034 v15 [n] 8 Solutions_Chapter_7[1].nb
à Problem 7.7
Solution
First we can verify that, in the the system below v n if n even w n = 0ifn odd
v[n] w[n] 2 2
1 n which yields w n = v n d2 n = Å2ÅÅÅ v n + -1 v n . Therefore, as you recall, 1 1 W w = ÅÅ2ÅÅ V w + ÅÅ2ÅÅÅ V w-p and therefore 1 1 W z = ÅÅ2ÅÅ V z + ÅÅ2ÅÅ V -z Applying this result it is easy to see that 1 1 Y z = G z ÅÅ2ÅÅ H z X z + ÅÅ2ÅÅ H -z X -z 1 1 = ÅÅÅÅ2 G z H z X z + ÅÅÅÅ2 G z H -z X -z à Problem 7.8
Solution
In this case we have a filter bank with two filters. Therefore M = 2 and forperfect reconstruction the filters have to be H z = h 0 + h -1 z G z = g 0 + g 1 z-1 with the condition 1 h 0 g 0 = ÅÅ2ÅÅ 1 h -1 g 1 = ÅÅÅÅ2 Then Let us see how to relate X z = Z x n with Y z = Z y n . Applying the result from the previ- ous problem we have Solutions_Chapter_7[1].nb 9
1 1 Y z = G z ÅÅ2ÅÅ H z X z + ÅÅ2ÅÅ H -z X -z + 1 1 + G -z ÅÅÅÅ2 H -z X z + ÅÅÅÅ2 H z X -z which becomes 1 1 Y z = Å2ÅÅÅ G z H z + G -z H -z X z + ÅÅ2ÅÅ G z H -z +G -z H z X -z Now let's see the two transfer functions X z Ø Y z and X -z Ø Y z with the perfect reconstruc- tion conditions above: Å1ÅÅÅ G z H z + G -z H -z = 2 1 -1 -1 ÅÅÅÅ2 g 0 + g 1 z h 0 + h -1 z + g 0 - g 1 z h 0 - h -1 z = = ÅÅÅÅ1 2 g 0 h 0 + g 1 h -1 + g 0 h -1 - g 0 h -1 z + g 1 h 0 - g 1 h 0 z-1 2 = 1 for all z 1 Å2ÅÅÅ G z H -z + G -z H z = 1 -1 -1 ÅÅÅÅ2 g 0 + g 1 z h 0 - h -1 z + g 0 - g 1 z h 0 + h -1 z = = ÅÅÅÅ1 2 g 0 h 0 - g 1 h -1 + -g 0 h -1 + g 0 h -1 z + g 1 h 0 - g 1 h 0 z-1 2 = 0 for all z
Therefore. as expected, Y z = X z and the filter bank perfectly reconstructs the input signal. à Problem 7.9
Solution
a) From the Problem 7.8, we can write Y z in terms of the input signal X z and its alias X -z . The aliasing comes from the downsampling operation. In terms of the DTFT we can write Y w = A w X w + B w X w-p where 1 1 A w = Å2ÅÅÅ G w H w + ÅÅ2ÅÅ G w-p H w-p 1 1 B w = ÅÅÅÅ2 G w H w-p + ÅÅÅÅ2 G w-p H w In our case the two prototype filters have frequency response H w = G w as shown below. 10 Solutions_Chapter_7[1].nb
H() 10 1 ( 0.55 )
0.45 0.55
Therefore:
2 2 ÅÅÅÅÅÅÅ10 w-0.55 p 2 + ÅÅÅÅÅÅÅ10 w-0.45 p 2 if 0.45 p < w < 0.55 p A w = p p 1 otherwise
A () 1
0.45 0.55
Analogously: - ÅÅÅ10ÅÅÅÅ 2 w -0.55 p w -0.45 p if 0.45 p< w < 0.55 p B w = p 0 otherwise and the maximum value is at w=≤ÅpÅÅÅ where the maxixmum is B ≤ ÅÅpÅÅ = 0.25, as shown below. 2 2 B()
0.45 0.55
b) For the given signal X w = 20 pd w + 2 pd w-0.2 p + 2 pd w+0.2 p - 3 pd w-0.7 p - 3 pd w+0.7 p , for -p § w < p Therefore the reconstructed signal becomes Solutions_Chapter_7[1].nb 11
Y w = A w X w + B w X w-p with A w , B w as above, and X w-p = 20 pd w-p + 2 pdw+ 0.8 p + 2 pd w- 0.8 p - 3 pd w+0.3 p - 3 pd w-0.3 p From the plot of A w and B w we can verify that A 0 = A ≤0.2 p = A ≤0.7 p = 1 B ≤p = B ≤0.8 p = A ≤0.3 p = 0 and therefore, for the given signal, y n = x n . à Problem 7.10
Solution
First let's see what is the transfer function (or the frequency response) of the system shown below. x[n] v[n] y[n] 2 Q(z) 2
Applying standard considerations we can see that 1 w w 1 w w Y w = Å2ÅÅÅ Q ÅÅ2ÅÅÅ V ÅÅ2ÅÅÅ + ÅÅ2ÅÅ Q ÅÅ2ÅÅÅ -p V ÅÅ2ÅÅÅ -p Also, from the upsampler, V w = X 2 w , which implies w V ÅÅ2ÅÅÅ = X w V w X 2 w X 2 X ÅÅÅÅÅ2 -p = ÅÅÅÅÅ2 -p = w- p = w using the periodicity of the DTFT. Therefore, subsituting into the expression for Y w we obtain 1 w w Y w = ÅÅ2ÅÅ Q ÅÅ2ÅÅÅ +Q ÅÅ2ÅÅÅ -p X w = Q0 w X w Therefore the impulse response q0 n = IDTFT Q0 w is the impulse response q n downsampled by two, ie q0 n = q 2 n In other words from the polyphase decomposition
2 -1 2 Q z = Q0 z +z Q1 z where 12 Solutions_Chapter_7[1].nb
Qk z = Z q 2 n + k
we can determine the transfer function Q0 z .
a) We want to determine the four transfer functions Yi z X j z for i, j = 0, 1. See each one separately: ÅÅÅÅÅÅÅÅÅÅÅÅÅY0 z = 0 : since, in this case X1 z 1 -1 -2 -3 1 1 2 3 Q z = G1 z H0 z = ÅÅ4ÅÅ 1 - z + z - z ÅÅ4ÅÅ 1 + z + z + z 1 -3 -1 3 = ÅÅÅÅÅÅÅ16 -z - z + z + z
and therefore Q0 z =0 since there are no even powers of z in Q z . ÅÅÅÅÅÅÅÅÅÅÅÅÅY1 z = 0 : since, in this case X0 z 1 -1 -2 -3 1 1 2 3 Q z = G0 z H1 z = ÅÅ4ÅÅ 1 + z + z + z ÅÅ4ÅÅ 1 - z + z - z 1 -3 -1 3 = ÅÅÅÅÅÅÅ16 z + z - z - z
and therefore Q0 z =0 since there are no even powers of z in Q z ÅÅÅÅÅÅÅÅÅÅÅÅÅY0 z = ÅÅÅÅÅÅÅ1 2 z-1 + 4 + 2 z : since, in this case X0 z 16 1 -1 -2 -3 1 1 2 3 Q z = G0 z H0 z = ÅÅ4ÅÅ 1 + z + z + z ÅÅ4ÅÅ 1 + z + z + z 1 -3 -2 -1 2 3 = ÅÅÅÅÅÅÅ16 z + 2 z + 3 z + 4 + 3 z + 2 z + z and the polyphase decomposition 1 -2 2 -1 1 -2 2 4 Q z = ÅÅÅ16ÅÅÅÅ 2 z + 4 + 2 z + z Å1ÅÅ6ÅÅÅÅ z + 3 + 3 z + z which yields
1 -1 Q0 z = Å1ÅÅ6ÅÅÅÅ 2 z + 4 +2 z . ÅÅÅÅÅÅÅÅÅÅÅÅÅY1 z = ÅÅÅÅÅÅÅ1 2 z-1 + 4 + 2 z : since, in this case X1 z 16 1 -1 -2 -3 1 1 2 3 Q z = G1 z H1 z = ÅÅ4ÅÅ 1 - z + z - z ÅÅ4ÅÅ 1 - z + z - z 1 -3 -2 -1 2 3 = ÅÅÅÅÅÅÅ16 -z + 2 z - 3 z + 4 - 3 z + 2 z - z and the polyphase decomposition 1 -2 2 -1 1 -2 2 4 Q z = ÅÅÅ16ÅÅÅÅ 2 z + 4 + 2 z + z Å1ÅÅ6ÅÅÅÅ -z - 3 - 3 z - z which yields
1 -1 Q0 z = Å1ÅÅ6ÅÅÅÅ 2 z + 4 + 2 z . b)