EECE 340 Homework, Spring 2005

EECE 340-Signals and Systems Homework #5 Problem # 1 An analog system has the impulse response given by ee, 120 A= , ) ( <0 Let a discrete-time system be obtained by sampling h,(4); i.e., having h(n) =h, (nT). Find and plot |H(a) . Problem # 2 The impulse response of a discrete-time system is given by 1, O<nsN-1 ion= Osn 0, otherwise Find and plot |H (@)| and arg(H! (a)) for N=5. Problem # 3 Find and sketch the output y(n) for an LTI discrete-time system with input x(n) = u(n) —u(n — 6) and impulse response: h(n) = a"u(n). Problem #4 Let A(n) = a"u(n) x(n) = b"u(n) Find the output y(n) = x(1)*h(n) fora #b. Repeat fora=b. Page 1 of 2 Problem #5 The impulse responses of two discrete-time LTI systems are given by h(n) = u(n) —u(n - 6) hy (n) =u(n+4)-u(n) a) Find the impulse response A(7) for the parallel combination of /,(”) and /,(). b) Find the impulse response /() for the cascaded combination of /, (7) and h, (n) . Problem # 6 Repeat Parts (a) and (b) in Problem 5 for the case where the systems impulse responses are given by h(n) = Ga)"u(n) h,(n) = (-jay"u(n) Problem #7 The unit step response for an LTI discrete-time system is given by s(1) = u(n)* h(a) (a) Express s(n) for a causal system as a summation of h(n). (b) Give the inverse relationship for A(n) in terms of s(n). (c) Find s(n) if h(n) = u(1)-u(n-6). (d) Find and sketch s(n) when h(n) =(-4)"u(n). Find lims(n). Problem # 8 Determine the conditions on the parameters of the following systems for stability: a) hA(n)=a"u(-n) b) h(n) = a" [u(n) —u(n~100)] c) A(n)=r" sin(na,T)u(n) d) A(n)= al" Page 2 of 2 0, 0 Herwrese— NF ne Ne! Hw) — s- 2 _ ( sJe7)” n=o0 <— _ yi =o , ' a fn VOUT ET EN) /— ev MT a a = 2 ay wt ( ae elu Vebe cus Ot =) _— te _QUNT —/ ae” Se ONY) Ain WT w HON) = 4h) — &h- €) Ah) = 2” un) HW <o ) y Chi =o a onek J O<n<5, Y(n)= 2 2 =a") a A k=o L=o / a?! __s oo _— "Zs 17 => alt _ "gr Zao kmo 6 = Ce. —/ 4-S vara a / p for @ = 8) YO= ATT = orsia, ny a C Ji) =o, WMO. &) 4 tn) = 4 (n)~ &€(n-~s) h, CN) = 40h pu) — “n). a) BN) = Fi Ch) +R CH) “| 2 -4F¢RNZ 5S Hh _D __ _ — n(n)= > |= nee , ~1<7<-! ba-4 #(n) = > / — 4 a ON] =-+ _I Ainj= > | =-n4r, [<Neg k=n-7 4ln)=o _ a a 4/9 7 0 5/9 9 “ 7 “ ; 1) . ° oa im 5 : ) - : , ” : tn : | g | ’ he : (n) = 4 > & i) hi Zz a. pr 7/9 7) S'Co) = (0) Sr) = Ro + RW) s(n-1) =f Us) + RU) +--+ Pon ~1) SCn) = RO) +2 4+-- + Rm) + Rin). > 4(n) = Sn) ~ SCh-t). S) A(n) = U(n)—~“(n- 6) SO) =D 1 =a 4) kL =o ) s ke =6 OK £5 7 AES J) Hh) = C4)'2cn) “a S(n) = (-4£ & _ /— (--L a+! 2/9 oy ee > a "=> Gb)" 9 =~. _ 7 =6 — ; fa}>l [a] b) A(n) = a [4207 ~ (ne tov) } Medags Stable _ 2) An) = Aten (2, F) et Cn) 2 bP = + pay, Yeo /—/r/ y A) #ln) = a =| “7 720