《信号与系统——连续时域分析运算》教学资源：Chapter 2 SOLUTIONS MANUAL to accompany.pdf_大学文库

3 vn xn dvn [] [] [ ] =− − 1 1 and y n d v n v n [] [] [ ] = +− 1 1 . Therefore we can rewrite the second equation as y n d x n d v n v n d x n d v n [] [] [ ] [ ] [] [ ] = −− ( ) + −= +− ( ) − 11 1 1 2 1 1 1 1 (1) = +− dxn d xn dvn ( )( −− − ) 1 1 2 1 [ ] [ ] [ ] 1 12 = +− dxn d xn d d vn ( ) −− − ( ) − 1 1 2 1 1 2 [] [ ] [ ] 1 11 2 From Eq. (1), y n d x n d v n [] [] [ ] −= −+− 1 11 2 ( ) − 1 1 2 , or equivalently, dyn d xn d d vn 1 1 2 1 1 2 [] [] [ ] −= −+ − 1 11 2 ( ) − . Therefore, yn dyn dxn d xn d d vn d xn d d vn [] [ ] [] [ ] [ ] [ ] [ ] + −= +− ( ) −− − ( ) − + −+ − ( ) − 11 1 2 1 1 2 1 2 1 1 2 1 1 11 2 11 2 = +− dxn xn 1 [ ] [ ], or y n d x n x n d y n 1 [] [] [ ] [ ] = + −− − 1 1 1 1. (d) From the figure shown below we obtain v[n] x[n] y[n] v[n–1] –1 d1 z –1 z –1 v[n–2] d 2 w[n] u[n] vn xn wn [ ] [ ] [ ], = − wn dvn d un [ ] [ ] [ ], = −+ 1 2 1 and u n v n x n [ ] [ ] [ ]. = −+ 2 From these equations we get w n d x n d x n d x n d w n d w n [] [] [ ] [ ] [ ] [ ] = + −+ − − −− − 21 2 1 2 1 2 1 2 . From the figure we also obtain y n v n w n x n w n w n [ ] [ ] [ ] [ ] [ ] [ ], = −+ = −+ − − 2 2 2 which yields dyn dxn dwn dwn 111 1 [ ] [ ] [ ] [ ], −= −+ −− − 1313 and d yn d xn d wn d wn 222 2 [ ] [ ] [ ] [ ], −= −+ −− − 2424 Therefore, yn dyn d yn xn dxn d xn [] [ ] [ ] [ ] [ ] [ ] + −+ − = − + −+ − 12 1 2 1 2 2 3 4 + + −+ − (wn dwn d wn wn dwn d wn [] [ ] [ ] [ ] [ ] [ ] ) − −+ −+ − ( ) 12 1 2 1 22 3 4 = −+ + − xn d xn dxn [ ] [] [ ] 2 1 2 1 or equivalently, yn d xn dxn xn dyn d yn [ ] [ ] [ ] [ ] [ ] [ ]. = + −+ − − −− − 21 1 2 12 1 2 2.3 (a) x n[ ] { }, = − 3 2 0 1 4 5 2 Hence, x n n [ ] { }, . − = − −≤≤ 25410 23 3 3 Thus, x n xn x n n ev[ ] ( [ ] [ ]) { / / / / }, , = + − = −≤≤ 1 2 5 2 3 2 2 1 2 3 2 5 2 3 3 and x n xn x n n od[ ] ( [ ] [ ]) { / / / / }, . = − − = − − − −≤≤ 1 2 12 72 2 0 2 72 12 3 3 (b) y n[] { } = −− 0 7 1 3 4 9 2 . Hence, y n n [ ] { }, . − =− − − ≤ ≤ 2 9 4 3 1 7 0 3 3 Thus, y n yn y n n ev[ ] ( [ ] [ ]) { / / }, , = + − =− − − − ≤ ≤ 1 2 1 8 5 2 3 5 2 8 1 3 3 and y n yn y n n od[ ] ( [ ] [ ]) { / / }, = − − = − − − −≤≤ 1 2 1 1 3 2 0 1 3 2 1 3 3

6 2.11 x n xn x n pcs N [ ] [ ] *[ ] = + ( ) 1 2 for 0 n N –1 ≤ ≤ , Since, x n xN n N [ ][ ] = − , it follows that x , 1 n N –1. pcs[ ] [ ] *[ ] n xn x N n = +− ( ) ≤ ≤ 1 2 For n = 0, x = Re{x[0]}. pcs[ ] [ ] *[ ] 0 00 1 2 = + (x x ) Similarly x n xn x n pca N [ ] [ ] *[ ] = − ( ) 1 2 = −− ( ) ≤ ≤ 1 2 xn x N n [ ] *[ ] , 1 n N –1. Hence, for n = 0, x = jIm{x[0]}. pca[ ] [ ] *[ ] 0 00 1 2 = − (x x ) 2.12 (a) Given x n n [] . =−∞ ∞ ∑ < ∞ Therefore, by Schwartz inequality, xn xn xn n nn [] [] [] . 2 =−∞ ∞ =−∞ ∞ =−∞ ∞ ∑ ∑∑ ≤         < ∞ (b) Consider x n n n otherwise [ ] /, , , . = ≥ { 1 1 0 The convergence of an infinite series can be shown via the integral test. Let a f x n = ( ), where f(x) is a continuous, positive and decreasing function for all x ≥ 1. Then the series an n= ∞ ∑ 1 and the integral f x dx ( ) 1 ∞ ∫ both converge or both diverge. For a n n = 1 / , f(x) = 1/n. But 1 0 1 1 x dx x ∞ ∞ ∫ = (ln . Hence, ) =∞− =∞ x n n n n [ ] =−∞ ∞ = ∞ ∑ ∑= 1 1 does not converge, and as a result, x[n] is not absolutely summable. To show {x[n]} is square-summable, we observe here a n n = 1 2 , and thus, f x x () . = 1 2 Now, 1 1 11 1 1 1 2 x 1 dx x ∞ ∞ ∫ = −    = − ∞ + = . Hence, 1 1 2 n n= ∞ ∑ converges, or in other words, x[n] = 1/n is square-summable. 2.13 See Problem 2.12, Part (b) solution. 2.14 x n n n n c 2[ ] 1 cos = , . π ≤ ≤∞ ω Now, cos . ωc n n n πn n       ≤ = π ∞ = ∞ ∑ ∑ 2 1 2 2 1 1 Since, 1 6 2 1 2 n= n ∞ ∑ = π , cos . ωc n n πn       ≤ = ∞ ∑ 2 1 1 6 Therefore x n 2[ ] is square-summable. Using integral test we now show that x n 2[ ] is not absolutely summable. cos cos cos cosint( ) ω ω ω ω c c c c x x dx x x x x x π = π ⋅ ⋅ ∞ ∞ ∫1 1 1 where cosint is the cosine integral function. Since cosωcx x dx π ∞ ∫1 diverges, cosωc n n = πn ∞ ∑ 1 also diverges

7 2.15 xn x n x n n ev od n 2 2 [] [] [] =−∞ ∞ =−∞ ∞ ∑ ∑ = + ( ) =++ =+ =−∞ ∞ =−∞ ∞ =−∞ ∞ =−∞ ∞ =−∞ ∞ ∑∑ ∑ ∑∑ x n x n x nx n x n x n ev n od n ev od n ev n od n 22 22 [] [] [] [] [] [] 2 as n= ∞ x nx n ev od ∞ ∑ = – [] [] 0 since x n x n ev od [ ] [ ] is an odd sequence. 2.16 x n kn N [ ] cos( / ), = ≤≤ 2π 0 n N –1. Hence, E kn N x n N = = − ∑cos ( / ) 2 0 1 2π = + ( ) = − ∑ 1 2 0 1 1 4 cos( / ) πkn N n N = + = − ∑ N n N kn N 2 1 2 0 1 cos( / ). 4π Let C kn N n N = = − ∑cos( / ), 4 0 1 π and S kn N n N = = − ∑sin( / ). 4 0 1 π Therefore C jS ej kn N n N + = = − ∑ 4 0 1 π / = − − = e e j k j kN 4 4 1 1 0 π π / . Thus, C C jS = + Re 0 . { } = As C C jS = + Re 0 { } = , it follows that Ex N = 2 . 2.17 (a) xn n 1[ ] [ ]. = µ Energy = µ2 2 [] . n 1 n n =−∞ ∞ =−∞ ∞ ∑ ∑= =∞ Average power = lim ( [ ]) lim lim . K n K K K n K K K n K K →∞ + =− →∞ = →∞ K = + = + ∑ ∑ = 1 2 1 1 2 1 1 2 1 1 2 2 2 0 µ (b) xn nn 2[ ] [ ]. = µ Energy = n n n n n ( µ[] . ) = =∞ =−∞ ∞ =−∞ ∞ ∑ ∑ 2 2 Aveerage power = lim ( [ ]) lim . K n K K K n K K n n K n →∞ =− →∞ = + = + ∑ ∑ = ∞ 1 2 1 1 2 1 2 2 0 µ (c) x n Aeo j no 3[] . = ω Energy = Ae A o j n n o n ωo =−∞ ∞ =−∞ ∞ ∫ = =∞ ∑ 2 2 . Average power = lim lim lim . K o j n n K K K o n K K K o o K A e K A K A K o A →∞ =− →∞ + =− →∞ = + = + ∑ ∑ = 1 2 1 1 2 1 2 2 1 2 2 2 ω 2 (d) xn A n M A e Ae o jn jn o o [ ] sin , = π +     = + 2 − φ 1 ω ω where ωo M= 2π , A A e o j = − 2 φ and A A e j 1 2 = − φ. From Part (c), energy = ∞ and average power = A A AA A o o 2 1 2 2 1 2 2 4 3 4 ++ = . 2.18 Now, µ[ ] , , , . n n n = ≥ <    1 0 0 0 Hence, µ[ ] , , , . −− = < ≥    n n n 1 1 0 0 0 Thus, x[n] = µ µ [ ] [ ]. n n + −− 1 2.19 (a) Consider a sequence defined by x n k k n [ ] [ ]. = =−∞ ∑δ

8 If n < 0 then k = 0 is not included in the sum and hence x[n] = 0, whereas for n ≥ 0 , k = 0 is included in the sum hence x[n] = 1 ∀ ≥ n0. Thus x n k n n n k n [] [] , , , , = = [ ]. ≥ <    = =−∞ ∑δ µ 1 0 0 0 (b) Since µ[ ] , , , , n n n = ≥ <    1 0 0 0 it follows that µ[ –] , , , . n n n 1 1 1 0 0 = ≥ ≤    Hence, µµ δ [n] – [ ] , , , , n [ ]. n n − = n = { ≠ 1 = 1 0 0 0 2.20 Now xn A n [ ] sin = + (ω φ 0 ). (a) Given x n[ ] = − −− { } 0 2 2 2 0 2 2 2 . The fundamental period is N = 4, hence ωo = π =π 28 4 / / . Next from x A [ ] sin( ) 0 0 = = φ we get φ = 0, and solving xA A [ ] sin( ) sin( / ) 1 4 = 4 2 π + = π =− φ we get A = –2. (b) Given x n[ ] = −− { } 22 2 2 . The fundamental period is N = 4, hence ω0 = 2π/4 = π/2. Next from x A [ ] sin( ) 0 2 = = φ and xA A [ ] sin( / ) cos( ) 12 2 = += = πφ φ it can be seen that A = 2 and = /4 is one solution satisfying the above equations. φ π (c) x n[ ] = − { } 3 3 . Here the fundamental period is N = 2, hence ω π 0 = . Next from x[0] = Asin( ) φ = 3 and x A A [ ] sin( ) sin( ) 1 3 = + =− =− φπ φ observe that A = 3 and = /2 φ π that A = 3 and φ = π/2 is one solution satisfying these two equations. (d) Given x n[] . . = − { } 0 15 0 15 , it follows that the fundamental period of x[n] is N = 4. Hence ωπ π 0 = = 24 2 / / . Solving x A [ ] sin( ) 0 0 = = φ we get φ = 0, and solving x A [ ] sin( / ) . 1 2 15 = = π , we get A = 1.5. 2.21 (a) ˜ [] . . xn e j n 1 0 4 = − π Here, ωo = π 0 4. . From Eq. (2.44a), we thus get N r r r o = π = π π = = 2 2 0 4 5 5 ω . for r =1. (b) xn n ˜ [ ] sin( . . ). 2 = π+ π 06 06 Here, ωo = π 0 6. . From Eq. (2.44a), we thus get N r r r o = π = π π = = 2 2 0 6 10 3 10 ω . for r = 3. (c) xn n n ˜ [ ] cos( . . ) sin( . ). 3 = π − π+ π 2 11 05 2 07 Here, ω1 = π 1 1. and ω2 = π 0 7. . From Eq. (2.44a), we thus get N r r r 1 1 1 1 1 2 2 1 1 20 11 = π = π π = ω . and N r r r 2 2 2 2 2 2 2 0 7 20 7 = π = π π = ω . . To be periodic

9 we must have N N 1 2 = . This implies, 20 11 20 7 1 2 r r = . This equality holds for r1 = 11 and r2 = 7, and hence N = N N 1 2 = = 20. (d) N r r 1 1 1 2 1 3 20 13 = π π = . and N r r 2 2 2 2 0 3 20 3 = π π = . . It follows from the results of Part (c), N = 20 with r1 = 13 and r2 = 3. (e) N r r 1 1 1 2 1 2 5 3 = π π = . , N r r 2 2 2 2 0 8 5 2 = π π = . and N N 3 2 = . Therefore, N N N N == = = 122 5 for r 1 = 3 and r2 = 2. (f) x n ˜ [ ] 6 =n modulo 6. Since x n ˜ [ ] 6 + = 6 (n+6) modulo 6 = n modulo 6 = x n ˜ [ ] 6 . Hence N = 6 is the fundamental period of the sequence x n ˜ [ ] 6 . 2.22 (a) ωo = π 0 14 . . Therefore, N r r r o = π = π π = = 2 2 0 14 100 7 100 ω . for r = 7. (b) ωo = π 0 24 . . Therefore, N r r r o = π = π π = = 2 2 0 24 25 3 25 ω . for r = 3. (c) ωo = π 0 34 . . Therefore, N r r r o = π = π π = = 2 2 0 34 100 17 100 ω . for r = 17. (d) ωo = π 0 75 . . Therefore, N r r r o = π = π π = = 2 2 0 75 8 3 8 ω . for r = 3. 2.23 x n x nT nT a o [ ] ( ) cos( ) = =Ω is a periodic sequence for all values of T satisfying ΩoTN r ⋅ =π2 for r and N taking integer values. Since, ΩoT rN = π2 / and r/N is a rational number, ΩoT must also be rational number. For Ωo = 18 and T = π/6, we get N = 2r/3. Hence, the smallest value of N = 3 for r = 3. 2.24 (a) xn n n n n n n [] [ ] [ ] [] [ ] [ ] [ ] = +− + + + −+ − + − 3 32 2 4 15 22 3 δ δ δδ δ δ (b) yn n n n n n n [] [ ] [ ] [] [ ] [ ] [ ] = + + +− + −+ − − − 7 2 13 4 19 22 3 δ δ δδ δ δ (c) wn n n n n n n [] [ ] [ ] [ ] [] [ ] [ ] =− + + + + + + − − + − 5 24 23 16 5 1 3 δ δ δ δδ δ 2.25 (a) For an input x n i[ ], i = 1, 2, the output is yn xn xn xn xn ii i i i [ ] [ ] [ ] [ ] [ ], = + −+ − + − αα α α 12 3 4 1 2 4 for i = 1, 2. Then, for an input x n Ax n Bx n [ ] [ ] [ ], = + 1 2 the output is y n A x n Bx n A x n Bx n [] [] [] [ ] [ ] = + α α 11 2 21 2 ( ) + −+ − ( 1 1 ) + −+ − α α 31 2 41 2 (Ax n Bx n Ax n Bx n [ ] [ ] [ ] [ ] 22 33 ) + −+ − ( ) = + −+ − + − A xn xn xn xn (αα α α 11 21 31 41 [ ] [ ] [ ] [ ] 12 4 ) + + −+ − + − B xn xn xn xn (αα α α 12 22 32 42 [ ] [ ] [ ] [ ] 12 4 ) = + Ay n By n 1 2 [ ] [ ]. Hence, the system is linear

(b) For an input xi[n ,i=l, 2, the output is yin]=boxi[n]+bixi[n-1]+baXi[n-2]+ayi[n-1]+ayi[n-2], i= 1, 2. Then, for an input x[n]=AXI[n]+ Bx2In], the output is y[n]=A(boxI[n]+biXI[n-1]+b2X1[n-2]+aiyI[n-1]+a2y1[n-2]) h-b(box21nJ+ bX2[n-1+b2x2In-2]+a1y2[n-1]+a2y2In-20)=Ayi[n]+By2In] ce, the system is linear (c)For an input xiIn], i=1, 2, the output is yin *i[n/L],n=0,tL,+2L,K Consider the input x[n]=AXI[n]+ Bx2[n], Then the output yIn] for n=0, tL, +2L,K given by yIn]= x[n/l]=AXIn/L]+ Bx2In/L]=Ayin]+By2ln]. For all other values of n, yIn]=A.0+B-0=0. Hence, the system is linear (d) For an input xi[n],i= l, 2, the output is yi[n]=xi[n/M]. Consider the input xn]=Ax,In]+ Bx2In], Then the output yIn]=AXIn/M]+Bx2ln/M=AyIIn]+ By2ln] Hence, the system is linear (e) For an input xi[n],i= l, 2, the output is yi[n] M2k=0XiLn-k. Then, for an input x(n =Ax[n]+Bx2[], the output is yi[n]=LEM-(Ax[n-k]+Bx2In-kJ) lica m 2k=o xi[n-K+M2k=0 x2In-k1=Ayi[n]+By2[n]. Hence, the system is (f) The first term on the RHs of Eq(2.58)is the output of an up-sampler. The second term on the rhs of Eq(2.58) is simply the output of an unit delay followed by an up- ereas he third term is the output of an unit adavance operator followed by an up-sampler We have shown in Part(c)that the up-sampler is a linear system. Moreover, the delay and the advance operators are linear systems. A cascade of two linear systems is linear and a linear combination of linear systems is also linear. Hence, the factor-of-2 interpolator is a linear system. (g) Following thearguments given in Part (f), it can be shown that the factor-of-3 interpolator is a linear system 226(a)ym]=n2x[n For an input xin] the output is yin]=n[],i=l, 2. Thus, for an input x3[n]= AxIn +Bx2[], the output y3[n] is given by y3[n]=n(AxI[n]+Bx2[n])=AyI[n]+By2[n] Hence the system is linear. Since there is no output before the input hence the system is causal. However, ly[n] being proportional to n, it is possible that a bounded input can result in an unbounded output. Let x[n]=1yn, then yIn]=n2. Hence yIn]→∞asn→o, hence not bibo stable Let yIn] be the output for an input x[n], and let yIIn] be the output for an input x,[n]. If X,In]=xIn-noI then y,n]=nx,[n]=n*xIn-nol. However, yln-nol=(n-no)-xn-nol Since y,[n]+yIn-nol, the system is not time-invariant

10 (b) For an input x n i[ ], i = 1, 2, the output is y n b x n bx n b x n ay n a y n i ii i i i [ ] [ ] [ ] [ ] [ ] [ ], = + −+ − + −+ − 01 2 1 2 1 2 1 2 i = 1, 2. Then, for an input x n A x n Bx n [ ] [ ] [ ], = + 1 2 the output is yn A b x n bx n b x n ay n a y n [] [] [ ] [ ] [ ] [ ] = + −+ − + −+ − ( 01 11 21 11 21 1 2 1 2 ) + + −+ − + −+ − Bb x n bx n b x n ay n a y n ( 02 12 22 12 22 [ ] [ ] [ ] [ ] [ ] 1 21 2 ) = + Ay n By n 1 2 [ ] [ ]. Hence, the system is linear. (c) For an input x n i[ ], i = 1, 2, the output is y n xnL n L L otherwise i i [] , [ / ], , , , , , =  =± ±   0 2 0 K Consider the input x n A x n Bx n [ ] [ ] [ ], = + 1 2 Then the output y[n] for n LL =± ± 0 2 ,, , K is given by y[n] = x n L A x n L Bx n L A y n By n [ / ] [ / ] [ / ] [] [] = + =+ 1 2 12 . For all other values of n, y n A B [] . = ⋅+ ⋅= 0 0 0 Hence, the system is linear. (d) For an input x n i[ ], i = 1, 2, the output is y n x n M i i [] [ / ] = . Consider the input x n Ax n Bx n [ ] [ ] [ ], = + 1 2 Then the output y n A x n M Bx n M A y n By n [ ] [ / ] [ / ] [ ] [ ]. = + =+ 1 2 12 Hence, the system is linear. (e) For an input x n i[ ], i = 1, 2, the output is y n M i i xn k k M [] [ ] = − = − ∑1 0 1 . Then, for an input x n A x n Bx n [ ] [ ] [ ], = + 1 2 the output is y n M i Ax n k Bx n k k M [] [ ] [ ] = −+ − ( ) = − ∑1 1 2 0 1 = −     + −     = + = − = − A ∑ ∑ M xn k B M x n k Ay n By n k M k 1 1 M 1 0 1 2 0 1 1 2 [ ] [ ] [ ] [ ]. Hence, the system is linear. (f) The first term on the RHS of Eq. (2.58) is the output of an up-sampler. The second term on the RHS of Eq. (2.58) is simply the output of an unit delay followed by an upsampler, whereas, the third term is the output of an unit adavance operator followed by an up-sampler We have shown in Part (c) that the up-sampler is a linear system. Moreover, the delay and the advance operators are linear systems. A cascade of two linear systems is linear and a linear combination of linear systems is also linear. Hence, the factor-of-2 interpolator is a linear system. (g) Following thearguments given in Part (f), it can be shown that the factor-of-3 interpolator is a linear system. 2.26 (a) y[n] = n2x[n]. For an input xi[n] the output is yi[n] = n2xi[n], i = 1, 2. Thus, for an input x3[n] = Ax1[n] + Bx2[n], the output y3[n] is given by y n n A x n Bx n A y n By n 3 2 12 12 [] [] [] [] [] = + ( ) = + . Hence the system is linear. Since there is no output before the input hence the system is causal. However, y n[ ] being proportional to n, it is possible that a bounded input can result in an unbounded output. Let x[n] = 1 ∀ n , then y[n] = n2. Hence y n[ ] →∞ →∞ as n , hence not BIBO stable. Let y[n] be the output for an input x[n], and let y1[n] be the output for an input x1[n]. If x n xn n 1 0 [] [ ] = − then y n n x n n x n n 1 2 1 2 0 [] [] [ ] = =− . However, y n n n n x n n [ ]( )[ ] − =− − 0 0 2 0 . Since y n y n n 1 0 [] [ ] ≠ − , the system is not time-invariant