Digital Signal Processing

Transcription

1 Dec./Jan. Fifth Semester B.E. Degree Examination Updates Digital Signal Processing Time: hrs. Max. Mars: ote:. Answer any FIVE full questions, selecting at least two questions from each part. PART A. a. Given the following x(n) : x(n) (n) + (n ) + (n ) i) Find the Fourier transform X(e jw ) and plot the X(e jw ) ii) Get the magnitude of the pont DFT of the first four samples of x(n) iii) Get the magnitude of the point of the first eight samples of x(n). Ans: Given x(n) (n) + (n ) + (n ) i) ( jw ) ( ) X e x n e n jw jwn + e + e ejw X(e jw ) ( Mars) ii) Magnitude of x(e j ) + + To nd the DFT of rst samples of x(n) X ( ) x ( n) j j n DFT n [ ] X( ) δ ( n) +δ ( n ) +δ( n ) n + + n n n n n n + + n w X() + + X() + ( j) + j X() + -Digital Signal Processing Dec.-Jan..indd 9// ::5 AM

2 Dec. /Jan. Digital Signal Processing Updates X() + j + ( ) j X() [ j j] X() [ ] iii) To nd points DFT of rst samples, 7 n [ ] X( ) δ ( n) +δ ( n ) +δ( n ) + + X() X().77 j.77 X() j X().9 + j.9 X() X(5).9 j.9 X(6) j X(7).77 + j.77 X() [,.,,.,,.,,.] n.77 j.77 j.77 j.77 πn. b. Find the point DFT of the sequence, x( n) 6+ sin, n. (6 Mars) πn Ans: x( n) 6+ sin, n The point DFT of x(n) is n X( ) x ( n),,, n π j jπ where e e j j The value of x(n) for n is tabulated below πn n x( n) 6+ sin Digital Signal Processing Dec.-Jan..indd 9// ::56 AM

3 Digital Signal Processing Dec./Jan. X( ) x( n) w n n x( ) + x() + x( ) + x( ) Updates X() X() 6 + 7( j) + 6( ) + 5(j) 6 7j 6 + 5j j X() 6 + 7( ) + 6 () + 5( ) X() 6 + 7(j) + 6( ) + 5( j) 6 + 7j 6 5j j X() [ j j]. c. Consider the sequence x( n) (,,,,) and x( n) (,,,,). Determine the sequence y(n) so that Y(K) X (K) X (K). where X (K) and X (K) are 5 point DFTs of x (n) and x (n) respectively. ( Mars) Ans: x (n) (,,,, ) x (n) (,,,, ) Y() X () X () To compute y(n), Using the property X() t X( K) x DFT (n) * x (n) y( n) x( n) * x( n) n 5 ( ) x ( m) x ( n m), n m 5 n x (m) x ((n m)) 5 y(n) y(n) ( ). a. n A sequence x( n) has an point DFT X(K). Compute the DFt of x (n) and x (n) in n 7 n n terms of X(K), for x ( n) n, x( n) n 5 5 n 7 6 n 7 (6 Mars) -Digital Signal Processing Dec.-Jan..indd 9// ::56 AM

4 Dec. /Jan. Digital Signal Processing Updates n Ans: x( n) n 7 [ ] n x ( n) n 5 n 7 [ ] n x ( n) n 5 6 n 7 [ ] It is evident from the sequence that x (n) x((n 5)) x (n) x((n )) Using circular time shift property x m (( n m) ) X ( ) 7 DFT n X( ) x( n) 7 e n x( ) + x() + x( ) + x( ) j π j j j j 7 j j + X() X() j. 6 X() X() j. X() Digital Signal Processing Dec.-Jan..indd 9// ::56 AM

5 Digital Signal Processing Dec./Jan. 5 5 X(5) j. 6 6 X(6) X(7) j. X() (, j.,, j.,, +j.,, +j.) Updates i) x (n) x((n 5)) Using circular time shift property 5 X () X() X () X() 5 X () X() ( + j.) X () X() 5 7 X () X() x() + j. X () 5 X (5) X(5) X(5) ( j.) X (6) 5 X (7) X(7) X(7) ( j.) X () [, + j.,, + j.,, j.,, j.] ii) x (n) x((n )) X () X() ; X () X() X () w X(). j X () 6 X () X(). + j X () X() X (5) X(5). j X (6) X (7) X(7). + j X () [,. j,,. + j,,. j,,. + j]. b. Consider a sequence x(n) (,,,, 5,,,,,, 7, ). Evaluate the following without explicitly computing X(K). i) DFT [DFT[DFT[x(n)]]]]. ii) X( K) K π j 6 iii) e X( K) ( Mars) K -Digital Signal Processing Dec.-Jan..indd 5 9// ::57 AM

6 Dec. /Jan. Digital Signal Processing Updates Ans: x(n) [ 5 7 ] i) DFT [x(n)] X() DFT [DFT (x(n))] [x(n)] DFT[DFT [DFT [DFT [x(n)]]]] [x(n)] [ 5 7 ] [ ] ii) X ( K) ; K n x( n) X( K) n K Let n x( ) X( K) K X( ) x ( ) 96 K jπ iii) e 6 X ( K) ; K jπn + x( ) n e X( K) K jπn K jπn 6 X( K) e 6 X( K) x ( n) K e (96, 6,,, 6,,,,,,, ). c. x(t) is an analog signal having a bandwidth of Hz. It is desired to compute the spectrum of this signal using M point DFT with a resolution better than or equal to 5 Hz. Determine the minimum sampling rate and the resulting resolution (M is an integer). ( Mars) Ans: The bandwidth of x(t) Hz the minimum resolution of sample is 5 Hz o. of samples samples 5 The minimum sampling rate is but, M the minimum sampling rate is samples, since 7 The resultant resolution.hz -Digital Signal Processing Dec.-Jan..indd 6 9// ::57 AM

7 Digital Signal Processing Dec./Jan.. a. Using overlap save method, compute y(n), of a FIR filter with impulse response h(n) (,, ) to an input x(n) (,,,,, 5, 6,,,,, ). Use only point circular convolution. Can the system exhibit linear phase? ( Mars) Ans: Given x(n) (,,,,, 5, 6,,,,, ) Updates h(n) (,, ) Given L (the length of convolution) L M, M, (Segment length) L M + 6 i) to compute y (n) (m )zeroes x (n) 5 h (n) h(n) x (n) zeroes y (n) x (n) * h (n) [ ] Discarding rst two values, we get y (n) ( ) ii) y (n) x (n) [ 5 6 ] h (n) [ ] y (n) x (n) * h (n) ( 5 5 7) Discarding rst two values Y (n) (5 7) iii) Y (n) x (n) (,,,,,,, ) h (n) (,,,,,,, ) Y (n) x (n) * h (n) [ 6 7 ] Y (n) ( ) y(n) (y (n), y (n), y (n)) ( ) y (n) y (n) y (n) -Digital Signal Processing Dec.-Jan..indd 7 9// ::57 AM

8 Dec. /Jan. Digital Signal Processing Updates. b. Show that the product of two complex numbers (a + jb) and (c + jd) can be performed with three real multiplications and five additions. ( Mars) Ans: (a + jb) (c + jd) ac + bd + j(bc + ad)...(a) Let ac bd bc + ad Let Product, P ac P bd P (a + c) (c + d) A (P P) + j(p + (P P )], 5...(B) total number of real products are three i.e. P, P, P and number of additions are fire as indicated in (B).. c. Bring out a comparison between linear convolution and circular convolution. ( Mars) Ans: Circular convolution Linear convolution y( n) x( n) * h( n) y( n) x ( n) *h( n) M ( m) ( ) x h n,m n used for periodic signals Circular convolution results in aliasing x( m) h( n m) m used for non periodic signals Linear convolution results can be obtained from circular convolution by appropriate padding zeroes. a. Using DIFFFT algorithm, compute of the sequence, x(n) (,,,,,,, ) If x (n) n(n ), compute x (K) without invoing FFT algorithm. Ans: Given, x(n) ( ) using DIF FFT, The basic butterfly is a j ( Mars) a + b j j b r (a b) r -Digital Signal Processing Dec.-Jan..indd 9// ::57 AM

9 Digital Signal Processing Dec./Jan. Input Output of stage Output of stage Output of nal stage x() X () 5 X () X() Updates x() X () X () 5 X() x() X () X () 7 7 X() x() X (6) X (6) 7 X(6) x() X () X () X() x(5) X (5) X (5) X(5) x(6) X () X () X() x(7) X (7) X (7) X(7) X(K) (,, 7,, 5,, 7, ) x (n) x((n )) X () X() X () x() 6 X () x() X () x() 7 X () x() 6 X () x() 5 X (5) x(5) X (6) x(6) 7 X (7) x(7) X () ( ). b. Compute the DFT of the sequence x(n) (,,, ) using the Goertzel algorithm. (6 Mars) Ans: Given x(n) (,,, ) According to Goertzd algorithm X() y (n) n where y (n) y (n ) x(n) ;,,, X() y () for, y (n) y (n ) x(n) for n, y () y ( ) x() ; assume y ( ) y () for n -Digital Signal Processing Dec.-Jan..indd 9 9// ::57 AM

10 Dec. /Jan. Digital Signal Processing Updates y () y () x() y () for n y () y () x() y () for n y () y () x() y () 6 for n y () y () x() y () 6 X() y () 6 Similarly, for y (n) jy (n ) x(n) for y (n) + y (n ) x() for y (n) + jy (n ) x(n) n n n n n y (n) y (n ) x(n) y (n) jy (n ) x(n) y (n) + y (n ) x(n) y (n) + j y (n ) x(n) y () y ( ) x() y () y () y () x() y () y () y () x() y () y () y () x() y () 6 y () y () x() y () y () jy ( ) x() y () y () jy () x() y () + j y () jy () x() y () xj y () jy () x() y () y () jy () x() y () y ()+y ( ) x() y () y ()+y () x() y () y () + y () x() y () y ()+y () x() y () y ()+y () x() y () X() 6 X() X() X() X() [6 ] y ()+jy ( ) x() y () y ()+jy () x() y () j y ()+jy () x() y () j y ()+jy () x() y () y ()+jy () x() y () PART B 5. a. Show that the bilinear transformation maps. i) The j axis in s plane onto the unit circle, z. ii) The left half s plane, Re(s) < inside the unit circle, z <. ( Mars) -Digital Signal Processing Dec.-Jan..indd 9// ::5 AM

11 Digital Signal Processing Dec./Jan. Ans: In bilinear transformation [ z ] [ + ] S T z S + j (A) Z re jw Substituting z in equation for S, we get jw S re T re jw + ( ) ( ) r cosω+ jsin ω T cosω+ jsin ω + Updates r rsinω + j T + r + r cosω + r + r cosω Comparing equation A and B r σ T + r + rcosω rsinω Ω T + r + rcosω From the above,. If r <, <, this means left hand side of the s plane is mapped inside the circle z. If r >, >, i.e, the right hand side of s plane is mapped outside circle z. If r,, i.e, imaginary axis in 's' domain is mapped to the circle of unit radius centered at z in z domain. I m (z) < > R e (z) 5. b. Design a digital low pass filter using the bilinear transformation method to satisfy the following characteristics: i) Monotoniv stopband and passband ii). db cut off frequency of.5 rad; iii) Magnitude down atleast 5dB at.75 rad. ( Mars) Ans: -Digital Signal Processing Dec.-Jan..indd 9// ::5 AM

12 Dec. /Jan. Digital Signal Processing Updates K P. db log H() P.5 S.75 db K S 5 db Fig: Magnitude frequency response of the specified lowpass digital IIR filter Step : Prewrp the band - edge frequencies P.5 rad and S.75 rad and T sec to get P and S. ωp.5π Ω p tan tan T ω.75π Ω S T S tan tan. Step : Choose a Butterworth filter to meet the monotonic passband and stopband requirement. Let use design a lowpass filter to meet the following specifications: K.dB log H jω ( ) p a P ( ) and log H jω K 5dB a S S order, K p K S log / Ω P log Ω S.9 Rounding off to the next larger integer, we get. Referring to normalized form Butterworth filter tables, we get H () s s + s+ s Let us determine the cutoff frequency C to meet the passband requirement presidents Ω P Ω c rad / sec Kp -Digital Signal Processing Dec.-Jan..indd 9// ::5 AM

13 Digital Signal Processing Dec./Jan. The prewarped analog lowpass filter Ha(s) is obtained by using lowpass - to - lowpass transformation to H (s). () () That is, H s H s a s s Ω C Updates s + s+ s s + + s s Step : Obtain H(z) by applying bilinear transformation (T ) to H a (s). () () That is, H z H s a z s z + z z z + z ( + z ). +.55z Specification of the design The frequency response of the digital lowpass filter is obtained by letting z e j in H(z). jω ( ) ( ) ( ) ω j ( + e ) That is, H e H ω. +.55e H ω ( ) ( ) Hence, log H().5. db and log H() db Difference equation realization () () ( ) Y z Let + z + z H() z X z.55z +. jω + cosω jsinω. +.55cos ω j.55sin ω + cosω + sin ω (.55cos ω+.) + (.55sin ω) Cross - multiplying, we get.55z - Y(z) +. Y(z) X(z) + z - x(z) + z - X(z) Taing inverse Z - transform on both the side and rearranging, we get y(n).75y(n ) +.9x(n) +.557x(n ) +.9x(n ) 5. c. Bring out a comparison between Butterworth filter and chebyshev filter. ( Mars) -Digital Signal Processing Dec.-Jan..indd 9// ::5 AM

14 Dec. /Jan. Digital Signal Processing Updates Ans: Difference between Butterworth and Chebyshev filters: Parameter Butter worth Chebyshev Frequency response Monotypically decreasing Ripples in pass band and monotomic in stop band Transition band Broader in given arrow for given Poles of Ha(S) Poles lie on circle Poles lie on ellipse Order for given specs Higher Lower 6. a. Transform s+ a H( s) ( s+ a) + b into a digital filter using impulse invariance technique. ( Mars) Ans: () Given H s s+ a ( ) s+ a + b s+ a ( s+ a+ jb)( s+ a jb) In partial fraction form () Hs A + B ( s+ a+ jb) ( s+ a jb) ( ) ( ) A s+ a+ jb H s s+ a ( s+ a jb) () s a jb s a jb a jb+ a jb + a jb+ a jb jb + Similarly B + + Hs + ( s+ a+ jb) ( s+ a jb) in impulse invariant transformation, s s e Z i sit + + H() z + e Z e Z at jbt at+ jbt 6. b. Let H ( s), for a nd order low pass Butterworth filter prototype. Determine the system s + s+ function for the digital bandpass filter using bilinear transformation. The cut off frequencies for the 5π 7π digital filter should lie at ω L and ω u. Tae T. ( Mars) Ans: -Digital Signal Processing Dec.-Jan..indd 9// ::5 AM

15 Digital Signal Processing Dec./Jan. () Hs s + s+ digital cut off frequencies are 5π ω L 7π ω n T Upon prewarping the edge frequencies considering T, ωl Ω L tan.767 T ωn Ω n tan. T a () () H s H s () Hs s +ΩnΩl s s( Ω u Ω l) s + s.559 s Updates Ha () s s + s s.559s 5 s s +.7s s + s.7s Using bilinear transformation () () z H z H s a s ;T T + z H a (s).s +.6s s +.s z z z+ z+ H() z z+ z+ z z 6. c. hat does linear phase do to the response of an input signal within the passband of the filter? hy choose an IIR filter instead of an FIR filter? ( Mars) Ans: A lter is said to have linear phase, if the phase response of the lter, i.e, phase Vs frequency is a linear function of frequency. On the passband of the filter, to an input signal a liner phase filter provide time shift of constant amount for all frequency components. Even if FIR filter provides linear phase response, it requires high orders and hence more hardwares than IIR filter. Therefore IIR filter we widely used which can approximate linear phase response. 7. Find the unit sample response of a symmetric FIR filter having a length of 9 samples. The desired jωα π frequency response is given by, Hα ( ω) e ω ωc, here ω C and ω ω c -Digital Signal Processing Dec.-Jan..indd 5 9// ::59 AM

16 Dec. /Jan. Digital Signal Processing Updates Ans: πn ω Hanning ( n) cos, n ( ). Also find H(z), linear constant coefficient difference equation and the frequency response H(e j ). Draw the structure of the filter. ( Mars) Given Hd() e j c > c c / πn ω Harning () n cos n 9, α a. The inverse DTFT of Hd() π jωn hd() n Hd( ) e d π π ω ω () π π o π ( n α) ( ) π sin π ωα j jωn e e dω n π π n α hd e d ω ; ; n π The unit sample response of the lter is n(n) hd(n) Hanning (n) n πn ω Hanning.5.5cos n π sin ( n ) πn.5.5cos ; n π () ( n ) h n π n.5.5cos ; n The coefcients of h(n) is tabulated below n hd(n) Han (n) h(n) hd(n) Han (n) / Digital Signal Processing Dec.-Jan..indd 6 9// ::59 AM

17 Digital Signal Processing Dec./Jan b. H() z () n h n z n Updates z.9z.75z.9z.6z.6z +.9z +.75z +.9z.6z () () 5 7 y z H() z.6 z + z +.9 z + z +.75 x z 7 5 c. The linear unit coefficient difference equation is h(n).6 [x(n ) + x(n 7)] +.9 [x(n ) + x(n 5)] +.75 x(n ) d. Since odd, h(n) h( n), iω jω H( e ) e h h() n cos n + ω n + ω n jω e h h n cos n () () ( ( )) ( () () () () () ) jω e h h cos h cos h cos h cos + ω+ ω+ ω+ ω ( ) jω e.75.9cos.7 cos ω+ ω e. The structure of the filter is x(n) z z z z z z z y(n) -Digital Signal Processing Dec.-Jan..indd 7 9// ::59 AM

18 Dec. /Jan. Digital Signal Processing Updates. a. Obtain the series and parallel form realization for a digital filter described by the system function, z z + z H () z. z z z+ ( Mars) z z + z Ans: Given H () z. z z z+ v (n) - z - H (z) x(n) v (n) -6 y (n) y (n) + y(n) y (n) z - - z - H (z) 6 H (z) Fig: Parallel realization of H(z). To obtain parallel realization, Dividing the numerator and the denominator by z -, we get () H z z + z z z z + z Let z - v. Then, the above equation becomes () H z v+ v v v v+ v A Bv+ C + + D v v+ v -Digital Signal Processing Dec.-Jan..indd 9// ::59 AM

19 Digital Signal Processing Dec./Jan. here A, B, C and D are the constants associated with the partial fraction expansion. These constants are found to have the following values. A, B, C 6 and D 6 () z 6 Thus, H z z z + z H z + H z + H z () () ()() Updates. b. Determine the parameters K m of the lattice filter corresponding to the FIR filter descirbed by, H(z) +.z +.z +.7z. (6 Mars) Ans: Given H(z) +. z +. z +.7z a (), a ()., a ()., a ().7 here m, for m m, m,...,, do m a m (m) a m(i) a m(m)a m(m i) a m (i) i m m if m, a ().7 if m, a () when m, and i a() a()a() a()..7. (.7) a() a()a() for m, a () a() Digital Signal Processing Dec.-Jan..indd 9 9// ::59 AM

20 June/July Fifth Semester B.E. Degree Examination Digital Signal Processing Time: hrs. Max. Mars: ote:. Answer any FIVE full questions, selecting at least two questions from each part. PART A. a. Dene DFT. Derive the relationship of DFT to Z transform. (5 Mars) Ans: Denition: The sampled discrete time fourier transform of a finite length discrete time signal in nown as discrete Fourier Transform (DFT). Consider a discrete time signal whose DTFT is X(w) x() n e n jwn upon sampling X (w) using total of equally spaced samples in the range jπn π i,e X( w) X X() x() n e n this is the equation of DFT. Relation to Z transform By definition Z transform of a sequence x(n) is n () x() n Z X z n Let Z e jw and sample X(e jw ) at equally spaced points on a circle () () () n () jπ X X z Z e x n Z x n e jπ n n jπ Z e jπ Z e, we get If the length of x(n) is less than or equal to, then the sequence can be recovered from its point DFT. Hence its Z transform is uniquely determined by its point DFT.. b. An analog signal is sampled at Hz and the DFT of 5 samples is computed. Determine the frequently spacing between the spectral samples of DFT. ( Mars) Ans: Sampling rate f s KHz o. of samples 5 Frequency spacing between spectral samples D f f s / 9.55 Hz -June July.indd // ::6 PM

21 Set - June Digital Signal Processing. c. Consider the nite length sequence x(n) (n) (n 5): Find i) The point DFT of x(n) (ii) The sequence y(n) that has a DFT y(k) jπ e X(K) where X(K) is the point DFT of x(n) (iii) The point sequence y(n) that has a DFT Y(K) X(K) (K) where X(K) is the point DFT of x(n) and (K) is the point DFT of u(n) u (n 6). ( Mars) Ans: Given x(n) (n) (n 5) i) point DFT of x(n) 9 n X( ) x( n) 9 n e π j [ ] n X( ) δ( n) δ( n 5) 9 n 9 n n n n X() w X() w 5 x X() X() X(9) X() ( ) π j ii) y( n) has DFT y( ) e X( ) here X() is DFT of x(n) x(n) (n) (n 5) π j y( ) e X( ) X( ) Using time shifting property y(n) x((n )) (n ) (n 7) iii) y(n) has pt DFT Y() X() () DFT X( ) x ( n) ; x( n) δ( n) δ( n 5) DFT ( ) w ( n) ; w( ) u( n) u( n 6) by using the property DFT x ( n) *w( n) X( ) ( ) y(n) x(n) * w(n) [(n) (n 5)] * [u(n) u(n 6)] -June July.indd // ::5 PM

22 Digital Signal Processing June Set - u(n) u(n 6) (n) + (n ) + (n ) + (n ) + (n ) + (n 5) y(n) [(n) (n 5)] * [(n) + (n ) + (n ) + (n ) + (n ) + (n 5) (n) + (n ) + (n ) + (n ) + (n ) + (n 5) (n 5) (n 6) (n 7) (n ) (n 9) (n ) y(n) [ ]. a. Determine the circular convolution of the sequences x(n) {,,, } and h (n) {,,, } using DFT and IDFT equations. ( Mars) Ans: x(n) {,,, } () () X x n n Similarly, () n H + + w + () () () ( )( ) Y XH () Y y n 7, 9,, () { }. b. Let X(K) be a point DFT of a length real sequence x(n). The rst samples of X(K) are given by: X(), X() + J, X() + j, X() J5, X() +, X(5) 6 + J, X(6) J, X(7). Determine the remaining samples of X(K). Also evaluate the following functions without computing the IDFJ. j j () () () () i) X ii) x 7 iii) x n iv) x n ( Mars) n n Ans: Since x(n) is a real sequence, the following symmetry condition must be satised: X () X* ( ), Since,, X () X*( ), -June July.indd // ::5 PM

23 Set - June Conjugate Symmetry: X() X*( ) Digital Signal Processing 7 Fig: Conjugate symmetry of X() for real x (n) Here, X () X*(6) + j X(9) X* (5) 6 j X () X*() j X() X* () + j5 X() X* () j X() X*() j The result is tabulated below: K X() K X() 7 + j + j + j 9 6 j j5 j + j + j j j 6 j j a. From the denition of point inverse DFT, we have n x() n X(), Since, we get n x() n X(), Letting n on both the sides of the above equation, we get x() X() X () + X ()... + X ( ).57 -June July.indd // ::5 PM

24 Digital Signal Processing June Set - b. Letting n in the expression for x(n), we get x X() π j K Xe () jπ Xe () X ()( ) Since, we get x 7 X.57 c. By denition: () ()( ) { ()} () () DFT x n X x n, X n () () n x n Letting on both the sides of the above equation, we get () () X x n n d. From Parseval's theorem: () X() x n n x n () X() n n n a. Consider a FIR lter with impulse response, h(n) {,,, }. If the input is x(n) {,,,,,,,,, 5, 6,,, }, using the overlap save method and point circular convolution. ( Mars) Ans: Given h(n) ( ) x (n) { 5 6 } the length of subsequence L and length of h (n), split x(n) into sequence of length ( + ) 5 h, ( 5 6 ) -June July.indd 5 // ::5 PM

25 Set - June Digital Signal Processing To nd y (n) h n x n points 5 points y n x n h n, 5,,,,,, 7 c To nd y (n) x n 5 last points y n x n h n, 7,,,, 5,, 6 c To nd y (n) x n 5 6 last points y n x n h n,, 7,,, 5,, c By discarding rst points of each individual output. y(n) 5 7 Discard discard 7 5 discard we will get y(n) ( ). b. hat are FFT algorithm? Prove the i) Symmetry and ii) Periodicity property of the twiddle factor. (6 Mars) Ans: Fast Fourier transforms are algorithms which are used for computing the discrete Fourier transform in a numerically efcient manner. There are two different FFT algorithm.. Decimation in time FFT. Decimation in frequency FFT Symmetry of Twiddle factor jπ e jπ ( a+ ) jπa jπa a+ π j a a+ a e e. e e Periodicity property: -June July.indd 6 // ::5 PM

26 Digital Signal Processing June Set - ( ) jπ a+ jπa jπa a+ jπ a a+ a e e. e e. c. How many complex multiplications and additions are required for computing 56 point DFT using FFT algorithms? ( Mars) Ans: Complex multiplications and additions required for computing 56 point DFT using FFT. 56 o. of complex addition required log 56 log o of complex multiplication required log log56. a. Find the DFT of the sequences x(n) {,,,,,,, } using the decimation in frequency FFT algorithm and draw the signal ow graph, Show the outputs for each stage. ( Mars) Ans: a a+b r b a b r The scale factors r are as follows: 5 j + j j + j 6 7 j + j Also, recall that: The ow diagram is an shown in Fig. Ex (a) -June July.indd 7 // ::5 PM

27 Set - June Digital Signal Processing To compute the output of rst stage X () + 5 X () X () + 5 X () X () + 5 X (5) X (6) + 5 X (7) To compute the output of the second stage X () X () + j X () 5 5 X () + j X () X (5) + j X (6) X (7) + j -June July.indd // ::5 PM

28 Digital Signal Processing June Set - To compute the output of the third stage X () X () + X () X () j + ( j) 5. j. X () X () + X () X () + j + ( + j).7 j. X () () X (5) X (5) j ( j).7 + j. X (6) X (6) X (7) X (7) + j ( + j) 5. + j. Since x (n) is a real sequence, we find that the symmetry condition: X () X* ( ) is satisfied.. b. Given x(n) {,,, }, nd x() using the Goertzel algorithm. (5 Mars) Ans: According to Goertzel algorithm. X () y (n) n Since, and, we get X () y () and y () n y ( n ) x() n Since, the recursive equation becomes y (n) + y (n ) x (n) Case (i): n y () + y (n ) x () Assuming y ( ), we get y () + y ( ) x () Case (ii): n y () + y () x () y () + y () Case (iii): n y () + y () x () y () y () Case (iv): n y () + y () x () y () + y () Case (v): n y () + y () x () y () y () Hence, X() y () -June July.indd 9 // ::5 PM

29 Set - June Digital Signal Processing. c. rite a note on chirp z transform algorithm. (5 Mars) Ans. The DFT of an point data sequence can be viewed as Z - transform of x(n) evaluated at equally spaced points of a unit circle. If we consider evaluation of X(Z) on other contours in the Z - plane, Let it be at points Z n Then X (z ) x( n) Z K ;,,,L Let V R e j X(z ) V K n ( n) gn ( ) V n This can be interpreted as convolution sum of g(n) with h(n) where h(n) n V The sequence h(n) j Re φ n if R, h(n) e jwn w nφ n j o e φ here we can see that frequency increases linearly with time. Such ind of signals are called chirp - signals. Chirp signals are used in Radar systems. Chirp Z transform has been implemented in hardware to computer DFT signals. jω 5. a. Given that ( ) jω Ans: Given ( ) PART B H e, determine the analog Butterworth low pass lter transfer function Ω (6 Mars) H e, Ω jω H( e ), ( 6 ) + 6Ω ( ) Ω + Ωc Ω + Ω + ΩC Ω Ω ; ΩC Ω C -June July.indd // ::5 PM

30 Digital Signal Processing June Set - () () () for, H S a s s Ω c ( S + S+ )( S+ ) ( S) + S + )( S + ) S S S H S H S lowpass to low pass transformation b. Design an analog Chebyshev lter with a maximum passband attenuation of.5 db at p rad/ sec and the stopband attenuation of db at S 5 rad/sec. ( Mars) Ans: log Ha(j).5dB db cut off frequency of normalised filter Kp log δ p.5 + p rad/s 5 Ω s.5 s 5 rad/sec K S db log S δ s.67 Ωp. Ωs.5 ( δ ) p.775 d.79 δ min filter order s 999 ( d) () Cosh.766 Cosh Since, -June July.indd // ::5 PM

31 Set - June Digital Signal Processing ( ) σ asin π b Ω b cos ( ) π b,...6 for, and (ie LHP of S plane) e get, H () S K ( S S)( S S )( S S ) and bo sin ce is odd.559 H () S s +.597s +.9s () () Ha S H s s s Ω p () H s ( s ) ( s ) ( s ) s s c. Compare Butterworth and Chebyshev lters. ( Mars) Ans. Difference between Butterworth and Chebyshev lters: Parameter Butter worth Chebyshev Frequency response Monotypically decreasing Ripples in pass band and monotomic in stop band Transition band Broader in given arrow for given Poles of Ha(S) Poles lie on circle Poles lie on ellipse Order for given specs Higher Lower 6. a. hat are the conditions to be satised while transforming an analog lter to a digital IIR lter? Explain how this is achieved in Bilinear transformation technique. (5 Mars) Ans: hile transforming an analog lter to a digital lter, for the system to be stable following conditions need to be satised.. The j axis in the 's' plane should map into the unit circle in the 'z' plane.. The left hand plane (LHP) of the 's' plane should map into the inside of the unit circle in the Z plane. In bilinear transformation S is replaced with z T + z putting Z re jwe and S + j -June July.indd // ::5 PM

32 Digital Signal Processing June Set - r σ T r + + r cosw rsinw Ω T r + + r cos w a. if r <, <, is, left hand side of 's' plane is mapped inside the circle z b. if r,, is, imaginary axis in mapped to the circle of unit radius centered at z in z domain hence the condition are satised 6. b. Design a Butterworth lter using the impulse invariance method for the following. H(e jw ). Specications: Tae T sec, H(e jw )..6 ( Mars) Ans: Given, T sec and. H(e jw ) w. H(e jw )..6 w pass band ripple p.. pass band edge wp. step band tolerance s. step band edge ws.6 w The analog frequencies corresponds above digital frequencies a calculated using Ω T T second p. s.6 p log ( p ).9 db s log s.97 db to design the equivalent analog filter Ha(s). p/ s/ [( ) ( ) ] log /.7 log( Ωp / Ωs) ( ) round off to next integer for, H ( s) S +.s + Ωp Ω c p/ ( ).755 -June July.indd // ::5 PM

33 Set - June Digital Signal Processing H ( S) H ( S) a S S Ω c s +.s + s s.755 ( s s.755) s ( s +.59) + (.5) To design an equivalent digital filter the transformation is at b e sinbtz ( s+ a) b + e cos btz + e z at at z.59 e sin(.5) z H() z e cos(.5) z + e z.5z.z +.5z 6. c. Determine H(z) for the given analog system function Hs () z transform. Ans: Given () Hs ( s+ a) ( + ) + s a b s+ a ( s+ a+ jb)( s+ a jb) ( s+ a) ( ) s+ a + b by using Matched (5 Mars) in matched z transform, z e +α z ( s ) () αt at at at z e z z cos bt + e ( ) ( ) a+ jb T a jb T wt, ( s+ a+ jb)( s+ a jb) Z e Z e z z at at Z z cos bt + e z Hz z z -June July.indd // ::5 PM

34 Digital Signal Processing June Set - Hdc z Hdc s () H() z () H() s () () at ( e ) z + s a cos bt e a + b at at ( + ) at ( a + b )( e ) a cos bt e Hdc z Hdc s at ( ) z e z Hz () z z cos bt + e at at at ( + ) at ( a + b )( e ) a cos bt e where 7. a. A z plane pole zero plot for a certain digital lter shown in Fig. Q7 (a). Determine the system function in the H() z ( + az )( + bz + bz ) ( + cz ) ( + dz + dz ) giving the numerical values for parameters a, b, c, d and d. Setch the direct form II and Cascade realizations of the system. ( Mars) Im {z} z P z / Z zeros 6 6 P Re {z} P Ans: From the pole zero plot at z there are three zeroes at z / there is pole o o j6 j6 and at z e and e another two poles P is at / o j6 P is at e i.e, at + j o j 6 P is at e i.e, at j -June July.indd 5 // ::5 PM

35 Set - June Digital Signal Processing system function is ( z+ ) z z j z j + ( )( )( ) Divide numerator and denominator by z we get ( ) + z z z + z ( )( ) One simplifying we get, ( )( ) H() z z z z + + a, b, b ( )( z z + + z ) c, d, d The direct form II and cascade realization is obtained as follows. Let H(z) H (z) H (z) x(n) + z z z H () z H () z + z + + z + z and the realization is y(n) / Z / + + H (z) H (z) / 7. b. A FIR lter is given by, () () ( ) ( ) ( ) and lattice structure. Ans: Given y() n x() n + x( n- ) + x( n- ) + x( n- 5 ) () () () Yz Xz z Xz Z Xz () Z Xz () recall the conversion formula for m m, m,..., m a m (m) () yn xn+ xn-+ xn-+ xn-. Draw the direct form 5 ( Mars) 5 () ( ) ( ) a i a m a m i m m m am i i m m since there are three zeroes, tae M taing Z transform on both sides -June July.indd 6 // ::5 PM

36 Digital Signal Processing June Set - for m ; a () () a a () a() a() a( ) (). a.675 ( ) a() a () a(). a ().6565 ( ) for m, a ().6565 () () ( ) () a a a for m, a. The lattice structure is y(n) x(n) Z Z Z here /. a. Design a FIR lter (low pass) with a desired frequency response jw ( ) jw Hd e e ; π π ω π ; ω π Use Hamming window with M 7. Also obtain the frequency response. Ans: By denition, the inverse DTFT of H d () is π jωn hd() n Hd() e d π π ω ω π ( ) ( ) π sin n jω jωn hd () n e e d,n π π π n ω ( Mars) π Also, hd () e d π π ω The impulse response of the FIR lter is -June July.indd 7 // ::5 PM

37 Set - June Digital Signal Processing h (n) h d (n) Ham (n), n 6 where π ω Ham () n.5.6 cos, n π.5.6 cos, n 6 6 Hence, the impulse response of the FIR lter for n 6 is () h n π ( ) ( n ) sin n πn.5.6 cos ( 6 ), n π πn.5.6 cos ( 6 ), n The table below shows the details of the computations for h (n). n h d (n) w Ham (n) h (n) h d (n) w Ham (n) Since is odd, the frequency response of the center symmetric FIR lter is given by H(e j ) H () H(e j ) H() ( ) j ( ) ω e h + h() n cos ω n n e j [h () + h () cos + h () cos + h () cos ] e j ( cos +.9 cos +.7 cos ) e j ( cos.9 cos +.6 cos ). b. Design a linear phase low pass FIR lter with 7 taps and cutoff frequency of C. rad. using the frequency sampling method. ( Mars) Ans: H d () π π 6π π π π Ideal response of linear phase low pass filter is π -June July.indd // ::5 PM

38 Digital Signal Processing June Set - j ( ) ω Hd() ω e <ω<.π. π<ω<π jω e <ω<.π otherwise The magnitude response is symmetric about and phase response is anti symmetric about. for. values of are, for. 5.9 are,,, 5 for 5.9 < < 6,, H(),,,,5, 6 Similarly 6π θ ω for,,, 7 6π ( 7) for,5,6 7 j6π 7 e ;, Hd() ω ;,,,5 j6 π( 7) 7 e ; 6 () IDFTofH () hn ( ) jπn H () + Re He () jπn H () Re He () + j6π j6πn Re 7 7 { e e 7 + } π + cos ( n ), n Filter coefficients are n h (n) June July.indd 9 // ::5 PM

39 December Fifth Semester B.E. Degree Examination Digital Signal Processing Time: hrs. Max. Mars: ote:. Answer any FIVE full questions, selecting at least two questions from each part. PART - A ; n. a. Find the -point DFT of x (n) if x (n) ; otherwise n Ans: X() x(n) K x(n) n { } X() x() + x() K + x() K K + + ; K,,... ( Mars). b. Two finite sequences x(n) [x(), x(), x(), x(), x()] and h(n) [h(), h(), h(), h()] have DFTs given by X(R) DFT {x(n)} {, J,, J}; H(R) DFT {h(n)} {, + J,, J}. Use the properties of DFT and find the following. i) X (R) DFT {h(), h() h(), h()} ii) X (R) DFT {y(n)} where y(n) x(n) h(n) iii) X (R) DFT{x(), h(), x(), h() x(), h(), x(), h()} ( Mars) Ans: Two nite sequences x(n) [x(), x(), x(), x()] and h(n) [h(), h(), h(), h()] have DFT X(R) [, j,, j] and H(R) [, + j,, j] Find the following. i) X (R) DFT {h(), h(), h(), h()} DFT {( ) n h(n)} e jπ π jπ ( ) ej e n n n n X (R) DFT { n h(n)]} H((K )) X (R) H((K )) -Digital signal process december.indd // :7: PM

40 Set - December [, j,, + j]. Digital Signal Processing ii) X (R) DFT {y(n) where y(n) x(n) h (n) y(n) x(n) h(n) DFT {y(n)} X() H() X (R) (, j,, j) (, + j,, j) [, + j, j] iii) X (R) DFT {x(), h(), x(), h(), x(), h(), x(), h()}, X ( R) x ( n) x ( n) + x ( n) Kn Kn Kn n n n 'even n' let n r for even 'n' and n r + for odd 'n', r r r r r x ( ) ( ) r + x r+ r r r ( + ) X ( R) x ( r) + x ( r+ ) K X() + H( );,,, X () X() + H() 'odd n' X ( ) X() + H()... X (7) X(7) + 7 H(7) X() + 7 H() X (R) [,. + j, j,,. + j, + j,. j]. a. Consider a length sequence defined for n, x(n) {,, 7,,,,,, 5,,, } with point DET given by X(R), R, JR 6 evaluate the following function without computing DFT, e (R) (5 Mars) R Ans: Consider the sequence x(n) {,, 7,,,,,, 5,,, } whose point DFT is X(R) πr j 6 evaluate e X( R) with out computing DFT. R X(R) is frequency domain signal. jπr 6 e X( R) x((n )) R jπr 6 e X( R) x((n )) R -Digital signal process december.indd // :7:7 PM

41 Digital Signal Processing December Set - { 5,,,,,, 7,,,,, } { 6,,, 6, 96,,,,,,, }. b. Determine x (n) x (n) x (n) for the sequences, x (n) e jn ; n 7; x (n) u (n) u (n 5). Setch all the sequences. Use time domain approach. Ans: Given X (n) e jn n 7 X (n) u(n) u(n 5) Determine X (n) X (n) X (n). Setch all the sequences ( Mars) x (n) x (n) n n x (n) * x(n) x(m)x (( )) n m n,,..., 7. m n x (m) x ((n m)) y(n) ans ( ) x()(n) n -Digital signal process december.indd // :7:7 PM

42 Set - December Digital Signal Processing. c. Show that: i) Real and even sequence has real DFT. ii) Multiplication of two DFT's in frequency domain corresponds to circular convolution in time domain. (7 Mars) Ans: i) Read and even sequence has leal DFT. DFT Let x(n) X(K) If x(n) is real and even x(n) x( n) n xn ( ) + x( n) ) Let x e (n) DFT {x e (n) / DFT x(n) + / DFT x (( n) / X() + / X(( )) / X() + / X * () / [X() + X * ()] Let X() a + jb X * () a jb DFT {x e (n)} / a + jb + a jb a which is purely real ii) Multiplication of two DFT's in frequency domain corresponds to circular convolution in time domain. DFT {h(n) * (x) (n)} DFT h( l)x( n l) ) l { } h( l) DFT x( n l )) l l Kl h( l) X( ) ( ) X h() l l X() H() Kl. a. Consider a FIR filter with impulse response h(n) {,,, } if the input is x(n) {,,,,,,,,, 5, 6,,,,, }, find the output y(n). Use overlap add method assuming the length of bloc is 7. (9 Mars) Ans: Impulse response h(n) {,,, } input x(n) {,,,,,,,,, 5, 6,,,,, } use overlap method. Length of bloc is 7. L m 6., bloc length 7 i) x (n) (,,,,,,, ) h(n) ( ) y (n) x (n) * h(n) [ 7 ] ii) x (n) [ 5 6 ] h (n) [ ] -Digital signal process december.indd // :7: PM

43 Digital Signal Processing December Set - y (n) x (n) * 6 h(n) [ ] iv) x (n) [,, ] h(n) [ ] y (n) x (n) * h(n) [ 6 7 ] to find y (n) y(n) b. rite a note on Chirp z transform. (6 Mars) Ans: Chip z transfer is used for evaluating Z transform of an M point sequence which lie on a spinal contour beginning at any arbitrary point in Z plane. The DFT can be considered as the Z transform on a sequence evaluated on a unit circle. By definition. ( ) K X Z x(n)z, K,,...,M n n K j π Let Z e,,,..., Let Z K be a point an the spinal centered about the origin. Z K AB K where A A o e j o B B o e j o i) B o <, the sequence spinal towards the origin. ii) B o >, the sequence spinal away from origin. iii) B o, the sequence will be in circle of radius A o iv) hen A o, B o, B e j /, X(Z ) will be DFT. n n XZ ( ) ( ) xna B o m n Let n / [n + (n ) ] n ( ) n n n XZ ( ) ( ) xna B B B Let us define g(n) x(n) A n B n / h(n) B / XZ ( ) ( ) ( ) gn hn,o M h ( ) n replace by n and n by. XZ ( ) ( ) ( ) n g h n,o n< M hn ( ) [ gn*hn ( ) ( )] hn ( ) I{z} A o B o < Re{z} -Digital signal process december.indd 5 // :7: PM

44 Set - x(n) December g(n) g(n) g (n) * h (n) X(Zn) h(n) Digital Signal Processing A n B n A n h(n) h(n) X(Zn) is the chirp Z transform. c. hat is in place computation? hat is the total number of complex additions and multiplications required for 5 point, if DFT is computed directly and if DFT is computed directly and if FFT is used? Also find the number of stages required and its memory requirement. (5 Mars) Ans: In place computation: hile calculating the FFT of a sequence, the algorithm is performed in stages. Each stage has butteries which operates on a pair of complex numbers and produce another pair of complex nos. Once output pair is calculated, input pair is no longer needed and the output can be can be stored in the same location of input pair. Therefore for an point DFT we need complex registers and real registers. The same registers are used for other stages also. This is called in place computation. Given 5. Direct DFT o. of addition required ( ) 66 o. of Multiplication required 6 FFT o. of addition log 6 o. of multiplication / log Total no. of stages required to calculate 5 pt FFT is 'seven' Totally real registers are required is real registers are required.. a. Derive DIT FFT algorithm for and draw the complete signal graph. ( Mars) Ans: In this approach the point DFT is successively decomposed into smaller DFTs. Because of this, the number of computations are reduced. The number of points is assumed as power of two i.e. p. In Decimation in time approach, point transform is divided into two / point transform, then each / point to / point transform and continue dividing till we get a point transform. If x(n) is an point sequence where is a power of, divide x(n) into two / point sequences where one composed of even indexed and other odd-indexed values. e now that X () () xnw n Decompose RHS into even-indexed and odd indexed values, n n we get () () + () X xn xn n n even n odd n let n r so every r, n even values if n r +, then every r, n odd values e can rewrite above eqn. as n r ( + ) () ( ) + ( + ) X xr xr r r r -Digital signal process december.indd 6 // :7: PM

45 Digital Signal Processing December Set - but r r r r r r () ( ) + ( + ) X xr xr G () + H, () Here G() + H() are two point sequences. G() and H() are periodic with period X() G () + H, () K G + + H +, Each of this point sequences can be further decimated into two point sequences. i.e G( K) ( ) r g r l ( l+ ) r g( l) + g( l+ ) l l l l g ( ) + g ( + ) l l A + B ; K l l () () also A() and B() are periodic with period K and G () A () + BK, ( ) K K A K+ + B, + K C () + DK, ( ) Similarly H() K C + + D, + e can continue this process till we get point DFTs. The nal signal ow diagram of radix, point decimation in time FFT algorithm is shown below, where each building bloc (specimen buttery) is as follows : Specimen buttery -Digital signal process december.indd 7 // :7: PM

46 Set - December Digital Signal Processing Signal ow diagram x() X() x() X() x() X() x(6) X() x() X() x(5) 5 X(5) x() X(6) x(7) 7 X(7). b. Find the IDFT of X(R) {, + j, j, j, + j, j, j} using inverse Radix DIT FFT algorithm. ( Mars) Ans: X() g() h() / x() X() g() h() w / x() X() g() w h() / x() X() X() w g() g() w h() h() / / x(6) x() X(5) w g(5) h(5) / x(5) X(6) w g(6) w h(6) / x() X(7) w g(7) w h(7) / x(7) -Digital signal process december.indd // :7: PM

47 Digital Signal Processing December Set - j 77 + j j.77 Output of First stage X() g() X() + j g() + hj X() j g() X() j g() j X() X(5) + j w w g() g(5) [( + j) ( + j)]x X(6) j w g(6) [ j j) j X(7) j w g(7) Output of first stage is [, + j,, j,,,, ] Output of second stage g() h() g() + j h() g() w h() g() j g() w h() h() g(5) h(5) g(6) g(7) Output of second stage is [,,,,,,, ] w w h(6) h(7) -Digital signal process december.indd 9 // :7:9 PM

48 Set - December Digital Signal Processing Output of Third stage g() x() / g() g() g() g() g(5) g(6) x() / x() / x(6) / x() / x(5) / x() / g(7) Final Output is [,,,,,,, ] PART - B x(7) / 5. a. Design a Chebyshev analog low pass filter that has db cut off frequency of rad/ sec and a stopband attenuation of 5 db or greater for all radian frequencies past 5 rad/ sec. ( Mars) Ans: Ω s 5 Ω s.5 Ωu the specification of normalized fitter is, s.5 log (j) O p db s 5 db s.5 p log Digital signal process december.indd // :7:9 PM

49 Digital Signal Processing December Set - δ p.95 + Ks 5 db log s s.56 Ωp K. Ω.5 Order s ( δ ) p d.56 δ s () d ( ) Cosh. Cosh to find H (s), a b +.65 K a Sin[( ) / ] a Sin [(K ) 6 π ] K b Cos [( ) / ] b Cos [( ) / 6 ]; for,,..., 6 for,, ie, for left half poles of H (), & are H(s) K ( s s )( s s )( s s ) ( s +.9 j.9)( s +.96)( s +. + j.9) K Since is odd K b o.559 s s +.9s H(s) s +.597s +.9s on applying low pass to low pass transformation Ha(s) H (s) s s / u s / 559. s s + 9. s K -Digital signal process december.indd // :7:9 PM

50 Set - December Digital Signal Processing Verification Let s j we get 559. Ha( jω ) Ω Ω + Ω+ ( ) Ha j Ω j 59.7 j ( Ω ) + ( 9.Ω Ω ) at log H(j) db at 5 log H(j) 5 db hence verified. 5. b. Derive the s to z plane transformation based on finite bacward difference method. Also show that entire left half s plane poles ar mapped inside the smaller circle of radius / centered at z / inside the unit the unit circle in the z plane. ( Mars) Ans: Refer (q. 7b, Dec. In bacward difference method, the differential equation of analog lter is approximated by equivalent difference equation. dy() t y( nt) y( nt T) dt T t nt Let y(n) y (nt) dy() t y( n) y( n ) dt T d() y t S sy() s dt y( n) y( n ) Z z Similarly T T y () z thus the analog domain to digital domain transformation can be achieved by letting s z Similarly s T where is the order of derivative z T H(z) Ha () s z s T z e have s T z st Let s + j we get z Ω j T z ( σ T) + ( ΩT) this is the necessary transformation. -Digital signal process december.indd // :7:9 PM

51 Digital Signal Processing December Set - from above we can see that, < implies z< letting, we get z j ΩT + jωt z ½ Ω j T ( ) z It shows image of j axis of s domain, which is a circle of radius.5 centered z.5 in z domain. This circle is in inside the unit circle. the LHP of j axis maps inside this circle. 6. a. Obtain the direct form II (canonic) and cascade realization of ( z )( z + 5z+ 6)( z ) H(z) ( z + 6z+ 5)( z 6z+ ) Ans: the cascade system should consist of two biquadratic sections, To obtain direct form II ( z )( z + 5z+ 6)( z ) H(z) ( z + 6z+ 5)( z 6z+ ) ( z + 5z+ 6)( z z+ ) ( + 5z + 6z )( z + z ) ( z + 6z+ 5)( z 6z+ ) ( + 6z + 5z )( 6z + z ) ( Mars) i) Cascade realization + 5z + 6z z + z H () () z, H z + 6z + 5z 6z + z H(z) H (z) H (z) X(z) Y(z) Z Z Z Z 5 6 H (z) H (z) ii) Direct form II realization ( + 5z + 6z )( z + z ) Hz () ( + 6z + 5z )( 6z + z ) Upon simplifying + z z 9z + z Hz () z + z + z -Digital signal process december.indd // :7:9 PM

52 Set - December Digital Signal Processing the direct form II realization X(z) Y(z) Z Z Z 9 Z 6. b. Given H(z) ( +.6z ) 5 Ans: i) Realize in direct form ii) Realize as a cascade of first order section only iii) As a cascade of st and nd order sections. i) Realize in direct form upon expansion, H(z) + z +.6 Z +.6 z +.6 z z 5 the direct form realization is x(z) y(n) ( Mars) Z Z Z.6 Z Z Z ii) Realization as a cascade of I order section x(n) Z Z.6 Z.6 Z.6 Z.6.6 y(n) -Digital signal process december.indd // :7: PM

53 Digital Signal Processing December Set - iii) Realization as cascade of I and II order systems. H(z) (+.6z ) 5 ( +.z +.6z ) ( +.z +.6Z ) ( +.6 z ) the realization is x(n) y(n) Z. Z Z..6 Z Z a. Using rectangular window technique, design a lowpass filter with passband gain of unity. Cut off frequency of Hz and woring at a sampling frequency of 5 KHz. The length of impulse response should be 7. ( Mars) Ans: fc Hz ormalized frequency fc fs. c f rad/sec c <.π rad Hd(w) elsewhere fc Hz for a rectangular window. o n 6 ( ) R n other wise hd(n) π π π Hd(w)e π.π jwn e dw.π jwn.π jwn dw e n n π j.π ej.πn e j.πn πnj sin. π n πn.π o for n ( ) π.π e dw π.π+.π.π the coefficients of hd(n) and h(n) are tabulated as n hd(n) R (n) h(n) hd (n) R (n) Digital signal process december.indd 5 // :7: PM

54 Set - December Digital Signal Processing Since is odd jw jw ( ) ( ) He H e h hn ( ) + Cosw n n e w ( Cos w +.65 Cos w +.7 Cos w) 7. b. ith necessary mathematical analysis, explain the frequency sampling technique of FIR filter design. ( Mars) Ans: In frequency sampling technique we sample the Hd(w) at equally spaced points in the interval to πk by replacing w by w for,..., w H() Hd(w) w w,,..., π Hd for is odd we must have H() real and H( ) H*(),,,... for is even, we must have H*() real, H( / ) and H( ) H*(),,,... h(n) is obtained by taing IDFT of H() jπn is, h(n) H( ) e,n,,..., (A) for is odd let H( ) H*(),,,... jπn h(n) H( o) + Re{H( ) e Similarly for is even jπn h(n) H( o) + Re{H( ) e -Digital signal process december.indd 6 // :7: PM

55 Digital Signal Processing December Set - upon taing z transform for eqn (A) j πn H(z) H( ) e Z n n j H ( ) π e Z n Z H ( ) jπ e Z the realization is as shown below. z x(n) Z n π j e Z () H () H y(n) π ( ) j Z e H( ). a. Design a digital filter H(z) that when used in A/D H(z) D/A structure, gives an equivalent analog filter with the following specifications: PB Ripple. db PB Edge : 5 Hz SB attenuation 5 db SB Edge : 75 Hz Sample rate : Hz The filter is to be designed by performing a bilinear transformation on an analog system function. Use Butterworth prototype. Also obtain the difference equation. (5 Mars) Ans: Given data p db f p 5 Hz; s 5 db f st 75 Hz f sampling Hz p f p s f st 5 Ωp p π.57 f sam p Ωs 5π.56 f sam Specification of equivalent dig. fitter ωp p T tan -Digital signal process december.indd 7 // :7: PM

56 Set - assume T December.57, 'p ( ) Digital Signal Processing.56 Similarly 's tan ( )..sdb log. pdb.9 log Ωs () Ωp for, the transfer function of normalized low pass butterwoath tter H (S) S + S+ Ω' p The cut of frequency c. p Ha(s) H () s s s Ω c s + s + s s + + s + s+ S S Applying bilinear transformation z H(z) Ha(S) S z + z z z + z ( ) + z H() z.+.55 Z. b. If Ha(s) ( s+ )( s+ ) ;find the corresponding H(z) using impulse invariance method for sampling frequency of 5 samples/ sec. (5 Mars) Ans: H a (s) ( ) )( ) Sampling frequency is 5 samples sec c c H a (s) s+ + s C ( ) H () a s s + s -Digital signal process december.indd // :7: PM

57 Digital Signal Processing December Set - C ( ) H () a s s + s H a (s) ( ) ( ) The sampling frequency F s 5 Hz Sampling period T F s 5 Impulse invariance transformation is given as s p PT e z s+ e. z s+ e. z H(z) e z.z 67. z H z) z e z.. z. z+. 5 -Digital signal process december.indd 9 // :7: PM

58 Fifth Semester B.E. Degree Examination Digital Signal Processing Time: hrs. Max. Mars:. (a) Let x p (n) be a periodic sequence with fundamental period. Consider the following DFTs. DFT ( ) ( ) DFT ( ) ( ) x X p x n X p hat is the relationship between X () and X ()? Sol. DFT x () n X () p n () p() n n n n xp() n w + xp() n + xp() n w n n n X x n w n n n ( + ) n n ( + ) p + p + + p( + ) n n n () () ( ) X x n x n w x n w () ( ) n n n () xp n xp n w w xp() n w w n n n () + () + n n K n K xp n xp n xp n n n n () + () + () K n p n ( ) () + + x n + + X K ( ) (b) Compute the -point DFT of the sequence x(n) an n Sol. X () () xn n n December - a n n n n b b,b n b, after differentiating Dec.indd // 9::9

59 Set - Dec n nb n ( b ) b ( b ) ( b ) Digital Signal Processing n nb n ( ( ) + ) ( b ) b b b Similarly n n ( ) { ( ) } + n ( ) ( ) ( ) ( ) w + n n ( ) a X () a n ; ( ) a, X () a, ( K ). (a) Given the point sequence () xn n n 7 Compute the DFT of the sequence x (n) using properties of DFT, n x ( n), n, 5 x 7 Sol. Given x() n [ ] x () n [ ] From the sequence it is clear that () ( ) x n x( n 5 ) Using circular time shifting property, if DFT {x(n)} X() m DFT{x(n m) } () ; X K ; K 7 5K DFT {x((n 5) } ( ) Dec.indd // 9::

60 Digital Signal Processing Dec. Set - given e now that, j,, j j 5 j,, + x(n) ( ) remember that 6 7 j j, + 7 X() () n X + w + w + w () x n w, n () + () + () + () + () x() 5 w + x() 6 w + x() 7 w x w x w x w x w x w + w + w + w w w, w w, w w and so on 9 () 6 () 6 9 () () 5 5 () 6 () 7 () X + w + w + w j. X + w + w + w X + w + w + w. j X + w + w + w X 5 + w + w + w +. j X 6 + w + w + w X 7 + w + w + w +. j X( K ) {,. j,,. j,,+. j,,+. j Let X () DFT{x (n)} 5 DFT x ( n 5) w X {( )} () X () X () X() 5 X() +. j X () X() 5 X () X() +. j X () X () 5 X (5) X (5). j X (6) X (6) Dec.indd // 9::5

61 Set - Dec Digital Signal Processing Ans. Ans. X (7) 5 X (7). j X () DFT {x (n)} [, +. j,, +. j,. j,,. j] (b) hat are the methods used to perform fast involution. Explain any one method giving all the steps involved to perform fast involution There are two basic methods to perform fast involution. This will be done when the data sequence is very long. These methods are (i) Overlap - add method (ii) Overlap - save method in both methods, the input signal is segmented into section of manageable length and then perform fast involution if each section and nally combine the output. Steps involved to perform overlap - add method is given below. Consider two sequences x(n) and h(n) having length (here is large integer) and M respectively. Let x(n) is subdivided into sections of length. (i) Decide the number of points for DFT to be computed. Let, L be the point, L M where M is length of h(n). h(n) is appended with zeroes. Let this be h (n). (ii) The input sequence x(n) is segmented into blocs of length such that, L M +, therefore each x' n, x' n segment is padded with M zeroes. Let each segment be () () (iii) The DFT of h(n) and x' () n, x' () n is obtained. (iv) The complex product Y' () H' () X' () is computed. (v) The IDFT of Y' () is obtained is, y' ( n) (vi) In a similar manner as in step (iv) and (v) we obtain Y' () n,y' () n (vii) y' () n y() n ; n L y' () n y () n ; n L (viii) The sequence y (n) is time shifted to produce y (n ) (ix) Last M points of y (n) and rst M points of y (n) are overlapped and added. (x) Repeat the steps until the end of data reached. Thus y(n) y (n) + y (n ) + y (n ) + where each addition term is overlap and add. (c) Given the sequences x(n) cos n x(n) h(n) y c (n) x(n) h (n) πn x(m) h ((n m)) m and h(n) n. Compute the point circular involution. Dec.indd // 9::5