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1、 Department of Computer Engineering 2005.12 信号与系统 奥本海姆第二版 习题解答1 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 2 Chapter 1 Answers 1.1 Converting from polar to Cartesian coordinates: 111cos222je 111c o s
2、 ()222je 2cos()sin()22jjje2c o s ()s i n ()22jjje 5 22jjjee 42 ( c o s ()s i n () )1442jjje 9 44122jjjee 9 44122jjjee 412jje 1.2 converting from Cartesian to polar coordinates: 055je, 22je , 233jje 213 22jje, 412jje , 2 221jje4(1)jje , 41 1j je1222 13j je1.3. (a) E=401 4tdte, P=0, because E (b) (2)4
3、 2( )jttxe, 2( )1tx.Therefore, E=2 2( )dttx=dt=, P=211limlim222( )TTTTTTdtdtTTtxlim1 1 T (c) 2( ) tx=cos(t). Therefore, E=2 3( )dttx=2cos( )dtt=, P=2111(2 )1limlim2222cos( )TTTTTTCOStdtdtTTt(d) 1 1 2nnu nx, 2 11 4nu nnx. Therefore, E=204 131 4nnnxP=0，because E7. (b) The signal xn is shifted by 4 to
4、the left. The shifted signal will be zero for n0. (c) The signal xn is flipped signal will be zero for n2. (d) The signal xn is flipped and the flipped signal is shifted by 2 to the right. The new Signal will be zero for n4. (e) The signal xn is flipped and the flipped and the flipped signal is shif
5、ted by 2 to the left. This new signal will be zero for n0. 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)
6、 +x(2-t) will be zero for t-2. (c) x(3t) is obtained by linearly compression x(t) by a factor of 3. Therefore, x(3t) will be zero for t3. (b) Since x1(t) is an odd signal, 2 vnxis zero for all values of t. (c) 11311 332211 22vnnnnnu nunxxx Therefore, 3 vnxis zero when n1,but y(t)=1 for t1. 1.41. (a)
7、 yn=2xn.Therefore, the system is time invariant. (b) yn=(2n-1)xn.This is not time-invariant because yn- N0(2n-1)2xn- N0. (c) yn=xn1+(-1)n+1+(-1)n-1=2xn.Therefore, the system is time invariant . 1.42.(a) Consider two system S1 and S2 connected in series .Assume that if x1(t) and x2(t) are the inputs
8、to S1then y1(t) and y2(t) are the outputs.respectively .Also,assume that if y1(t) and y2(t) are the input to S2 ,then z1(t) and z2(t) are the outputs, respectively . Since S1 is linear ,we may write 11212,s ax tbxtay tbyt where a and b are constants. Since S2 is also linear ,we may write 21212,s ay
9、tbytaz tbzt We may therefore conclude that )()()()(212121tbtatbtazzxxss Therefore ,the series combination of S1 and S2 is linear. Since S1 is time invariant, we may write 11010sxtTytT and 21010sytTztT Therefore, 1 21010s s x tTz tT Therefore, the series combination of S1 and S2 is time invariant. (b
10、) 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. 00. ( )( ).00x ty t0( )( )( )( )0x tx ty ty t1 1-1/2e-t/ u（t） 1 - 0 t u（t） 1 1/2 - 0 1/2e-t/ t Figure s3.18 15 (c) Let us name the o
11、utput of system 1 as wn and the output of system 2 as zn .Then 11 2 2 212224y nznw nw nw n 241121nxnxnx The overall system is linear and time-invariant. 1.43. (a) We have )(tytxs Since S is time-invariant. )(TtyTtxs Now if x (t) is periodic with period T. xt=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 tim
