山东大学：《生物医学信号处理 Biomedical Signal Processing》精品课程教学资源（试卷与答案）A卷-2019_答案

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 1.(10 pts)Solution: 1+2:+22 The ystem funtions(日)1-05e05“(4o.5s90-s可 S0, 4(-） (5pts) Poles:.5:Since the system is causal,the ROC is> 3.(15pts)Solution: Find the impulse response of the system,hin]. (a)The impulse response of the Linear-Phase FIR system is hln-26[n]+36[n-1]-6[n-2]+6[n-3]+8[n-4]-6[n-5]+36[m-6]+26[n-7] （） 8/ H阳o3-可2+益+各. The system function of the Linear-Phase FIR system is He)=2+3z1-22+3+4+5+326+2=7 (3pts) Taling the iaverserr: (5 pts) (b)The difference equation relating input x[n]and output y[n]: (2pts) b)(5 pts)Find the output of this system,yn].for the input m=2x+3n-刂-x对n-2]+可n-3]+m-4-对n-]+3xn-6+2n-7刀 We use the eigenfunction of the the inputxn: (c)The signal flow graph of direct form structure for the FIR linear-phase system is y[n]=H( x 1+2ee四+e24=四 H(e儿Aa= ,121=4 =1-0.5ee吗-0.5e4啊-1+0.57+0.53+ Fna,儿=He儿a网ee2n=eean=4 3+ 3+1 (5 pts) 网 (8pts) 2.(15 pts)Solution: (d)Since the impulse response is symmetric at n=3.5,h[7-n]=hin] eros:,poles:o-1;ROC:(for right-sided h (3pts) ∠H(eo）=-3.5o+p 531 20 The group delay of the system is: 的 + )H(e)=22_- 9 3 8[He】=-品[He〗=-品35o+py=35 (2pts) c,a-小sa+-a(4回 (5pts) 4.(20 pts)Solution: (c)The system is stable,since the ROC()include the unit circle. (2 pts) (a)(5pts)The corresponding impulse response of) 强 53 (eedo(o)edo(o)et-udo (d)H()= 22 =H.()H() 第1项共2

2018-2019 2 数字信号处理（双语） （A）答案与评分细则 1 2 1．(10 pts) Solution: a) The system function is ( ) 1 2 1 2 1+2 1 0.5 -0.5 z z H z z z − − − − + = − ( ) ( )( ) 2 1 1 1 1+ 1+0.5 1 − − − = − z z z Poles: 1 2 = = − 1, 0.5 p p z z ; Since the system is causal, the ROC is : z 1 Find the impulse response of the system, h[n]. ( ) ( ) ( )( ) 2 1 1 1 1 1 1 8 1+ 3 3 = 2 , 1 1+0.5 1 1+0.5 1 − − − − − = − + +  − − z H z z z z z z (5 pts) b) (5 pts) Find the output of this system, y[n], for the input ( /2) [ ] = j n x n e  . We use the eigenfunction of the the input ( /2) [ ] = j n x n e  :   ( ) ( ) /2 / 2 ( ) = = j j n z e y n H z e   ( ) ( ) ( ) ( ) ( ) / 2 /2 2 /2 /2 2 /2 1+2 1 2 1 4 ( ) 1 0.5 0.5 1 0.5 0.5 3 − − = − − + − − − = = = − − + + + j j j z e j j e e j j H z e e j j      Finaly,   ( ) ( ) /2 / 2 ( ) = = j j n z e y n H z  e  4 4 ( / 2) 1 3 3 = j n n j j j e j − −  + + + = (5 pts) 2．(15 pts) Solution: (a) zeros: -3, 1/2; poles: ∞, -1/3; ROC: 1 3    z (for right-sided h[n] ) (3pts) (b) ( ) 1 1 1 5 3 20 2 2 9 3 = = + 1 1 2 1 1 3 3 − − − + − − + + z z H z z z z , since ROC: 1 3    z ,         9 20 1 = 1 - 2 3 3 n h n n n u n     + − +     (5pts) (c) The system is stable, since the ROC ( 1 3    z ) include the unit circle. (2 pts) (d) ( ) ( ) 1 1 1 1 1 1 5 3 1 1 3 1 2 2 2 = 1 1 1 1 3 3 − − − − − −   + − + −     =   +   +   z z z z H z z z z 1 1 1 1 1 1 3 3 1 2 1 1 3 − − − −     +     = −         +   z z z z min ( ) ( ) = H z H z ap So, ( ) 1 min 1 =3 1 2 H z z   −   −   , ( ) 1 1 1 1 3 = 1 1 3 ap z H z z z − − −     +       +   (5 pts) 3．(15 pts) Solution: (a) The impulse response of the Linear-Phase FIR system is h[n]=2δ[n]+3δ[n-1]-δ[n-2]+δ[n-3]+δ[n-4]-δ[n-5]+3δ[n-6]+2δ[n-7] The system function of the Linear-Phase FIR system is 1 2 3 4 5 6 7 ( ) 2+3 + + + +3 +2 − − − − − − − H z z z z z z z z = − (3pts) (b) The difference equation relating input x[n] and output y[n]: (2pts) y n x n x n x n x n x n x n x n x n [ ] 2 [ ] 3 [ 1] [ 2] [ 3] [ 4] [ 5] 3 [ 6] 2 [ 7] = + − − − + − + − − − + − + − (c) The signal flow graph of direct form structure for the FIR linear-phase system is (8pts) (d) Since the impulse response is symmetric at n=3.5 , h[7-n] = h[n]  = ( ) - + 3.5 j H e    The group delay of the system is:         ( ) = − arg ( )  j j d grd H e H e d    = − - + = 3.5 3.5  d d    (2pts) 4．(20 pts) Solution: (a) (5pts) The corresponding impulse response h n diff   of ( ) j H e diff  :   1 ( ) 2 − =  j j n diff diff h n H e e d       ( ) 1 / 2 2 − − =  j M j n j e e d        ( ) 1 ( / 2) 2 − − =  j n M j e d       ( ) ( ) 1 1 ( / 2) 2 / 2 − − = −  j n M j de j n M      ( ) ( ) ( ) ( ) 1 1 /2 /2 2 / 2 − − − − = − −  j n M j n M e e d n M         

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) -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) -ooe (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)