Homework #1 Solution

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1 February 7, 4 Department of Electrical and Computer Engineering University of Wisconsin Madison ECE 734 VLSI Array Structures for Digital Signal Processing Homework # Solution Due: February 6, 4 in class. This homework covers weeks,, and 3 materials including reference materials and notes.. You must do the homework by yourself. No collaborations are allowed. Late homework will receive a 5% penalty per day. There are total points. This homework is worth % of your overall grades. Homework solution will be posted after the class when the homework is due. Hence, NO LATE HOMEWORK SUBMISSION WILL BE ACCEPTED. You must either type the solution or write it neatly. Each homework submitted must be stapled and with a cover page. On the cover page, you must specify the homework #, your last and first name. Failure to follow instruction will result in 5 point penalty of homework grade. CC problems: A problem with CC marked behind the points assignment is a completion credit only problem. As long as you show reasonable effort, you will receive full credit. Many of these questions are open-ended and hence may not have absolutely correct answers exist. The grader will be instructed to grade each part independently.. ( points, Convolution) A Töplitz matrix like the one shown below is defined by a sequence {t k, 3 k 3} that form the first column and the first row of this matrix. It has been said that the product of a Töplitz matrix and a vector can be realized using a convolution operation. t t t t3 x y t t t t x y t t t t x3 y3 t 3 t t t x4 y4 (a) (5 points) Show that {y i } can be obtained by convolving two sequences. Give explicitly these sequences, and explicitly specify which part of the results of the convolution corresponding to the desired y i ; i where h k t -k. Hence, {y i } may be computed as y xt xh i k + i k k i k k k the convolution of {x,, x 4 } and {t 3, t,, t -3 }. The output of the convolution contains 4+7- elements. y i 's corresponds to the 4 th to the 7 th elements of the outputs. (b) (5 points) Since convolution can be realized with fast Fourier transform. Using high level language pseudo-code write a program to implement the a Töplitz matrix-vector product operation by calling a FFT subroutine. Assume that the syntax as follows: X FFT(N,x) where N is the N-point FFT to be computed and is also the length of the output sequence X. x is the input sequence. If its length is shorter than N, it will be padded with s. We may also define IFFT routine calling sequence as: x IFFT(N,X). Page of

2 February 7, 4 Since the convolution in time domain multiplication in frequency domain, we can implement the convolution as follow. Let h(n) t(-n) p(n) x(n)*h(n) Since the output of the convolution of {x,, x 4 } and {t 3, t,, t -3 }contains element, and the FFT require the input to be power of, N should be 6 to obtain the complete result. P(f) FFT(6,x) FFT(6,h) p(t) IFFT(6,Y) Therefore, y i p(n) where i to 4 and n 4 to 7. An example is given in hws4conv.m. ( points) In the class note, we give a vector inner product implementation example (ref. Intro.ppt). Refer to the dependence graph, note that all the n multiplications, a(k)*x(k), k n, can be computed simultaneously provided n hardware multipliers are available. However, it is not clear if these n multiplication results can be added together at the same time for n >. Given a full adder module as shown below, sketch the logic schematic of a logic circuit that implement a special adder that can add four 4-bit unsigned binary numbers in parallel. For convenience, denote the binary representation of these four numbers as: A 3 A A A, B 3 B B B, C 3 C C C, D 3 D D D. X Y C out C in S This can be implement using an adder tree. 3. ( points) Graph transpose theorem An important graph-theoretic theorem known as the graph transpose theorem states the following: The transfer function of a single-input, single output signal flow graph remains unchanged if (i) the direction of all arcs are reversed, and (ii) the position of inputs and outputs are inter-changed. Recall the direct form-i, nd order IIR digital filter shown in the digitalfilter.ppt note. (a) (5 points) Apply the graph transpose theorem to the SFG of the direct form I, and draw the resulting SFG. Page of

3 February 7, 4 y(n) b() u(n) x(n) z - b() a() z - z - b() a() z - (b) (5 points) Derive the transfer function of the SFG you obtained in part (a). Show that it is identical to the transfer function of the original SFG. Note that the transfer function of the original SFG is b b z b z Bz H() z () + () + () ( ) + a() z + a() z Az ( ) Now the time domain equations corresponding to the transposed SFG are: un ( ) a() un ( ) a() un ( ) + xn ( ) yn ( ) b() un ( ) + b() un ( ) + b() un ( ) Now apply z-transform on both equations, one obtains U() z a() z U( z) a() z U() z + X() z Hence a z a z U z AzUz X z, + () + () () ( ) ( ) () Y z b U z b z U z b z U z b b z b z U z BzUz Y() z BzUz ( ) ( ) Bz () Htranspose() z ( ) X() z AzUz ( ) ( ) A( z ) H z ( ) () () + () ( ) + () () () + () + () () ( ) ( ) 4. (5 points) DFT Refer to the DFT algorithm in the intro.ppt notes. A pseudo code is given below: /* cos(θ(n,k))cos( π*n*k/n), sin(θ(n,k))sin( π*n*k/n), /* n, k N, are assumed to have been computed in advance, /* and can be loaded into the memory. So the computation cost is not /* included in this problem For k:n-, For n:n-, C(n,k)cos(-*pi*n*k/N); S(n,k)sin(-*pi*n*k/N); End End /* Here is DFT computation /* Note that X(k) is a complex number where i squart root of /* mcount# of real multiplications, acount # of real additions mcount, acount, Page 3 of 3

4 February 7, 4 For k:n-, X(k), For n:n-, X(k)X(k)+C(n,k)*x(n)+i*S(n,k)*x(n), mcountmcount+, acountacount+, End End (a) (3 points, CC) Turn in the source code of a sequential program in a high level programming language of your choice to implement the DFT algorithm. hws4dft.m. This is only a Matlab example. Other High level languages may also be used. (b) ( points) Give an eight-point DFT example to illustrate the correctness of the computation results. For convenience, report the results by filling the table below: n, k x(n) X(k) i i i i i i i (c) ( points) Give the mcount and acount result of this 8-point example, as well as a function of N. mcount acount 8. In general, mcount acount N. (d) (3 points) Observe a symmetry property exhibited in the result from part (ii) that if x(n) is a real-valued sequence, then some X(k) s are complex conjugate each other. State this property formally (together with the range of k) and prove it for the general case of N. Symmetric property of DFT: If {x(n); n N-} is a real-valued sequence, for k N/-, X(N-k) X * (k). The proof is as follows: N N j πnn ( k) jπnk * X( N k) xn ( )exp xn ( )exp X ( k) n N n N (e) (3 points, CC) Modify the program you implemented in part (i) so that you can compute the same results. Turn in the source code of the program. hws4dft.m. This is only a Matlab example. Other High level languages may also be used. (f) ( points) Repeat part iii) for program in part iv). mcount acount 8. In general, mcount acount N(N/+). Page 4 of 4

5 February 7, 4 5. ( points) (a) (5 points) Refer to the direct form-ii realization in the class notes digitalfilter.ppt. The time domain equations that describes the nd order IIR filter are: wn ( ) a() wn ( ) a() wn ( ) + xn ( ) yn ( ) b() wn ( ) + b() wn ( ) + b() wn ( ) Denote two state variables v (n) w(n ), v (n) w(n ). Derive the state space representation of this digital filter v( n+ ) A v( n) + b xn ( ) T yn ( ) cv( n) + d xn ( ) by specifying the A, b, c and d matrices (or vectors). You must give derivations. Answer alone will not receive credit. The state space equations are: v( n + ) a() v( n) a() v( n) + x( n) v ( n + ) v( n) yn ( ) b() wn ( ) + b() wn ( ) + b() wn ( ) Hence ( ) ( ) ( ) [ ] v or a() v( n + ) b() a() v( n) a() v( n) + xn ( ) + b() v( n) + b() v ( n) b() b() a() v ( n) + b() b() a() v ( n) + b() xn ( ) b() b() a() b() b() a() ( n) + b() xn ( ) a() a() b() b() a() A,,, and d b() b c b() b() a() a() ( n) x( n) v + (b) ( points) Suppose that the two complex conjugate roots of the characteristic polynomial A ( z) z + a() z + a() are r cosθ ± j r sinθ. Represent a(), a() in terms of r and θ. jθ jθ ( z re )( z re ) z r cosθ z + r z + a() z + a() By comparison of coefficients, one has: a() -r cosq, a() r. (c) (5 points) Consider the orthogonal state space realization (r > ) u( n+ ) cosθ sinθ u( n) β r xn ( ) ( n) xn ( ) u( n ) sinθ cosθ + + u( n) β R u ß + u( n) T yn ( ) [ γ γ ] + δ xn ( ) ( n) + δ xn ( ) u ( n)?u Find its transfer function H(z) Y(z)/X(z). Express the result as a function of r, θ, β, β, γ, γ, and δ, and find b(), b(), and b() as a function of these parameters too. (**) Page 5 of 5

6 February 7, 4 [ γ γ ] rsinθ z rcosθ β δ T T (?ß cos ) (?ß cos ( ) sin ) δz + δr θ z+ δ r r θ + γβ γβ r θ () + () + () z rcosθ rsinθ β Y() z T H() z? ( zi R) ß+ + δ X() z z rcosθ z + r b z b z b z + a() z+ a() z rcosθ + rz Comparing coefficient, b(), b(), and b() can be easily found. (d) (5 points) Suppose that the IIR filter can be described by a difference equation y(n).96*y(n-).64*y(n-) + x(n) Represent this IIR filter in a orthogonal state space representation (**). Assume that β and β. a() -r cosq -.96, and a() r.64. Hence r.8, r cosq -.96/(-).48, and cosq.6. Now, b() d, b() b(). Therefore,? T ß δr cosθ. Thus,? T ß ()(.8)(.6). 96. In addition, T δ r?ß rcos θ + ( γ β γβ ) rsin θ ()(.64) (.96)(.48) + ( γ β γβ )(.8)(.8) Hence we have a set of equations: β γ β γ + β γ β γ.96.8 Since b, b, it can be solved that g.96, and g -.8. (e) (3 points) Use induction method, prove that for the R matrix defined in eq. (**), cos( kθ ) sin ( kθ ) ( ) ( ) R k r k sin kθ cos kθ The case of k is true by definition of R. Now suppose that above expression is true for a particular value of k. We want to examine R k+ as follows: R k+ R r r k k+ k+ R cos cos cos sin Thus, by induction, it is proved. cos( kθ ) sin ( kθ ) ( ) ( ) ( ) ( ) cos θ sin θ k r r ( ) ( ) sin kθ cos kθ sin θ cos θ ( kθ ) cos( θ ) sin ( kθ ) sin ( θ ) cos( kθ ) sin ( θ ) sin ( kθ ) cos( θ ) ( kθ ) sin ( θ ) + sin ( kθ ) cos( θ ) cos( kθ ) cos( θ ) sin ( kθ ) sin ( θ ) (( k + ) θ ) sin ( k + ) θ ) (( k + ) θ ) cos( ( k + ) θ ) 6. (5 points) Quantization Error Consider a first order IIR filter Page 6 of 6

7 February 7, 4 yn ( ) a yn ( ) + xn ( ) a <, xn ( ) <, y(), n,,... The input data sequence {x(n)} is sampled and quantized to b-bit per sample. In order to custom-design a hardware MAC unit and data path to implement this digital filter, we want to determine the appropriate register length in this problem. (a) (7 points) First, we want to find the maximum absolute value y(n) may reach. That is called the dynamic range of the output y(n). Show that lim yn ( ) n a Proof: Thus, lim yn ( ). n a ( ) yn ( ) ay( n ) + xn ( ) a ay( n ) + xn ( ) + xn ( ) n n n k ay() + a x() + + ax( n ) + xn ( ) axn ( k) k k k k n n n k k k a axn ( k) a xn ( k) a a For fixed-point arithmetic implementation, this upper bound specifies how many bits are needed in the integer portion so that there is no overflow during computation. (b) (8 points) Next, consider a fixed-point arithmetic implementation. Since {x(n)} has only b significant factional binary bits, there is no need to retain more than b bits on the final results. However, intermediate results may retain more than b bits so that the quantization error can be reduced. To analyze the effect of quantization, denote Q[y] to be a quantized quantity of a précised value of variable y such that Q[y] y + ε where ε is a uniformly distributed random variable varying between ± l where l is the number of fractional bits of Q[y]. The quantization errors ε for different values of y are assumed to be i.i.d. i) ( points) Show that E{ε}, Var.{ε} l /3. denote a -l-, Var.{e)} a x a x a a dx a 3 a l ii) ( points) For the n th iteration of the IIR filter, computed value of y(n) subject to two types of quantization error: the quantization error inherited from the previous iterations that accompanying y(n ), as well as additional quantization error imposed after the MAC operation. For convenience, write yn ˆ( ) Qayn ˆ( ) + xn ( ) a yn ˆ( ) + xn ( ) + ε( n) yn ( ) + δ( n) [ ] n. Page 7 of 7

8 February 7, 4 where δ(n) is the accumulated quantization error on y(n), ε(n) is the uniform quantization error described in part (i), and yn ˆ() is the computed value of y(n), taking into account all the accumulated quantization errors. Show that δ( n) a δ( n ) + ε( n) ˆ ( ˆ + + ε ) ( + ) ( ˆ( ) ( ) ) ε( ) δ( ) ε( ) δ( n) yn ( ) yn ( ) a y( n ) xn ( ) ( n) a y( n ) xn ( ) a yn yn + n a n + n iii) ( points) Moreover, use an analysis similar to part (a), deduce a worst case upper bound for δ(n): n l k δ( n) a δ( n ) + ε( n) a ε( n k) a k Hence if one decides that d (n) should be -b, then the desired quantization length l can be solved as follows: iv) ( points) Assume that m a m. Solve for a lower bound of l such that the maximum accumulated quantization error max. δ(n) b : Since this implies l Equivalently, one has ( a ) l a b b m b l b + m -.. This leads to the condition 7. ( points, CC) (a) ( points CC) Refer to the qr734.m matlab m-file posted in the ece734 website. Develop a high level language program (in C, C++, visual basic, etc. any high level language) to perform the QR factorization with Givens rotation of the following 4 3 matrix: A Give the R matrix of the result, the numbers of multiplication, addition and division performed. Derive a general formula of the numbers of multiply, addition and division for a real-valued M N matrix. Note M needs not be equal to N. hws4qr.m. In general, the number of divisions the number of elements MN N( N + )/ M N; to be annihilated in the matrix M( M )/ M N. Page 8 of 8

9 February 7, 4 The plane rotation will be applied to two row vectors at a time. For a pair of n vectors, there will be 4n multiplications and n additions. Thus, the number of multiplications and additions can be computed as follows: For M N, # of multiplications N N( N + ) N( N + )(N + ) ( M i)4( N i+ ) 4 MN( N + ) 4( M + N+ ) i M + N N + N( N + )(3M N ) 4 N( N + ) M The number of addition is half of multiplication N(N+)(3M-N-)/3. For M < N, M # of multiplications ( M i) 4( N i + ) i The number of addition is half of multiplication (b) (5 points, CC) Refer to the Kalman filter power point notes statisticaldsp.ppt, the input u(n), output y(n) are both scalar, the state vector x(n) is a m vector, the state transition matrix A is a m m matrix. The state vector covariance matrix P is m m, the Kalman gain vector k is m, the input covariance σ u is a scalar, and so is the disturbance covariance σ v. Compute the number of MAC operations that are needed to update from xˆ( n n ) to xn ˆ( + n+ ). From xn ˆ( + n+ ) and k(n+) equation, it can be seen that the total number of MAC operation 4m(m+). (c) (5 points, CC) First, solve for the Yule-Walker equation: Given r(), r() r( ), r() r( ), solve for a(), and σ. r() r( ) σ r() r() a() r()+a()r(-) s () r()+a()r() () from (), a() -r()/r() -----(3) from () and (3), s r()-[ [r()r(-)]/r()] r()-r ()/r() Next, evaluate {b ij } in terms of variables to the left. r() r() r() b b r() r() r() a() a() b b r() r() r() b3 b 3 Page 9 of 9

10 February 7, 4 r() + a() r() r() + a() r() r() + a() r() a() r() + r() b a() r() + r() b a() r() + r() b Therefore, b b 3, b b, b 3 b -----(4) Next, post-multiply to both sides of above equation to obtain the desired format. Compute the values of κ and q. Note that means that entry should be equal to. r() r() r() b b q κ κ r() r() r() a() a() b b κ κ r() r() r() b3 b 3 q r() r( ) σ from r() r() a(), one has r() r( ) a() σ r() r() a(). Therefore, σ r() r() r() r() r() r() a () a() κ κ r() r() r() σ b b b q κ κ b b κ κ b3 σ b3 b 3 q By comparison of coefficients, we have b s b 3, b b. b 3 b r()a()+r() r() - r ()/r(). Moreover, b k b. Hence, k -b /b, and q b -kb b + b /b 3 b b b 3 Page of

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