山东大学2018-2019学年2学期数字信号处理（双语）课程试卷(A)答案与评分细则 cosπ(n-M/2）sinπ(n-M2) (n-M/2)(n-M2) -<n< 5r-em）0.5-(-my H(）= 1-e-a+b 1-e-a- (b)(3pts)Anti-symmetry of h os(M-n-M/2)sin(M-n-M2) Since>0,(-e)°→0.(-ee-r）°→0 (M-n-MB2)(M-n-M/2) 0.5T 05T05l-e:1-ee-） .osap-川_n-a.-osa-B+nza-g=-r问.mce He)-e-e- -er1-ee-y】 (M2-m）x(M2-mj (n-M2) π(n-M/2) h问=csxa-AM2_smxa-Mp T(1-eacos(bT)z-） (7pts) (m-M/2)x(n-M2) -oo<n<oo,ie.har [M-n]=-hse [n] 1-2e cos(bT)ROC:>e (c)2pts)Since h=hw[w[n,hw[M-=-h[可，wM-可=wn,we get h[M- 6.(15 pts)Solution: (a)xsc]4159879878J (7 pts) n]=-h[n].hin]is anti-symmetry. b)xn=015987987841 (6 pts) (d)(2pts)If M9,is typeIV FIR generalized linear-phase system. (c)x3L [n]is not equal to xac[n]. (2pts) (e)(2pts)If M10,is type IlI FIR generalized linear-phase system. 7.(10 pts)Solution: (a)The flow graph is from decimation-in-time FFT algorithm. (Ipts) (f)(6pts)M-9,sinceis type IV FIR generalized linear-phase system when M=9,and (b)The input should be placed into the A[r]in bit-reversed order. A0Fx0hAI]=4:A2]=x[2:A[3=6: can be high-pass filter which is necessary for differentiator.If M10,is type A[4]-x[1]A[5]-x[5]:A[6]-x[3]:A[7]-x[7] The output X[k]should be extracted from the Dfr]in sequential order. III FIR generalized linear-phase system,andcannot be high-pass filter. Xk=Dkk==0,1,…,7. (4pts) 的 (c)First,find the 8-pointDFT of x[n]=,n0,1,.,N=8. 5.(15 pts)Solution: Hc(s)=-s+a =05 +05 x[内=2aw-2gw*=2Wgw-上间 (s+a旷+6s+a+乃s+a-b 1-g可=86k-] Since the system is causal,the ROC of Hc(s):Re(sp-a.>0 Then,a sketch of the sequence Dfr=X[k]is shown below 怨 Find the inverse Laplace transform of Hc(s),we get, h.(0=0.5eu0+0.5ey0 (5pts) hin]is got by sampling he(t):n]=Th (nT)=0.5Ten]+0.5Teun] (3pts) 01234567了 Find the z-transform of hin],we get: (5pts) 第2页共2页2018-2019 2 数字信号处理（双语） （A）答案与评分细则 2 2 ( ) ( ) ( ) ( ) 2 cos 2 sin 2 , 2 2 − − = − −     − − n M n M n n M n M    (b)(3pts)Anti-symmetry of h n diff   :   ( ) ( ) ( ) ( ) 2 cos 2 sin 2 2 2 − − − − − = − − − − − diff M n M M n M h M n M n M M n M    ( ) ( ) ( ) ( ) 2 cos 2 sin 2 2 2 − − = − − − M n M n M n M n    ( ) ( ) ( ) ( ) 2 cos 2 sin 2 + 2 2 − − = − − − n M n M n M n M    = , − h n diff   since   ( ) ( ) ( ) ( ) 2 cos 2 sin 2 , 2 2 − − = − −     − − diff n M n M h n n n M n M    , i.e. h M n h n diff diff  − − =   (c) (2pts)Since h n h n w n  = diff    , h M n h n diff diff  − − =   , w[M – n] =w[n], we get h[M – n] = -h[n]. h[n] is anti-symmetry . (d)(2pts)If M=9, h n  is type IV FIR generalized linear-phase system. (e) (2pts)If M=10, h n  is type III FIR generalized linear-phase system. (f) (6pts)M=9, since h n  is type IV FIR generalized linear-phase system when M=9, and h n  can be high-pass filter which is necessary for differentiator. If M=10, h n  is type III FIR generalized linear-phase system, and h n  cannot be high-pass filter. 5．(15 pts) Solution: 2 2 0.5 0.5 ( ) ( ) + = = + + + + + + − C s a H s s a b s a jb s a jb Since the system is causal, the ROC of Hc(s): Re(s)> - a, a>0 Find the inverse Laplace transform of Hc(s), we get, ( ) ( ) ( ) 0.5 ( ) 0.5 ( ) − + − − = + a jb t a jb t c h t e u t e u t (5pts) h[n] is got by sampling hc(t) : ( ) ( ) [ ] ( ) 0.5 [ ] 0.5 [ ] − + − − = = + a jb nT a jb nT c h n Th nT Te u n Te u n (3pts) Find the z-transform of h[n], we get: ( ) ( ( ) ) ( ) ( ) ( ( ) ) ( ) 1 1 1 1 0.5 1 0.5 1 ( ) 1 1   − − − + − − − + − − − − − − = + − − a jb T a jb T a jb T a jb T T z e T z e H z e z e z Since a>0, ( ) ( ) 1 0  z e − − +a jb T → , ( ) ( ) 1 0  z e − − −a jb T → ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 1 1 1 1 1 1 0.5 1 +1 0.5 0.5 ( ) = 1 1 1 1 − − − + − − − + − − − − − + − − − − − − = + − − − − a jb T a jb T a jb T a jb T a jb T a jb T T e z e z T T H z e z e z e z e z ( ) 1 1 2 2 1 cos( ) , : 1 2 cos( ) + − − − − − − − − =  − aT aT aT aT T e bT z ROC z e e bT z e z (7pts) 6．(15 pts) Solution: (a) x3C[n]=[ 4 1 5 9 8 7 9 8 7 8] (7 pts) (b) x3L[n]=[ 0 1 5 9 8 7 9 8 7 8 4 ] (6 pts) (c) x3L [n] is not equal to x3C[n]. (2pts) 7．(10 pts) Solution： (a) The flow graph is from decimation-in-time FFT algorithm. (1pts) (b) The input should be placed into the A[r] in bit-reversed order. A[0]=x[0]; A[1]=x[4]; A[2]=x[2]; A[3]=x[6]; A[4]=x[1]; A[5]=x[5]; A[6]=x[3]; A[7]=x[7]; The output X[k] should be extracted from the D[r] in sequential order. X[k]=D[k], k= = 0, 1, ... ,7. (4pts) (c) First, find the 8-pointDFT of x[n] = −n WN , n= 0,1, ... ,7, N=8.   1 7 8 8 0 0 [ ] = − − = = =   N nk n nk N n n X k x n W W W ( ) 7 1 0 − = =  n k N n W ( ) ( ) 1 1 8 1 8 [ 1] 1 − − − = = − − k k W k W  Then, a sketch of the sequence D[r]=X[k] is shown below. (5pts)