[CONTINUOUS-TIME](#page-10-0) SYSTEM ANALYSIS USING THE LAPLACE TRANSFORM

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PROBLEMS

• 3.1-1 Find the energy of the signals depicted in Fig. P3.1-1.
• 3.1-2 Find the power of the signals illustrated in Fig. P3.1-2.
• 3.1-3 Show that the power of a signal ^Dej(2π/N0)ⁿ is |D| 2. Hence, show that the power of a signal ^x[n] =%N0−^{1 r}=^{0 D}rejr(2π/N0)ⁿ is Px ⁼%N0−^{1 r}=⁰ |Dr| 2. Use the fact that

$$\sum_{k=0}^{N_0-1} e^{j(r-m)2\pi k/N_0} = \begin{cases} N_0 & r=m\\ 0 & \text{otherwise} \end{cases}$$

• 3.1-4 (a) Determine even and odd components of the signal ^x[n] = (0.8)nu[n].
  • (b) Show that the energy of x[n] is the sum of energies of its odd and even components found in part (a).
  • (c) Generalize the result in part (b) for any finite energy signal.
• 3.1-5 (a) If xe[n] and xo[n] are the even and the odd components of causal energy signal x[n], then determine Exe and Ex⁰ , and show that Exe +Ex⁰ = Ex.
  • (b) Show that the cross-energy of xe and xo is zero, that is,

$$\sum_{n=-\infty}^{\infty} x_e[n]x_o[n] = 0$$

3.1-6 Define

$$x[n] = \begin{cases} \left(\frac{1}{3}\right)^n & n \ge 0\\ A^n & n < 0 \end{cases}$$

• (a) Determine the energy Ex and power Px of ^x[n] if ^{A = 1} 2 .
• (b) Determine the energy Ex and power Px of x[n] if A = 1.
• (c) Determine the energy Ex and power Px of x[n] if A = 2.
• 3.1-7 Determine the energy Ex and power Px of the complex DT signal ^x[n] = Re* 3(ejπ/4)ⁿ +
• 3.2-1 If the energy of a signal x[n] is Ex, then find the energy of the following:
  • (a) x[−n]
  • (b) x[n−m]
  • (c) x[m−n]
  • (d) Kx[n] (m integer and K constant)
• 3.2-2 If the power of a periodic signal x[n] is Px, find and comment on the powers and the rms values of the following:
  • (a) −x[n]
  • (b) x[−n]
  • (c) x[n−m] (m integer)
  • (d) cx[n]
  • (e) x[m−n] (m integer)

Problems 315

Figure P3.1-2

• 3.2-3 Letting ↓ identify n = 0, define the nonzero values of signal x[n] as [−1, 2,−3, 4,−5, 4,−3, ↓ 2,−1].

  • (a) Using vector form, represent signal y[n] = x[−3n + 2]. Be sure to identify the n = 0 element.
  • (b) Using vector form, represent signal z[n] = x[n/2 − 3]. Be sure to identify the n = 0 element.

• 3.2-4 Letting ↓ identify n = 0, define the nonzero values of signal x[n] as [−1, ↓ 2,

  • −3, 4,−5, 4,−3, 2,−1]. (a) Determine the energy Ex and power Px of the signal x[n].
  • (b) Using vector form, represent signal y[n] = x[2(n + 2)]. Be sure to identify the n = 0 element.
  • (c) Using vector form, represent signal z[n] = ^x[−n−^{6 3} ]. Be sure to identify the n = 0 element.

• 3.2-5 Letting ↓ identify the n = 0 value, describe a 4-periodic signal w[n] using vector notation as

  • [· · · , 1, 2, 3, 4, ↓ 1, 2, 3, 4, 1, 2, 3, 4,···].
  • (a) Determine the energy Ex and power Px of signal x[n] = w[2n].
  • (b) Determine the energy Ey and power Py of signal ^y[n] = ^w[2^{− n} 3 ].

• 3.2-6 Let DT signal x[n] have values [1, 2, 3, 4, 5, 6] for 0 ≤ n ≤ 5 and let DT signal y[n] have values [5, 0, 0, 3, 0, 0, 1] for 0 ≤ n ≤ 6. Both signals are zero outside the ranges given. Further, define a 6-periodic replication of y[n] as ^y˜[n] = %^{∞ k}=−∞ y[n−6k] .

  • (a) Determine the energy Ex and power Px of signal x[n].
  • (b) Determine the smallest-magnitude integers N1, N2, and N³ such that y[n] = x[ N1n ^N2 +N3].
  • (c) Determine the energy Ey˜ and power Py˜ of y˜[n].

• 3.2-7 For the signal shown in Fig. P3.1-1b, sketch the following signals:

  • (a) x[−n]
  • (b) x[n+6]
  • (c) x[n−6]
  • (d) x[3n]
  • (e) x ’ n (
  • 3 (f) x[3−n]

• 3.2-8 Repeat Prob. 3.2-7 for the signal depicted in Fig. P3.1-1c.

• 3.2-9 Letting ↓ identify n = 0, consider a DT signal x[n] whose nonzero values are given as x[n]=[1, −3, 2, 2, 3, −2, −1, 1, 2, −3, ↓ 3, 3, −2, 1, −3, 2, 3, −1]. Accurately sketch y[n] = x[−1 − 2n] and z[n] = x[−2 + n/3] over −5 ≤ n ≤ 4.

• 3.2-10 Define

$$x[n] = \begin{cases} \left(\frac{1}{2}\right)^n & n \ge 0\\ 0 & n < 0 \end{cases}$$

Determine and locate the two largest non-zero values of:

• (a) ya[n] = x[2n]
• (b) yb[n] = x[n/3]
• (c) yc[n] = x[3n+1]
• (d) yd[n] = x[−2n+5]
• (e) ye[n] = x[−(n+8)/2]
• 3.3-1 Sketch, and find the power of, the following signals:
  • (a) (1)ⁿ
  • (b) (−1)ⁿ
  • (c) u[n]
  • (d) (−1)nu[n]
  • (e) cos’^{π 3 n+ π} 6 (
• 3.3-2 Show that
  • (a) δ[n] +δ[n−1] = u[n] −u[n−2]
  • (b) 2n−¹ sin ^πn 3 ^u[n]= ^{1 2 2n} sin ^πn 3 u[n−1]
  • (c) ⁿ(n−1)γ nu[n] = ⁿ(n−1)γ nu[n−2]
  • (d) (u[n] +(−1)nu[n])sinπ*ⁿ* = 0 for all n

(e)

$$(u[n] + (-1)^{n+1}u[n])\cos(\frac{\pi n}{2}) = 0$$

for all n

• 3.3-3 Sketch the following signals:

  • (a) u[n−2] −u[n−6]
    • (b) n{u[n] −u[n−7]}
    • (c) (n−2){u[n−2] −u[n−6]}
    • (d) (−n+8){u[n−6] −u[n−9]}
    • (e) (n−2){u[n−2]−u[n−6]}+(−n+8){u[n− 6] −u[n−9]}

• 3.3-4 Describe each of the signals in Fig. P3.1-1 by a single expression valid for all n.

• 3.3-5 Why are DT signals of the form zⁿ so important to the study of LTID systems?

• 3.3-6 Explain the similarities and differences between the Kronecker delta function δ[n] and the Dirac delta function δ(t).

• 3.3-7 The following signals are in the form eλn. Express them in the form γ ⁿ:

  • (a) e−0.5*ⁿ*
  • (b) e0.5*ⁿ*
  • (c) e−jπ*ⁿ*
  • (d) ejπ*ⁿ*

In each case show the locations of λ and γ in the complex plane. Verify that an exponential is growing if γ lies outside the unit circle (or if λ lies in the RHP), is decaying if γ lies within the unit circle (or if λ lies in the LHP), and has a constant amplitude if γ lies on the unit circle (or if λ lies on the imaginary axis).

• 3.3-8 Express the following signals, which are in the form eλn, in the form γ ⁿ:

  • (a) e−(1+jπ )ⁿ
  • (b) e−(1−jπ )ⁿ
  • (c) e(1+jπ )ⁿ
  • (d) e(1−jπ )ⁿ
  • (e) e−[1+j(π/3)]ⁿ
  • (f) e[1−j(π/3)]ⁿ

• 3.3-9 The concepts of even and odd functions for discrete-time signals are identical to those of the continuous-time signals discussed in Sec. 1.5. Using these concepts, find and sketch the odd and the even components of the following:

  • (a) u[n]
  • (b) nu[n]
  • (c) sinπ*ⁿ* 4
  • (d) cosπ*ⁿ* 4

• 3.4-1 A cash register output y[n] represents the total cost of n items rung up by a cashier. The input x[n] is the cost of the nth item.

  • (a) Write the difference equation relating y[n] to x[n].
  • (b) Realize this system using a time-delay element.

• 3.4-2 Let p[n] be the population of a certain country at the beginning of the nth year. The birth and death rates of the population during any year are 3.3 and 1.3%, respectively. If i[n] is the total number of immigrants entering the country during the nth year, write the difference equation relating p[n + 1], p[n], and i[n]. Assume that the immigrants enter the country throughout the year at a uniform rate.

• 3.4-3 A moving average is used to detect a trend of a rapidly fluctuating variable, such as the stock market average. A variable may fluctuate (up and down) daily, masking its long-term (secular) trend. We can discern the long-term trend by smoothing or averaging the past N values of the variable. For the stock market average, we may consider a 5-day moving average y[n] to be the mean of the past 5 days’ market closing values x[n], x[n−1],…, x[n−4].

  • (a) Write the difference equation relating y[n] to the input x[n].
  • (b) Use time-delay elements to realize the 5-day moving-average filter.

• 3.4-4 The digital integrator in Ex. 3.9 is specified by

$$y[n] - y[n-1] = Tx[n]$$

If an input u[n] is applied to such an integrator, show that the output is (n + 1)Tu[n], which approaches the desired ramp nTu[n] as T → 0.

3.4-5 Approximate the following second-order differential equation with a difference equation.

$$\frac{d^2y(t)}{dt^2} + a_1 \frac{dy(t)}{dt} + a_0 y(t) = x(t)$$

• 3.4-6 Letting ↓ identify n = 0, define the nonzero values of signal g[n] in vector form as [1, 2, 3, 4, 5, 4, 3, ↓ 2, 1]. The impulse response of an LTID system is defined in terms of g[n] as h[n] = g[−2n−1].

  • (a) Express the nonzero values of h[n] in vector form, taking care to identify the n = 0 point.
  • (b) Write a constant-coefficient linear difference equation (input x[n] and output y[n]) that has impulse response h[n].
  • (c) Show that the system is both linear and time-invariant.
  • (d) Determine, if possible, whether the system is BIBO-stable.
  • (e) Determine, if possible, whether the system is memoryless.
  • (f) Determine, if possible, whether the system is causal.

• 3.4-7 An LTID system has an impulse response function h[n] = u[−(5−n)/3].

  • (a) Using an accurate sketch or vector representation, graphically depict h[n].

• (b) Determine, if possible, whether the system is BIBO-stable.

• (c) Determine, if possible, whether the system is memoryless.

• (d) Determine, if possible, whether the system is causal.

• 3.4-8 The voltage at the nth node of a resistive ladder in Fig. P3.4-8 is v[n], (n = 0, 1, 2,…,N). Show that v[n] satisfies the second-order difference equation

$$v[n+2] - Av[n+1] + v[n] = 0 \quad A = 2 + \frac{1}{a}$$

[Hint: Consider the node equation at the nth node with voltage v[n].]

• 3.4-9 Determine whether each of the following statements is true or false. If the statement is false, demonstrate by proof or example why the statement is false. If the statement is true, explain why.

  • (a) A discrete-time signal with finite power cannot be an energy signal.

• (b) A discrete-time signal with infinite energy must be a power signal.

• (c) The system described by y[n] = (n+1)x[n] is causal.

• (d) The system described by y[n − 1] = x[n] is causal.

• (e) If an energy signal x[n] has energy E, then the energy of x[an] is E/|a|.

• 3.4-10 A linear time-invariant system produces output y1[n] in response to input x1[n], as shown in Fig. P3.4-10. Determine and sketch the output y2[n] that results when input x2[n] is applied to the same system.

• 3.4-11 A system is described by

$$y[n] = \frac{1}{2} \sum_{k=-\infty}^{\infty} x[k] (\delta[n-k] + \delta[n+k])$$

• (a) Explain what this system does.
• (b) Is the system BIBO-stable? Justify your answer.
• (c) Is the system linear? Justify your answer.

Figure P3.4-8

Figure P3.4-10

• (d) Is the system memoryless? Justify your answer.
• (e) Is the system causal? Justify your answer.
• (f) Is the system time-invariant? Justify your answer.
• 3.4-12 A discrete-time system is given by

$$y[n+1] = \frac{x[n]}{x[n+1]}$$

• (a) Is the system BIBO-stable? Justify your answer.
• (b) Is the system memoryless? Justify your answer.
• (c) Is the system causal? Justify your answer.
• 3.4-13 Explain why the continuous-time system y(t) = x(2t) is always invertible and yet the corresponding discrete-time system y[n] = x[2n] is not invertible.
• 3.4-14 Consider the input–output relationships of two similar discrete-time systems:

$$y_1[n] = \sin\left(\frac{\pi}{2}n + 1\right)x[n]$$

and

$$y_2[n] = \sin\left(\frac{\pi}{2}(n+1)\right) x[n]$$

Explain why x[n] can be recovered from y1[n] yet x[n] cannot be recovered from y2[n].

• 3.4-15 Consider a system that multiplies a given input by a ramp function, r[n]. That is, y[n] = x[n]r[n].
  • (a) Is the system BIBO-stable? Justify your answer.
  • (b) Is the system linear? Justify your answer.
  • (c) Is the system memoryless? Justify your answer.
  • (d) Is the system causal? Justify your answer.
  • (e) Is the system time-invariant? Justify your answer.
• 3.4-16 A jet-powered car is filmed using a camera operating at 60 frames per second. Let variable n designate the film frame, where n = 0 corresponds to engine ignition (film before ignition is discarded). By analyzing each frame of the film, it is possible to determine the car position x[n], measured in meters, from the original starting position x[0] = 0.

From physics, we know that velocity is the time derivative of position:

$$v(t) = \frac{d}{dt}x(t)$$

Furthermore, we know that acceleration is the time derivative of velocity:

$$a(t) = \frac{d}{dt}v(t)$$

We can estimate the car velocity from the film data by using a simple difference equation v[n] = k(x[n] −x[n−1]).

• (a) Determine the appropriate constant k to ensure v[n] has units of meters per second.
• (b) Determine a standard-form constant coefficient difference equation that outputs an estimate of acceleration, a[n], using an input of position, x[n]. Identify the advantages and shortcomings of estimating acceleration a(t) with a[n]. What is the impulse response h[n] for this system?
• 3.5-1 An LTID system is described by a constant coefficient linear difference equation 2y[n] + 2y[n−1] = x[n−1].
  • (a) Express this system in standard advance operator form.
  • (b) Using recursion, determine the first 5 values of the system impulse response h[n].
  • (c) Using recursion, determine the first 5 values of the system zero-state response to input x[n] = 2u[n].
  • (d) Using recursion, determine for (0 ≤ n ≤ 4) the system zero-input response if y[−1] = 1.
• 3.5-2 Solve recursively (first three terms only):
  • (a) y[n+1] −0.5y[n] = 0, with y[−1] = 10
  • (b) y[n + 1] + 2y[n] = x[n + 1], with x[n] = ^e−nu[n] and ^y[−1] = ⁰
• 3.5-3 Solve the following equation recursively (first three terms only):

$$y[n] - 0.6y[n-1] - 0.16y[n-2] = 0$$

with

$$y[-1] = -25, y[-2] = 0.$$

3.5-4 Solve recursively the second-order difference Eq. (3.6) for sales estimate (first three terms only), assuming y[−1] = y[−2] = 0 and x[n] = 100u[n].

3.5-5 Solve the following equation recursively (first three terms only):

$$y[n+2] + 3y[n+1] + 2y[n] = x[n+2] + 3x[n+1] + 3x[n]$$

with ^x[n] = (3)nu[n], ^y[−1] = 3, and ^y[−2] = ²

3.5-6 Repeat Prob. 3.5-5 for

$$y[n] + 2y[n-1] + y[n-2] = 2x[n] - x[n-1]$$

with ^x[n] = (3)−nu[n], ^y[−1] = 2, and ^y[−2] = 3.

• 3.6-1 Given y0[−1] = 3 and y0[−2]=−1, determine the closed-form expression of the zero-input response y0[n] of an LTID system described by the equation ^y[n]+ ^{1 6 y}[n−1]− ^{1 6 y}[n−2] = ^{1 3} x[n] +2 ³ x[n−2].
• 3.6-2 Solve

$$y[n+2] + 3y[n+1] + 2y[n] = 0$$

if

$$y[-1] = 0$$

and $y[-2] = 1$ .

3.6-3 Solve

$$y[n+2] + 2y[n+1] + y[n] = 0$$

if

$$y[-1] = 1$$

and $y[-2] = 1$ .

3.6-4 Solve

$$y[n+2] - 2y[n+1] + 2y[n] = 0$$

if y[−1] = 1 and y[−2] = 0.

3.6-5 For the general Nth-order difference Eq. (3.16), letting

a¹ = a² =···= aN−¹ = 0

results in a general causal Nth-order LTI nonrecursive difference equation

$$y[n] = b_0 x[n] + b_1 x[n-1] + \cdots + b_N x[n-N]$$

Show that the characteristic roots for this system are zero—hence, that the zero-input response is zero. Consequently, the total response consists of the zero-state component only.

3.6-6 Leonardo Pisano Fibonacci, a famous thirteenthcentury mathematician, generated the sequence of integers

$$\{0,1,1,2,3,5,8,13,21,34,\dots\}$$

while addressing, oddly enough, a problem involving rabbit reproduction. An element of the Fibonacci sequence is the sum of the previous two.

• (a) Find the constant-coefficient difference equation whose zero-input response f[n] with auxiliary conditions f[1] = 0 and f[2] = 1 is a Fibonacci sequence. Given f[n] is the system output, what is the system input?

• (b) What are the characteristic roots of this system? Is the system stable?

• (c) Designating 0 and 1 as the first and second Fibonacci numbers, determine the fiftieth Fibonacci number. Determine the one thousandth Fibonacci number.

• 3.6-7 Find v[n], the voltage at the nth node of the resistive ladder depicted in Fig. P3.4-8, if V = 100 volts and a = 2. [Hint 1: Consider the node equation at the nth node with voltage v[n]. Hint 2: See Prob. 3.4-8 for the equation for v[n]. The auxiliary conditions are v[0] = 100 and v[N] = 0.]

• 3.6-8 Consider the discrete-time system y[n] + y[n − ¹] + 0.25y[^{n − 2}] = ^√3x[^{n − 8}]. Find the zero input response, y0[n], if y0[−1] = 1 and y0[1] = 1.

• 3.6-9 Provide a standard-form polynomial Q(X) such that Q(E){y[n]} = x[n] corresponds to a marginally stable third-order LTID system and Q(D){y(t)} = x(t) corresponds to a stable third-order LTIC system.

• 3.7-1 Find the unit impulse response h[n] of systems specified by the following equations: (a) y[n+1] +2y[n] = x[n] (b) y[n] +2y[n−1] = x[n]

• 3.7-2 Determine the unit impulse response h[n] of the following systems. In each case, use recursion to verify the n = 3 value of the closed-form expression of h[n].

  • (a) (E² +1){y[n]} = (E +0.5){x[n]}
  • (b) y[n] −y[n−1] +0.25y[n−2] = x[n]
  • (c) ^y[n] − ^{1 6 y}[n−1] − ^{1 6 y}[n−2] = ^{1 3} x[n−2]

• (d) ^y[n] + ^{1 6 y}[n−1] − ^{1 6 y}[n−2] = ^{1 3} x[n]

• (e) ^y[n] + ^{1 4} y[n−2] = x[n]

• (f) (E^{2 − 4 9} ){y[n]} = (E^{2 +}1){x[n]}

• (g) (E^{2 − 1 4} )(^{E + 1 2} ){y[n]} = *^E*3{x[n]}

• (h) (^{E − 1 2} )2{y[n]} = ^x[n]

• 3.7-3 Consider a DT system with input x[n] and output y[n] described by the difference equation

$$4y[n+1] + y[n-1] = 8x[n+1] + 8x[n]$$

• (a) What is the order of this system?
• (b) Determine the characteristic mode(s) of the system.
• (c) Determine a closed-form expression for the system’s impulse response h[n].
• 3.7-4 Repeat Prob. 3.7-3 for a system described by the difference equation

$$y[n+3] - \frac{3}{10}y[n+2] - \frac{1}{10}y[n+1] = 2x[n+1]$$

3.7-5 Repeat Prob. 3.7-1 for

$$(E2 - 6E + 9)y[n] = Ex[n]$$

3.7-6 Repeat Prob. 3.7-1 for

$$y[n] - 6y[n-1] + 25y[n-2] = 2x[n] - 4x[n-1]$$

3.7-7 (a) For the general Nth-order difference Eq. (3.16), letting

$$a_0 = a_1 = a_2 = \cdots = a_{N-1} = 0$$

results in a general causal Nth-order LTI nonrecursive difference equation

$$y[n] = \sum_{i=0}^{N} b_i x[n-i]$$

Find the impulse response h[n] for this system. [Hint: The characteristic equation for this case is ^{γ n =} 0. Hence, all the characteristic roots are zero. In this case, yc[n] = 0, and the approach in Sec. 3.7 does not work. Use a direct method to find h[n] by realizing that h[n] is the response to unit impulse input.]

(b) Find the impulse response of a nonrecursive LTID system described by the equation

$$y[n] = 3x[n] - 5x[n-1] - 2x[n-3]$$

Observe that the impulse response has only a finite (N) number of nonzero elements. For this reason, such systems are called finite-impulse response (FIR) systems. For a general recursive case [Eq. (3.20)], the impulse response has an infinite number of nonzero elements, and such systems are called infinite-impulse response (IIR) systems.

3.8-1 The convolution ^y[n] = ^{5 2n} u[n+5] ^∗ (3nu[−n−2]) can be represented as

$$y[n] = \begin{cases} C_1(\gamma_1)^n & n < N \\ C_2(\gamma_2)^n & n \ge N \end{cases}$$

Using the graphical convolution procedure, determine constants C1, C2, γ1, γ2, and N.

• 3.8-2 Use the graphical convolution procedure to determine the following:
  • (a) ya[n] = u[n] ∗ (u[n − 5] − u[n − 9] + (0.5)(n−8) u[n−9])
  • (b) *^y*b[n] = ( ^{1 2} )|n^{| ∗ u}[−n+5]
• 3.8-3 Let ^x[n] = (0.5)ⁿ (u[n+4] −u[n−4]) be input into an LTID system with an impulse response given by

$$h[n] = \begin{cases} 2 & \text{[(}n \text{ mod } 6) < 4 \text{]} \text{ and } [n \ge 0] \\ 0 & \text{otherwise} \end{cases}$$

Recall, (n mod p) is the remainder of the division n/p. The system is described according to the difference equation y[n]−y[n−6] = 2x[n]+ 2x[n−1] +2x[n−2] +2x[n−3].

• (a) Determine the six characteristic roots (γ¹ through γ6) of the system.
• (b) Determine the value of y[10], the zero-state output of system h[n] in response to x[n] at time n = 10. Express your result in decimal form to at least three decimal places (e.g., y[10] = 3.142).
• 3.8-4 An LTID system has impulse response h[n] = (0.5)(n+3) (u[n] −u[n+6]). A 6-periodic DT

input signal x[n] is given by

$$x[n] = \begin{cases} 1 & n = 0, \pm 3, \pm 6, \pm 9, \pm 12, \dots \\ 2 & n = 1, 1 \pm 6, 1 \pm 12, \dots \\ 3 & n = 2, 2 \pm 6, 2 \pm 12, \dots \\ 0 & \text{otherwise} \end{cases}$$

• (a) Is system h[n] causal? Mathematically justify your answer.
• (b) Determine the value of y[12], the zero-state output of system h[n] in response to x[n] at time n = 12. Express your result in decimal form to at least three decimal places (e.g., y[12] = 1.234).
• 3.8-5 Find the (zero-state) response y[n] of an LTID system whose unit impulse response is

$$h[n] = (-2)^n u[n-1]$$

and the input is ^x[n] = ^e−nu[^{n + 1}]. Find your answer by computing the convolution sum and also by using Table 3.1.

3.8-6 Find the (zero-state) response y[n] of an LTID system if the input is ^x[n] = ³n−1u[n+2], and

$$h[n] = \frac{1}{2} [\delta[n-2] - (-2)^{n+1}] u[n-3]$$

3.8-7 Find the (zero-state) response y[n] of an LTID system if the input ^x[n] = (3)n+2u[n+1], and

$$h[n] = [(2)^{n-2} + 3(-5)^{n+2}]u[n-1]$$

3.8-8 Find the (zero-state) response y[n] of an LTID system if the input ^x[n] = (3)−n+2u[n+3], and

$$h[n] = 3(n-2)(2)^{n-3}u[n-4]$$

3.8-9 Find the (zero-state) response y[n] of an LTID system if its input ^x[n] = (2)nu[n−1], and

$$h[n] = (3)^n \cos\left(\frac{\pi}{3}n - 0.5\right) u[n]$$

Find your answer using only Table 3.1.

• 3.8-10 Consider an LTID system (“system 1”) described by (^{E − 1 2} ){y[n]} = x[n].

  • (a) Determine the impulse response h1[n] for system 1. Simplify your answer.
  • (b) Determine the step response s[n] for system 1 (the step response is the output in response to a unit step input). Simplify your answer.

• (c) Determine the impulse response hcascade[n] of system 1 cascaded with an LTID system with impulse response h2[n]=−3u[n−13]. Simplify your answer.

• 3.8-11 Derive the results in entries 1, 2, and 3 in Table 3.1. [Hint: You may need to use the information in Sec. B.8-3.]

• 3.8-12 Derive the results in entries 4, 5, and 6 in Table 3.1.

• 3.8-13 Derive the results in entries 7 and 8 in Table 3.1. [Hint: You may need to use the information in Sec. B.8-3.]

• 3.8-14 Derive the results in entries 9 and 11 in Table 3.1. [Hint: You may need to use the information in Sec. B.8-3.]

• 3.8-15 Find the total response of a system specified by the equation

$$y[n+1] + 2y[n] = x[n+1]$$

if ^y[−1] = 10, and the input ^x[n] = ^e−nu[n].

• 3.8-16 Find an LTID system (zero-state) response if its impulse response ^h[n] = (0.5)nu[n], and the input x[n] is (a) 2nu[n] (b) 2n−3u[n]
  • (c) 2nu[n−2]

[Hint: You may need to use the convolution shift property of Eq. (3.32).]

3.8-17 For a system specified by equation

$$y[n] = x[n] - 2x[n-1]$$

Find the system response to input x[n] = u[n]. What is the order of the system? What type of system (recursive or nonrecursive) is this? Is the knowledge of initial condition(s) necessary to find the system response? Explain.

• 3.8-18 (a) A discrete-time LTI system is shown in Fig. P3.8-18. Express the overall impulse response of the system, h[n], in terms of h1[n], h2[n], h3[n], h4[n], and h5[n].
  • (b) Two LTID systems in cascade have impulse response h1[n] and h2[n], respectively. Show that if h1[n] = (0.9)nu[n] − 0.5(0.9)n−1u[^{n − 1}] and ^h2[n] = (0.5)nu[n] − 0.9(0.5)n−1u[^{n − 1}], the cascade system is an identity system.

Figure P3.8-18

3.8-19 (a) Show that for a causal system, Eq. (3.37) can also be expressed as

$$g[n] = \sum_{k=0}^{n} h[n-k]$$

• (b) How would the expressions in part (a) change if the system is not causal?
• 3.8-20 An LTID system with input x[n] and output y[n] has impulse response h[n] = 2(u[n + 2] − u[n − 3]).
  • (a) Write a constant-coefficient linear difference equation that has the given impulse response. [Hint: First express h[n] in terms of delta functions δ[n].]
  • (b) Using graphical convolution, determine the zero-state output of this system in response to the anticausal input ^x[n] = ²nu[−n]. A simplified closed-form solution is required.
• 3.8-21 Consider three LTID systems: system 1 has impulse response h1[n]=[ ↓ 2, −3, 4], system 2 has impulse response h2[n]=[ ↓ 0, 0, −6, − 9, 3], and system 3 is an identity system (output equals input).
  • (a) Determine the overall impulse response h[n] if system 1 is connected in cascade with a parallel connection of systems 2 and 3.
  • (b) For input x[n] = u[−n], determine the zero-state response yzsr[n] of system 2.
• 3.8-22 In the savings account problem described in Ex. 3.6, a person deposits $500 at the beginning of every month, starting at n = 0 with the exception at n = 4, when instead of depositing $500, she withdraws $1000. Find y[n] if the interest rate is 1% per month (r = 0.01).

3.8-23 To pay off a loan of M dollars in N number of payments using a fixed monthly payment of P dollars, show that

$$P = \frac{rM}{1 - (1 + r)^{-N}}$$

where r is the interest rate per dollar per month. [Hint: This problem can be modeled by Eq. (3.3) with the payments of P dollars starting at n = 1. The problem can be approached in two ways. First, consider the loan as the initial condition y0[0]=−M, and the input x[n] = Pu[n − 1]. The loan balance is the sum of the zero-input component (due to the initial condition) and the zero-state component h[n] ∗ x[n]. Second, consider the loan as an input −M at n = 0 along with the input due to payments. The loan balance is now exclusively a zero-state component h[n] ∗ x[n]. Because the loan is paid off in N payments, set y[N] = 0.]

• 3.8-24 A person receives an automobile loan of $10,000 from a bank at the interest rate of 1.5% per month. His monthly payment is $500, with the first payment due one month after he receives the loan. Compute the number of payments required to pay off the loan. Note that the last payment may not be exactly $500. [Hint: Follow the procedure in Prob. 3.8-23 to determine the balance y[n]. To determine N, the number of payments, set y[N] = 0. In general, N will not be an integer. The number of payments K is the largest integer ≤ N. The residual payment is |y[K]|.]
• 3.8-25 Letting ↓ identify the n = 0 values, use the sliding-tape method to determine the following:
  • (a) y^a = [ ↓ 2, 3,−2,−3] ∗ [−10, ↓ 0,−5] (b) y^b = [2, ↓ −1, 3,−2] ∗ [−1,−4, 1, ↓ −2] (c) y^c = [ ↓ 0, 0, 3, 2, 1, 2, 3]∗[2, 3,−2, ↓ 1]
  • (d) y^d = [5, 0, 0, ↓ −2, 8] ∗ [−1, 1, ↓ 3, 3,−2, 3]
  • (e) y^e = ([1, ↓ −1]∗[ ↓ 1,−1]) ∗ ([ ↓ 1,−1]∗[1, ↓ −1])
  • (f) y^f = ([2, ↓ −1]∗[ ↓ 1,−2]) ∗ ([ ↓ 1,−2]∗[2, ↓ −1]) Outside the values shown, assume all signals are zero.

324 CHAPTER 3 TIME-DOMAIN ANALYSIS OF DISCRETE-TIME SYSTEMS

• 3.8-26 Let ↓ identify the n = 0 and consider DT signals x[n] and h[n] whose nonzero values are given as x[n]=[1, 2, 3, ↓ 4, 5] and h[n] = [−2, −1, 1, ↓ 2]. Use DT convolution (any method) to determine y[n] = (2x[n − 30]) ∗ −3 ² h[n−10] . Express your result in vector notation, making sure to indicate the time index of the leftmost (nonzero) element.

• 3.8-27 Using the sliding-tape algorithm, show that (a) u[n] ∗ u[n] = (n+1)u[n] (b) (u[n] −u[n−m])∗u[n] = (n+1)u[n]−(n− m+1)u[n−m]

• 3.8-28 Using the sliding-tape algorithm, find x[n] ∗g[n] for the signals shown in Fig. P3.8-28.

• 3.8-29 Repeat Prob. 3.8-28 for the signals shown in Fig. P3.8-29.

• 3.8-30 Repeat Prob. 3.8-28 for the signals shown in Fig. P3.8-30.

• 3.8-31 Letting ↓ identify n = 0, define the nonzero values of signal x[n] as [1, 2, ↓ 2]. Similarly, define the non-zero values of signal y[n] as [3, 4, 6, 6, 11, ↓ 2,−2]. Using the sliding-tape algorithm as the basis for your work, determine the signal h[n] so that y[n] = x[n] ∗ h[n].

• 3.8-32 The convolution sum in Eq. (3.33) can be expressed in a matrix form as ^{y =} Hx, where ^y is a column vector containing ^y[0], ^y[1],… ^y[n]; ^x is a column vector containing x[0], x[1],… x[n];

Figure P3.8-30

and H is a lower triangular matrix defined as

	⎡ h[0]	0	0		0	⎤
	h[1] ⎢	h[0]	0		0	⎥
H =	⎢ ⎢					⎥ ⎥
	⎣ h[n]	h[n−1]		···	h[0]	⎦

Knowing h[n] and the output y[n], we can determine the input ^x[n] according to ^{x = H}−1y. This operation is the reverse of convolution and is known as deconvolution. Moreover, knowing x[n] and y[n], we can determine h[n]. This can be done by expressing the foregoing matrix equation as n + 1 simultaneous equations in terms of n + 1 unknowns h[0], h[1], … , h[n]. These equations can readily be solved iteratively. Thus, we can synthesize a system that yields a certain output y[n] for a given input x[n].

• (a) Design a system (i.e., determine h[n]) that will yield the output sequence (8, 12, 14, 15, 15.5, 15.75, …) for the input sequence (1, 1, 1, 1, 1, 1, …).
• (b) For a system with the impulse response sequence (1, 2, 4, …), the output sequence was (1, 7/3, 43/9, …). Determine the input sequence.
• 3.8-33 A second-order LTID system has zero-input response

$$y_0[n] = [3, 2\frac{1}{3}, 2\frac{1}{9}, 2\frac{1}{27}, \dots]$$

=

$$\sum_{k=0}^{\infty} \left\{ 2 + \left(\frac{1}{3}\right)^k \right\} \delta[n-k]$$

• (a) Determine the characteristic equation of this system, a0γ ² +a1γ +a² = 0.
• (b) Find a bounded, causal input with infinite duration that would cause a strong response from this system. Justify your choice.

Figure P3.8-34

• (c) Find a bounded, causal input with infinite duration that would cause a weak response from this system. Justify your choice.
• 3.8-34 An LTID filter has an impulse response function given by h1[n] = δ[n + 2] − δ[n − 2]. A second LTID system has an impulse response function given by h2[n] = n(u[n+4] −u[n−4]).
  • (a) Carefully sketch the functions h1[n] and h2[n] over (−10 ≤ n ≤ 10).
  • (b) Assume that the two systems are connected in parallel, as shown in Fig. P3.8-34a. Determine the impulse response hp[n] for the parallel system in terms of h1[n] and h2[n]. Sketch hp[n] over (−10 ≤ n ≤ 10).
  • (c) Assume that the two systems are connected in cascade, as shown in Fig. P3.8-34b. Determine the impulse response hs[n] for the cascade system in terms of h1[n] and h2[n]. Sketch hs[n] over (−10 ≤ n ≤ 10).
• 3.8-35 This problem investigates an interesting application of discrete-time convolution: the expansion of certain polynomial expressions.
  • (a) By hand, expand (z3+z2+z+1)2. Compare the coefficients to [1, 1, 1, 1]∗[1, 1, 1, 1].
  • (b) Formulate a relationship between discretetime convolution and the expansion of constant-coefficient polynomial expressions.
  • (c) Use convolution to expand (z−⁴ − 2z−³ + 3z−2)4.
  • (d) Use convolution to expand (z⁵ +2z⁴ +3z² + 5)2(z−⁴ −5z−² +13).
• 3.8-36 Joe likes coffee, and he drinks his coffee according to a very particular routine. He begins by adding two teaspoons of sugar to his mug, which he then fills to the brim with hot coffee. He drinks 2/3 of the mug’s contents, adds another two teaspoons of sugar, and tops the mug off with steaming hot coffee. This refill procedure

continues, sometimes for many, many cups of coffee. Joe has noted that his coffee tends to taste sweeter with the number of refills.

Let independent variable n designate the coffee refill number. In this way, n = 0 indicates the first cup of coffee, n = 1 is the first refill, and so forth. Let x[n] represent the sugar (measured in teaspoons) added into the system (a coffee mug) on refill n. Let y[n] designate the amount of sugar (again, teaspoons) contained in the mug on refill n.

• (a) The sugar (teaspoons) in Joe’s coffee can be represented using a standard second-order constant coefficient difference equation y[n] + a1y[n − 1] + a2y[n − 2] = b0x[n] + b1x[n − 1] + b2x[n − 2]. Determine the constants a1, a2, b0, b1, and b2.
• (b) Determine x[n], the driving function to this system.
• (c) Solve the difference equation for y[n]. This requires finding the total solution. Joe always starts with a clean mug from the dishwasher, so y[−1] (the sugar content before the first cup) is zero.
• (d) Determine the steady-state value of y[n]. That is, what is y[n] as n → ∞? If possible, suggest a way of modifying x[n] so that the sugar content of Joe’s coffee remains a constant for all nonnegative n.
• 3.8-37 A system is called complex if a real-valued input can produce a complex-valued output. Consider a causal complex system described by a first-order constant coefficient linear difference equation:

(jE +0.5)y[n] = (−5E)x[n]

• (a) Determine the impulse response function h[n] for this system.

• (b) Given input x[n] = u[n − 5] and initial condition y0[−1] = j, determine the system’s total output y[n] for n ≥ 0.

• 3.8-38 A discrete-time LTI system has impulse response function h[n] = n(u[n − 2] − u[n+2]).

  • (a) Carefully sketch the function h[n] over (−5 ≤ n ≤ 5).
  • (b) Determine the difference equation representation of this system, using y[n] to designate the output and x[n] to designate the input.

• 3.8-39 Consider three discrete-time signals: x[n], y[n], and z[n]. Denoting convolution as ∗, identify the expression(s) that is(are) equivalent to x[n](y[n] ∗ z[n]):

  • (a) (x[n] ∗ y[n])z[n] (b) (x[n]y[n]) ∗ (x[n]z[n])
  • (c) (x[n]y[n]) ∗ z[n]
  • (d) none of the above
  • Justify your answer!

• 3.8-40 A causal system with input x[n] and output y[n] is described by

$$y[n] - ny[n-1] = x[n]$$

• (a) By recursion, determine the first six nonzero values of h[n], the response to x[n] = δ[n]. Do you think this system is BIBO-stable? Why?

• (b) Compute yR[4] recursively from yR[n] − nyR[n−1] = x[n], assuming all initial conditions are zero and x[n] = u[n]. The subscript R is only used to emphasize a recursive solution.

• (c) Define yC[n] = x[n]∗h[n]. Using x[n] = u[n] and h[n] from part (a), compute yC[4]. The subscript C is only used to emphasize a convolution solution.

• (d) In this chapter, both recursion and convolution are presented as potential methods to compute the zero-state response (ZSR) of a discrete-time system. Comparing parts (b) and (c), we see that yR[4] = yC[4]. Why are the two results not the same? Which method, if any, yields the correct ZSR value?

• 3.9-1 In Sec. 3.9-1 we showed that for BIBO stability in an LTID system, it is sufficient for its impulse response h[n] to satisfy Eq. (3.43). Show that this is also a necessary condition for the system to be BIBO-stable. In other words, show that if Eq. (3.43) is not satisfied, there exists a bounded input that produces unbounded output. [Hint: Assume that a system exists for which h[n] violates Eq. (3.43), yet its output is bounded for every bounded input. Establish the contradiction in this statement by considering an input x[n] defined by x[n¹ − m] = 1 when h[m] > 0 and x[n1−m]=−1 when h[m]<0, where n¹ is some fixed integer.]

• 3.9-2 Each of the following equations specifies an LTID system. Determine whether each of these systems is BIBO-stable or -unstable. Determine also whether each is asymptotically stable, unstable, or marginally stable.

  • (a) y[n + 2] + 0.6y[n + 1] − 0.16y[n] = x[n + 1] −2x[n]
  • (b) y[n] + 3y[n − 1] + 2y[n − 2] = x[n − 1] + 2x[n−2]
  • (c) (E −1)^{2 E + 1} 2 y[n] = x[n]
  • (d) y[n] +2y[n−1] +0.96y[n−2] = x[n]
  • (e) y[n]+y[n−1]−2y[n−2] = x[n]+2x[n−1]
  • (f) (E² −1)(E² +1)y[n] = x[n]

• 3.9-3 Consider two LTIC systems in cascade, as illustrated in Fig. 3.29. The impulse response of the system ^S1 is ^h1[n] = ^{2n u}[n] and the impulse response of the system S² is h2[n] = δ[n] − 2δ[n−1]. Is the cascaded system asymptotically stable or unstable? Determine the BIBO stability of the composite system.

• 3.9-4 Figure P3.9-4 locates the characteristic roots of ten causal, LTID systems, labeled A through J. Each system has only two roots and is described using operator notation as Q(E)y[n] = P(E)x[n]. All plots are drawn to scale, with the unit circle shown for reference. For each of the following parts, identify all the answers that are correct.

  • (a) Identify all systems that are unstable.
  • (b) Assuming all systems have P(E) = E2, identify all systems that are real. Recall that a real system always generates a real-valued response to a real-valued input.
  • (c) Identify all systems that support oscillatory natural modes.

• (d) Identify all systems that have at least one mode whose envelop decays at a rate of 2−n.

• (e) Identify all systems that have only one mode.

• 3.9-5 A discrete-time LTI system has impulse response given by

$$h[n] = \delta[n] + \left(\frac{1}{3}\right)^n u[n-1]$$

• (a) Is the system stable? Is the system causal? Justify your answers.
• (b) Plot the signal x[n] = u[n−3] −u[n+3].
• (c) Determine the system’s zero-state response y[n] to the input x[n] = u[n − 3] − u[n + 3]. Plot y[n] over (−10 ≤ n ≤ 10).
• 3.9-6 An LTID system has an impulse response given by

$$h[n] = \left(\frac{1}{2}\right)^{|n|}$$

• (a) Is the system causal? Justify your answer.
• (b) Compute %^{∞ n}=−∞ |h[n]|. Is this system BIBO-stable?
• (c) Compute the energy and power of input signal x[n] = 3u[n−5].
• (d) Using input x[n] = 3u[n − 5], determine the zero-state response of this system at time n = 10. That is, determine yzsr[10].
• 3.10-1 Determine a constant coefficient linear difference equation that describes a system for which the input ^x[n] = ²( ^{1 3} )nu[−^{n − 4}] causes resonance.
• 3.10-2 If one exists, determine a real input x[n] that will cause resonance in the causal LTID system described by (E² +1){y[n]} = (E+0.5){x[n]}. If no such input exists, explain why not.

Figure P3.9-4

• 3.10-3 Consider two lowpass LTID systems, one with infinite-duration impulse response h1[n] = ⁻(0.5)nu[n] and the other with finite-duration impulse response h2[n] = 2(u[n] − u[n − 4]). Which system (1, 2, both, or neither) would more efficiently transmit a binary communication signal? Carefully justify your result.
• 3.11-1 Write a MATLAB program that recursively computes and then plots the solution to ^y[n] − ^{1 3 y}[^{n − 1}] + ^{1 2} y[n − 2] = x[n] for (0 ≤ n ≤ 100) given x[n] = δ[n] + u[n − 50] and y[−2] = y[−1] = 2.
• 3.11-2 Consider the discrete-time function f[n] = ^e−n/⁵ cos(πn/5)u[n]. Section 3.11 uses anonymous functions in describing DT signals.
  • f = @(n) exp(-n/5).*cos(pi*n/5).*(n>=0);

While this anonymous function operates correctly for a downsampling operation such as f[2n], it does not operate correctly for an upsampling operation, such as f[n/2]. Modify the anonymous function f so that it also correctly accommodates upsampling operations. Test your code by computing and plotting f(n/2) over (−10 ≤ n ≤ 10).

• 3.11-3 Write MATLAB code to compute and plot the DT convolutions of Prob. 3.8-25.
• 3.11-4 An indecisive student contemplates whether he should stay home or take his final exam, which is being held 2 miles away. Starting at home, the student travels half the distance to the exam location before changing his mind. The student turns around and travels half the distance between his current location and his home before changing his mind again. This process of changing direction and traveling half the remaining distance continues until the student either reaches a destination or dies from exhaustion.
  • (a) Determine a suitable difference equation description of this system.
  • (b) Use MATLAB to simulate the difference equation in part (a). Where does the student end up as n → ∞? How does your answer change if the student goes two-thirds the way each time, rather than halfway?
  • (c) Determine a closed-form solution to the equation in part (a). Use this solution to verify the results in part (b).

3.11-5 The cross-correlation function between x[n] and y[n] is given as

$$r_{xy}[k] = \sum_{n=-\infty}^{\infty} x[n]y[n-k]$$

Notice that rxy[k] is quite similar to the convolution sum. The independent variable k corresponds to the relative shift between the two inputs.

• (a) Express rxy[k] in terms of convolution. Is rxy[k] = ryx[k]?
• (b) Cross-correlation is said to indicate similarity between two signals. Do you agree? Why or why not?
• (c) If x[n] and y[n] are both finite duration, MATLAB’s conv command is well suited to compute rxy[k]. Write a MATLAB function that computes the cross-correlation function using the conv command. Four vectors are passed to the function (x, y, nx, and ny) corresponding to the inputs x[n], y[n], and their respective time vectors. Notice that x and y are not necessarily the same length. Two outputs should be created (rxy and k) corresponding to rxy[k] and its shift vector.
• (d) Test your code from part (c) using x[n] = u[n − 5] − u[n − 10] over (0 ≤ n = nx ≤ 20) and y[n] = u[−n−15]−u[−n−10]+δ[n− 2] over (−20 ≤ n = ny ≤ 10). Plot the result rxy as a function of the shift vector k. What shift k gives the largest magnitude of rxy[k]? Does this make sense?
• 3.11-6 Suppose a vector x exists in the MATLAB workspace, corresponding to a finite-duration DT signal x[n]
  • (a) Write a MATLAB function that, when passed vector x, computes and returns Ex, the energy of x[n].
  • (b) Write a MATLAB function that, when passed vector x, computes and returns Px, the power of x[n]. Assume that x[n] is periodic and that vector x contains data for an integer number of periods of x[n].
• 3.11-7 A causal N-point max filter assigns y[n] to the maximum of {x[n],…, x[n−(N −1)]}.
  • (a) Write a MATLAB function that performs N-point max filtering on a length-M input vector x. The two function inputs are vector x and scalar N. To create the length-M output

vector y, initially pad the input vector with N − 1 zeros. The MATLAB command max may be helpful.

• (b) Test your filter and MATLAB code by filtering a length-45 input defined as x[n] = cos(πn/5) + δ[n − 30] − δ[n − 35]. Separately plot the results for N = 4, N = 8, and N = 12. Comment on the filter behavior.
• 3.11-8 A causal N-point min filter assigns y[n] to the minimum of {x[n],…, x[n−(N −1)]}.
  • (a) Write a MATLAB function that performs N-point min filtering on a length-M input vector x. The two function inputs are vector x and scalar N. To create the length-M output vector y, initially pad the input vector with N − 1 zeros. The MATLAB command min may be helpful.
  • (b) Test your filter and MATLAB code by filtering a length-45 input defined as x[n] = cos(πn/5) + δ[n − 30] − δ[n − 35]. Separately plot the results for N = 4,N = 8, and N = 12. Comment on the filter behavior.
• 3.11-9 A causal N-point median filter assigns y[n] to the median of {x[n],…, x[n−(N −1)]}. The median is found by sorting sequence {x[n],…, x[n − (N − 1)]} and choosing the middle value (odd N) or the average of the two middle values (even N).
  • (a) Write a MATLAB function that performs N-point median filtering on a length-M input vector x. The two function inputs are vector x and scalar N. To create the length-M output vector y, initially pad the input vector with N − 1 zeros. The MATLAB command sort or median may be helpful.
  • (b) Test your filter and MATLAB code by filtering a length-45 input defined as x[n] = cos(πn/5) + δ[n − 30] − δ[n − 35]. Separately plot the results for N = 4, N = 8, and N = 12. Comment on the filter behavior.
• 3.11-10 Recall that y[n] = x[n/N] represents an upsample by N operation. An interpolation filter replaces the inserted zeros with more realistic values. A linear interpolation filter has impulse response

$$h[n] = \sum_{k=-(N-1)}^{N-1} \left(1 - \left|\frac{k}{N}\right|\right) \delta(n-k)$$

• (a) Determine a constant coefficient difference equation that has impulse response h[n].
• (b) The impulse response h[n] is noncausal. What is the smallest time shift necessary to make the filter causal? What is the effect of this shift on the behavior of the filter?
• (c) Write a MATLAB function that will compute the parameters necessary to implement an interpolation filter using MATLAB’s filter command. That is, your function should output filter vectors b and a given an input scalar N.
• (d) Test your filter and MATLAB code. To do this, create x[n] = cos(n) for (0 ≤ n ≤ 9). Upsample x[n] by N = 10 to create a new signal xup[n]. Design the corresponding N = 10 linear interpolation filter, filter xup[n] to produce y[n], and plot the results.
• 3.11-11 A causal N-point moving-average filter has impulse response h[n] = (u[n] − u[n − N])/N.
  • (a) Determine a constant-coefficient difference equation that has impulse response h[n].
  • (b) Write a MATLAB function that will compute the parameters necessary to implement an N-point moving-average filter using MATLAB’s filter command. That is, your function should output filter vectors b and a given a scalar input N.
  • (c) Test your filter and MATLAB code by filtering a length-45 input defined as x[n] = cos(πn/5) + δ[n − 30] − δ[n − 35]. Separately plot the results for N = 4, N = 8, and N = 12. Comment on the filter behavior.
  • (d) Problem 3.11-10 introduces linear interpolation filters, for use following an upsample by N operation. Within a scale factor, show that a cascade of two N-point moving-average filters is equivalent to the linear interpolation filter. What is the scale factor difference? Test this idea with MAT-LAB. Create x[n] = cos(n) for (0 ≤ n ≤ 9). Upsample x[n] by N = 10 to create a new signal xup[n]. Design an N = 10 moving-average filter. Filter xup[n] twice and scale to produce y[n]. Plot the results. Does the output from the cascaded pair of moving-average filters linearly interpolate the upsampled data?

CONTINUOUS-TIME SYSTEM ANALYSIS USING THE LAPLACE TRANSFORM

Because of the linearity (superposition) property of linear time-invariant systems, we can find the response of these systems by breaking the input x(t) into several components and then summing the system response to all the components of x(t). We have already used this procedure in time-domain analysis, in which the input x(t) is broken into impulsive components. In the frequency-domain analysis developed in this chapter, we break up the input x(t) into exponentials of the form est, where the parameter s is the complex frequency of the signal est, as explained in Sec. 1.4-3. This method offers an insight into the system behavior complementary to that seen in the time-domain analysis. In fact, the time-domain and the frequency-domain methods are duals of each other.

The tool that makes it possible to represent arbitrary input x(t) in terms of exponential components is the Laplace transform, which is discussed in the following section.

4.1 THE LAPLACE TRANSFORM

For a signal x(t), its Laplace transform X(s) is defined by

$$X(s) = \int_{-\infty}^{\infty} x(t)e^{-st} dt$$

\n(4.1)

The signal x(t) is said to be the inverse Laplace transform of X(s). It can be shown that

$$x(t) = \frac{1}{2\pi j} \int_{c-j\infty}^{c+j\infty} X(s)e^{st} ds$$

\n(4.2)

where c is a constant chosen to ensure the convergence of the integral in Eq. (4.1), as explained later. See also [1].

This pair of equations is known as the bilateral Laplace transform pair, where X(s) is the direct Laplace transform of x(t) and x(t) is the inverse Laplace transform of X(s). Symbolically,

^X(s) ^{= L}[x(t)] and ^x(t) ^{= L−1} [X(s)]

Note that

CHAPTER

4

$$\mathcal{L}^{-1}\{\mathcal{L}[x(t)]\} = x(t) \qquad \text{and} \qquad \mathcal{L}\{\mathcal{L}^{-1}[X(s)]\} = X(s)$$

It is also common practice to use a bidirectional arrow to indicate a Laplace transform pair, as follows:

$$x(t) \Longleftrightarrow X(s)$$

The Laplace transform, defined in this way, can handle signals existing over the entire time interval from −∞ to ∞ (causal and noncausal signals). For this reason it is called the bilateral (or two-sided) Laplace transform. Later we shall consider a special case—the unilateral or one-sided Laplace transform—which can handle only causal signals.

LINEARITY OF THE LAPLACE TRANSFORM

We now prove that the Laplace transform is a linear operator by showing that the principle of superposition holds, implying that if

$$x_1(t) \Longleftrightarrow X_1(s)$$

and $x_2(t) \Longleftrightarrow X_2(s)$

then

$$a_1x_1(t) + a_2x_2(t) \Longleftrightarrow a_1X_1(s) + a_2X_2(s)$$

The proof is simple. By definition,

$$\mathcal{L}[a_1x_1(t) + a_2x_2(t)] = \int_{-\infty}^{\infty} [a_1x_1(t) + a_2x_2(t)]e^{-st} dt$$

\n

$$= a_1 \int_{-\infty}^{\infty} x_1(t)e^{-st} dt + a_2 \int_{-\infty}^{\infty} x_2(t)e^{-st} dt$$

\n

$$= a_1X_1(s) + a_2X_2(s)$$

\n(4.3)

This result can be extended to any finite sum.

THE REGION OF CONVERGENCE (ROC)

The region of convergence (ROC), also called the region of existence, for the Laplace transform, X(s), is the set of values of s (the region in the complex plane) for which the integral in Eq. (4.1) converges. This concept will become clear in the following example.

EXAMPLE 4.1 Laplace Transform and ROC of a Causal Exponential

For a signal x(t) = e−atu(t), find the Laplace transform X(s) and its ROC.

By definition,

$$X(s) = \int_{-\infty}^{\infty} e^{-at} u(t) e^{-st} dt$$

Because u(t) = 0 for t < 0 and u(t) = 1 for t ≥ 0,

$$X(s) = \int_0^\infty e^{-at} e^{-st} dt = \int_0^\infty e^{-(s+a)t} dt = -\frac{1}{s+a} e^{-(s+a)t} \Big|_0^\infty \tag{4.4}$$

332 CHAPTER 4 CONTINUOUS-TIME SYSTEM ANALYSIS

Note that s is complex and as t → ∞, the term e−(s+a)^t does not necessarily vanish. Here we recall that for a complex number z = α +jβ,

$$e^{-zt} = e^{-(\alpha+j\beta)t} = e^{-\alpha t}e^{-j\beta t}$$

Now |e−jβ*^t* | = 1 regardless of the value of βt. Therefore, as t → ∞, e−zt → 0 only if α > 0, and e−zt → ∞ if α < 0. Thus,

$$\lim_{t \to \infty} e^{-zt} = \begin{cases} 0 & \text{Re } z > 0 \\ \infty & \text{Re } z < 0 \end{cases}$$

(4.5)

Clearly,

$$\lim_{t \to \infty} e^{-(s+a)t} = \begin{cases} 0 & \text{Re}(s+a) > 0\\ \infty & \text{Re}(s+a) < 0 \end{cases}$$

Use of this result in Eq. (4.4) yields

$$X(s) = \frac{1}{s+a} \qquad \text{Re}(s+a) > 0$$ $$e^{-at}u(t) \Longleftrightarrow \frac{1}{s+a} \qquad \text{Re } s > -a \tag{4.6}$$

or

The ROC of

$$X(s)$$

is Re $s > -a$ , as shown in the shaded area in Fig. 4.1a. This fact means that the integral defining $X(s)$ in Eq. (4.4) exists only for the values of $s$ in the shaded region in Fig. 4.1a. For other values of $s$ , the integral in Eq. (4.4) does not converge. For this reason, the shaded region is called the ROC (or the region of existence) for $X(s)$ .

Figure 4.1 Signals (a) e−atu(t) and (b) −e−atu(−t) have the same Laplace transform but different regions of convergence.

REGION OF CONVERGENCE FOR FINITE-DURATION SIGNALS

A finite-duration signal xf(t) is a signal that is nonzero only for t¹ ≤ t ≤ t2, where both t¹ and t² are finite numbers and t² > t1. For a finite-duration, absolutely integrable signal, the ROC is the entire s plane. This is clear from the fact that if xf(t) is absolutely integrable and a finite-duration signal, then x(t)e−σ*^t* is also absolutely integrable for any value of σ because the integration is over the finite range of t only. Hence, the Laplace transform of such a signal converges for every value of s. This means that the ROC of a general signal x(t) remains unaffected by the addition of any absolutely integrable, finite-duration signal xf(t) to x(t). In other words, if R represents the ROC of a signal x(t), then the ROC of a signal x(t)+xf(t) is also R.

ROLE OF THE REGION OF CONVERGENCE

The ROC is required for evaluating the inverse Laplace transform x(t) from X(s), as defined by Eq. (4.2). The operation of finding the inverse transform requires an integration in the complex plane, which needs some explanation. The path of integration is along c+jω, with ω varying from −∞ to ∞. † Moreover, the path of integration must lie in the ROC (or existence) for X(s). For the signal e−atu(t), this is possible if c > −a. One possible path of integration is shown (dotted) in Fig. 4.1a. Thus, to obtain x(t) from X(s), the integration in Eq. (4.2) is performed along this path. When we integrate [1/(s + a)]est along this path, the result is e−atu(t). Such integration in the complex plane requires a background in the theory of functions of complex variables. We can avoid this integration by compiling a table of Laplace transforms (Table 4.1), where the Laplace transform pairs are tabulated for a variety of signals. To find the inverse Laplace transform of, say, 1/(s + a), instead of using the complex integral of Eq. (4.2), we look up the table and find the inverse Laplace transform to be e−atu(t) (assuming that the ROC is Re s > −a). Although the table given here is rather short, it comprises the functions of most practical interest. A more comprehensive table appears in Doetsch [2].

THE UNILATERAL LAPLACE TRANSFORM

To understand the need for defining unilateral transform, let us find the Laplace transform of signal x(t) illustrated in Fig. 4.1b:

$$x(t) = -e^{-at}u(-t)$$

The Laplace transform of this signal is

$$X(s) = \int_{-\infty}^{\infty} -e^{-at}u(-t)e^{-st}dt$$

Because u(−t) = 1 for t < 0 and u(−t) = 0 for t > 0,

$$X(s) = \int_{-\infty}^{0} -e^{-at}e^{-st} dt = -\int_{-\infty}^{0} e^{-(s+a)t} dt = \frac{1}{s+a}e^{-(s+a)t} \Big|_{-\infty}^{0}$$

^† The discussion about the path of convergence is rather complicated, requiring the concepts of contour integration and understanding of the theory of complex variables. For this reason, the discussion here is somewhat simplified.

No.	x(t)	X(s)
1	δ(t)	1
2	u(t)	1 s
3	tu(t)	1 s2
4	nu(t) t	n! sn+1
5	eλt u(t)	1 s−λ
6	teλt u(t)	1 (s−λ)2
7	neλt t u(t)	n! (s−λ)n+1
8a	cos bt u(t)	s s2 +b2
8b	sin bt u(t)	b s2 +b2
9a	e−at cos bt u(t)	s+a (s+a)2 +b2
9b	e−atsin bt u(t)	b (s+a)2 +b2
10a	re−at cos(bt +θ )u(t)	(r cos θ )s +(ar cos θ −brsin θ ) s2 +2as +(a2 +b2)
10b	re−at cos(bt +θ )u(t)	0.5rejθ 0.5re−jθ s+a−jb + s +a+jb
10c	re−at cos(bt +θ )u(t)	As+B s2 +2as+c
	A2c+B2 −2ABa r = c−a2
	Aa−B θ = tan−1 √ c−a2 A
	b = √ c−a2
10d	e−at ! B−Aa Acos bt + sin bt u(t) b	As+B s2 +2as+c
	b = √ c−a2

TABLE 4.1 Select (Unilateral) Laplace Transform Pairs

Equation (4.5) shows that

lim ^t→−∞e−(s+a)^{t =} 0 Re (s+a) < ⁰

Hence,

$$X(s) = \frac{1}{s+a} \qquad \text{Re } s < -a$$

The signal −e−atu(−t) and its ROC (Re s < −a) are depicted in Fig. 4.1b. Note that the Laplace transforms for the signals e−atu(t) and −e−atu(−t) are identical except for their regions of convergence. Therefore, for a given X(s), there may be more than one inverse transform, depending on the ROC. In other words, unless the ROC is specified, there is no one-to-one correspondence between X(s) and x(t). This fact increases the complexity in using the Laplace transform. The complexity is the result of trying to handle causal as well as noncausal signals. If we restrict all our signals to the causal type, such an ambiguity does not arise. There is only one inverse transform of X(s) = 1/(s + a), namely, e−atu(t). To find x(t) from X(s), we need not even specify the ROC. In summary, if all signals are restricted to the causal type, then, for a given X(s), there is only one inverse transform x(t). †

The unilateral Laplace transform is a special case of the bilateral Laplace transform in which all signals are restricted to being causal; consequently, the limits of integration for the integral in Eq. (4.1) can be taken from 0 to ∞. Therefore, the unilateral Laplace transform X(s) of a signal x(t) is defined as

$$X(s) = \int_{0^{-}}^{\infty} x(t)e^{-st} dt$$

(4.7)

We choose 0⁻ (rather than 0⁺ used in some texts) as the lower limit of integration. This convention not only ensures inclusion of an impulse function at t = 0, but also allows us to use initial conditions at 0⁻ (rather than at 0+) in the solution of differential equations via the Laplace transform. In practice, we are likely to know the initial conditions before the input is applied (at 0−), not after the input is applied (at 0+). Indeed, the very meaning of the term “initial conditions” implies conditions at t = 0⁻ (conditions before the input is applied). Detailed analysis of desirability of using t = 0⁻ appears in Sec. 4.3.

The unilateral Laplace transform simplifies the system analysis problem considerably because of its uniqueness property, which says that for a given X(s), there is a unique inverse transform. But there is a price for this simplification: we cannot analyze noncausal systems or use noncausal inputs. However, in most practical problems, this restriction is of little consequence. For this reason, we shall first consider the unilateral Laplace transform and its application to system analysis. (The bilateral Laplace transform is discussed later, in Sec. 4.11.)

Basically there is no difference between the unilateral and the bilateral Laplace transform. The unilateral transform is the bilateral transform that deals with a subclass of signals starting at t = 0 (causal signals). Therefore, the expression [Eq. (4.2)] for the inverse Laplace transform remains unchanged. In practice, the term Laplace transform means the unilateral Laplace transform.

^† Actually, ^X(s) specifies ^x(t) within a null function ⁿ(t), which has the property that the area under ^|n(t)^| 2 is zero over any finite interval 0 to t (t > 0) (Lerch’s theorem). For example, if two functions are identical everywhere except at finite number of points, they differ by a null function.

EXISTENCE OF THE LAPLACE TRANSFORM

The variable s in the Laplace transform is complex in general, and it can be expressed as s =σ +jω. By definition,

$$X(s) = \int_{0^-}^{\infty} x(t)e^{-st} dt = \int_{0^-}^{\infty} [x(t)e^{-\sigma t}]e^{-j\omega t} dt$$

Because |ejω*^t* | = 1, the integral on the right-hand side of this equation converges if

$$\int_{0^{-}}^{\infty} \left| x(t)e^{-\sigma t} \right| dt < \infty \tag{4.8}$$

Hence the existence of the Laplace transform is guaranteed if the integral in Eq. (4.8) is finite for some value of σ. Any signal that grows no faster than an exponential signal Meσ0*^t* for some M and σ⁰ satisfies the condition of Eq. (4.8). Thus, if for some M and σ0,

$$|x(t)| \le Me^{\sigma_0 t} \tag{4.9}$$

we can choose σ>σ⁰ to satisfy Eq. (4.8).† The signal et 2 , in contrast, grows at a rate faster than e^σ0*^t* , and consequently is not Laplace–transformable.‡ Fortunately such signals (which are not Laplace–transformable) are of little consequence from either a practical or a theoretical viewpoint. If σ⁰ is the smallest value of σ for which the integral in Eq. (4.8) is finite, σ⁰ is called the abscissa of convergence and the ROC of X(s) is Re s > σ0. The abscissa of convergence for e−atu(t) is −a (the ROC is Re s > −a).

EXAMPLE 4.2 Bilateral Laplace Transform of Common Causal Signals

Determine the Laplace transform of the following: (a) δ(t), (b) u(t), and (c) cos ω0t u(t).