Signals and Systems 2Ed Haykin Solutions Manual PDF

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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 p...

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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

l.43

π 2 y ( t ) =  3 cos  200t + ---    6  2 π = 9 cos  200t + ---  6

9 π = --- cos  400t + --- 1  2 3 9 (a) DC component = --2 9 π (b) Sinusoidal component = --- cos  400t + ---  2 3 9 Amplitude = --2

1

200 Fundamental frequency = --------- Hz π

1.44

The RMS value of sinusoidal x(t) is A ⁄ 2 . Hence, the average power of x(t) in a 1-ohm 2

resistor is ( A ⁄ 2 ) = A2/2.

1.45

Let N denote the fundamental period of x[N]. which is defined by 2π N = -----Ω The average power of x[n] is therefore N -1

1 2 P = ---- ∑ x [ n ] N 1 = ---N

n=0 N -1

∑A

n=0 2 N -1

A = -----N

1.46

2

2 2πn cos  ---------- + φ  N 

∑ cos

n=0

2

 2πn ---------- + φ  N 

The energy of the raised cosine pulse is E =

π⁄ω

1

∫–π ⁄ ω --4- ( cos ( ωt ) + 1 )

2

dt

1 π⁄ω 2 = --- ∫ ( cos ( ωt ) + 2 cos ( ωt ) + 1 ) dt 2 0 1 π ⁄ ω 1 1 --- cos ( 2ωt ) + --- + 2 cos ( ωt ) + 1 dt = --- ∫  2 0 2 2 1 3 π = ---  ---  ---- = 3π ⁄ 4ω 2  2  ω

1.47

The signal x(t) is even; its total energy is therefore 5 2

E = 2 ∫ x ( t ) dt 0

2

4

5

= 2 ∫ ( 1 ) dt + 2 ∫ ( 5 – t ) dt 2

0

2

4

5

1 4 3 = 2 [ t ] t=0 + 2 – --- ( 5 – t ) 3

t=4

2 26 = 8 + --- = -----3 3

1.48

(a) The differentiator output is  1  y ( t ) =  –1   0

for – 5 < t < – 4 for 4 < t < 5 otherwise

(b) The energy of y(t) is E =

–4

∫–5

5

( 1 ) dt + ∫ ( – 1 ) dt 2

2

4

= 1+1 = 2

1.49

The output of the integrator is t

y ( t ) = A ∫ τ dτ = At

for 0 ≤ t ≤ T

0

Hence the energy of y(t) is E =

1.50

T

∫0

2 3

A T A t dt = ------------3 2 2

(a) x(5t) 1.0

-1

-0.8

(b)

0

0.8

1

t

25

t

x(0.2t) 1.0

-25

-20

0

20

3

1.51 x(10t - 5)

1.0

0

1.52

0.1

0.5

0.9

1.0

t

(a) x(t) 1 -1

1

2

t

3

-1 y(t - 1)

-1

1

2

t

3

-1 x(t)y(t - 1) 1 1 -1

2

t

3

-1

4

1.52

(b) x(t + 1) x(t - 1) 1

1 -1

1

2

3

4

t

-1 y(-t)y(-t) 1 -2

-1

1

2

3

4

3

4

t

-1 x(t - 1)y(-t) 1 t -2

-1

1

2

-1

1.52

(c)

-2 -1

1

2

3

3

4

3

4

t

-1

-2

-1 1

2

t

x(t + 1)y(t - 2)

-2

-1

1

2

5

t

1.52

(d) x(t) 1

-3

-2

-1

1

2

t

3

-1

y(1/2t + 1) 6

-5

-4 -3

-2

-1 1

2

4

6

t

-1.0

x(t - 1)y(-t) 1 t -3

1.52

-2

-1 -1

1

2

3

(e) x(t) 1 -4

-3

-2

-1 1

2

3

t

-1

y(2 - t)

1 -4

-3

-2

2

3

t

-1

x(t)y(2 - t) -1 1

2

3

-1

6

t

1.52

(f) x(t) 1

-2

-1

1

t

2

-1

y(t/2 + 1) 1.0

-5

-3

-2

-1

-6

1

1

2

t

3

-1.0

x(2t)y(1/2t + 1) +1 -0.5 1

-1

2

t

-1

1.52

(g) x(4 - t) 1 -7

-6

-5 -4

-3

t

-2 -1

y(t)

-2

-1

1

2

4

t

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

-3

-2

-1

1

2

3

7

t

1.53

We may represent x(t) as the superposition of 4 rectangular pulses as follows: g1(t) 1 1

2

11

2

3

4

t

g2(t) 1 3

4

t

g3(t) 1 1

2

3

4

t

g4(t) 1 1

2

3

4

t

0

To generate g1(t) from the prescribed g(t), we let g 1 ( t ) = g ( at – b ) where a and b are to be determined. The width of pulse g(t) is 2, whereas the width of pulse g1(t) is 4. We therefore need to expand g(t) by a factor of 2, which, in turn, requires that we choose 1 a = --2 The mid-point of g(t) is at t = 0, whereas the mid-point of g1(t) is at t = 2. Hence, we must choose b to satisfy the condition a(2) – b = 0 or 1 b = 2a = 2  --- = 1  2 1 Hence, g 1 ( t ) = g  --- t – 1 2  Proceeding in a similar manner, we find that 2 5 g 2 ( t ) = g  --- t – ---  3 3 g3 ( t ) = g ( t – 3 ) g 4 ( t ) = g ( 2t – 7 ) Accordingly, we may express the staircase signal x(t) in terms of the rectangular pulse g(t) as follows:

8

2 5 1 x ( t ) = g  --- t – 1 + g  --- t – --- + g ( t – 3 ) + g ( 2t – 7 )  3 3 2 

1.54

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

0

1

t

2

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

0

1

2

-1

t

3 -1

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

-3

2

3

t

0 -1

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

-1

0

1

2

t

3

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

-3

-2

-1

0

1

t

2

9

1.55

We may generate x(t) as the superposition of 3 rectangular pulses as follows: g1(t) 1 -4

-2

0

2

4

2

4

t

g2(t) 1 -4

-2

0

t

g3(t) 1 -4

-2

0

2

4

t

All three pulses, g1(t), g2(t), and g3(t), are symmetrically positioned around the origin: 1. g1(t) is exactly the same as g(t). 2. g2(t) is an expanded version of g(t) by a factor of 3. 3. g3(t) is an expanded version of g(t) by a factor of 4. Hence, it follows that g1 ( t ) = g ( t ) 1 g 2 ( t ) = g  --- t 3  1 g 3 ( t ) = g  --- t 4  That is, 1 1 x ( t ) = g ( t ) + g  --- t + g  --- t 3  4  1.56

(a) x[2n]

o 2

o o 0

-1

1

n

(b) x[3n - 1] o

2 o1 o -1

0

1

10

n

1.56

(c) y[1 - n] o1

o

o

o

o

-4

-3

-2

-1 0 -1

o 1

2

3

4

5

o

o

o

o

2

3

4

5

o

o

o

o

n

(d) y[2 - 2n] o

o

-3

-2

o

o1 o 1

-1 -1

n

(e) x[n - 2] + y[n + 2] o

o

4 o

o3 o

2 o o

1 -7 o

-6 o

-5 o

-4 o

-3

o -2

-1

0

1

2

3

4

o

o

o

5

6

7

8

o

o

o

5

6

7

n

o

(f) x[2n] + y[n - 4] o -5

-4

-3

1 o

-2

2

3

-1 o

o

o

o

o -1

11

o

o

o 4

n

1.56

(g) x[n + 2]y[n - 2]

o

-5

-4

-3

-2 o

-1

1

o

n

1

o

o

o

o

o2 3 o

o

(h) x[3 - n]y[-n] 3

o o

2

o -2

o

o

1 o -3

o

o -1

1

2

o 7

o 8

n

3

4

5

6

3

4 o

5 o

6 o

n

4 o

5 o

6 o

n

(i) x[-n] y[-n] o

3 o

2 o

o -5

o -4

-3

-2

-1

1 1

o

2

-1 o -2

o

-3

o

(j) x[n]y[-2-n] o

3 2 1

o -6

o -5

o -4

-2 o

-1

1

o

2

3

-3 o

-1 o -2 -3

12

o o

1.56

(k) x[n + 2]y[6-n] 3

o

2 o 1 -8 o

-7 o

-6 o

-5

-4

-3 o

o o

1.57

o -2

-1

1

o 2

o 3

o 4

o 5

o 6

n

-1 -2 -3

(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 = 2π seconds (h) Nonperiodic (i) Periodic Fundamental period = 15 samples

1.58

The fundamental period of the sinusoidal signal x[n] is N = 10. Hence the angular frequency of x[n] is 2πm m: integer Ω = ----------N The smallest value of Ω is attained with m = 1. Hence, 2π π Ω = ------ = --- radians/cycle 10 5

13

1.59

The amplitude of complex signal x(t) is defined by 2

2

xR( t ) + xI ( t ) =

2

2

2

2

A cos ( ωt + φ ) + A sin ( ωt + φ ) 2

2

= A cos ( ωt + φ ) + sin ( ωt + φ ) = A 1.60

Real part of x(t) is αt

Re { x ( t ) } = Ae cos ( ωt ) Imaginary part of x(t) is αt

Im { x ( t ) } = Ae sin ( ωt )

1.61

We are given     x(t ) =     

t ∆ ∆ --- for – --- ≤ t ≤ --∆ 2 2 ∆ 1 for t ≥ --2 ∆ 2 for t < – --2

The waveform of x(t) is as follows x(t)

1 1 2 -∆/2 ∆/2

0

- 12

14

t

The output of a differentiator in response to x(t) has the corresponding waveform: y(t) 1 δ(t - 1 ) 2 2

1/∆

t

∆/2

0

-∆/2

1 δ(t + ∆ ) 2 2

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 at t = ∆/2. 3. An impulse of strength -1/2 at t = -∆/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: d lim y ( t ) = lim ----- x ( t ) ∆→0 ∆ → 0 dt = δ(t )

1.62

We are given a triangular pulse of total duration ∆ and unit area, which is symmetrical about the origin: x(t) 2/∆ slope = -4/∆2

slope = 4/∆2

area = 1

-∆/2

∆/2

0

15

t

(a) Applying x(t) to a differentiator, we get an output y(t) depicted as follows: y(t) 4/∆2

area = 2/∆

∆/2

t

-∆/2 area = 2/∆ -4/∆2

(b) As the triangular pulse duration ∆ approaches zero, the differentiator output approaches the combination of two impulse functions described as follows: • An impulse of positive infinite strength at t = 0-. • An impulse of negative infinite strength at t = 0+. (c) The total area under the differentiator output y(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 -∞. 1.63

From Fig. P.1.63 we observe the following: x1 ( t ) = x2 ( t ) = x3 ( t ) = x ( t ) x4 ( t ) = y3 ( t ) Hence, we may write y 1 ( t ) = x ( t )x ( t – 1 )

(1)

y2 ( t ) = x ( t )

(2)

y 4 ( t ) = cos ( y 3 ( t ) ) = cos ( 1 + 2x ( t ) )

(3)

The overall system output is y ( t ) = y1 ( t ) + y2 ( t ) – y4 ( t )

(4)

Substituting Eqs. (1) to (3) into (4): y ( t ) = x ( t )x ( t – 1 ) + x ( t ) – cos ( 1 + 2x ( t ) )

(5)

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

16

1.64 (a) (b) (c) (d) (e) (f) (g) (h) (i) (j) (k) (l)

1.65

Memoryless ✓ ✓ ✓ x x x ✓ x x ✓ ✓ ✓

Stable ✓ ✓ ✓ ✓ ✓ ✓ ✓ ✓ ✓ ✓ ✓ ✓

Causal ✓ ✓ ✓ ✓ x ✓ x ✓ x ✓ ✓ ✓

Linear x ✓ x ✓ ✓ ✓ x ✓ ✓ ✓ ✓ x

Time-invariant ✓ ✓ ✓ ✓ ✓ ✓ ✓ ✓ ✓ ✓ ✓ ✓

We are given y [ n ] = a0 x [ n ] + a1 x [ n – 1 ] + a2 x [ n – 2 ] + a3 x [ n – 3 ] Let k

S { x(n)} = x(n – k ) We may then rewrite Eq. (1) in the equivalent form 1

2

3

y [ n ] = a0 x [ n ] + a1 S { x [ n ] } + a2 S { x [ n ] } + a3 S { x [ n ] } 1

2

3

= ( a0 + a1 S + a2 S + a3 S ) { x [ n ] } = H { x[n]} where 1

2

H = a0 + a1 S + a2 S + a3 S

.

3

.

(a) Cascade implementation of operator H: x[n]

a0

S

S

.

S

a2

a1

a3

Σ y[n]

17

(1)

(b) Parallel implementation of operator H:

x[n]

1.66

a0

.. .

S1

a1 Σ

S2

a2

S3

a3

y[n]

Using the given input-output relation: y [ n ] = a0 x [ n ] + a1 x [ n – 1 ] + a2 x [ n – 2 ] + a3 x [ n – 3 ] we may write y [ n ] = a0 x [ n ] + a1 x [ n – 1 ] + a2 x [ n – 2 ] + a3 x [ n – 3 ] ≤ a0 x [ n ] + a1 x [ n – 1 ] + a2 x [ n – 2 ] + a3 x [ n – 3 ] ≤ a0 M x + a1 M x + a2 M x + a3 M x = ( a 0 + a 1 + a 2 + a 3 )M x where M x = x ( n ) . 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:

.

x(n + k)

ak

.

x[n] Sk

x(n - l) Sl

a0

al

Σ y[n] l{x[n]}

= x[n - l], we have the input-output relation That is, with S y [ n ] = a0 x [ n ] + ak x [ n + k ] + al x [ n – l ] This system is noncausal by virtue of the term akx[n + k]. The system has memory by virtue of the term alx[n - l].

18

1.69

(a) The operator H relating the output y[n] to the input x[n] is 1

H = 1+S +S where

2

k

S { x[n]} = x[n – k ]

for integer k

(b) The inverse operator Hinvis correspondingly defined by inv 1 H = -------------------------1 2 1+S +S Cascade implementation of the operator H is described in Fig. 1. Correspondingly, feedback implementation of the inverse operator Hinvis described in Fig. 2 x[n]

.

.

S

S

Σ

Fig. 1 Operator H

y[n]

y[n] +

Σ

.

S

.

x[n]

S

Fig. 2 Inverse Operator Hinv Figure 2 follows directly from the relation: x[n] = y[n] – x[n – 1] – x[n – 2] 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 n

0 S H = HS where

n0

for integer n0

(1)

n

S 0 { x [ n ] } = x [ n – n0 ] and 1

2

H = 1+S +S We first note that n

n

1

n

n +1

2

S 0H = S 0(1 + S + S ) = S 0+S 0 Next we note that HS

n0

1

+S

2

= (1 + S + S )S

n0 + 2

(2)

n0

19

n

1+n

2+n

0 0 = S 0+S +S (3) From Eqs. (2) and (3) we immediately see that Eq. (1) is indeed satisfied. Hence, the system described in Problem ...