信号与系统练习与解答-英文.pdf
信号与系统奥本海姆英文版课后答案chapter10
∞
−n
n =−∞
∑ (1/ 4)
∞
z −n zn
n =−3
∑ (1/ 4)
= ∑ (1/ 4) − n +3 z n −3
n=0
= (1/ 64) z −3 /(1 − 4 z ), z < 1/ 4 = (1/16) z −4 /(1 − (1/ 4) z −1 ), z < 1/ 4
The Fourier transform does not exist because the ROC includes the unit circle. (g) Consider x1 ( z ) = 2n μ[ − n].
X ( z) =
n =−∞
jw
r> 1 . The second summation
2
10.2
∑ ( 5 ) u[n − 3]z
n
∞
1
z −3 ∞ 1 n − n ]∑ ( ) z =[ 125 n = 0 5
∞ 1 = ∑ ( )n z − n n =3 5 z −3 1 ] =[ 125 1 − 1 z −1 5 −n
X ( Z ) = ∑ x[n]z − n
n =∞ ∞
−5
= ∑ (−1) n z − n
n =0
∞
= 1/(1 + z −1 ), z > 1
The Fourier transform does not exist because the ROC does not include the unit circle (d) For x [n]= ( 1 ) n +1 μ[n + 3],
x( z ) =
英文版《信号与系统》第九章习题解答
2 H (s ) = s (s + 4 )
Re{s} > 0
The system is unstable.
15
Chapter 9 9.33 A causal LTI system H (s ) =
x(t ) = e
t
Problem Solution
s +1 s 2 + 2s + 2
, ∞ < t < +∞
Determine and sketch the response y (t )
2 X (s ) = 2 s 1
1 < Re{s} < 1 Re{s} > 1
2(s + 1) (s + 1)(s 1) s 2 + 2s + 2
s +1 H (s ) = 2 s + 2s + 2 Y (s ) = X (s )H (s ) =
2 2 4 y (t ) = + e t u ( t ) + e t (cos t )u (t ) + e t (sin t )u (t ) 5 5 5
17
Chapter 9
Problem Solution
9.35 Consider a causal LTI system with the input x(t ) and output y (t ) . (b) Is this system stable?
2
②
1
③
1
④
2
σ
① Re{s} < 2 anticausal , unstable ② -2 < Re{s} < 1 noncausal , unstable
英文版《信号与系统》第三章习题解答
ck
1 e jk 2
sin
k / 2 k
e jk / 2
bk
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jk
2 1
dx1t
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ck
1
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4t
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Chapter 3
Problem Solution
0
ak
1
jk
2
jk
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k is even k is odd
(c) x t
ake jk0t
ak 0 , k is odd
for each of the following inputs :
(a) xt t n n
(b) xt 1n t n n
(c) xt is the periodic wave depicted in Figure P3.34
1/ 2 1 xt
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1
2
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Chapter 3
(c) ak 0 , k is odd xt is even harmonic
Could T be the fundamental period for such a signal?
(d) If one of two things happens,T is fundamental period.
xt xt T / 2
Chapter 3
Problem Solution
(b) Solution 1 xt x1t x1t 1
xt
x1 t FSbk
1
2 1 0 1 2 3
ak bk 1 e jk0t0 0 t0 1
1
x1 t
1
ak bk 1 e jk
英文版《信号与系统》第二章习题解答
Problems Solution
dx(t ) ∵ → 3 y (t ) + e 2t u (t ) dt
2e 3 h(t 1) = e 2t u (t )
the impulse 1 3 2(t +1) u (t + 1) response h(t ) = 2 e e
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Chapter 2
∫
t
∞
h (τ )d τ =
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t<0,t>2
sin (πτ )d τ =
1
∫
t
1
0
π
cos πτ | 0t
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=
π
(1 cos π t )[u (t ) u (t 2 )]
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Chapter 2
y (t ) = 2
Problems Solution
π
2
(1 cos π (t 1 ))[u (t 1 ) u (t 3 )] (1 cos π (t 3 ))[u (t 3 ) u (t 5 )]
∞
+∞
2
t
y (t ) = ∫ e (t τ 2 )u (t τ 2 )x(τ )dτ
∞
+∞
h(t ) = e
(t 2 )
u (t 2)
h(t τ )
(b ) y(t ) = x(t ) h(t ) = x′(t ) h (1) (t )
= h (1) (t + 1) h (1) (t 2)
0
e 3τ dτ
1 e 3(t 3 ) = 3 e 3(t 5 ) e 3(t 3 ) = 3
x(t τ )
t 3
信号与系统(习题课)
∴ y(t) = e-3t + t e-3t = (1+ t) e-3t
by wky
习题 3-6 (1)
已知系统的微分方程为 y’’(t) +5 y’(t) + 4 y(t) =2 f ’(t) + 5f(t), t >0; 初始状态y(0-) =1，y’(0-) =5， 求系统的零输入响应yx(t)。 解：系统特征方程为 s2+5s+4=0 , 解得特征根 s1=-1, s2=-4
特解 （强迫响应）
比较：完全响应=零输入响应 + 零状态响应 = e－t + (1 - 1/2e－t -1/2e－3t)
by wky
习题 3-4
已知微分方程为 y’(t) + 3 y (t) = f(t)，t >0; y(0) =1，
求系统的固有响应(齐次解) yh(t)、强迫响应 (特解) yp(t)和完全响应(全解) y(t) 解：系统特征方程为 s+3=0,
f(t)
f(-t)
2
2
1
1
-3 -2 -1 0 1 2 3 t -3 -2 -1 0 1 2 3 t
2 f(t+2)
f(-3t)
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-3 -2 -1 0 1 2 3 t -3 -2 -1 0 1 2 3 t by wky
2-10 已知信号波形, 绘出下列信号波形
f(t)
f(-t)
2
2
1
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-3 -2 -1 0 1 2 3 t -3 -2 -1 0 1 2 3 t
信号与系统奥本海姆英文版课后答案chapter
Chapter 1 Answers1.1 Converting from polar to Cartesian coordinates: 1.2 converting from Cartesian to polar coordinates:55j=, 22j e π-=,233jj eπ--=212je π--=, 41j j π+=, ()2221jj eπ-=-4(1)j j eπ-=, 411j jeπ+=- 12e π-= 1.3. (a) E ∞=4014tdt e∞-=⎰, P ∞=0, because E ∞<∞ (b) (2)42()j t t x eπ+=, 2()1t x =.Therefore, E ∞=22()dt t x +∞-∞⎰=dt +∞-∞⎰=∞,P ∞=211limlim222()TTTTT T dt dt TTt x --→∞→∞==⎰⎰lim11T →∞=(c) 2()t x =cos(t). Therefore, E∞=23()dt t x +∞-∞⎰=2cos()dt t +∞-∞⎰=∞,P ∞=2111(2)1lim lim 2222cos()TTTT T T COS t dt dt T Tt --→∞→∞+==⎰⎰(d)1[][]12nn u n x =⎛⎫⎪⎝⎭,2[]11[]4nu n n x =⎛⎫ ⎪⎝⎭. Therefore, E ∞=24131[]4nn n x +∞∞-∞===⎛⎫∑∑⎪⎝⎭P ∞=0，because E ∞<∞. (e) 2[]n x =()28n j e ππ-+, 22[]n x =1. therefore, E ∞=22[]n x +∞-∞∑=∞,P ∞=211limlim1122121[]NNN N n Nn NN N n x →∞→∞=-=-==++∑∑.(f) 3[]n x =cos 4nπ⎛⎫ ⎪⎝⎭. Therefore, E ∞=23[]n x +∞-∞∑=2cos()4n π+∞-∞∑=2cos()4n π+∞-∞∑,P ∞=1limcos 214nNN n NN π→∞=-=+⎛⎫∑ ⎪⎝⎭1cos()112lim ()2122NN n Nn N π→∞=-+=+∑ 1.4. (a) The signal x[n] is shifted by 3 to the right. The shifted signal will be zero for n<1, And n>7. (b) The signal x[n] is shifted by 4 to the left. The shifted signal will be zero for n<-6. And n>0. (c) The signal x[n] is flipped signal will be zero for n<-1 and n>2.(d) The signal x[n] is flipped and the flipped signal is shifted by 2 to the right. The new Signal will be zero for n<-2 and n>4.(e) The signal x[n] is flipped and the flipped and the flipped signal is shifted by 2 to the left. This new signal will be zero for n<-6 and n>0.1.5. (a) x(1-t) is obtained by flipping x(t) and shifting the flipped signal by 1 to the right. Therefore, x (1-t) will be zero for t>-2. (b) From (a), we know that x(1-t) is zero for t>-2. Similarly, x(2-t) is zero for t>-1, Therefore, x (1-t) +x(2-t) will be zero for t>-2. (c) x(3t) is obtained by linearly compression x(t) by a factor of3. Therefore, x(3t) will be zero for t<1. (d) x(t/3) is obtained by linearly compression x(t) by a factor of 3. Therefore, x(3t) will be zero for t<9.1.6 (a) x 1(t ) is not periodic because it is zero for t<0. (b) x 2[n ]=1 for all n. Therefore, it is periodic with a fundamental period of 1. (c) x 3[n1.7. (a)v ε[4])n --Therefore,()1[]vn xεis zero for 1[]n x >3. (b) Since x 1(t ) is an odd signal, ()2[]vn x εis zero for all values of t.(c) (){}11311[][][][3][3]221122v nnn n n u n u n x x x ε-⎡⎤⎢⎥=+-=----⎢⎥⎢⎥⎣⎦⎛⎫⎛⎫ ⎪ ⎪⎝⎭⎝⎭Therefore, ()3[]vn x εis zero whenn <3 and when n →∞.(d)()1554411()(()())(2)(2)22vttt t t u t u t x x x ee ε-⎡⎤=+-=---+⎣⎦Therefore,()4()vt x εis zero only whent →∞.1.8. (a) ()01{()}22cos(0)tt t x eπℜ=-=+l (b) ()02{()}cos()cos(32)cos(3)cos(30)4t t t t t x e ππℜ+==+l(c) ()3{()}sin(3)sin(32t t t t t x e e ππ--ℜ=+=+l(d) ()224{()}sin(100)sin(100)cos(1002t t t t t t t x e e e ππ---ℜ=-=+=+l1.9. (a) 1()t xis a periodic complex exponential.(b) 2()t x is a complex exponential multiplied by a decaying exponential. Therefore,2()t x is not periodic.（c ）3[]n x is a periodic signal. 3[]n x =7j n e π=j n e π.3[]n x is a complex exponential with a fundamental period of22ππ=. (d) 4[]n x is a periodic signal. The fundamental period is given by N=m(23/5ππ)=10().3m By choosing m=3. We obtain the fundamental period to be 10.(e) 5[]n x is not periodic. 5[]n x is a complex exponential with 0w =3/5. We cannot find any integer msuch that m(02wπ ) is also an integer. Therefore, 5[]n xis not periodic.1.10. x (t )=2cos(10t ＋1)-sin(4t-1)Period of first term in the RHS =2105ππ=.Period of first term in the RHS =242ππ= .Therefore, the overall signal is periodic with a period which the least commonmultiple of the periods of the first and second terms. This is equal to π . 1.11.x[n] = 1+74j n e π?25j n e πPeriod of first term in the RHS =1. Period of second term in the RHS =⎪⎭⎫ ⎝⎛7/42π=7 （when m=2）Period of second term in the RHS =⎪⎭⎫ ⎝⎛5/22ππ=5 (when m=1)Therefore, the overall signal x[n] is periodic with a period which is the least common Multiple of the periods of the three terms inn x[n].This is equal to 35.1.12. The signal x[n] is as shown in figure S1.12. x[n] can be obtained by flipping u[n] and thenShifting the flipped signal by 3 to the right. Therefore, x[n]=u[-n+3]. This implies that 1.13y (t)= ⎰∞-tdt x )(τ =dt t))2()2((--+⎰∞-τδτδ=⎪⎩⎪⎨⎧>≤≤--<2,022,12,0,t t tTherefore ⎰-==∞224dt E=2x 1[n-2]+ 5x 1[n-3] + 2x 1[n-4] The input-output relationship for S isy[n]=2x[n-2]+ 5x [n-3] + 2x [n-4](b) The input-output relationship does not change if the order in which S 1and S 2 are connected series reversed. . We can easily prove this assuming that S 1 follows S 2. In this case , the signal x 1[n], which is the input to S 1 is the same as y 2[n].Therefore y 1[n] =2x 1[n]+ 4x 1[n-1]= 2y 2[n]+4 y 2[n-1]=2( x 2[n-2]+21 x 2[n-3] )+4(x 2[n-3]+21x 2[n-4]) =2 x 2[n-2]+5x 2[n-3]+ 2 x 2[n-4]The input-output relationship for S is once againy[n]=2x[n-2]+ 5x [n-3] + 2x [n-4]1.16 (a)The system is not memory less because y[n] depends on past values of x[n].(b)The output of the system will be y[n]= ]2[][-n n δδ=0(c)From the result of part (b), we may conclude that the system output is always zero for inputs of the form ][k n -δ, k ∈ ?. Therefore , the system is not invertible .1.17 (a) The system is not causal because the output y(t) at some time may depend on future values of x(t). For instance , y(-π)=x(0).(b) Consider two arbitrary inputs x 1(t)and x 2(t).x 1(t) →y 1(t)= x 1(sin(t)) x 2(t) → y 2(t)= x 2(sin(t))Let x 3(t) be a linear combination of x 1(t) and x 2(t).That is , x 3(t)=a x 1(t)+b x 2(t)Where a and b are arbitrary scalars .If x 3(t) is the input to the given system ,then the corresponding output y 3(t) is y 3(t)= x 3( sin(t))=a x 1(sin(t))+ x 2(sin(t))=a y 1(t)+ by 2(t)Therefore , the system is linear.1.18.(a) Consider two arbitrary inputs x 1[n]and x 2[n].x 1[n] → y 1[n] =][001k x n n n n k ∑+-=x 2[n ] → y 2[n] =][02k x n n n n k ∑+-=Let x 3[n] be a linear combination of x 1[n] and x 2[n]. That is :x 3[n]= ax 1[n]+b x 2[n]where a and b are arbitrary scalars. If x 3[n] is the input to the given system, then the corresponding outputy 3[n] is y 3[n]=][03k x n n n n k ∑+-==])[][(2100k bx k ax n n n n k +∑+-==a ][001k x n n n n k ∑+-=+b ][02k x n n n n k ∑+-== ay 1[n]+b y 2[n]Therefore the system is linear.(b) Consider an arbitrary input x 1[n].Lety 1[n] =][01k x n n n n k ∑+-=be the corresponding output .Consider a second input x 2[n] obtained by shifting x 1[n] in time:x 2[n]= x 1[n-n 1]The output corresponding to this input isy 2[n]=][02k x n n n n k ∑+-== ]n [1100-∑+-=k x n n n n k = ][01011k x n n n n n n k ∑+---=Also note that y 1[n- n 1]=][01011k x n n n n n n k ∑+---=.Therefore , y 2[n]= y 1[n- n 1] This implies that the system is time-invariant.(c) If ][n x <B, then y[n]≤(2 n 0+1)B. Therefore ,C ≤(2 n 0+1)B.1.19 (a) (i) Consider two arbitrary inputs x 1(t) and x 2(t). x 1(t) → y 1(t)= t 2x 1(t-1)x 2(t) → y 2(t)= t 2x 2(t-1)Let x 3(t) be a linear combination of x 1(t) and x 2(t).That is x 3(t)=a x 1(t)+b x 2(t)where a and b are arbitrary scalars. If x 3(t) is the input to the given system, then the corresponding output y 3(t) is y 3(t)= t 2x 3 (t-1)= t 2(ax 1(t-1)+b x 2(t-1))= ay 1(t)+b y 2(t)Therefore , the system is linear.(ii) Consider an arbitrary inputs x 1(t).Let y 1(t)= t 2x 1(t-1)be the corresponding output .Consider a second input x 2(t) obtained by shifting x 1(t) in time:x 2(t)= x 1(t-t 0)The output corresponding to this input is y2(t)= t2x2(t-1)= t2x1(t- 1- t)Also note that y1(t-t)= (t-t)2x1(t- 1- t)≠y2(t)Therefore the system is not time-invariant.(b) (i) Consider two arbitrary inputs x1[n]and x2[n]. x1[n] →y1[n] = x12[n-2]x2[n ] →y2[n] = x22[n-2].Let x3(t) be a linear combination of x1[n]and x2[n].That is x3[n]= ax1[n]+b x2[n]where a and b are arbitrary scalars. If x3[n] is the input to the given system, then the corresponding outputy3[n] is y3[n] = x32[n-2]=(a x1[n-2] +b x2[n-2])2=a2x12[n-2]+b2x22[n-2]+2ab x1[n-2] x2[n-2] ≠ay1[n]+b y2[n]Therefore the system is not linear.(ii) Consider an arbitrary input x1[n]. Let y1[n] = x12[n-2]be the corresponding output .Consider a second input x2[n] obtained by shifting x1[n] in time:x 2[n]= x1[n- n]The output corresponding to this input isy 2[n] = x22[n-2].= x12[n-2- n]Also note that y1[n- n]= x12[n-2- n]Therefore , y2[n]= y1[n- n]This implies that the system is time-invariant.(c) (i) Consider two arbitrary inputs x1[n]and x2[n].x 1[n] →y1[n] = x1[n+1]- x1[n-1]x2[n ]→y2[n] = x2[n+1 ]- x2[n -1]Let x3[n] be a linear combination of x1[n] and x2[n]. That is :x3[n]= ax1[n]+b x2[n]where a and b are arbitrary scalars. If x3[n] is the input to the given system, then thecorresponding output y3[n] is y3[n]= x3[n+1]- x3[n-1]=a x1[n+1]+b x2[n +1]-a x1[n-1]-b x2[n -1]=a(x1[n+1]- x1[n-1])+b(x2[n +1]- x2[n -1])= ay1[n]+b y2[n]Therefore the system is linear.(ii) Consider an arbitrary input x1[n].Let y1[n]= x1[n+1]- x1[n-1]be the corresponding output .Consider a second input x2[n] obtained by shifting x1[n] in time: x2[n]=x 1[n-n]The output corresponding to this input isy 2[n]= x2[n +1]- x2[n -1]= x1[n+1- n]- x1[n-1- n]Also note that y1[n-n]= x1[n+1- n]- x1[n-1- n]Therefore , y2[n]= y1[n-n]This implies that the system is time-invariant.(d) (i) Consider two arbitrary inputs x1(t) and x2(t).x 1(t) →y1(t)= dO{}(t)x1x 2(t) →y2(t)= {}(t)x2dOLet x 3(t) be a linear combination of x 1(t) and x 2(t).That is x 3(t)=a x 1(t)+b x 2(t)where a and b are arbitrary scalars. If x 3(t) is the input to the given system, then the corresponding output y 3(t) is y 3(t)= d O {}(t) x 3={}(t) x b +(t) ax 21d O=a d O {}(t) x 1+b {}(t) x 2d O = ay 1(t)+b y 2(t)Therefore the system is linear.(ii) Consider an arbitrary inputs x 1(t).Lety 1(t)= d O {}(t) x 1=2)(x -(t) x 11t -be the corresponding output .Consider a second input x 2(t) obtained by shifting x 1(t) in time:x 2(t)= x 1(t-t 0)The output corresponding to this input isy 2(t)= {}(t) x 2d O =2)(x -(t) x 22t -=2)(x -)t -(t x 0101t t --Also note that y 1(t-t 0)= 2)(x -)t -(t x 0101t t --≠ y 2(t)Therefore the system is not time-invariant. 1.20 (a) Givenx )(t =jt e 2 y(t)=tj e 3x )(t =jte2- y(t)=tj e3-Since the system liner+=tj et x 21(2/1)(jt e 2-) )(1t y =1/2(tj e3+tj e3-)Thereforex 1(t)=cos(2t))(1t y =cos(3t)(b) we know thatx 2(t)=cos(2(t-1/2))= (j e -jte 2+jejt e 2-)/2Using the linearity property, we may once again write x 1(t)=21( j e -jt e 2+j e jt e 2-))(1t y =(je-jt e 3+je jt e 3-)= cos(3t-1)Therefore,x 1(t)=cos(2(t-1/2)))(1t y =cos(3t-1)1.21.The signals are sketched in figure S1.21.(f) Not period.1.26 (a) periodic, period=7.(b) Not period.(c) periodic, period=8.(d) x[n]=(1/2)[cos(3πn/4+cos(πn/4)). periodic, period=8. (e) periodic, period=16. 1.27 (a) Linear, stable(b) Not period. (c) Linear(d) Linear, causal, stable(e) Time invariant, linear, causal, stable (f) Linear, stable(g) Time invariant, linear, causal 1.28 (a) Linear, stable(b) Time invariant, linear, causal, stable (c)Memoryless, linear, causal (d) Linear, stable (e) Linear, stable(f) Memoryless, linear, causal, stable (g) Linear, stable1.29 (a) Consider two inputs to the system such that[][][]{}111.S e x n y n x n −−→=ℜand [][][]{}221.Se x n y n x n −−→=ℜNow consider a third inputx 3[n]= x2[n]+x 1[n]. The corresponding system outputWill be [][]{}[][]{}[]{}[]{}[][]33121212e e e e y n x n x n x n x n x n y n y n ==+=+=+ℜℜℜℜtherefore, we may conclude that the system is additive Let us now assume that inputs to the system such that andNow consider a third input x 3 [n]= x 2 [n]+ x 1 [n]. The corresponding system output Will betherefore, we may conclude that the system is additive (b) (i) Consider two inputs to the system such that()()()()211111Sdx t x t y t x t dt ⎡⎤−−→=⎢⎥⎣⎦and ()()()()222211S dx t x t y t x t dt ⎡⎤−−→=⎢⎥⎣⎦Now consider a third inputx 3[t]= x2[t]+x 1[t]. The corresponding system outputWill betherefore, we may conclude that the system is not additiveNow consider a third input x 4 [t]= a x 1 [t]. The corresponding system output Will beTherefore, the system is homogeneous.(ii) This system is not additive. Consider the fowling example .Let δ[n]=2δ[n+2]+ 2δ[n+1]+2δ[n] andx2[n]=δ[n+1]+ 2δ[n+1]+ 3δ[n]. The corresponding outputs evaluated at n=0 areNow consider a third input x 3 [n]= x 2 [n]+ x 1 [n].= 3δ[n+2]+4δ[n+1]+5δ[n]The corresponding outputs evaluated at n=0 is y 3[0]=15/4. Gnarly, y 3[0]≠]0[][21y y n +.ThisTherefore, the system is homogenous.1.30 (a) Invertible. Inverse system y(t)=x(t+4)(b)Non invertible. The signals x(t) and x 1(t)=x(t)+2πgive the same output (c) δ[n] and 2δ[n] give the same output d) Invertible. Inverse system; y(t)=dx(t)/dt(e) Invertible. Inverse system y(n)=x(n+1) for n ≥0 and y[n]=x[n] for n<0 (f) Non invertible. x (n) and –x(n) give the same result (g)Invertible. Inverse system y(n)=x(1-n) (h) Invertible. Inverse system y(t)=dx(t)/dt(i) Invertible. Inverse system y(n) = x(n)-(1/2)x[n-1] (j) Non invertible. If x(t) is any constant, then y(t)=0 (k) δ[n] and 2δ[n] result in y[n]=0 (l) Invertible. Inverse system: y(t)=x(t/2)(m) Non invertible x 1 [n]= δ[n]+ δ[n-1]and x 2 [n]= δ[n] give y[n]= δ[n] (n) Invertible. Inverse system: y[n]=x[2n]1.31 (a) Note that x 2[t]= x 1 [t]- x 1 [t-2]. Therefore, using linearity we get y 2 (t)= y 1 (t)- y 1 (t-2).this is shown in Figure S1.31(b)Note that x3 (t)= x1 [t]+ x1 [t+1]. .Therefore, using linearity we get Y3 (t)= y1 (t)+ y1 (t+2). this is2220201.34. (a) ConsiderIf x[n] is odd, x[n] +x [-n] =0. Therefore, the given summation evaluates to zero. (b) Let y[n] =x 1[n]x 2[n] .Theny [-n] =x 1[-n] x 2[-n] =-x 1[n]x 2[n] =-y[n]. This implies that y[n] is odd.(c)ConsiderUsing the result of part (b), we know that x e [n]x o [n] is an odd signal .Therefore, using {}1[][0][][]n n x n x x n x n ∞∞=-∞==++-∑∑22[][]e o n n n n x x ∞∞=-∞=-∞=+∑∑222[][][]e on n n n n n x x x ∞∞∞=-∞=-∞=-∞==+∑∑∑the result of part (a) we may conclude thatTherefore,(d)ConsiderAgain, since x e (t) x o (t) is odd,Therefore,1.35. We want to find the smallest N 0 such that m(2π /N) N 0 =2πk or N 0 =kN/m,where k is an integer, then N must be a multiple of m/k and m/k must be an integer .this implies that m/k is a divisor of both m and N .Also, if we want the smallest possible N 0, then m/k should be the GCD of m and N. Therefore, N 0=N/gcd(m,N).1.36．(a)If x[n] is periodic 0(),0..2/j n N T o ewhere T ωωπ+= This implies that 022o T kNT k T T Nππ=⇒==a rational number . (b)T/T 0 =p/q then x[n] =2(/)j n p q e π,The fundamental period is q/gcd(p,q) and the fundmental frequencyis(c) p/gcd(p,q) periods of x(t) are needed .1.37.(a) From the definition of ().xy t φWe have (b) Note from part(a) that()().xx xx t t φφ=-This implies that ()xy t φis even .Therefore,the odd part of ().xx t φis zero.(c) Here, ()().xy xx t t T φφ=-and ()().yy xx t t φφ= 1.38.(a) We know that /22(2)().t t δδ=V V Therefore This implies that(b)The plot are as shown in Figure s3.18.1.39 We have Also,2[][]0eon n n x x ∞=-∞=∑222[][][].e on n n n n n xx x ∞∞∞=-∞=-∞=-∞==+∑∑∑2220()()()2()().eoet dt t dt t dt t t dt x x x x x ∞∞∞∞-∞-∞-∞-∞=++⎰⎰⎰⎰0()()0.et t dt x x ∞-∞=⎰222()()().e ot dt t dt t dt xx x ∞∞∞-∞-∞-∞=+⎰⎰⎰0022gcd(,)gcd(,)gcd(,)gcd(,).T pp q p q p q p q q p q p pωωππ===/21lim (2)lim ().2t t δδ→∞→∞=V V V V(b) y(t)=x 2(t) is such a systerm . (c) No.For example,consider y(t) ()()ty t x d ττ-∞=⎰with ()()(1).x t u t u t =--Then x(t)=0for t>1,but y(t)=1 for t>1.1.41. (a) y[n]=2x[n].Therefore, the system is time invariant.(b) y[n]=(2n-1)x[n].This is not time-invariant because y[n- N 0]≠(2n-1)2x [n- N 0]. (c) y[n]=x[n]{1+(-1)n +1+(-1)n-1}=2x[n].Therefore, the system is time invariant .1.42.(a) Consider two system S 1 and S 2 connected in series .Assume that if x 1(t) and x 2(t) arethe inputs to S 1..then y 1(t) and y 2(t) are the outputs.respectively .Also,assume thatif y 1(t) and y 2(t) are the input to S 2 ,then z 1(t) and z 2(t) are the outputs, respectively . Since S 1 is linear ,we may writewhere a and b are constants. Since S 2 is also linear ,we may write We may therefore conclude thatTherefore ,the series combination of S 1 and S 2 is linear. Since S 1 is time invariant, we may write andTherefore,Therefore, the series combination of S 1 and S 2 is time invariant.(b) False, Let y(t)=x(t)+1 and z(t)=y(t)-1.These corresponds to two nonlinear systems. If these systems are connected in series ,then z(t)=x(t) which is a linear system.(c) Let us name the output of system 1 as w[n] and the output of system 2 as z[n] .Then The overall system is linear and time-invariant. 1.43. (a) We haveSince S is time-invariant.Now if x (t) is periodic with period T. x{t}=x(t-T). Therefore, we may conclude that y(t)=y(t-T).This implies that y(t) is also periodic with T .A similar argument may be made in discrete time . (b)1.44 (a) Assumption : If x(t)=0 for t<t 0 ,then y(t)=0 for t< t 0.To prove That : The system is causal.Let us consider an arbitrary signal x 1(t) .Let us consider another signal x 2(t) which is the same as x 1(t)fort< t 0. But for t> t 0 , x 2(t) ≠x 1(t),Since the system is linear, Since ()()120x t x t -=for t< t 0 ,by our assumption =()()120y t y t -=for t< t 0 .This implies that()()12y t y t =for t< t 0 . In other words, t he output is not affected by input values for 0t t ≥. Therefore, thesystem is causal .Assumption: the system is causal . To prove that :If x(t)=0 for t< t 0 .then y(t)=0 for t< t 0 .Let us assume that the signal x(t)=0 for t< t 0 .Then we may express x(t) as ()()12()x t x t x t =-, Where()()12x t x t = for t< t 0 . the system is linear .the output to x(t) will be()()12()y t y t y t =-.Now ,since the system is causal . ()()12y t y t = for t< t 0 .implies that ()()12y t y t = for t< t 0 .Therefore y(t)=0 for t< t 0 .(b) Consider y(t)=x(t)x(t+1) .Now , x(t)=0 for t< t 0 implies that y(t)=0 for t< t 0 .Note that the system is nonlinear and non-causal .(c) Consider y(t)=x(t)+1. the system is nonlinear and causal .This does not satisfy the condition of part(a). (d) Assumption: the system is invertible. To prove that :y[n]=0 for all n only if x[n]=0 for all n . Consider[]0[]x n y n =→. Since the system is linear :2[]02[]x n y n =→.Since the input has not changed in the two above equations ,we require that y[n]= 2y[n].This implies that y[n]=0. Since we have assumed that the system is invertible , only one input could have led to this particular output .That input must be x[n]=0 .Assumption: y[n]=0 for all n if x[n]=0 for all n . To prove that : The system is invertible . Suppose that andSince the system is linear ,By the original assumption ,we must conclude that 12[][]x n x n =.That is ,any particular y 1[n] can be produced that by only one distinct input x 1[n] .Therefore , the system isinvertible.(e) y[n]=x 2[n].1.45. (a) Consider ,and()222()()s hx x t y t t φ→=.Now, consider ()()()312x t ax t bx t =+. The corresponding system output will be Therefore, S is linear .Now ,consider x 4(t)=x 1(t-T).The corresponding system output will beClearly, y 4(t)≠ y 1(t-T).Therefore ,the system is not time-invariant.The system is definitely not causal because the output at any time depends on future values of the input signal x(t).(b) The system will then be linear ,time invariant and non-causal.1.46. The plots are in Figure S1.46.1.47.(a) The overall response of the system of Figure P1.47.(a)=(the response of the system to x[n]+x 1[n])-the response of the system to x 1[n]=(Response of a linear system L to x[n]+x 1[n]+ zero input response of S)- (Response of a linear system L to x 1[n]+zero input response of S)=( (Response of a linear system L to x[n]).。
信号与系统习题集英文版
信号与系统习题集第一部分：时域分析 一、填空题 1. =---)3()()2(t t u et δ()。
2. The unit step response )(t g is the zero-state response when the input signal is ( ).3. Given two continuous – time signals x(t) and h(t), if their convolution is denoted by y(t), then the convolution of )1(-t x and )1(+t h is ( ).4. The convolution =+-)(*)(21t t t t x δ( ).5. The unit impulse response )(t h is the zero-state response when the input signal is ( ).6. A continuous – time LTI system is stable if its unit impulse response satisfies the condition: ( ).7. A continuous – time LTI system can be completely determined by its ( ).8. =⎰∞∞-(t)dt 2sin 2 δt t ( ).9. Given two sequences }1,2,2,1{][=n x and }5,6,3{][=n h , their convolution=][*][n h n x ( ).10. Given three LTI systems S1, S2 and S3, their unit impulse responses are )(1t h ,)(2t h and )(3t h respectively. Now, construct an LTI system S using these threesystems: S1 parallelinterconnected by S2, then series interconnected by S3. the unit impulse response of the system S is ( ).11. It is known that the zero-stat response of a system to the input signal x(t) is⎰∞-=td x t y ττ)()(, then the unit impulse response h(t) is ( ).12. The complete response of an LTI system can be expressed as a sum of its zero-state response and its ( ) response.13. It is known that the unit step response of an LTI system is )(2t u et-, then the unit impulseresponse h(t) is ( ).14. =++-=⎰∞dt t t t t x ))1()1((2sin )(0δδπ( ).15. We can build a continuous-time LTI system using the following three basic operations: ( ) , ( ), and ( ). 16. The zero-state response of an LTI system to the input signal )1()()(--=t u t u t xis )1()(--t s t s , where s(t) is the unit step response of the system, then the unit impulse response h(t) is ( ).17. The block diagram of a continuous-time LTI system is illustrated in the following figure. The differential equation describing the input-output relationship of the system is ( )。
[信号与系统][第二版][奥本海姆][课后答案][英文][文字版]
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Signals & Systems
(Second Edition)
—Learning Instructions
(Exercises Answers)
Department of Computer Engineering 2005.12
Contents
Chapter 1 ······················································· 2 Chapter 2 ······················································· 17 Chapter 3 ······················································· 35 Chapter 4 ······················································· 62 Chapter 5 ······················································· 83 Chapter 6 ······················································· 109 Chapter 7 ······················································· 119 Chapter 8 ······················································· 132 Chapter 9 ······················································· 140 Chapter 10 ·······················································160