One important use of inverse systems is in situations in which one wishes to remove

Let x[n] = 8[n] + 28[n- 1] - 8[n- 3] and h[n] = 28[n + 1] + 28[n- 1]. Compute and plot each of the following convolutions: (a) YI [n] = x[n] * h[n] (b) Y2[n] = x[n + 2] * h[n] (c) Y3 [n] = x[n] * h[n + 2]

Consider the signal ( ) n-1 h[n] = ~ {u[n + 3] - u[n - 10]}. Express A and B in terms of n so that the following equation holds: h[n - k] = 2 ' - - . { ( _!_ )n- k- I A < k < B 0, elsewhere

Consider an input x[n] and a unit impulse response h[n] given by x[n] = (U-2 u[n- 2], h[n] = u[n + 2]. Determine and plot the output y[n] = x[n] * h[n].

Compute and plot y[n] = x[n] * h[n], where x[n] = { ~ h[n] = { ~: 3:Sn:S8 otherwise ' 4 ::; n ::; 15 otherwise

Let x[n] = { ~ h[n] = { ~: 3:Sn:S8 otherwise ' 4 ::; n ::; 15 otherwise xn [ ] = an n = { 1, 0 ::; n ::; 9 d h[ ] { 1, 0 ::; n ::; N 0, elsewhere 0, elsewhere ' Chap. 2 where N ::; 9 is an integer. Determine the value of N, given that y[n] = x[n] * h[n] and y[4] = 5, y[l4] = 0.

Compute and plot the convolution y[n] = x[n] * h[n], where x[n] = (~ r u[-n- I] and h[n] = u[n- 1].

A linear systemS has the relationship CIJ y[n] = L x[k]g[n - 2k] k=-c;IO between its input x[n] and its output y[n], where g[n] = u[n] - u[n- 4]. (a) Determine y[n] when x[n] = o[n- 1]. (b) Determine y[n] when x[n] = o[n- 2]. (c) IsS LTI? (d) Determine y[n] when x[n] = u[n].

Determine and sketch the convolution of the following two signals: 2.9. Let { t + 1, 0 ::; t ::; 1 x(t) = 2 - t, 1 < t ::; 2 , 0, elsewhere h(t) = o(t + 2) + 2o(t + 1).

Let { t + 1, 0 ::; t ::; 1 x(t) = 2 - t, 1 < t ::; 2 , 0, elsewhere h(t) = o(t + 2) + 2o(t + 1). h(t) = e21 u(-t+4)+e-21 u(t-5). Determine A and B such that { -2(t-T) e ' h(t- T) = 0, e2(t-T), 2.10. Suppose that r

Let x(t) = u(t- 3)- u(t- 5) and h(t) = e- 3 t u(t). (a) Compute y(t) = x(t) * h(t). (b) Compute g(t) = (dx(t)ldt) * h(t). (c) How is g(t) related to y(t)?

Let 00 y(t) = e-tu(t) * .:L set- 3k). k= -00 Show that y(t) = Ae-t for 0 ::; t < 3, and determine the value of A.

Consider a discrete-time system S1 with impulse response \(h[n]=\left(\frac{1}{5}\right)^{n} u[n]\) (a) Find the integer A such that h[n] - Ah[n- 1] = \(\delta\)[n]. (b) Using the result from part (a), determine the impulse response g[n] of an LTI system S2 which is the inverse system of S1.

Which of the following impulse responses correspond(s) to stable LTI systems? (a) h1(t) = e-(1-2j)ru(t) (b) h2(t) = e-r cos(2t)u(t)

Which of the following impulse responses correspond(s) to stable LTI systems? (a) \( h_{1}[n]=n \cos \left(\frac{\pi}{4} n\right) u[n]\) (b) \(h_{2}[n]=3^{n} u[-n+10]\)

For each of the following statements, determine whether it is true or false: (a) If x[n] = 0 for n < N 1 and h[n] = 0 for n < N2, then x[n] * h[n] = 0 for n < N1 + N2. (b) If y[n] = x[n] * h[n], then y[n- 1] = x[n- 1] * h[n- 1]. (c) If y(t) = x(t) * h(t), then y( -t) = x( -t) * h( -t). (d) If x(t) = 0 for t > T1 and h(t) = 0 for t > T2, then x(t) * h(t) = 0 for t > T1 + T2.

Consider an LTI system whose input x(t) and output y(t) are related by the differential equation d dt y(t) + 4 y(t) = x(t). The system also satisfies the condition of initial rest. (a) If x(t) = e<-l+3j)tu(t), what is y(t)? (P2.17-1) (b) Note that CRe{x(t)} will satisfy eq. (P2.17-1) with CRe{y(t)}. Determine the output y(t) of the LTI system if x(t) = e -r cos(3t)u(t)

Consider a causal LTI system whose input x[n] and output y[n] are related by the difference equation 1 y[n] = 4 y[n - 1] + x[n]. Determine y[n] if x[n] = o[n - 1].

Consider the cascade of the following two systems 5 1 and 52, as depicted in Figure P2.19: S 1 : causal LTI, 1 w[n] = 2w[n- 1] + x[n]; S2 : causal LTI, y[n] = ay[n- 1] + f3w[n]. The difference equation relating x[n] and y[n] is: 1 3 y[n] = - S y[n - 2] + 4 y[n - 1] + x[n]. (a) Determine a and f3. (b) Show the impulse response of the cascade connection of S1 and S2

Compute the convolution y[n] = x[n] * h[n] of the following pairs of signals: (a) x[n] = a 11 u[n], } a = f3 h[n] = {3 11 u[n], (b) x[n] = h[n] = anu[n] (c) x[n] = (- ~ )" u[n - 4] h[n] = 4"uC2 - n] (d) x[n] and h[n] are as in Figure P2.21.

For each of the following pairs of waveforms, use the convolution integral to find the response y(t) of the LTI system with impulse response h(t) to the input x(t). Sketch your results. (a) x(t) = e-ar u(t)} (Do this both when a = f3 and when a = f3 .) (b) x(t) = u(t) - 2u(t - 2) + u(t - 5) h(t) = e2 t u(l - t) (c) x(t) and h(t) are as in Figure P2.22(a). (d) x(t) and h(t) are as in Figure P2.22(b). (e) x(t) and h(t) are as in Figure P2.22(c).

Let h(t) be the triangular pulse shown in Figure P2.23(a), and let x(t) be the impulse train depicted in Figure P2.23(b ). That is, +x x(t) = L, 8(t - kT). k=-X Determine and sketch y(t) = x(t) * h(t) for the following values ofT: (a) T = 4 (b) T = 2 (c) T = 3/2 (d) T = I

Consider the cascade interconnection of three causal LTI systems, illustrated in Figure P2.24(a). The impulse response h2 [n] is h2[n] = u[n] - u[n - 2], and the overall impulse response is as shown in Figure P2.24(b ). x[n] y[n] (a) -1 0 1 2 3 4 5 6 7 n (b) Figure P2.24 (a) Find the impulse response h1 [n]. (b) Find the response of the overall system to the input x[n] = o[n] - o[n- 1].

Let the signal y[n] = x[n] * h[n], where \(x[n]=3^{n} u[-n-1]+\left(\frac{1}{3}\right)^{n} u[n]\) and \(h[n]=\left(\frac{1}{4}\right)^{n} u[n+3]\) (a) Determine y[n] without utilizing the distributive property of convolution. (b) Determine y[n] utilizing the distributive property of convolution.

Consider the evaluation of y[n] = XJ [n] * x2[n] * x3[n], where x1 [n] = (0.5Yu[n], x2[n] = u[n + 3], and x3[n] = B[n]- B[n- 1]. (a) Evaluate the convolution x 1 [n] * x2 [n]. (b) Convolve the result of part (a) with x3[n] in order to evaluate y[n]. (c) Evaluate the convolution x2 [n] * x3 [n]. (d) Convolve the result of part (c) with x 1 [n] in order to evaluate y[n].

We define the area under a continuous-time signal v(t) as I +oo A\. = -oo v(t) dt. Show that if y(t) = x(t) * h(t), then

The following are the impulse responses of discrete-time LTI systems. Determine whether each system is causal and/or stable. Justify your answers. (a) h[n] = (~)nu[n] (b) h[n] = (0.8)nu[n + 2] (c) h[n] = (~)nu[-n] (d) h[n] = (5)nu[3 - n] (e) h[n] = (- ~)nu[n] + (1.01) 11 u[n- 1] (f) h[n] = (- ~)nu[n] + (l.Ol)n u[l - n] (g) h[n] = n(~)' u[n- 1]

The following are the impulse responses of continuous-time LTI systems. Determine whether each system is causal and/or stable. Justify your answers. (a) h(t) = e-4r u(t- 2) (b) h(t) = e- 6t u(3 - t) (c) h(t) = e-2r u(t + 50) (d) h(t) = e21 u( -1 - t) (e) h(t) = e-61tl (0 h(t) = te-t u(t) (g) h(t) = (2e-t - e(t-IOO)!IOO)u(t)

Consider the LTI system initially at rest and described by the difference equation y[n] + 2y[n - 1] = x[n] + 2x[n - 2]. Find the response of this system to the input depicted in Figure P2.31 by solving the difference equation recursively. x[n]

Consider the difference equation 1 y[n] - 2 y[n- 1] = x[n], (P2.32-1) and suppose that x[n] = (~ )" u[n]. (P2.32-2) Assume that the solution y[n] consists of the sum of a particular solution Yp[n] to eq. (P2.32-l) and a homogeneous solution Yh[n] satisfying the equation 1 Yh[n]- 2Yh[n- 1] = 0. (a) Verify that the homogeneous solution is given by Yh[n] = AG)" (b) Let us consider obtaining a particular solution Yp[n] such that 1 (1 )n Yp[n]- 2 Yp[n- 1] = 3 u[n]. By assuming that Yp[n] is of the form B( * )'1 for n 2: 0, and substituting this in the above difference equation, determine. the value of B. (c) Suppose that the LTI system described by eq. (P2.32-1) and initially at rest has as its input the signal specified by eq. (P2.32-2). Since x[n] = 0 for n < 0, we have that y[n] = 0 for n < 0. Also, from parts (a) and (b) we have that y[n] has the form (1)11 (1)11 y[n] =A "2 + B "3 for n :::::: 0. In order to solve for the unknown constant A, we must specify a value for y[n] for some n 2: 0. Use the condition of initial rest and eqs. (P2.32-1) and (P2.32-2) to determine y[O]. From this value determine the constant A. The result of this calculation yields the solution to the difference equation (P2.32-1) under the condition of initial rest, when the input is given by eq. (P2.32-2).

Consider a system whose input x(t) and output y(t) satisfy the first-order differential equation dy(t) -----;[( + 2y(t) = x(t). (P2.33-1) The system also satisfies the condition of initial rest. (a) (i) Determine the system output y 1 (t) when the input is x 1 (t) = e3t u(t). (ii) Determine the system output y2(t) when the input is x2(t) = e2t u(t). (iii) Determine the system output y3(t) when the input is x3(t) = ae3t u(t) + {3e2tu(t), where a and {3 are real numbers. Show that y3(t) = ay1 (t) + {3 Y2(t). (iv) Now let x 1 (t) and x 2(t) be arbitrary signals such that (b) (i) (ii) (iii) x 1(t) = 0, fort< t1, x2(t) = 0, fort < t2. Letting Y1 (t) be the system output for input x 1 (t), y2(t) be the system output for input x2(t), and y3(t) be the system output for x3(t) = ax1 (t) + {3x2(t), show that y3(t) = ay1 (t) + {3 Y2(t). We may therefore conclude that the system under consideration is linear. Determine the system output y 1 (t) when the input is x 1 (t) = K e2t u(t). Determine the system output y2(t) when the input is x2(t) = K e2(t-T) u(t - T). Show that Y2(t) = Y1 (t - T). Now let x1 (t) be an arbitrary signal such that x 1 (t) = 0 fort < t0. Letting Y! (t) be the system output for input x 1 (t) and y2(t) be the system output for x 2(t) = x 1 (t - T), show that Y2(t) = Y! (t - T). We may therefore conclude that the system under consideration is time invariant. In conjunction with the result derived in part (a), we conclude that the given system is LTI. Since this system satisfies the condition of initial rest, it is causal as well.

The initial rest assumption corresponds to a zero-valued auxiliary condition being imposed at a time determined in accordance with the input signal. In this problem we show that if the auxiliary condition used is nonzero or if it is always applied at a fixed time (regardless of the input signal) the corresponding system cannot be LTI. Consider a system whose input x(t) and output y(t) satisfy the first-order differential equation (P2.33-1). (a) Given the auxiliary condition y(1) = 1, use a counterexample to show that the system is not linear. (b) Given the auxiliary condition y(l) = 1, use a counterexample to show that the system is not time invariant. (c) Given the auxiliary condition y(1) = 1, show that the system is incrementally linear. (d) Given the auxiliary condition y(l) = 0, show that the system is linear but not time invariant. (e) Given the auxiliary condition y(O) + y(4) = 0, show that the system is linear but not time invariant.

In the previous problem we saw that application of an auxiliary condition at a fixed time (regardless of the input signal) leads to the corresponding system being not time-invariant. In this problem, we explore the effect of fixed auxiliary conditions on the causality of a system. Consider a system whose input x(t) and output y(t) satisfy the first-order differential equation (P2.33-1 ). Assume that the auxiliary condition associated with the differential equation is y(O) = 0. Determine the output of the system for each of the following two inputs: (a) x 1 (t) = 0, for all t { 0 t<-1 (b) X 2 ( t) = } : t > _ 1 Observe that if y 1 (t) is the output for input x 1 (t) and y2(t) is the output for input x 2(t), then y1 (t) and y2(t) are not identical fort < -1, even though x 1 (t) and x 2(t) are identical fort < - 1. Use this observation as the basis of an argument to conclude that the given system is not causal.

Consider a discrete-time system whose input x[n] and output y[n] are related by y(n] = (~ )y[n- I] + x[n]. (a) Show that if this system satisfies the condition of initial rest (i.e., if x[n] = 0 for n < n0 , then y[n] = 0 for n < n0 ), then it is linear and time invariant. (b) Show that if this system does not satisfy the condition of initial rest, but instead uses the auxiliary condition y[O] = 0, it is not causal. [Hint: Use an approach similar to that used in Problem 2.35.]

Consider a system whose input and output are related by the first-order differential equation (P2.33-l ). Assume that the system satisfies the condition of final rest [i. e., if x(t) = 0 fort > t0 , then y(t) = 0 fort > t0 ]. Show that this system is not causal. [Hint: Consider two inputs to the system, x1 (t) = 0 and x2(t) = er(u(t)- u(t- 1)), which result in outputs y 1 (t) and y2(t), respectively. Then show that y 1 (t) =I= y2(t) fort< 0.]

Draw block diagram representations for causal LTI systems described by the following difference equations: (a) y[n] = *y[n- 1] + ~x[n] (b) y[n] = *y[n- 1] + x[n- 1]

Draw block diagram representations for causal LTI systems described by the following differential equations: (a) y(t) = -( ~) dy(t)ldt + 4x(t) (b) dy(t)ldt + -3y(t) = x(t)

Consider the signal x[n] = a 11 u[n]. (a) Sketch the signal g[n] = x[n] - ax[n - 1]. Figure P2.40 (b) Use the result of part (a) in conjunction with properties of convolution in order to determine a sequence h[n] such that ( 1 ) 11 x[n] * h[n] = 2 {u[n + 2] - u[n - 2]}.

Suppose that the signal x(t) = u(t + 0.5) - u(t - 0.5) is convolved with the signal (a) Determine a value of w 0 which ensures that y(O) = 0, where y(t) = x(t) * h(t). (b) Is your answer to the previous part unique?

One of the important properties of convolution, in both continuous and discrete time, is the associativity property. In this problem, we will check and illustrate this property. (a) Prove the equality [x(t) * h(t)] * g(t) = x(t) * [h(t) * g(t)] (P2.43-l) by showing that both sides of eq. (P2.43-l) equal I +oo I +oo -oo -oo x( r)h(u)g(t - T - u) dr du. (b) Consider two LTI systems with the unit sample responses hdn] and h2 [n] shown in Figure P2.43(a). These two systems are cascaded as shown in Figure P2.43(b). Let x[n] = u[n]. (i) Compute y[n] by first computing w[n] = x[n] * h1 [n] and then computing y[n] = w[n] * h2[n]; that is, y[n] = [x[n] * h1 [n]] * h2[n]. (ii) Now find y[n] by first convolving h1 [n] and h2[n] to obtain g[n] h1 [n] * h2[n] and then convolving x[n] with g[n] to obtain y[n] x[n] * [hi [n] * h2[n]]. The answers to (i) and (ii) should be identical, illustrating the associativity property of discrete-time convolution. (c) Consider the cascade of two LTI systems as in Figure P2.43(b), where in this case h1 [n] = sin 8n and and where the input is x[n] = B[n] - aB[n- 1]. Determine the output y[n]. (Hint: The use of the associative and commutative properties of convolution should greatly facilitate the solution.)

(a) If x(t) = 0, Jtl > TJ, and h(t) = 0, JtJ > T2, then x(t) * h(t) = 0, ltl > T3 for some positive number T3 . Express T3 in terms of T1 and T2 . (b) A discrete-time LTI system has input x[n], impulse response h[n], and output y[n]. If h[n] is known to be zero everywhere outside the interval No :s; n :s; N1 and x[n] is known to be zero everywhere outside the interval N2 :s; n :s; N3, then the output y[n] is constrained to be zero everywhere, except on some interval N4 :s; n :s; Ns. (i) Determine N4 and N5 in terms of N0 , N1, N2 , and N3 . (ii) If the interval N0 :s; n :s; N1 is of length Mh, N2 :s; n :s; N3 is of length Mx, and N4 :s; n :s; N5 is of length My, express My in terms of M11 and Mx. (c) Consider a discrete-time LTI system with the property that if the input x[n] = 0 for all n 2: 10, then the output y[n] = 0 for all n 2: 15. What condition must h[n], the impulse response of the system, satisfy for this to be true? (d) Consider an LTI system with impulse response in Figure P2.44. Over what interval must we know x(t) in order to determine y(O)?

(a) Show that if the response of an LTI system to x(t) is the output y(t), then the response of the system to x'(t) = dx(t) dt is y'(t). Do this problem in three different ways: (i) Directly from the properties of linearity and time invariance and the fact that '( ) _ 1 . x(t) - x(t - h) X t - Ill h . h---'>0 (ii) By differentiating the convolution integral. (iii) By examining the system in Figure P2.45. x(t) y(t) Figure P2.45 (b) Demonstrate the validity of the following relationships: (i) y'(t) = x(t) * h'(t) (ii) y(t) = cCxx(r)dr) * h'(t) = Cx[x'(r) * h(r)Jdr = x'(t) * cf~xh(r)dr) [Hint: These are easily done using block diagrams as in (iii) of part (a) and the fact that UJ (t) * U-] (t) = O(t).] (c) An LTI system has the response y(t) = sin w0t to input x(t) = e-St u(t). Use the result of part (a) to aid in determining the impulse response of this system. (d) Let s(t) be the unit step response of a continuous-time LTI system. Use part (b) to deduce that the response y(t) to the input x(t) is (P2.45-1) Show also that (P2.45-2) (e) Use eq. (P2.45-1) to determine the response of an LTI system with step response s(t) = (e- 3 '- 2e- 2 ' + l)u(t) to the input x(t) = e' u(t). (0 Let s[n] be the unit step response of a discrete-time LTI system. What are the discrete-time counterparts of eqs. (P2.45-l) and (P2.45-2)?

Consider an LTI systemS and a signal x(t) = 2e- 3t u(t - 1). If x(t) ~ y(t) and dx(t) dt ~ -3y(t) + e- 2 tu(t), determine the impulse response h(t) of S.

We are given a certain linear time-invariant system with impulse response h0(t). We are told that when the input is x0(t) the output is y0(t), which is sketched in Figure P2.47. We are then given the following set of inputs to linear time-invariant systems with the indicated impulse responses: Input x(t) (a) x(t) = 2xo(t) (b) x(t) = x0(t)- xo(t- 2) (c) x(t) = xo(t - 2) (d) x(t) = x0( -t) (e) x(t) = x0( -t) (0 x(t) = xb(t) Impulse response h(t) h(t) = h0(t) h(t) = ho(t) h(t) = h0(t + 1) h(t) = ho(t) h(t) = h0(-t) h(t) = hb(t) [Here xb(t) and hb(t) denote the first derivatives of x0(t) and h0(t), respectively.] 0 2 Figure P2.47 In each of these cases, determine whether or not we have enough information to determine the output y(t) when the input is x(t) and the system has impulse response h(t). If it is possible to determine y(t), provide an accurate sketch of it with numerical values clearly indicated on the graph.

Determine whether each of the following statements concerning LTI systems is true or false. Justify your answers. (a) If h(t) is the impulse response of an LTI system and h(t) is periodic and nonzero, the system is unstable. (b) The inverse of a causal LTI system is always causal. (c) If lh[n]l :::::; K for each n, where K is a given number, then the LTI system with h[n] as its impulse response is stable. (d) If a discrete-time LTI system has an impulse response h[n] of finite duration, the system is stable. (e) If an LTI system is causal, it is stable. (f) The cascade of a non causal LTI system with a causal one is necessarily noncausal. (g) A continuous-time LTI system is stable if and only if its step response s(t) is absolutely integrable-that is, if and only if (h) A discrete-time LTI system is causal if and only if its step response s[n] is zero for n < 0.

In the text, we showed that if h[n] is absolutely summable, i.e., if +oc L, lh[kJI < oo, k= -oc then the LTI system with impulse response h[n] is stable. This means that absolute summability is a sufficient condition for stability. In this problem, we shall show that it is also a necessary condition. Consider an LTI system with impulse response h[n] that is not absolutely summable; that is, +oc L, ih[kJI = oo. k= -oc (a) Suppose that the input to this system is { 0, x[n] = h[-n] lh[-nll' if h[ -n] = 0 if h[ -n] =I= 0 Does this input signal represent a bounded input? If so, what is the smallest number B such that lx[n]l :::::; B for all n? (b) Calculate the output at n = 0 for this particular choice of input. Does the result prove the contention that absolute summability is a necessary condition for stability? (c) In a similar fashion, show that a continuous-time LTI system is stable if and only if its impulse response is absolutely integrable.

In the text, we saw that the overall input-output relationship of the cascade of two LTI systems does not depend on the order in which they are cascaded. This fact, known as the commutativity property, depends on both the linearity and the time in variance of both systems. In this problem, we illustrate the point. (a) Consider two discrete-time systems A and B, where system A is an LTI system with unit sample response h[n] = (l/2)"u[n]. System B, on the other hand, is linear but time varying. Specifically, if the input to system B is w[n], its output is x[n] z[n] = nw[n]. Show that the commutativity property does not hold for these two systems by computing the impulse responses of the cascade combinations in Figures P2.5l(a) and P2.51(b), respectively (b) Suppose that we replace system B in each of the interconnected systems of Figure P2.51 by the system with the following relationship between its input w[n] and output z[n]: z[n] = w[n] + 2. Repeat the calculations of part (a) in this case.

Consider a discrete-time LTI system with unit sample response h[n] = (n + l)a"u[n], where Ia I < 1. Show that the step response of this system is [ 1 a n a 1) n] ] s[n] = (a _ 1)2 - (a _ 1 )2 a + (a _ 1 ) (n + a u[n . (Hint: Note that

(a) Consider the homogeneous differential equation Show that if s0 is a solution of the equation N p(s) = L aksk = 0, k=O (P2.53-1) (P2.53-2) then Aesot is a solution of eq. (P2.53-1 ), where A is an arbitrary complex constant. (b) The polynomial p(s) in eq. (P2.53-2) can be factored in terms of its roots SJ, ... , Sr as where the Si are the distinct solutions of eq. (P2.53-2) and the ui are their multiplicities-that is, the number of times each root appears as a solution of the equation. Note that CT I + U2 + . + CT r = N. In general, if ui > 1, then not only is Aesit a solution of eq. (P2.53-1), but so is Atj e1 J, as long as j is an integer greater than or equal to zero and less than or equal to ui - 1. To illustrate this, show that if Ui = 2, then Atesit is a solution of eq. (P2.53-l). [Hint: Show that if sis an arbitrary complex number, Thus, the most general solution of eq. (P2.53-l) is r fT 1 -] ~ ~ Atjes;t LL I.J ' i= I j=O where the Ai.i are arbitrary complex constants. (c) Solve the following homogeneous differential equations with the specified auxiliary conditions: (i) d~i~~n + 3d~~n + 2y(t) = 0, y(O) = 0, y'(O) = 2 (ii) dd~y) + 3 d;;~n + 2y(t) = o, y(O) = I, y'(O) = -1 (iii) d~i~;n + 3 d~;~n + 2y(t) = 0, y(O) = 0, y'(O) = 0 (iv) d~i~y) + 2 d:;~n + y(t) = 0, y(O) = 1, y'(O) = 1 (v) dd~~n + d~i~;'l - d~;~n - y(t) = 0, y(O) = 1, y'(O) = 1, y"(O) = -2 (vi) y~n + 2 dyUJ + 5y(t) = 0 v(O) = 1 y'(O) =

(a) Consider the homogeneous difference equation N Laky[n- k] = 0, k=O Show that if zo is a solution of the equation N Lakz-k = 0, k=O (P2.54-1) (P2.54-2) then Azg is a solution of eq. (P2.54-l ), where A is an arbitrary constant. (b) As it is more convenient for the moment to work with polynomials that have only nonnegative powers of z, consider the equation obtained by multiplying both sides of eq. (P2.54-2) by zN: N p(z) = L akzN-k = 0. k=O The polynomial p(z) can be factored as p(z) = ao(z - ZJ yr 1 (z - z,.)u', where the z1, , z,. are the distinct roots of p(z). Show that if y[n] = nzn-l, then aky[n- k] = dp(z) zn-N + (n- N)p(z)zn-N-l. k=O dz (P2.54-3) Use this fact to show that if (Ji = 2, then both Az? and Bnz?- 1 are solutions of eq. (P2.54-1), where A and Bare arbitrary complex constants. More generally, one can use this same procedure to show that if O"i > 1, then n! zn-r r!(n- r)! is a solution of eq. (P2.54-l) for r = 0, 1, ... , ui- 1.7 (c) Solve the following homogeneous difference equations with the specified auxiliary conditions: (i) y[n] + ~y[n- 1] + iy[n- 2] = 0; y[O] = 1, y[ -1] = -6 (ii) y[n] - 2y[n- 1] + y[n - 2] = 0; y[O] = 1, y[l] = 0 (iii) y[n] - 2y[n- 1] + y[n- 2] = 0; y[O] = 1, y[10] = 21 (iv) y[n] - f!- y[n- 1] + ~y[n- 2] = 0; y[O] = 0, y[ -1] = 1

In the text we described one method for solving linear constant-coefficient difference equations, and another method for doing this was illustrated in Problem 2.30. If the assumption of initial rest is made so that the system described by the difference equation is LTI and causal, then, in principle, we can determine the unit impulse response h[n] using either of these procedures. In Chapter 5, we describe another method that allows us to determine h[n] in a more elegant way. In this problem we describe yet another approach, which basically shows that h[n] can be determined by solving the homogeneous equation with appropriate initial conditions. (a) Consider the system initially at rest and described by the equation 1 y[n]- ly[n- 1] = x[n]. (P2.55-1) Assuming that x[n] = S[n], what is y[O]? What equation does h[n] satisfy for n 2:: 1, and with what auxiliary condition? Solve this equation to obtain a closed-form expression for h[n]. (b) Consider next the LTI system initially at rest and described by the difference equation 1 y[n]- ly[n- 1] = x[n] + 2x[n- 1]. (P 2.55-2) This system is depicted in Figure P2.55(a) as a cascade of two LTI systems that are initially at rest. Because of the properties of LTI systems, we can reverse the order of the systems in the cascade to obtain an alternative representation of the same overall system, as illustrated in Figure P2.55(b ). From this fact, use the result of part (a) to determine the impulse response for the system described by eq. (P2.55-2). (c) Consider again the system of part (a), with h[n] denoting its impulse response. Show, by verifying that eq. (P2.55-3) satisfies the difference equation (P2.55- 1), that the response y[ n] to an arbitrary input x[ n] is in fact given by the convolution sum +oc y[n] = L h[n - m]x[m]. (P2.55-3) 7Here, weareusingfactorialnotation-thatis, k! = k(k - l)(k- 2) ... (2)(1), whereO! is defined to be 1. x[n] z[n] = x[n] + 2x[n-1] y[n] 1 ....,._~~ y[n]- 2y[n-1] = z[n] z[n] (a) x[n] ...... w[n]- ~w[n-1] = x[n] w[n] y[n] = w[n] + 2w[n-1] ~ y[n] (b) Figure P2.55 (d) Consider the LTI system initially at rest and described by the difference equation N ~ aky[n - k] = x[n]. k=O (P2.55-4) Assuming that a0 =I= 0, what is y[O] if x[n] = B[n]? Using this result, specify the homogeneous equation and initial conditions that the impulse response of the system must satisfy. Consider next the causal LTI system described by the difference equation N M ~ aky[n - k] = ~ bkx[n - k]. (P2.55-5) k=O k=O Express the impulse response of this system in terms of that for the LTI system described by eq. (P2.55-4). (e) There is an alternative method for determining the impulse response of the LTI system described by eq. (P2.55-5). Specifically, given the condition of initial rest, i.e., in this case, y[-N] = y[-N + 1] = ... = y[ -1] = 0, solve eq. (P2.55-5) recursively when x[n] = B[n] in order to determine y[O], ... , y[M]. What equation does h[n] satisfy for n 2:: M? What are the appropriate initial conditions for this equation? (0 Using either of the methods outlined in parts (d) and (e), find the impulse responses of the causal LTI systems described by the following equations: (i) y[n] - y[n - 2] = x[n] (ii) y[n] - y[n - 2] = x[n] + 2x[n - 1] (iii) y[n] - y[n - 2] = 2x[n] - 3x[n - 4] (iv) y[n] - ( ]3!2)y[n- 1] + ~y[n- 2] = x[n]

In this problem, we consider a procedure that is the continuous-time counterpart of the technique developed in Problem 2.55. Again, we will see that the problem of determining the impulse response h(t) for t > 0 for an LTI system initially at rest and described by a linear constant-coefficient differential equation reduces to the problem of solving the homogeneous equation with appropriate initial conditions. (a) Consider the LTI system initially at rest and described by the differential equation dy(t) dt + 2y(t) = x(t). (P2.56-l) Suppose that x(t) = o(t). In order to determine the value of y(t) immediately after the application of the unit impulse, consider integrating eq. (P2.56-1) from t = o- to t = o+ (i.e., from "just before" to "just after" the application of the impulse). This yields o+ o+ y(O+)- y(O-) + 2 fa_ y(T)dT = fa_ D(T)dT = 1. (P2.56-2) Since the system is initially at rest and x(t) = 0 fort < 0, y(O-) = 0. To satisfy eq. (P2.56-2) we must have y(O+) = 1. Thus, since x(t) = 0 for t > 0, the impulse response of our system is the solution of the homogeneous differential equation with initial condition dy(t) + 2y(t) = 0 dt Solve this differential equation to obtain the impulse response h(t) for the system. Check your result by showing that J +CX) y(t) = -oc h(t - T)X( T) dT satisfies eq. (P2.56-l) for any input x(t). (b) To generalize the preceding argument, consider an LTI system initially at rest and described by the differential equation N dky(t) .L,ak-k- = x(t) k=O dt (P2.56-3) with x(t) = o(t). Assume the condition of initial rest, which, since x(t) = 0 for t < 0, implies that - dy - dN-1 y - y(O ) = d t (0 ) = .. . = d tN -I (0 ) = 0. (P2.56-4) Integrate both sides of eq. (P2.56-3) once from t = o- to t = o+, and use eq. (P2.56-4) and an argument similar to that used in part (a) to show that the resulting equation is satisfied with + dy + y(O ) = dt (0 ) = ... (P2.56-5a) and (P2.56-5b) Consequently, the system's impulse response fort> 0 can be obtained by solving the homogeneous equation ~ dky(t) - 0 Lak--- k=O dtk with initial conditions given by eqs. (P2.56-5). (c) Consider now the causal LTI system described by the differential equation ~ dky(t) _ ~ b dkx(t) Lak k -Lk k" k=O dt k=O dt (P2.56-6) Express the impulse response of this system in terms of that for the system of part (b). (Hint: Examine Figure P2.56.) (d) Apply the procedures outlined in parts (b) and (c) to find the impulse responses for the LTI systems initially at rest and described by the following differential equations: (i) d~i~~t) + 3 d~~n + 2y(t) = x(t) (ii) d~i~~t) + 2 d;;~tl + 2y(t) = x(t) (e) Use the results of parts (b) and (c) to deduce that if M ~ N in eq. (P2.56-6), then the impulse response h(t) will contain singularity terms concentrated at t = 0. In particular, h(t) will contain a term of the form M-N L lXrUr(t), r=O where the a,. are constants and the ur(t) are the singularity functions defined in Section 2.5. (f) Find the impulse responses of the causal LTI systems described by the following differential equations: (i) d;;~n + 2y(t) = 3 d~~;n + x(t) (ii) l('-y~t) + 5 dy(t) + 6y(t) = d'x(t) + 2 x~f) + 4 dx(t) + 3x(t)

Consider a causal LTI systemS whose input x[n] and output y[n] are related by the difference equation y[n] = -ay[n- 1] + box[n] + b1x[n- 1]. (a) Verify that S may be considered a cascade connection of two causal LTI systems sl and s2 with the following input-output relationship: sl : Yl [n] = box I [n] + bl X) [n- 1], s2 : Y2[n] = -ay2[n- 1] + X2[n]. (b) Draw a block diagram representation of S 1 (c) Draw a block diagram representation of S2 (d) Draw a block diagram representation of S as a cascade connection of the block diagram representation of S 1 followed by the block diagram representation of S2. (e) Draw a block diagram representation of S as a cascade connection of the block diagram representation of s2 followed by the block diagram representation of S1 (f) Show that the two unit-delay elements in the block diagram representation of S obtained in part (e) may be collapsed into one unit-delay element. The resulting block diagram is referred to as a Direct Form II realization of S, while the block diagrams obtained in parts (d) and (e) are referred to as Direct Form I realizations of S.

Consider a causal LTI systemS whose input x[n] and output y[n] are related by the difference equation 2y[n] - y[n - 1] + y[n - 3] = x[n] - 5x[n- 4]. (a) Verify that S may be considered a cascade connection of two causal LTI systems S 1 and S2 with the following input-output relationship: S1 : 2yl [n] = X1 [n] - 5XJ [n- 4], 1 1 S2 : Y2[n] = 2 Y2[n- 1] - 2 Y2[n- 3] + x2[n]. (b) Draw a block diagram representation of S1 (c) Draw a block diagram representation of S2 . (d) Draw a block diagram representation of S as a cascade connection of the block diagram representation of S 1 followed by the block diagram representation of S2. (e) Draw a block diagram representation of S as a cascade connection of the block diagram representation of s2 followed by the block diagram representation of S1 (f) Show that the four delay elements in the block diagram representation of S obtained in part (e) may be collapsed to three. The resulting block diagram is referred to as a Direct Form II realization of S, while the block diagrams obtained in parts (d) and (e) are referred to as Direct Form I realizations of S.

Consider a causal LTI system 5 whose input x(t) and output y(t) are related by the differential equation dv(t) dx(t) a1-d + aov(t) = box(t) + b1 -d-. t - t (a) Show that and express the constants A, B, and C in terms of the constants a 0 , a 1, b0 , and b1 (b) Show that 5 may be considered a cascade connection of the following two causal LTI systems: sl : )'J(I) ~ Bx\(t) + c Lyx(T)dT, s, : y,(t) ~ A r /'( T) dT + x,(r). (c) Draw a block diagram representation of 51 (d) Draw a block diagram representation of 52. (e) Draw a block diagram representation of 5 as a cascade connection of the block diagram representation of 51 followed by the block diagram representation of 52. (f) Draw a block diagram representation of 5 as a cascade connection of the block diagram representation of 52 followed by the block diagram of representation 5 1 (g) Show that the two integrators in your answer to part (f) may be collapsed into one. The resulting block diagram is referred to as a Direct Form II realization of 5, while the block diagrams obtained in parts (e) and (f) are referred to as Direct Form I realizations of 5.

(a) In the circuit shown in Figure P2.6l(a), x(t) is the input voltage. The voltage y(t) across the capacitor is considered to be the system output. x(t) L=1H 1 C =1F y(t) l (a) Figure P2.61 a (i) Determine the differential equation relating x(t) and y(t). (ii) Show that the homogeneous solution of the differential equation from part (i) has the form K1 eiw 1 t + K2e.iw21 Specify the values of w 1 and w 2 . (iii) Show that, since the voltage and current are restricted to be real, the natural response of the system is sinusoidal. (b) In the circuit shown in Figure P2.61 (b), x(t) is the input voltage. The voltage y(t) across the capacitor is considered to be the system output. R = 10 x(t) + C =1F + (b) Figure P2.61 b (i) Determine the differential equation relating x(t) and y(t). (ii) Show that the natural response of this system has the form K e-ar, and specify the value of a. (c) In the circuit shown in Figure P2.61(c), x(t) is the input voltage. The voltage y(t) across the capacitor is considered to be the system output. R = 2!1 L = 1H r (c) Figure P2.61 c (i) Determine the differential equation relating x(t) and y(t). (ii) Show that the homogeneous solution of the differential equation from part (i) has the form e- 0 '{K1 ei2' + K2e-i 2 '}, and specify the value of a. (iii) Show that, since the voltage and current are restricted to be real, the natural response of the system is a decaying sinusoid.

(a) In the mechanical system shown in Figure P2.62(a), the force x(t) applied to the mass represents the input, while the displacement y(t) of the mass represents the output. Determine the differential equation relating x(t) and y(t). Show that the natural response of this system is periodic. (b) Consider Figure P2.62(b ), in which the force x(t) is the input and the velocity y(t) is the output. The mass of the car ism, while the coefficient of kinetic friction is p. Show that the natural response of tqis system decays with increasing time. (c) In the mechanical system shown in Figure P2.62(c), the force x(t) applied to the mass represents the input, while the displacement y(t) of the mass represents the output. (i) Determine the differential equation relating x(t) and y(t). 165 m = 1,000 Kg p=0.1 N-s/m (ii) Show that the homogeneous solution of the differential equation from part (i) has the form e-at{K1 ejt + K2e-jt}, and specify the value of a. (iii) Show that, since the force and displacement are restricted to be real, the natural response of the system is a decaying sinusoid.

A $100,000 mortgage is to be retired by equal monthly payments of D dollars. Interest, compounded monthly, is charged at the rate of 12% per annum on the unpaid balance; for example, after the first month, the total debt equals $100,000 + ;~ )$100,000 = $101,000. The problem is to determine D such that after a specified time the mortgage is paid in full, leaving a net balance of zero. (a) To set up the problem, let y[n] denote the unpaid balance after the nth monthly payment. Assume that the principal is borrowed in month 0 and monthly payments begin in month 1. Show that y[n] satisfies the difference equation y[n] - yy[n - 1] = - D n 2: 1 (P2.63-1) with initial condition y[O] = $100,000, where y is a constant. Determine y. (b) Solve the difference equation of part (a) to determine y[n] for n 2: 0. (Hint: The particular solution of eq. (P2.63-1) is a constant Y. Find the value of Y, and express y[n] for n 2: 1 as the sum of particular and homogeneous solutions. Determine the unknown constant in the homogeneous solution by directly calculating y[l] from eq. (P2.63-1) and comparing it to your solution.) (c) If the mortgage is to be retired in 30 years after 360 monthly payments of D dollars, determine the appropriate value of D. (d) What is the total payment to the bank over the 30-year period? (e) Why do banks make loans?

One important use of inverse systems is in situations in which one wishes to remove distortions of some type. A good example of this is the problem of removing echoes from acoustic signals. For example, if an auditorium has a perceptible echo, then an initial acoustic impulse will be followed by attenuated versions of the sound at regularly spaced intervals. Consequently, an often -used model for this phenomenon is an LTI system with an impulse response consisting of a train of impulses, i.e., h(t) = ~ hko(t - kT). (P2.64-1) k=O Here the echoes occur T seconds apart, and hk represents the gain factor on the kth echo resulting from an initial acoustic impulse. (a) Suppose that x(t) represents the original acoustic signal (the music produced by an orchestra, for example) and that y(t) = x(t) * h(t) is the actual signal that is heard if no processing is done to remove the echoes. In order to remove the distortion introduced by the echoes, assume that a microphone is used to sense y(t) and that the resulting signal is transduced into an electrical signal. We will also use y(t) to denote this signal, as it represents the electrical equivalent of the acoustic signal, and we can go from one to the other via acoustic-electrical conversion systems. The important point to note is that the system with impulse response given by eq. (P2.64-1) is invertible. Therefore, we can find an LTI system with impulse response g(t) such that y(t) * g(t) = x(t), and thus, by processing the electrical signal y(t) in this fashion and then converting back to an acoustic signal, we can remove the troublesome echoes. The required impulse response g(t) is also an impulse train: g(t) = L gk8(t- kT). k=O Determine the algebraic equations that the successive gk must satisfy, and solve these equations for go, g 1, and g2 in terms of hk (b) Suppose that h0 = 1, h 1 = 1/2, and hi = 0 for all i ~ 2. What is g(t) in this case? (c) A good model for the generation of echoes is illustrated in Figure P2.64. Hence, each successive echo represents a fed-back version of y(t), delayed by T seconds and scaled by a. Typically, 0 < a < 1, as successive echoes are attenuated. x(t) _, ... + y(t) a Delay """- T Figure P2. 64 (i) What is the impulse response of this system? (Assume initial rest, i.e., y(t) = 0 fort < 0 if x(t) = 0 fort < 0.) (ii) Show that the system is stable if 0 < a < 1 and unstable if a > 1. (iii) What is g(t) in this case? Construct a realization of the inverse system using adders, coefficient multipliers, and T-second delay elements. (d) Although we have phrased the preceding discussion in terms of continuous-time systems because of the application we have been considering, the same general ideas hold in discrete time. That is, the LTI system with impulse response h[n] = L, hk8[n - kN] k=O is invertible and has as its inverse an LTI system with impulse response g[n] = L, gk8[n- kN]. It is not difficult to check that the gk satisfy the same algebraic equations as in part (a). Consider now the discrete-time LTI system with impulse response h[n] = L 8[n- kN]. k=-% This system is not invertible. Find two inputs that produce the same output.

In Problem 1.45, we introduced and examined some of the basic properties of correlation functions for continuous-time signals. The discrete-time counterpart of the correlation function has essentially the same properties as those in continuous time, and both are extremely important in numerous applications (as is discussed in Problems 2.66 and 2.67). In this problem, we introduce the discrete-time correlation function and examine several more of its properties. Let x[n] and y[n] be two real-valued discrete-time signals. The autocorrelation functions yy[n] of x[n] and y[n], respectively, are defined by the expressions +ex; yy[n] = L y[m + n]y[m], m=-oc and the cross-correlation functions are given by +rye xy[n] = L x[m + n]y[m] m= -oc and +oo yx[n] = L y[m + n]x[m]. m= -oo As in continuous time, these functions possess certain symmetry properties. Specifically, yy[n] are even functions, while xy[n] = yx[- n]. (a) Compute the autocorrelation sequences for the signals Xt [n], x2[n], x3[n], and x4 [n] depicted in Figure P2.65. (b) Compute the cross-correlation sequences x;x1 [n], i # j, i, j = 1, 2, 3, 4, for xi[n], i = 1, 2, 3, 4, as shown in Figure P2.65. (c) Let x[n] be the input to an LTI system with unit sample response h[n], and let the corresponding output be y[n]. Find expressions for xy[n] and yy[n] in terms of yy[n] can be viewed as the output of LTI systems with 4>xAn] as the input. (Do this by explicitly specifying the impulse response of each of the two systems.) (d) Let h[n] = x 1 [n] in Figure P2.65, and let y[n] be the output of the LTI system with impulse response h[n] when the input x[n] also equals x 1 [n]. Calculate 4>x_v[n] and 4>_v_v[n] using the results of part (c).

Let h 1 (t), h2(t), and h3(t), as sketched in Figure P2.66, be the impulse responses of three LTI systems. These three signals are known as Walsh functions and are of considerable practical importance because they can be easily generated by digital logic circuitry and because multiplication by each of them can be implemented in a simple fashion by a polarity-reversing switch. -1 Figure P2.66 (a) Determine and sketch a choice for x 1 (t), a continuous-time signal with the following properties: (i) x 1 (t) is real. (ii) x 1 (t) = 0 fort < 0. (iii) \xi (t)\ ::; 1 fort ~ 0. (iv) y 1 (t) = x 1 (t) * h(t) is as large as possible at t = 4. (b) Repeat part (a) for x2(t) and x3(t) by making Y2(t) = x2(t) * h2(t) and y3(t) = x3(t) * h3(t) each as large as possible at t = 4. (c) What is the value of Yij(l) = Xi(t) * h j(l), i =I= j at time t = 4 for i, j = 1, 2, 3? The system with impulse response hi(t) is known as the matched filter for the signal Xi(t) because the impulse response is tuned to xi(t) in order to produce the maximum output signal. In the next problem, we relate the concept of a matched filter to that of the correlation function for continuous-time signals.

The cross-correlation function between two continuous-time real signals x(t) and y(t) is I c/>xy(t) = J_~ x(t + T)y(T)dT. (P2.67-l) The autocorrelation function of a signal x(t) is obtained by setting y(t) = x(t) in eq. (P2.67-1): c/>xx(t) = r: X(t + T)X(T)dT. (a) Compute the autocorrelation function for each of the two signals x 1 (t) and x2(t) depicted in Figure P2.67(a). (b) Let x(t) be a given signal, and assume that x(t) is of finite duration-i.e., that x(t) = 0 for t < 0 and t > T. Find the impulse response of an LTI system so that xxU - T) is the output if x(t) is the input. (c) The system determined in part (b) is a matched filter for the signal x(t). That this definition of a matched filter is identical to the one introduced in Problem 2.66 can be seen from the following: Let x(t) be as in part (b), and let y(t) denote the response to x(t) of an LTI system with real impulse response h(t). Assume that h(t) = 0 for t < 0 and fort> T. Show that the choice for h(t) that maximizes y(T), subject to the constraint that JoT h2(t)dt = M, a fixed positive number, (P2.67-2) is a scalar multiple of the impulse response determined in part (b). [Hint: Schwartz's inequality states that [ b ] I /2 [ b ] 1/2 r u(t)v(t)dt ,; L u2 (t)dt L v 2 (t)dt for any two signals u(t) and v(t). Use this to obtain a bound on y(T).] (d) The constraint given by eq. (P2.67-2) simply provides a scaling to the impulse response, as increasing M merely changes the scalar multiplier mentioned in part (c). Thus, we see that the particular choice for h(t) in parts (b) and (c) is matched to the signal x(t) to produce maximum output. This is an extremely important property in a number of applications, as we will now indicate. In communication problems, one often wishes to transmit one of a small number of possible pieces of information. For example, if a complex message is encoded into a sequence of binary digits, we can imagine a system that transmits the information bit by bit. Each bit can then be transmitted by sending one signal, say, x0(t) , if the bit is a 0, or a different signal x 1 (t) if a 1 is to be communicated. In this case, the receiving system for these signals must be capable of recognizing whether x0(t) or x1 (t) has been received. Intuitively, what makes sense is to have two systems in the receiver, one tuned to x0(t) and one tuned to x 1 (t), where, by "tuned," we mean that the system gives a large output after the signal to which it is tuned is received. The property of producing a large output when a particular signal is received is exactly what the matched filter possesses. In practice, there is always distortion and interference in the transmission and reception processes. Consequently, we want to maximize the difference between the response of a matched filter to the input to which it is matched and the response of the filter to one of the other signals that can be transmitted. To illustrate this point, consider the two signals x0 (t) and XI (t) depicted in Figure P2.67(b). Let L0 denote the matched filter for x0(t), and let L1 denote the matched filter for x 1 (t). (i) Sketch the responses of Lo to x0(t) and x1 (t). Do the same for L1 (ii) Compare the values of these responses at t = 4. How might you modify x0 (t) so that the receiver would have an even easier job of distinguishing between x0 (t) and x1 (t) in that the response of L0 to XI (t) and L1 to x0 (t) would both be zero at t = 4?

Another application in which matched filters and correlation functions play an important role is radar systems. The underlying principle of radar is that an electromagnetic pulse transmitted at a target will be reflected by the target and will subsequently return to the sender with a delay proportional to the distance to the target. Ideally, the received signal will simply be a shifted and possibly scaled version of the original transmitted signal. Let p(t) be the original pulse that is sent out. Show that c/>pp(O) = max c/>pp(t). t That is, c/>pp(O) is the largest value taken by c/>pp(t). Use this equation to deduce that, if the waveform that comes back to the sender is x(t) = a p(t - to), where a is a positive constant, then c/>xp(to) = max c/>xp(t). t (Hint: Use Schwartz's inequality.) Thus, the way in which simple radar ranging systems work is based on using a matched filter for the transmitted waveform p(t) and noting the time at which the output of this system reaches its maximum value.

In Section 2.5, we characterized the unit doublet through the equation J +oo x(t) * u 1(t) = -oo x(t- r)u 1(r)dr = x'(t) (P2.69-1) for any signal x(t). From this equation, we derived the relationship J +oo -oo g(r)u 1(r)dr = -g'(O). (P2.69-2) (a) Show that eq. (P2.69-2) is an equivalent characterization of u1 (t) by showing that eq. (P2.69-2) implies eq. (P2.69-1). [Hint: Fix t, and define the signal g( r) = x(t - r).] Thus, we have seen that characterizing the unit impulse or unit doublet by how it behaves under convolution is equivalent to characterizing how it behaves under integration when multiplied by an arbitrary signal g(t). In fact, as indicated in Section 2.5, the equivalence of these operational definitions holds for all signals and, in particular, for all singularity functions. (b) Let f(t) be a given signal. Show that f(t)ul (t) = f(O)ui (t)- f'(O)o(t) by showing that both functions have the same operational definitions. (c) What is the value of Find an expression for f(t)u2(t) analogous to that in part (b) for f(t)u 1 (t).

In this chapter, we have used several properties and ideas that greatly facilitate the analysis of LTI systems. Among these are two that we wish to examine a bit more closely. As we will see, in certain very special cases one must be careful in using these properties, which otherwise hold without qualification. (a) One of the basic and most important properties of convolution (in both con tinuous and discrete time) is associativity. That is, if x(t), h(t), and g(t) are three signals, then x(t) * [g(t) * h(t)] = [x(t) * g(t)] * h(t) = [x(t) * h(t)] * g(t). (P2.71-l) This relationship holds as long as all three expressions are well defined and finite. As that is usually the case in practice, we will in general use the associativity property without comments or assumptions. However, there are some cases in which it does not hold. For example, consider the system depicted in Figure P2.71, with h(t) = u1 (t) and g(t) = u(t). Compute the response of this system to the input x(t) = 1 for all t. x(t) -8--~E]----. y(t) x(t) -G~--~~ y(t) Figure P2. 71 Do this in the three different ways suggested by eq. (P2.71-l) and by the figure: (i) By first convolving the two impulse responses and then convolving the result with x(t). (ii) By first convolving x(t) with u1 (t) and then convolving the result with u(t). (iii) By first convolving x(t) with u(t) and then convolving the result with u1 (t). (b) Repeat part (a) for and (c) Do the same for x(t) = e-t h(t) = e -t u(t), g(t) = u1 (t) + o(t). x[n] = (U h[n] = (4 )" u[n], 1 g[n] = o[n] - 2o[n- 1]. 175 Thus, in general, the associativity property of convolution holds if and only if the three expressions in eq. (P2.71-1) make sense (i.e., if and only if their interpretations in terms of LTI systems are meaningful). For example, in part (a) differentiating a constant and then integrating makes sense, but the process of integrating the constant from t = -oo and then differentiating does not, and it is only in such cases that associativity breaks down. Closely related to the foregoing discussion is an issue involving inverse systems. Consider the LTI system with impulse response h(t) = u(t). As we saw in part (a), there are inputs-specifically, x(t) = nonzero constant-for which the output of this system is infinite, and thus, it is meaningless to consider the question of inverting such outputs to recover the input. However, if we limit ourselves to inputs that do yield finite outputs, that is, inputs which satisfy (P2.71-2) then the system is invertible, and the LTI system with impulse response u1 (t) is its inverse. (d) Show that the LTI system with impulse response u1 (t) is not invertible. (Hint: Find two different inputs that both yield zero output for all time.) However, show that the system is invertible if we limit ourselves to inputs that satisfy eq. (P2.71-2). [Hint: In Problem 1.44, we showed that an LTI system is invertible if no input other than x(t) = 0 yields an output that is zero for all time; are there two inputs x(t) that satisfy eq. (P2.71-2) and that yield identically zero responses when convolved with u1 (t)?] What we have illustrated in this problem is the following: (1) If x(t), h(t), and g(t) are three signals, and if x(t) * g(t), x(t) * h(t), and h(t) * g(t) are all well defined and finite, then the associativity property, eq. (P2. 71-1 ), holds. (2) Let h(t) be the impulse response of an LTI system, and suppose that the impulse response g(t) of a second system has the property h(t) * g(t) = o(t). (P2.71-3) Then, from (1), for all inputs x(t) for which x(t) * h(t) and x(t) * g(t) are both well defined and finite, the two cascades of systems depicted in Figure P2. 71 act as the identity system, and thus, the two LTI systems can be regarded as inverses of one another. For example, if h(t) = u(t) and g(t) = u1 (t), then, as long as we restrict ourselves to inputs satisfying eq. (P2.71-2), we can regard these two systems as inverses. Therefore, we see that the associativity property of eq. (P2.71-1) and the definition ofLTI inverses as given in eq. (P2.71-3) are valid, as long as all convolutions that are involved are finite. As this is certainly the case in any realistic problem, we will in general use these properties without comment or qualification. Note that, although we have phrased most of our discussion in terms of continuous-time signals and systems, the same points can also be made in discrete time [as should be evident from part (c)].

Let 8L1(t) denote the rectangular pulse of height -k for 0 < t ::; Ll. Verify that d 1 dt oLl(t) = K [o(t) - o(t - L1)].

Show by induction that tk-1 u_k(t) = (k _ 1 )! u(t) fork = 1, 2, 3 ...