EECE 340 Homework, Spring 2001

EECE 340 — Signals and Systems Homework # 1 Consider the following voltage signal across a 50 Q resistance. 11 ately SO= 2rel Determine the energy dissipated by jt) across the resistance. -hetst Hint: rect(t)=3 2 2 0, elsewhere 1.2 Consider the signal f(t) =e" for allt. Apply the time-domain integration formulas for energy and average power to determine the nature of the signal f(t). Consider the following system: sW=TIOl=4 f f@dr tut). Determine the impulse response A(t) of the system and the system output g(t) when the input 13 fis given by f(t) =e u(t). 1.4 Consider the system shown in the figure below. t >) JO dt. = ——~> a(t) SAC) In this system, f(t) is the input signal and g(t) is the output signal. Determine the impulse response, /i(t), of the system and the system response, g(t), for an input f(t)=0(t-tq), where d(t) is the unit impulse function. Conclude whether the system is time-invariant or time-varying. Page 1 of 2 1.5 Consider an LTI system with impulse response /(t) given by A(t) = e*u(-t). Determine the output g/t) of this system when the input f(t) = u(t) and use g(t) to determine whether the system is or is not physically realizable. Explain. 1.6 Consider the following system: BS =TIPO]=2f(O +e). Determine whether this system is linear or non-linear and time-invariant or time-varying. 1.7 Consider the system represented by the following input-output transformation: gs) =T/(M) =e" FO Examine the linearity, time-invariance and causality of this system. 18 ue h(t) =e"u(t) | sO, For the above LTI system, determine the energy or average power, whichever applies, of the system output g(t), when f(t) = u(t). Use time-domain analysis. Page 2 of 2 EECE 340 — Signals and Systems Homework # 1 — Solution 1.1 The voltage signal S(O = 2rect( Ty is across a 50Q resistance. Using , erect rect(t) = 2 2 0, elsewhere then IP yy S(t) = 2rect( _f2,0<¢s or ~ 0, elsewhere T E, at dt = Ah 27 Joules 50 50. 25 1.2. Consider the signal fW=e", allt E= ferar=-Le" =1(e?-e*)=4e" 0 2 2° le 9 2 lyr or —l|e -e P, = lim— a edt = lim 2| ) Ta0 T T30 T T -T T = lim =-lim 2— = tim = 0 Toa 2T Taw QT T2027 => /(®) is neither an energy nor a power signal 1.3. The following system is considered: s@=T[S(t)]=4 [ f@dr+uO A(t) =T[ S(t) ]=4 [st@ar+u) For? <0, A(t) =0 For?20, A(t) =4u()+uO=Fa &) For f(t) =e"u(t), the system output is: g=4 | f@dr+uy t 4 Jevar +u(t)= -26*|' +u(t) Qa 2(1-e* u(t) +u(e) 3-2e*' u(t) 1.4 The system shown in the figure below is considered. t cA) my fdr -— 2) The system impulse response is the system output, denoted by A(t), when the input f(t) = d(t). Thus, A(t) = fetodr = Jo(e)de 0, t<0 -{ t>0 =u(t) For f(t) =6d(t-1,), 1 a= fets(r-1,)dr =e" [o(r-1)ae 0 =e“u(t-t,) Obviously, the system, which is formed by the cascaded connection of a multiplier and integrator is a time-varying system although the integrator is time-invariant. 1.5 The LTI system with impulse response h(t) given by h(t) =e"u(-t) is considered. The output g(t) of this system when the input f(t) = u(t) can be obtained using convolution. A gy eu(-r) _ > t 0 ; Ly py 1 =5( 2, -0)=<e* 1; 1 Fort20,g(t)= fevde=se"| => an t<0 =>g()= 1 -, t20 2 The system input is f(t) = u(t). So, it applied for t 2 0. But, the system produces an output for t < 0; i.e., before the input is applied. Hence, the system is not physically realizable. 1.6 The following system is considered: BO=TILO]= 2/04 ue). Linearity: Tla AMt+a,40]= 2a, fi()+ 2a, f,() +u(t) #aT[fO]+a7[40] = 2a, fi(t)+u(t)+2a, f,(t)+u(t) = 2a, f(O+2a, f+ 2u(y So, the system is non-linear. Time-invariance: T[ f(t-t) |= 2s (¢-%) +4 (1) # g(t-4) =2f (1-4) +u(t-4) Hence, the system is time-varying. 1.7 The system represented by the following input-output transformation is considered: s@)=TPMO]=e" SO Linearity: Tla fO+a,f4,0]=47[A0]+47[40]??? Ta fOrtaAOl=e"[ahOr+ahO] sae’ fOt+ae"f0 =al[f£O]+a7[40] So, the system is linear. Causality: a()=e S(t) Hence, the system output at time 4, depends on the input at time 4). => the system is causal Time-invariance: g(t-%) =e" £(t-4) #7 [F(t-%) |= eS (t-4) Hence, the system is time-varying. 1.8 ee A(t) = e'u(t) | 8 When the input to the system is //)=u(t), then the output is g(f)=(/-e"Ju(t). This can be verified using convolution. It is obvious from the plot of g(t) that g(t) has an « energy. Hence g(t) is not an energy signal. g(t) is a power signal if it has a finite non-zero average power. As shown in the computations done below, g(t) is therefore a power signal. P, = tim> [(1-e*) dt = unt] fos e #4] wv Tae T Ta0 T i T