《信号与系统教学资料》信号与系统奥本海姆英文版课后答案chapter10.pdfVIP

Chapter 10 Answers 10.1 (a)The given summation may be written as ∑∞ 1 ( 1 r ?1)ne? jwn 2n=?1 2 ,by replacing z with re jw . If r 1 ,then 2 1 r ?1 1 2 And the function within the summation grows towards infinity with increasing n.. Also , the summation dose not converge. But if r 1 ,then the summation converges. 2 (b) The given summation may be written as ∑∞ 1(1 r?1)ne? jwn , by replacing z with re jw . If r 1 , then 2r 1 2n=?1 2 2 And the function within the summation grows towards infinity with increasing n ,. Also , the summation dose not converge. 1 But if r , then the summation converges. 2 ∑( c )The summation may be written as ∞ r ?n + (?r)?n e? jwn . by replacing z with re jw . If n=0 2 r1, then the function within the summation grows towards infinity with increasing n ,. Also , the summation dose not converge. But if r 1, then the summation converges. ∑ ∑(a) The summation may be written as ∞ (1 r?1)n cos(π n / 4)e? jwn + 0 (1 r)?n cos(π n / 4)e? jwn n=0 2 2n=?∞ by replacing z with re jw . The first summation converges for r 1 . The second summation 2 converge for r2. Therefore, the sum of these two summations converges for 1/2r2 . 10.2 Using eq.910.3). ∑ ∑X (z) = ∞ (1)n u[n ? 3]z?n = ∞ (1)n z?n 5n=?∞ n=3 5 ∑= [ z?3 ] ∞ (1)n z?n 125 n=0 5 = [ z?3 ] 125 1? 1 1 z ?1 , z 1 5 5 10.18. (a) using the analysis of example10.18,we may show that H (Z ) = 1? 6Z ?1 + 8Z ?2 1 ? 2 3 Z ?2 + 1 9Z ?2 Since h(z)=y(z)/x(z), we may write Y (Z )[1? 2 Z ?1 + 1 Z ?2 ] = X (Z )[1? 6Z ?18Z ?2 ] 39 Taking the inverse z-transform we obtain y[n] ? 2 y[n ?1] + 1 [n ? 2] = x[n] ? 6x[n ?1] + 8x[n ? 2] 39 (b) H (Z) has only two poles, these are both at z=1/3. Since the system is causal, the ROC of H (Z) will be the form z 1/3. Since the ROC includes the unit circle, the system is Stable. 10.19. (a) The unilateral z-transform is ∑x(Z ) = ∞ ( 1 )nu[n + 5]z?n N =0 4 ∑= ∞ ( 1 )n z?n n=0 4 1 1? (1 )z?1 , z 1 4 4 (b) The unilateral z-transform is 178 ∞ ∑x(z) = (σ[n + 3] + σ[n] + 2n[?n])z?n n=0