ENEE630 ADSP RECITATION 5 w/ solution Ver

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1 ENEE630 ADSP RECITATION 5 w/ solution Ver20209 Consider the structures shown in Fig RI, with input transforms and filter responses as indicated Sketch the quantities Y 0 (e jω ) and Y (e jω ) Figure RI: Comment: For a down-sampled signal, it s still possible to recover the original signal (not necessarily low-pass) using filters and multirate building blocks as long as there is no aliasing (H in this problem) 2 For each case shown in the Fig RI2, prove or disprove whether the left system is equivalent to the right system? Assume,L,K are all integers larger than Assume the input, output, intermediate signals are x(n), y(n), and u(n), respectively (a) (FD) U(z) = X(z ), Y 2 (z) = k=0 U(z/ W k ) = k=0 X(zWk ) = X(z) (TD) u(n) = x( n ) if n is a multiple of, and u(n) = 0 otherwise Then, y 2(n) = u(n) = x(n) (b) (FD) U(z) = k=0 X(z/ W k ), Y 2(z) = U(z ) = k=0 X(zWk ) { x(n) n is multiple of (TD)u(n) = x(n), y 2 (n) = u( n ) = x(n)onlyifnisamultipleof,ie, y 2(n) = 0 otherwise

2 Figure RI2: (c) (FD) U(z) = X(z K ), Y 2 (z) = U(z L ) = X(z KL ) (TD) u(n) = x( n K ) only if n is a multiple of K, and y 2(n) = u( n L ) only if n is a multiple of L Zero otherwise Hence, y 2 (n) = x( n ) only when n is a multiple of KL = K (d) (FD) U(z) = K KL K K m=0 k=0 X(z/K e k=0 X(z/K WK k), Y 2(z) = 2πm 2πk j j K ) = Ke = K K t=0 X(z /K WK t ) (TD) u(n) = x(kn), y 2 (n) = u(n) = x(kn) K m=0 m=0 U(z/ W m ) K k=0 X(z/K W k+m K ) (e) (FD) Left: Y (z) = k=0 X(zL/ W k ); Right: Y 2(z) = K K k=0 X(z L/ WK k ) (TD) Left: y 2 (n) = x(n L ) only when n is a multiple of L; right: y 2(n) = x(n L ) only when n is a multiple of KL Note: To prove/disprove the equivalence, you can prove either in time domain or in freq domain, no need to do both You are recommended to try all on your own (except (d) freq domain) to familiar yourself with the derivation 3 Simplify the following systems in Fig I3 Figure RI3: The simplified systems are as follows Note: The basic relationships from the previous problem and Nobel identities are applied For part (d), the trick is z = z 3 z 2 and z = z 3 z 2 2

3 4 In this problem, the term polyphase components stands for the Type components with = 2 (a) Let H(z) represent an FIR filter of length 0 with impulseresponsecoefficients h(n) = (/2) n for 0 n 9 and zero otherwise Find the polyphase components E 0 (z) and E (z) (b) Let H(z) be IIR with h(n) = (/2) n u(n)+(/3) n u(n 3) Find the polyphase components E 0 (z) and E (z) Give simplified, closed form expressions (Hint: +x = x x 2 ) (c) Let H(z) = /( 2Rcosθz + R 2 z 2 ), with R > 0 and θ real This is a system with a pair of complex conjugate poles at Re ±jθ Find the polyphase components E 0 (z) and E (z) (a) E 0 (z) = + (/2) 2 z + (/2) 4 z 2 + (/2) 6 z 3 + (/2) 8 z 4, E (z) = /2 + (/2) 3 z + (/2) 5 z 2 +(/2) 7 z 3 +(/2) 9 z 4 (b) H(z) = (/3) 4 z 2 (/9)z, and E (z) = (c) H(z) = + (/3)3 z 3 = +(/2)z + (/3)3 z 3 +(/3) 4 z 4 (/2)z (/3)z (/4)z 2 /2 + (/3)3 z (/4)z (/9)z 2Rcosθz +R 2 z 2 = +R 2 z (+R 2 z ) 2 4R 2 cos 2 θz, and E (z) = (/9)z 2 Hence, E 0 (z) = (+2Rcosθz +R 2 z 2 ) (/4)z + ( 2Rcosθz +R 2 z 2 )(+2Rcosθz +R 2 z 2 ) Hence, E 0(z) = 2Rcosθ (+R 2 z ) 2 4R 2 cos 2 θz 5 A uniformdft analysis bank (Type) is shown in Fig RI5(a), wherew is the IDFT matrix, ie, the (m,n)-th entry is W mn with indices m, n starting from 0 The transfer function from input port x(n) to output port x k (n) is denoted by H k (z) Answer the following questions (a) (b) Figure RI5: (a) Prove H k (z) = H 0 (zw k ) for 0 k Given H 0(e jω ) in Fig RI5(b), sketch H (e jω ) when = 2, and H (e jω ), H 2 (e jω ) when = 3 (b) = 4 Assume E 0 (z) = +z, E (z) = +2z, E 2 (z) = 2+z, and E 3 (z) = 05+z 3

4 Find numerical values of these filter coefficients for H k (z), 0 k 3 (c) = 2 Let H 0 (z) = + 2z + 4z 2 + 2z 3 + z 4, and let H (z) = H 0 ( z) Draw an implementation for the pair [H 0 (z),h (z)] in the form of a uniform DFT analysis bank, explicitly showing the polyphase components, the 2 2 IDFT box, and other relevant details (a) H 0 (z) = m=0 z m E m (z ), H k (z) = m=0 z m E m (z )W mk = m=0 (zwk ) m E m ((zw k ) ) = H 0 (zw k ) H k(e jω ) = H 0 (e j(ω 2πk/) ) (b) H 0 (z) = +z + 2z z 3 +z 4 +2z 5 +z 6 +z 7 H (z) = +jz 2z 2 05jz 3 +z 4 +2jz 5 z 6 jz 7 H 2 (z) = z +2z 2 05z 3 +z 4 2z 5 +z 6 z 7 H 3 (z) = jz 2z 2 +05jz 3 +z 4 2jz 5 z 6 +jz 7 (j = ) (c) Note that W2 =, then they can be implemented together using one DFT analysis bank E 0 (z) = +4z +z 2, E (z) = 2+2z Note: W = e j 2π, eg, W 2 =, W 4 = j 2 W (with entry W ij ) is called the IDFT matrix, whereas W (with entry Wij ) is called the DFT matrix Note that the index starts from 0 and goes up to One important property is that W W = I For instance, when = 2, W and W happen to be the same [ ], and hence using the property, [ ][ ] [ 2 0 = 0 2 ], which you will encounter very often when studying the QF bank 3 Type 2 DFT bank (synthesis bank) is found in Problem # in Homework #2 Similar relationship ( shifted version ) can be found between transfer functions 4

5 4 Given the DFT bank structure, we can expect transfer functions have relationship (a); on the other hand, given the transfer functions have the relationship, we can implement it by the DFT structure Part (b) and part (c) of this problem show the two aspects 5 For H 0 (z), since all weights are, it is just the type polyphase representation, and hence H 0 (z) = m=0 z m E m (z ) 7 Prove or disprove: the systems are linear time-invariant (LTI) (a) NO U(z) = H(z) Figure RI7: k=0 X(z/ W k ), Y(z) = U(z ) = H(z ) k=0 U(z/ W k ) = k=0 X(zWk ) k=0 H(z/ W k )X(z) = (b)yesu(z) = H(z)X(z ),andy(z) = [ k=0 H(z/ W ]X(z) k ) Actually, the effective transfer function is just the polyphase component E 0 (z) An alternative way is using the equivalence Note: A system is LTI if and only if it can be represented by H(z), ie, Y(z) = H(z)X(z) 8 Prove that the block diagram shown in Fig RI8(a) is a perfect reconstruction system Based on this, answer the following questions (No knowledge of QF bank is needed though We will briefly revisit this problem in next recitation): Figure RI8: (a) What is the transfer function H (z) for the block diagram in Fig RI8(b)? (b) Suppose that we are to design a system with the desired system function H(z) = +2z + 3z 2 + 4z 3 + 5z 4 + 6z 5 + 7z 6 in a very high-speed environment If the speed of the devices used to implement the system is three times slower than the required operating speed, please devise a scheme that can implement H(z) under the constraint of the device (Hint: implement z 2 H(z)) 5

6 (FD)U 0 (z) = V (z) = k=0 X(z/ W k ),V 0(z) = k=0 z W k X(zWk ); Ingeneral, U m(z) = X(zWk ) As a result, Y(z) = k=0 z m W mk m=0 k=0 W mk k=0 X(zWk ); U (z) = k=0 z / W k X(z/ W k ), k=0 z m/ W mk X(z/ W k ),V m(z) = m=0 z ( m) V m (z) = z ( ) m=0 W mk = 0 when k =,2,, we get H(z) = z ( ) or X(zWk ) Use the fact that m=0 W mk = when k = 0 and (TD) u 0 (n) = x(n),v 0 (n) = u 0 (n/) = x(n) only if mod (n,) = 0 Let w 0 (n) = v 0 (n + ), then w 0 (n) = x 0 (n + ) only if mod (n,) = and 0 otherwise u (n) = x(n ),v (n) = u (n/) = x(n ) only if mod (n,) = 0 Let w (n) = v (n +2), then w (n) = x(n +)onlyif mod (n,) = 2and0otherwise Ingeneral, w m (n) = x(n +) only if mod (n,) = m y(n) = m=0 w m(n) = x(n +) (a) z ( ) G(z ) (b) R 0 (z) = 3+6z,R (z) = 2+5z,R 2 (z) = +4z +7z 2 9 ConsidertheQFbankinFigRI9, whereh 0 (z) = +2z +3z 2 +z 3, andh (z) = H 0 ( z) Figure RI9: (a) Design an FIR synthesis bank F 0 (z),f (z) such that the system is alias-free (b) Design an IIR synthesis bank F 0 (z),f (z) such that the system is perfect reconstruction Is this bank stable? (By default, we assume it is causal) (c) Design a stable IIR synthesis bank F 0 (z),f (z) such that the system is alias-free and amplitude-distortion-free (Hint: a +z +az is an all-pass filter) 6

7 E 0 (z) = +3z, E (z) = 2+z F 0 (z) = z R 0 (z 2 )+R (z 2 ), F (z) = F 0 ( z) = z R 0 (z 2 ) R (z 2 ) (a) R 0 (z) = E (z), R (z) = E 0 (z) T(z) = 2z E 0 (z 2 )E (z 2 ) (b) R 0 (z) = /E 0 (z), R (z) = /E (z) T(z) = 2z However, check that R 0 (z) is not stable (c) Let Ẽ0(z) = 3 + z, then E 0(z) Ẽ 0 (z) is an all-pass filter R 0(z) = Ẽ 0 (z), R (z) = T(z) = 2z E 0 (z 2 ) Ẽ 0 (z 2 ) Note: E 0(z) Ẽ 0 (z)e (z) When designing a QF bank, we usually use the polyphase decomposition as in this problem In this problem, since the IDFT and DFT matrix cancel out each other, we get a structure similar to Fig RI8(b) with each branch being E m (z)r m (z) To make it alias-free, we require all branches should be the same, ie, E 0 (z)r 0 (z) = E (z)r (z) If we further desire a system that is amplitudedistortion-freeorpr, thennotonlyallbranchesshouldbethesame, butalso E m (z)r m (z) = const (for amp-distortion-free) or E m (z)r m (z) = cz n 0 (for PR) We will discuss more in the next recitation 2 In general, if A(z) = a 0 +a z +a 2 z 2 ++a n z n, then define Ã(z) = a n +a n z + a n 2 z a 0 z n (reverse the order and take conjugate), and Ã(z) A(z) is an all-pass filter For a+z example, +az is the -order real-coefficient all-pass filter 0 Consider a 4-channel filter bank which is alias-free If we know part of the matrix P(z) as P(z) = 3z 2z 2 z 3 write down the entire matrix P(z) What is the distortion function T(z)?, Since the matrix should be pseudo-circulant, 3z z 2 2z 2z 2 3z z 2 P(z) = z 3 2z 2 3z 3z 2 z 3 2z 2 7

8 2z 7 ] The distortion function can be calculated using the first row, ie, T(z) = z 3 [+3z 5 +z 0 + Problem RI8 revisited First find the transfer function for Fig RI(a) and (b), Then, given the desired H(z) = +2z +3z 2 +4z 3 +5z 4 +6z 5 +7z 6, how to implement it on a hardware platform which is 3 times slower than needed? (Hint: using pseudo-circulant representation) Figure RI: (a) P(z) = G(z) (b) P (z) = G(z), H(z) = z ( ) (c) Implement z 2 H(z) P(z) =, H (z) = z ( ) G(z ) G(z) P 0 (z) P (z) P 2 (z) z P 2 (z) P 0 (z) P (z) z P (z) z P 2 (z) P 0 (z) 7z 2,P (z) = 2+5z,P 2 (z) = 3+6z, where P 0(z) = +4z + 8

9 2 Consider the following -channel multirate system H0 (z) C0 (z) F0 (z) H (z) C (z) F (z) H (z) C (z) F (z) Assume throughout that the functions F k (z), H k (z), and C k (z) are rational and stable (a)supposeh k (z)andf k (z)aresuchthatthesystemisalias-freeinabsenceofchanneldistortion C k (z), ie when C k (z) is unity for all k Now with C k (z) present, find a modified set of synthesis filters F k (z) to retain alias-free property (b) Assume C k (z) has no zeros on the unit circle and has the form of C k (z) = A k(z)b k (z) D k (z) where A k (z) has all the zeros inside the unit circle and B k (z) has them outside Repeat part (a) by replacing alias-free condition with alias-free and amplitude distortion free condition (c) Now assume that C k (z) has no zeros on or outside the unit circle, repeat part (a) by replacing alias-free with perfect reconstruction (a) G k (z) = F k (z) i=0,i k C i(z ) 9

10 (b) G k (z) = F k (z) D k(z ) A k (z ) all-pass filter (c) G k (z) = F k (z)/c k (z ) B i (z ) B k (z ) i=0,i k B, where B i (z ) k (z ) is chosen such that B k(z ) B k (z ) is an 0