Signals and Systems 2Ed - Haykin - Solutions Manual

Signals and Systems 2Ed - Haykin - Solutions Manual

(Parte 1 de 2)

CHAPTER 1 1.1 to 1.41 - part of text

1.42 (a) Periodic: Fundamental period = 0.5s

(b) Nonperiodic

(c) Periodic Fundamental period = 3s

(d) Periodic Fundamental period = 2 samples

(e) Nonperiodic

(f) Periodic: Fundamental period = 10 samples

(g) Nonperiodic (h) Nonperiodic

(i) Periodic: Fundamental period = 1 sample


(a)DC component =

(b)Sinusoidal component = Amplitude =

Fundamental frequency =

1.44The RMS value of sinusoidalx(t) is. Hence, the average power ofx(t) in a 1-ohm resistor is =A2/2.

1.45LetN denote the fundamental period ofx[N]. which is defined by

The average power ofx[n] is therefore

1.46The energy of the raised cosine pulse is

1.47The signalx(t) is even; its total energy is therefore pi ---------Hz cos cos

1.48(a)The differentiator output is

(b)The energy ofy(t) is


1.49The output of the integrator is Hence the energy ofy(t) is

1.50 (a)

-1-0.8 0 0.8 1t
-25-20 0 20 25t
00.1 0.5 0.9 1.0

1.52 (a)

1.0 t

-12 3

y(t- 1)

t-11 2 3

1.52 (b)

1.52 (c) x(t- 1)

12 3 4-1
-2-1 1 2 3 4
-2-1 1 2 3 4
-11 2 3

-2 -1

12 3 4t
-2-1 1 2 3 4

1.52 (d)

1.52 (e)

y(1/2t+ 1)
-3-2 -1 1 2 3
1 2 4 6
-3-2 -1 1 2 3
6-5 -4 -3 -2 -1
12 3
-4-3 -2 -1
-4-3 -2 -1 1 2 3
12 3

1.52 (f)

-2-1 1 2
1 1 2 3

t t

-3 -2 -1.0
-7-6 -5 -4 -3 -2

x(4 -t)

-2-1 1 2 4

t t

-3-2 -1 1 2 3

x(4 -t)y(t) = 0

1.53We may representx(t) as the superposition of 4 rectangular pulses as follows:

To generateg1(t) from the prescribedg(t), we let

wherea andb are to be determined. The width of pulseg(t) is 2, whereas the width of pulseg1(t) is 4. We therefore need to expandg(t) by a factor of 2, which, in turn, requires that we choose

The mid-point of g(t)i sa t t = 0, whereas the mid-point of g1(t)i sa t t = 2. Hence, we must chooseb to satisfy the condition

Hence, Proceeding in a similar manner, we find that

Accordingly, we may express the staircase signal x(t) in terms of the rectangular pulse g(t) as follows:

12 3 4
12 3 4
12 3 4
12 3 4

1.54 (a) (b)


(d) (e)

01 2

x(t) =u(t) -u(t - 2)

01 2-1 3

x(t) =u(t + 1) - 2u(t) +u(t - 1)

12 3-3

x(t) =-u(t + 3) + 2u(t +1) -2u(t - 1) +u(t - 3) t x(t) =r(t + 1) -r(t) +r(t - 2)

-2-1 0 1 2 3

x(t) =r(t + 2) -r(t + 1) -r(t - 1)+r(t - 2) 1

-3 -2 -1 0 1 2

1.55We may generatex(t) as the superposition of 3 rectangular pulses as follows:

Hence, it follows that

That is, 1.56 (a)


-4-2 0 2 4
-4-2 0 2 4
-4- 2 0 2 4
-10 1

o o2 n

o o

-10 1

x[3n- 1]

1.56 (c)

(d) (e)

(f) o o o o o o

-4-3 -2 -1 01
23 4 5

y[1 -n] o o o o o

-3 -2 -1
123 4 5

y[2 - 2n] o oo o o o o o o o

-7-6 -5 -4 -3
-2-1 0 1 2 3 4 5 6 7 8

x[n - 2] +y[n + 2]

o o o o o o o o

-5-4 -3 -2 2 3
-14 5 6 7

o o 1 x[2n] +y[n - 4]

1.56 (g)

(h) (i)

(j) noo o o o o

-5-4 -3 -2 -1 1

x[n + 2]y[n - 2] o o o o o o o

-3-2 -1 1 2 3 4 5 6 7 8

o o o o o o

-5-4 -3 -2 -1
12 3 4 5 6

o o o o

-6-5 -4 -3
-2-1 1 2 3 4 5 6

1.56 (k)

1.57 (a) Periodic Fundamental period = 15 samples

(b) Periodic Fundamental period = 30 samples

(c) Nonperiodic

(d) Periodic Fundamental period = 2 samples

(e) Nonperiodic (f) Nonperiodic

(g) Periodic Fundamental period = 2pi seconds

(h) Nonperiodic

(i) Periodic Fundamental period = 15 samples

1.58 The fundamental period of the sinusoidal signal x[n]i s N = 10. Hence the angular frequency ofx[n] is m: integer

The smallest value of is attained withm = 1. Hence, radians/cycle o o o o o

12 3 4 5 6-8 -7 -6 -5 -4 -3

1.59The amplitude of complex signalx(t) is defined by

1.60Real part ofx(t) is Imaginary part ofx(t) is

1.61We are given

The waveform ofx(t) is as follows xt()

The output of a differentiator in response tox(t) has the corresponding waveform:

y(t) consists of the following components:

1. Rectangular pulse of duration ∆ and amplitude 1/∆ centred on the origin; the area under this pulse is unity. 2.An impulse of strength 1/2 att =∆/2. 3.An impulse of strength -1/2 att = -∆/2.

As the duration∆ is permitted to approach zero, the impulses (1/2)δ(t-∆/2) and -(1/2)δ(t+∆/2) coincide and therefore cancel each other. At the same time, the rectangular pulse of unit area (i.e., component 1) approaches a unit impulse at t = 0. We may thus state that in the limit:

1.62 We are given a triangular pulse of total duration ∆ and unit area, which is symmetrical about the origin:

12δ(t -)1
12δ(t +)∆


slope = -4/∆2
area = 1

(a)Applyingx(t) to a differentiator, we get an outputy(t) depicted as follows:

(b) As the triangular pulse duration ∆ approaches zero, the differentiator output approaches the combination of two impulse functions described as follows:

•An impulse of negative infinite strength att = 0+ .

(c)The total area under the differentiator outputy(t) is equal to (2/∆) + (-2/∆) = 0.

In light of the results presented in parts (a), (b), and (c) of this problem, we may now make the following statement: When the unit impulse δ(t) is differentiated with respect to time t, the resulting output consists of a pair of impulses located at t =0 - and t =0 +, whose respective strengths are +∞ and -∞.

1.63From Fig. P.1.63 we observe the following:

Hence, we may write (1)

The overall system output is (4)

Substituting Eqs. (1) to (3) into (4): (5)

Equation (5) describes the operator H that defines the output y(t) in terms of the input x(t).

area = 2/∆

1.65We are given (1)


We may then rewrite Eq. (1) in the equivalent form

where (a)Cascade implementation of operatorH:

(b)Parallel implementation of operatorH:

1.66Using the given input-output relation: we may write

where . Hence, provided that Mx is finite, the absolute value of the output will always be finite. This assumes that the coefficients a0, a1, a2, a3 have finite values of their own. It follows therefore that the system described by the operator H of Problem 1.65 is stable.

1.67 The memory of the discrete-time described in Problem 1.65 extends 3 time units into the past.

1.68 It is indeed possible for a noncausal system to possess memory. Consider, for example, the system illustrated below:

That is, withSl{x[n]} =x[n -l], we have the input-output relation

This system is noncausal by virtue of the term akx[n + k]. The system has memory by virtue of the termalx[n -l].

SlSk ak a0 al

1.69(a)The operatorH relating the outputy[n] to the inputx[n] is

for integerk

where (b)The inverse operatorHinvis correspondingly defined by

Cascade implementation of the operator H is described in Fig. 1. Correspondingly, feedback implementation of the inverse operatorHinvis described in Fig. 2

Figure 2 follows directly from the relation:

1.70 For the discrete-time system (i.e., the operator H) described in Problem 1.65 to be timeinvariant, the following relation must hold

for integern0(1)


and We first note that

(2) Next we note that

Fig. 1 Operator H

Fig. 2 Inverse Operator Hinv

From Eqs. (2) and (3) we immediately see that Eq. (1) is indeed satisfied. Hence, the system described in Problem 1.65 is time-invariant.

1.71(a)It is indeed possible for a time-variant system to be linear. (b) Consider, for example, the resistance-capacitance circuit where the resistive component is time variant, as described here:

This circuit, consisting of the series combination of the resistor R(t) and capacitor C,i s time variant because ofR(t).

Doubling the input v1(t) results in doubling the output v2(t). Hence, the property of homogeneity is satisfied.

Moreover, if


where ,k = 1,2,...,N

Hence, the property of superposition is also satisfied. We therefore conclude that the time-varying circuit of Fig. P1.71 is indeed linear.

1.72We are given thepth power law device: (1)


Hence the system described by Eq. (1) is nonlinear.

This system is both linear and time invariant. Consider another discrete-time system described by the operatorH2:

which is also both linear and time invariant. The system H1 is causal, but the second systemH2 is noncausal.

1.74 The system configuration shown in Fig. 1.56(a) is simpler than the system configuration shown in Fig. 1.56(b). They both involve the same number of multipliers and summer. however, Fig. 1.56(b) requires N replicas of the operator H, whereas Fig. 1.56(a) requires a single operatorH for its implementation.

1.75(a)All three systems •have memory because of an integrating action performed on the input,

•are causal because (in each case) the output does not appear before the input, and

• are time-invariant.

(b) H1 is noncausal because the output appears before the input. The input-output relation of H1 is representative of a differentiating action, which by itself is memoryless. However, the duration of the output is twice as long as that of the input. This suggests that H1 may consist of a differentiator in parallel with a storage device, followed by a combiner. On this basis,H1 may be viewed as a time-invariant system with memory.

System H2 is causal because the output does not appear before the input. The duration of the output is longer than that of the input. This suggests that H2 must have memory. It is time-invariant.

System H3 is noncausal because the output appears before the input. Part of the output, extending from t =- 1t o t = +1, is due to a differentiating action performed on the input; this action is memoryless. The rectangular pulse, appearing in the output from t =+ 1t o t = +3, may be due to a pulse generator that is triggered by the termination of the input. On this basis,H3 would have to be viewed as time-varying.

Finally, the output of H4 is exactly the same as the input, except for an attenuation by a factor of 1/2. Hence,H4 is a causal, memoryless, and time-invariant system.

1.76 H1 is representative of an integrator, and therefore has memory. It is causal because the output does not appear before the input. It is time-invariant.

H2 is noncausal because the output appears at t = 0, one time unit before the delayed input at t = +1. It has memory because of the integrating action performed on the input. But, how do we explain the constant level of +1 at the front end of the output, extending from t =0t o t = +1? Since the system is noncausal, and therefore operating in a non real-time fashion, this constant level of duration 1 time unit may be inserted into the output by artificial means. On this basis,H2 may be viewed as time-varying.

H3 is causal because the output does not appear before the input. It has memory because of the integrating action performed on the input from t =1t o t = 2. The constant level appearing at the back end of the output, from t =2t o t = 3, may be explained by the presence of a strong device connected in parallel with the integrator. On this basis, H3 is time-invariant.

Consider next the input x(t) depicted in Fig. P1.76b. This input may be decomposed into the sum of two rectangular pulses, as shown here:

01 2t 0 1 2t 0 1 2t
2 12
01 2t0 1 2t
01 2t

The rectangular pulse of unit amplitude and unit duration at the front end of y2(t)i s inserted in an off-line manner by artificial means

1.7(a)The response of the LTI discrete-time system to the inputδ[n-1] is as follows: (b)The response of the system to the input 2δ[n] -δ[n - 2] is as follows

-1 01
01 2 t0 1 2 3t 0 1 2 3t

o o o

-1 1 3

o o o o o o

-2-1 2 3 5 6

(c) The input given in Fig. P1.77b may be decomposed into the sum of 3 impulse functions: δ[n + 1], -δ[n], and 2δ[n - 1]. The response of the system to these three components is given in the following table:

Thus, the total responsey[n] of the system is as shown here:

Advanced Problems 1.78(a)The energy of the signal x(t) is defined by

Substituting into this formula yields

Timen δ[n + 1]-δ[n]2δ[n - 1]Total response

-1 +1 -2
+1 -1 +4

o o o o

-3-2 -1 1 32 4 5 6

Withxe(t) even andxo(t) odd, it follows that the productxe(t)xo(t) is odd, as shown by Hence,

Accordingly, Eq. (1) reduces to (b)For a discrete-time signalx[n],, we may similarly write

With it follows that xe t() xo t()() td xe t() xo t() tx e t() xo t() td

xe n[] xo n[] xe n[] xo n[] n=0

Accordingly, Eq. (2) reduces to

1.79(a)From Fig. P1.79, (1)

Eliminatingi2(t) between Eqs. (1) and (2):

Rearranging terms: (4)

(b) Comparing Eqs. (4) with Eq. (1.108) for the MEMS as presented in the text, we may derive the following analogy:

MEMS of Fig. 1.64LRC circuit of Fig. P1.79


1.80 (a) As the pulse duration ∆ approaches zero, the area under the pulse x∆(t) remains equal to unity, and the amplitude of the pulse approaches infinity.

(b) The limiting form of the pulse x∆(t) violates the even-function property of the unit impulse:

Substitutingt +T0 fort into Eq. (1) and then using Eq. (2), we may write

1.82(a)For, we have

Att =∆/2, we have

Sincex∆(t) is even, then


For this area to equal unity, we require (c)



∆ = 1 ∆ = 0.5

∆ = 0.25

∆ = 0.125 yt() x τ() τd

xt() tx τ() τd

The definite integral, representing theinitial condition, is a constant. With differentiation as the operation of interest, we may also write

Clearly, the value ofx(t) is unaffected by the value assumed by the initial condition

It would therefore be wrong to say that differentiation and integration are the inverse of each other. To illustrate the meaning of this statement, consider the two following two waveforms that differ from each other by a constant value for:

Yet,, as illustrated below:

(b)For Fig. P1.83(a):

ForR/L large, we approximately have

Equivalently, we have a differentiator described by , large xt() td xt() td


slope =aslope =a

yt() L R

For Fig. P1.83(b):

ForR/L small, we approximately have

Equivalently, we have an integrator described by small

(c)Consider the following two scenarios describing theLR circuits of Fig. P1.83 •The input x(t) consists of a voltage source with an average value equal to zero.

•The input x(t) includes a dc componentE (exemplified by a battery). These are two different input conditions. Yet for large R/L, the differentiator of Fig. P1.83(a) produces the same output. On the other hand, for small R/L the integrator of Fig. P1.83(b) produces different outputs. Clearly, on this basis it would be wrong to say that these twoLR circuits are the inverse of each other.

This input-output relation satisfies the following two conditions:

• Homogeneity: If the input x(t) is scaled by an arbitrary factor a, the output y(t) will be scaled by the same factor.

where ,k = 1,2

Hence, the system of Fig. P1.84 is linear.

(b)For the impulse input ,

Eq. (1) yields the corresponding output yt() L R

For, Eq. (1) yields

Recognizing that, the system of Fig. P1.84 is time-variant.

The output is nonlinear as the system violates both the homogeneity and superposition properties:

•Letx(t) be scaled by the factora. The corresponding value of the output is


(b)For the impulse input ,

Eq. (1) yields yt() 2pi f ctk x τ() τd ya t() 2pi f ctk a x τ() τd

For the delayed impulse input, Eq. (1) yields

Recognizing that, it follows that the system is time-variant.

1.86For the square-law device

, the input

yields the output

The outputy(t) contains the following components:

•DC component of amplitude

1.87The cubic-law device

, in response to the input,

, produces the output

The outputy(t) consists of two components:

•Sinusoidal component of frequencyω, amplitudeA3 and phaseφ •Sinusoidal component of frequency 3ω, amplitudeA3/2, and phase 3φ

To extract the component with frequency 3ω (i.e., the third harmonic), we need to use a band-pass filter centered on 3ω and a pass-band narrow enough to suppress the fundamental component of frequencyω.

From the analysis presented here we infer that, in order to generate the pth harmonic in response to a sinusoidal component of frequency ω, we require the use of two subsystems:

•Nonlinear device defined by

,p = 2,3,4,... •Narrowband filter centered on the frequencypω.

1.8(a)Following the solution to Example 1.21, we start with the pair of inputs:

The corresponding outputs are respectively given by

The response to the input is given by

As. We also note that

Hence, with , we find that the impulse response of the system is ωn t ∆ ωn t ∆ ωnt ωn∆ ωnt ωn∆ lim= lim=

e αt–

Since att = 0, Eq. (1) reduces to (2)

(b) Using Euler’s formula, we can write

The step response can therefore be rewritten as Again, the impulse response in this case can be obtained as

whereα1 =α -αn andα2 =α +αn. 1.89Building on the solution described in Fig. 1.69, we may relabel Fig. P1.89 as follows

(Parte 1 de 2)