# Continuous.and.Discrete.time.Signals.and.Systems.(Mandal.&.Asif).Solutions.chap13

July 17, 2017 | Author: Ricardo Gálmez | Category: Sine, Analysis, Mathematical Relations, Geometry, Applied Mathematics

#### Description

Chapter 13: The z-Transform Problem 13.1 (i) x1[k ] = 0.5 k +1 u[k + 5]

1 1 − 0.5 z −1

Z 0.5 k u[k ] ←⎯→

By definition,

Z Using the time shifting property, 0.5 k +5 u[k + 5] ←⎯→

Z 0.5 k +1 u[k + 5] ←⎯→

which implies that (ii) x 2 [k ] = (k + 2)0.5

ROC : z > 0.5

z5 1 − 0.5 z −1

0.5 −4 z 5 1 − 0.5 z −1

ROC : z > 0.5 , z ≠ ∞ .

k

1 1 − 0.5 z −1

Z 0.5 k u[k ] ←⎯→

By definition,

ROC : z > 0.5

Z − 0.5 −k u[−k − 1] = −2 k u[−k − 1] ←⎯→

and

ROC : z > 0.5 , z ≠ ∞

1 1 − 2 z −1

ROC : z < 2.

Applying the linearity property, Z 0.5 k u[k ] + 0.5 −k u[ −k − 1] ←⎯→

Z 0.5 ←⎯→ k

or,

1 1 − −1 1 − 0 .5 z 1 − 2 z −1

− 1.5 z −1

(1 − 0.5z ) (1 − 2 z ) −1

−1

ROC : [ z > 0.5] ∩ [ z < 2],

ROC : 0.5 < z < 2.

Applying the frequency differentiation property Z k 0.5 ←⎯→ −z k

d − 1.5 z −1 dz 1 − 0.5 z −1 1 − 2 z −1

(

)(

)

ROC : 0.5 < z < 2.

We have just proved that Z 2 × 0.5 ←⎯→ k

3z −1

(1 − 0.5z ) (1 − 2 z ) −1

−1

ROC : 0.5 < z < 2.

Using the linearity property Z (k + 2)0.5 ←⎯→ k

3 z −1

(1 − 0.5z ) (1 − 2 z ) −1

−1

+z

d 1.5 z −1 dz 1 − 0.5 z −1 1 − 2 z −1

(

)(

)

ROC : 0.5 < z < 2.

2

Chapter 13

(iii) x3 [k ] = k + 2 × 0.5 Consider

k +2

k 0 .5

⎧ k × 0.5 k =⎨ −k ⎩ − k × 0 .5

k

k ≥0

= k × 0.5 k U [k ] − k × 2 k U [− k − 1] .

k 0.5

(1 − 0.5z )

−1 2

0 . 5 z −1

2 z −1

ROC : z < 2

(1 − 2 z )

−1 2

2 z −1

+

ROC : 0.5 < z < 2 .

(1 − 0.5 z ) (1 − 2 z ) −1 2

−1 2

Using the time shifting property

k + 2 0.5

k +2

Z ←⎯→

0 .5 z

2z

+

(1 − 0.5z ) (1 − 2 z ) −1 2

−1 2

ROC : 0.5 < z < 2

(iv) x 4 [k ] = 3 k +1 cos( π3 k − π4 )u[−k + 5] Expressing

x4 [k ] = 3k +1 cos( π3 k ) cos( π4 ) u[−k + 5] − 3 k +1 sin( π3 k ) sin( π4 ) u[ −k + 5] x4 [k ] =

or, x4 [k ] =

or,

x4 [k ] =

or,

3 2 2

3k +1 2

[cos( π3 k ) − sin( π3 k )]u[−k + 5]

3k +1 ⎡ e jπk / 3 + e − jπk / 3 e jπk / 3 − e − jπk / 3 ⎤ − ⎢ ⎥ u[−k + 5] 2 2j 2 ⎣ ⎦

(1 + j1)(3e jπ / 3 )

k

u[ −k + 5] +

Z − α k u[ −k − 1] ←⎯→

We know that

Z − α k u[−k + 5] ←⎯→

or, Hence, X 4 ( z ) =

37 2 2

(1 − j1)(3e − jπ / 3 )

z −6 1 − αz −1

( z / α) −6 1 − αz −1

k

u[−k + 5] .

ROC : z < α ROC : z < α

ROC : z < α .

) 37 ( ze jπ / 3 ) −6 ( ) + 1 − j 1 1 − 3e jπ / 3 z −1 2 2 1 − 3e − jπ / 3 z −1

(1 + j1) ( ze

− jπ / 3 −6

2 2

1 1 − αz −1

Z − α k −6 u[−(k − 6) − 1] ←⎯→

or,

3

ROC : z < 3.

Problem 13.2

(i)

The sequence can be expressed as

x1[k ] = δ[k − 10] + δ[k − 11] + 2δ[ k − 12] + 2δ[k − 15],

Solutions

3

which has the z-transform X 1 ( z ) = z −10 + z −11 + 2 z −12 + 2 z −15 , ROC: entire z-plane except z = 0. (ii)

By definition, X 2 ( z) =

∑ ⎢⎣3

−k +2

k =−∞

4 ⎤ u[k ] + ∑ mδ [k − m]⎥ z − k m =1 ⎦

= ∑ 3− k + 2 z − k + k =0

∑ {δ [k − 1] + 2δ [k − 2] + 3δ [k − 3] + 4δ [k − 4]} z

−k

k =−∞

= 9∑ (3 z ) − k + z −1 + 2 z −2 + 3 z −3 + 4 z −4

z ≠0

= 1−(39z )−1 + z −1 + 2 z −2 + 3 z −3 + 4 z −4

z ≠ 0, z > 13

k =0

= 1−91 z −1 + z −1 + 2 z −2 + 3 z −3 + 4 z −4

ROC: z > 13

3

(iii) Expressing the sequence as x3 [k ] = sin( π5k + π3 )u[k ] = ⎡⎣sin( π5k ) cos( π3 ) + cos( π5k ) sin( π3 ) ⎤⎦ u[k ] = ⎡⎣ 12 sin( π5k ) +

3 2

cos( π5k ) ⎤⎦ u[k ]

the z-transform is given by

{⎡⎣

X 3 ( z) = = = = ≈

(iv)

1 2

sin( π5k ) +

3 2

1 2

sin( π5 ) z −1 + 1 − 2 cos( π5 ) z −1 + z −2

1 2

sin( π5 ) z −1 +

3 2

}

cos( π5k ) ⎤⎦ u[k ] =

{1 − cos(

3 2

π

5

1 2

{[sin( π )] u[k ]} + k 5

1 − cos( π5 ) z −1 1 − 2 cos( π5 ) z −1 + z −2

3 2

{[cos( π )] u[k ]} k 5

ROC: z > 1

) z −1 }

1 − 2 cos( π5 ) z −1 + z −2 3 2

+ 12 sin( π5 ) z −1 − 23 cos( π5 ) z −1 1 − 2 cos( π5 ) z −1 + z −2

0.866 − 0.407 z −1 1 − 1.618 z −1 + z −2

ROC: z > 1

We know that Z (0.5e jπ / 5 ) k u[k ] ←⎯ →

Z → (0.5e− jπ / 5 ) k u[k ] ←⎯

and

1 1 − 0.5e jπ / 5 z −1

1 1 − 0.5e − jπ / 5 z −1

ROC : z > 0.5

ROC : z > 0.5

Adding the two transform pairs after multiplying respectively with exp(jπ/3) and exp(−jπ/3), gives Z e jπ / 3 (0.5e jπ / 5 ) k u[k ] − e − jπ / 3 (0.5e − jπ / 5 ) k u[k ] ←⎯ →

or, (v)

Z 2 j × 0.5k sin( π5k + π3 )u[ k ] ←⎯ →

By definition

e jπ / 3 e − jπ / 3 − 1 − 0.5e jπ / 5 z −1 1 − 0.5e − jπ / 5 z −1

e jπ / 3 e − jπ / 3 − 1 − 0.5e jπ / 5 z −1 1 − 0.5e − jπ / 5 z −1

ROC : z > 0.5

ROC : z > 0.5 .

4

Chapter 13 ∞

k =0

k =0

X 5 ( z ) = ∑ ku[k ]z − k = ∑ kz − k = ( z −z1)2

ROC: z > 1

Problem 13.3

In parts (i)-(vii), the sequences are all causal (right hand sided), and hence the ROC is the outside of the circle with radius equal to the magnitude of the pole furthermost from the origin in the z-plane. z z − 0.9 z + 0.2

(i) X 1 ( z ) =

2

By partial fraction expansion, we obtain X1 ( z) =

z z 1 ⎤ z ⎞ ⎡ 1 ⎛ z = = 10 z ⎢ − = 10 ⎜ − ⎟ ⎥ z − 0.9 z + 0.2 ( z − 0.5)( z − 0.4) ⎣ z − 0.5 z − 0.4 ⎦ ⎝ z − 0.5 z − 0.4 ⎠ 2

(

Therefore, (ii) X 2 ( z ) =

)

(

)

x1[k ] = 10 0.5k u[k ] − 0.4k u[ k ] = 10 0.5k − 0.4k u[k ] .

z z − 2.1z + 0.2 2

By partial fraction expansion, X 2 ( z) =

z z 1 ⎤ 10 ⎛ z z ⎞ ⎡ 1 = = 1.91 z ⎢ − = 19 ⎜ − ⎟ ⎥ z − 2.1z + 0.2 ( z − 2)( z − 0.1) ⎣ z − 2 z − 0.1 ⎦ ⎝ z − 2 z − 0.1 ⎠ 2

(

(iii) X 3 ( z ) =

)

(

)

k k k k 10 . x2 [ k ] = 10 19 2 u[ k ] − 0.1 u[ k ] = 19 2 − 0.1 u[ k ]

Therefore,

z2 + 2 ( z − 0.3)( z + 0.4)( z − 0.7)

By partial fraction expansion, X 3 ( z) =

k3 z2 + 2 k1 k2 ≡ + + ( z − 0.3)( z + 0.4)( z − 0.7) z − 0.3 z + 0.4 z − 0.7

⎡ ⎤ z2 + 2 2.09 k1 = ⎢ = = −7.4643 ⎥ ( z 0.4)( z 0.7) 0.28 + − − ⎣ ⎦ z =0.3

where

⎡ ⎤ z2 + 2 2.16 = = 2.8052 k2 = ⎢ ⎥ ⎣ ( z − 0.3)( z − 0.7) ⎦ z =−0.4 0.77

⎡ ⎤ z2 + 2 2.49 k3 = ⎢ = = 5.6591 ⎥ ⎣ ( z − 0.3)( z + 0.4) ⎦ z =0.7 0.44

In other words, X 3 ( z) =

z2 + 2 7.4643 2.8052 5.6591 ⎡ 7.4643z 2.8052 z 5.6591z ⎤ =− + + = z −1 ⎢ − + + ( z − 0.3)( z + 0.4)( z − 0.7) z − 0.3 z + 0.4 z − 0.7 z + 0.4 z − 0.7 ⎥⎦ ⎣ z − 0.3 =P( z )

where Therefore,

p[k ] = z −1{P( z )} = [−7.4643 × 0.3k + 2.8052 × (−0.4) k + 5.6591× 0.7 k ]u[k ] .

Solutions

x3 [k ] = z −1{z −1 P ( z )} = p[k − 1] = ⎡⎣ −7.4643 × 0.3k −1 + 2.8052 × (−0.4) k −1 + 5.6591× 0.7 k −1 ⎤⎦ u[k − 1] = ⎡⎣ −24.881× 0.3k − 7.0130 × (−0.4) k + 8.0844 × 0.7 k ⎤⎦ u[k − 1]. (iv) X 4 ( z ) =

z2 + 2 ( z − 0.3)( z + 0.4) 2

By partial fraction expansion,

k3 X 4 ( z) k k2 k4 z2 + 2 = ≡ 1+ + + 2 z z ( z − 0.3)( z + 0.4) z z − 0.3 z + 0.4 ( z + 0.4) 2 where k1 =

lim ⎡ ⎤ 2 z2 + 2 × z⎥ = = −41.6667 , ⎢ 2 2 z → 0 ⎣ z ( z − 0.3)( z + 0.4) ⎦ (−0.3)(0.4)

k2 =

lim ⎡ lim ⎡ z 2 + 2 ⎤ ⎤ 2.09 z2 + 2 ( 0 . 3 ) × z − = = = 14.2177 ⎢ ⎥ ⎢ 2 2⎥ 2 z → 0.3 ⎣ z ( z − 0.3)( z + 0.4) ⎦ z → 0.3 ⎣ z ( z + 0.4) ⎦ (0.3)(0.7)

⎡d lim ⎡ d z 2 + 2 ⎤ z2 + 2 2⎤ z × + = ( 0 . 4 ) ⎢ ⎥ ⎢ ⎥ z → −0.4 ⎣ dz z ( z − 0.3)( z + 0.4) 2 ⎦ z → −0.4 ⎣ dz z ( z − 0.3) ⎦ lim ⎡ z2 + 2 z2 + 2 ⎤ 2z = − − 2 ⎢ ⎥ = 27.4490 z → −0.4 ⎣ z ( z − 0.3) z ( z − 0.3) z ( z − 0.3) 2 ⎦ lim

k3 =

k4 =

⎡ lim ⎡ z 2 + 2 ⎤ z2 + 2 2.16 2⎤ ( 0 . 4 ) × z + = = 7.7143 ⎢ ⎥ ⎢ ⎥= 2 z → −0.4 ⎣ z ( z − 0.3)( z + 0.4) ⎦ z → −0.4 ⎣ z ( z − 0.3) ⎦ (−0.4)(−0.7) lim

k3 k2 k4 z −1 X 4 ( z ) ≡ k1 + + + 1 − 0.3 z −1 1 + 0.4 z −1 (1 + 0.4 z −1 ) 2

Hence,

Assuming right hand sequences and taking the inverse z-transform, we get

x4 [k ] = −41.6667δ [k ] + 14.2177 × 0.3k u[ k ] + 27.4490 × (−0.4) k u[k ] − 19.2858k (−0.4)k u[k ] = −41.6667δ [k ] + ⎡⎣14.2177 × 0.3k − (19.2858k − 27.4490 ) × (−0.4) k ⎤⎦ u[k ] (v) X 5 ( z ) =

4 z −1 4z = 2 −1 −2 z − 5z + 6 1 − 5z + 6 z

By partial fraction expansion,

X 5 ( z) k k 4 = ≡ 1 + 2 z ( z − 2)( z − 3) z − 2 z − 3 where

k1 =

lim ⎡ ⎤ 4 4 × ( z − 2)⎥ = = −4 , ⎢ z → 2 ⎣ ( z − 2)( z − 3) ( − 1) ⎦

and

k2 =

lim ⎡ ⎤ 4 × ( z − 3) ⎥ = 4 . ⎢ z → 3 ⎣ ( z − 2)( z − 3) ⎦

5

6

Chapter 13

X 5 ( z) ≡

Hence,

−4 4 + −1 1 − 2z 1 − 3z −1

Assuming right hand sequences and taking the inverse z-transform, we get

x5 [k ] ≡ −4 × 2k u[k ] + 4 × 3k u[k ] = 4 ( 3k − 2k ) u[k ] .

4 z −2 4 z −2 2 z 2 = = − z −2 × 2 = − z −2 × P ( z ) (vi) X 6 ( z ) = −1 −1 10 − 6( z + z ) −(6 z − 10 + 6 z ) 3 z − 5z / 3 + 1 3 Applying the linearity and time shifting property of the z-transform, x6 [k ] can be expressed as

x6 [k ] = − 23 p[k − 2] where

{ p[k ]} = P( z ) =

z z − 5z / 3 + 1 2

In order to calculate p[ k ] , the transform pair in the Entry 12 of Table 13.1 can be used.

z Az + B . = 2 z − 5 z / 3 + 1 z + 2γ z + α 2

P( z ) =

2

A= 0, B =1,γ =−5/ 6,α =1

k The function p[ k ] can be expressed as, p[k ] = r ⋅ α sin ( Ω 0 k + θ ) ⋅ u[k ] where

r=

A2α 2 + B 2 − 2 ABγ

α 2 −γ 2

θ = tan −1

(

A α 2 −γ 2 B − Aγ

=

1 1− 25 / 36

) = tan

−1

=

1 11/ 36

≈ 1.809 , Ω0 = cos −1 ( −αγ ) = cos −1 ( 56 ) = 0.5857 , and

( 0) = 0 .

In other words, p[k ] = 1.809sin ( 0.5857 k ) u[ k ] , and the function x6 [k ] can be expressed as

p[k ] = − 23 × 1.809sin ( 0.5857(k − 2) ) u[k − 2]

= 1.206sin ( 0.5857 k − 67.11630 ) u[k − 2] ≡ −1.11108δ [k ] − 0.6667δ [k − 1] − 1.206sin ( 0.5857 k − 67.11630 ) u[ k ]

(vii) X 7 ( z ) =

2 z −2

(1 − 4 z ) (1 − 2 z ) −1

2

−1

=

2z ( z − 4) 2 ( z − 2)

By partial fraction expansion, X 7 ( z) 2 0 .5 1 0.5 = ≡− + + 2 2 z z − 4 ( z − 4) z−2 ( z − 4) ( z − 2)

Assuming right hand sequences and taking the inverse z-transform, we get

x7 [k ] = −0.5 × 4k u[k ] + 0.25k × 4k u[k ] + 0.5 × 2k u[k ] = ⎡⎣(0.25k − 0.5) × 4k + 0.5 × 2k ⎤⎦ u[k ]. The first 10 sample values of each x[k]’s are shown in MATLAB Program 13.3.

Solutions

Program 13.3: MATLAB program for Problem 13.3. % MATLAB Code to calculate x[k]’s in Problem 13.3. This code % generates first 10 samples of each time-domain function. % k=0:9 ; % % part (i) x1=10*((0.5).^k-(0.4).^k) ; % x1 = [0, 1, 0.9, 0.61, 0.369, 0.2101, 0.11529, 0.0617, 0.0325, 0.0169] % part (ii) x2=(10/19)*(2.^k-(0.1).^k) ; % x2 = [0 1.00 2.10 4.2100 8.4210 16.8421 269.4737]

33.6842

%part (iii) k=[0:10] ; p1 = (0.3).^k ; p2 = (-0.4).^k; p3 = (0.7).^k; x3 = (-24.881*p1-7.0130*p2+8.0844*p3).*(k>=1) ; % x3 = [0 1.0 0.60 2.55 1.56 1.3701 0.9043 0.2275]

67.3684

0.6718

134.7368

0.4598

0.3276

%part (iv) k=[0:10] ; x4 = 14.2177*((0.3).^k) - (19.2858*k-27.4490).*((-0.4).^k); x4(1) = x4(1)-41.6667; %x4 = [0, 1, -0.5, 2.33, -1.157, 0.741 -0.351, 0.179, -0.0822, 0.0386, 0.0173] %part (v) k=[0:10] ; x5 = 4*((3.^k) - (2.^k)); %x5 = 0 4 20 76 260

844

2660

8236

25220

76684

232100

%part (vi) k=[0:10] ; x6 = -1.206*sin(0.5857*k-0.5857*2); x6(1) = x6(1)- x6(1); % =1.11108; x6(2) = x6(2)- x6(2); % = 0.6666561; x6 % x6= [0, 0, 0, -0.6666, -1.1111, -1.1851, -0.864, -0.2550, 0.439, 0.9867, 1.2055] %part (vii) k=[0:10] ; x7 = (0.25*k-0.5).*4.^k ; x7 = x7 + 0.5*(2.^k); % x7= [0 0 2 20 136

784

4128

20544

98432

459008

2097664]

7

8

Chapter 13

Problem 13.4

(i) X 1 ( z ) =

z z − 0.9 z + 0.2 2

z 2 − 0.9 z + 0.2

z −1 + 0.9 z −2 + 0.61z −3 + 0.369 z −4 + 0.2101z −5 z z ∓ 0.9 ± 0.2 z −1 + 0.9 − 0.2 z −1 ± 0.9 ∓ 0.81z −1 + 0.18 z − 2 + 0.61z −1 − 0.18 z − 2 ± 0.61z −1 ∓ 0.549 z − 2 ± 0.122 z −3 + 0.369 z − 2 − 0.122 z −3 ± 0.369 z − 2 ∓ 0.3321z −3 ± 0.0738 z − 4 + 0.2101z −3 − 0.0738 z − 4 ± 0.2101z −3 ∓ 0.1891z −4 ± 0.0420

Hence, X 1 (z ) =

z = z −1 + 0.9 z − 2 + 0.61z −3 + 0.369 z −4 + 0.2101z −5 + z − 0.9 z + 0.2 2

Taking the inverse transform gives the following values for the first five samples of x1[k]

x1[0] = 0, x1[1] = 1, x1[2] = 0.9, x1[3] = 0.61, x1[4] = 0.369, x1[5] = 0.2101, Note that the above values are consistent with those in Problem 13.3(i). (ii) X 2 ( z ) =

z z − 2.1z + 0.2 2

z 2 − 2.1z + 0.2

z −1 + 2.1z −2 + 4.21z −3 + 8.421z −4 + 16.8421z −5 z z ∓ 2.1 ± 0.2 z −1 + 2.1 − 0.2 z −1 ± 2.1 ∓ 4.41z −1 ± 0.42 z − 2 + 4.21z −1 − 0.42 z − 2 ± 4.21z −1 ∓ 8.841z − 2 ± 0.842 z −3 + 8.421z − 2 − 0.842 z −3 ± 8.421z − 2 ∓ 17.6841z −3 ± 1.6841z − 4 + 16.8421z −3 − 1.6841z − 4 ± 16.8421z −3 ∓ 35.3684 z − 4 ± 3.36842 z −5

Hence, X 2 (z ) =

z = z −1 + 2.1z − 2 + 4.21z −3 + 8.421z − 4 + 16.8421z −5 + z − 2.1z + 0.2 2

Taking the inverse transform gives the following values for the first five samples of x2[k]

.

Solutions

x2 [0] = 0, x2 [1] = 1, x2 [2] = 2.1, x2 [3] = 4.21, x2 [4] = 8.421, x2 [5] = 16.8421,

9

.

Note that the above values are consistent with those in Problem 13.3(ii). (iii) X 3 ( z ) =

z2 + 2 z2 + 2 = 3 ( z − 0.3)( z + 0.4)( z − 0.7) z − 0.6 z 2 − 0.19 z + 0.084

By power division, we get

z 3 − 0.6 z 2 − 0.19 z + 0.084

z −1 + 0.6 z −2 + 2.55 z −3 + 1.560 z −4 + 1.3685 z −5 z 2 + 0z + 2 z 2 ∓ 0.6 z ∓ 0.19 ± 0.084 z −1 + 0.6 z + 2.19 − 0.084 z −1 ± 0.6 z ∓ 0.36 ∓ 0.114 z −1 ± 0.048 z −2 + 2.55 + 0.030 z −1 − 0.048 z − 2 ± 2.55 ∓ 1.530 z −1 ∓ 0.4845 z − 2 ± 0.2142 z −3 + 1.560 z −1 + 0.4325 z − 2 − 0.2142 z −3 ± 1.560 z −1 ∓ 0.9360 z − 2 ∓ 0.2964 z −3 ± 0.1310 z − 4 + 1.3685 z −2 + 0.0822 z −3 − 0.1310 z − 4 ± 1.3685 z −2 ∓ 0.8211z −3 ∓ 0.2600 z − 4 ± 0.1150 z −5

Hence, X 3 (z ) =

z2 + 2 = z −1 + 0.6 z −2 + 2.55 z −3 + 1.560 z − 4 + 1.3685 z −5 + ( z − 0.3)( z + 0.4)( z − 0.7)

Taking the inverse transform gives the following values for the first five samples of x3[k]

x3 [0] = 0, x3 [1] = 1, x3 [2] = 0.6, x3 [3] = 2.55, x3 [4] = 1.560, x3 [5] = 1.3685, Note that the above values are consistent with those in Problem 13.3(iii). (iv) X 4 (z ) =

z2 + 2 z2 + 2 = ( z − 0.3)( z + 0.4) 2 z 3 + 0.5 z 2 − 0.08 z − 0.048

By power division, we obtain

.

10

Chapter 13

z 3 + 0.5 z 2 − 0.08 z − 0.048

z −1 − 0.5 z −2 + 2.33 z −3 − 1.157 z −4 + 0.7409 z −5 z 2 + 0z + 2 z 2 ± 0.5 z ∓ 0.08 ∓ 0.048 z −1 − 0.5 z + 2.08 + 0.048 z −1 ∓ 0.5 z ∓ 0.25 ± 0.040 z −1 ± 0.024 z − 2 + 2.33 + 0.008 z −1 − 0.024 z − 2 ± 2.33 ± 1.165 z −1 ∓ 0.1864 z −2 ∓ 0.1118 z −3 − 1.157 z −1 + 0.1624 z − 2 + 0.1118 z −3 ∓ 1.157 z −1 ∓ 0.5785 z − 2 ± 0.0926 z −3 ± 0.0555 z − 4 + 0.7409 z − 2 + 0.0192 z −3 + 0.0555 z − 4 ± 0.7409 z − 2 ± 0.3705 z −3 ∓ 0.0593 z − 4 ∓ 0.0356 z −5

Hence, X 4 (z ) =

z2 + 2 = z −1 − 0.5 z − 2 + 2.33 z −3 − 1.157 z − 4 + 0.7409 z −5 − + 2 ( z − 0.3)( z + 0.4)

Taking the inverse transform gives the following values for the first five samples of x4[k]

x4 [0] = 0, x4 [1] = 1, x4 [2] = −0.5, x4 [3] = 2.33, x4 [4] = −1.157, x4 [5] = 0.7409, Note that the above values are consistent with those in Problem 13.3(iv). (v) X 5 ( z ) =

4 z −1 4z = 2 −1 −2 1 − 5z + 6 z z − 5z + 6

By power division, we get

z 2 − 5z + 6

4 z −1 + 20 z −2 + 76 z −3 + 260 z −4 + 844 z −5 4z 4 z ∓ 20 ± 24 z −1 + 20 − 24 z −1 ± 20 ∓ 100 z −1 ± 120 z − 2 + 76 z −1 − 120 z − 2 ± 76 z −1 ∓ 380 z −2 ± 456 z −3 + 260 z − 2 − 456 z −3 ± 260 z − 2 ∓ 1300 z −3 ± 1560 z − 4 + 844 z −3 − 1560 z − 4 ± 844 z −3 ∓ 4220 z − 4 ± 5064 z −5

Hence, X 5 (z ) =

4 z −1 = 4 z −1 + 20 z −2 + 76 z −3 + 260 z −4 + 844 z −5 + (1 − 5 z −1 + 6 z − 2 )

Taking the inverse transform gives the following values for the first five samples of x5[k]

x5 [0] = 0, x5 [1] = 4, x5 [2] = 20, x5 [3] = 76, x5 [4] = 260, x5 [5] = 844,

.

.

Solutions Note that the above values are consistent with those in Problem 13.3(v). (vi) X 6 ( z ) =

4 z −2 − 4 z −1 = 10 − 6( z1 + z −1 ) 6 z 2 − 10 z + 6

By power division, we get, −5 −6 62 −7 − 23 z −3 − 109 z −4 − 32 − 70 − 243 z 27 z 81 z

−4 z −1

6 z − 10 z + 6 2

−4 z −1 + 203 z −2 − 4 z −3 − 203 z −2 + 4 z −3 −3 − 203 z −2 + 100 − 203 z −4 9 z

− 649 z −3 + 203 z −4 −4 − 649 z −3 + 320 − 649 z −5 27 z −4 − 140 + 649 z −5 27 z −4 −5 −6 − 140 + 700 − 140 27 z 81 z 27 z −5 −6 − 124 + 140 81 z 27 z −5 −6 −7 − 124 + 620 + 124 81 z 243 z 81 z

Hence,

X 6 ( z) =

4 z −2 − 4 z −1 −6 32 −5 62 = = − 23 z −3 − 109 z −4 − 27 z − 70 − 243 z −7 − ...... 81 z 10 − 6( z1 + z −1 ) 6 z 2 − 10 z + 6

Taking the inverse transform gives the following values for the first five samples of x6 [k ] .

x6 [0] = 0, x6 [1] = 0, x6 [2] = 0, x6 [3] = − 32 , x6 [4] = − 109 , x6 [5] = − 32 27 , 62 x6 [6] = − 70 81 , x6 [7] = − 243 ,

Note that the above values are consistent with those in Problem 13.3(vi). (vii) X 7 ( z ) =

2 z −2

(1 − 4 z ) (1 − 2 z ) −1

2

By power division, we get,

−1

=

2z 2z = 3 2 2 ( z − 4) ( z − 2) z − 10 z + 32 z − 32

.

11

12

Chapter 13 2z −2 +20z −3 + 136 z −4 + 784 z −5 + 4128 z −6

z 3 − 10 z 2 + 32 z − 32

2z 2 z − 20 + 64 z −1 − 64 z −2 20 − 64 z −1 + 64 z −2 20 − 200 z −1 + 640 z −2 − 640 z −3 136z −1 − 576 z −2 + 640 z −3 136 z −1 − 1360 z −2 + 4352 z −3 − 4352 z −4 784 z −2 − 3712 z −3 + 4352 z −4 784 z −2 − 7840 z −3 + 25088 z −4 − 25088 z −5 4128 z −3 − 20736 z −4 + 25088 z −5 4128 z −3 − 41280 z −4 + 132096 z −5 − 132096 z −6

Hence, X 7 ( z ) =

2 z −2

(1 − 4 z ) (1 − 2 z ) −1 2

−1

= 2z −2 +20z −3 + 136 z −4 + 784 z −5 + 4128 z −6 + ......

Taking the inverse transform gives the following values for the first five samples of x7 [k ] .

x7 [0] = 0, x7 [1] = 0, x7 [2] = 2, x7 [3] = 20, x7 [4] = 136, x7 [5] = 784, x7 [6] = 4128,

.

Note that the above values are consistent with those in Problem 13.3(vii).

Problem 13.5

(a) We know that (Entry 4 of Table 13.1, with α = α e Z (αe jΩ0 ) k U [ k ] ←⎯→

Z (αe − jΩ0 ) k U [ k ] ←⎯→

and

± j Ω0

)

1 1 − αe jΩ0 z −1 1 1 − αe − jΩ0 z −1

ROC : z > α ROC : z > α

Adding the two transform pairs after multiplying with re jθ and re − jθ , respectively, gives jθ

re (α e

jΩ0 k

) U [k ] − re

− jθ

(α e

re jθ re − jθ ) U [k ] ←⎯→ − 1 − α e jΩ0 z −1 1 − α e − jΩ0 z −1

− j Ω0 k

Z

j ( Ω k +θ ) Z − e − j ( Ω0 k +θ ) ⎤⎦ u[k ] ←⎯ → or, rα k ⎡⎣e 0

re jθ (1 − α e − jΩ0 z −1 ) − re − jθ (1 − α e jΩ0 z −1 ) (1 − α e jΩ0 z −1 )(1 − α e − jΩ0 z −1 )

(

)

re jθ − re − jθ + α r e j ( Ω0 −θ ) − e − j ( Ω0 −θ ) z −1

2 jr × α sin(Ω0 k + θ )u[k ] ←⎯→ k

Z

(

)

1 − α e jΩ0 z −1 + α e − jΩ0 z −1 + α 2 z −2

or,

=

ROC : z > α

r × 2 j sin θ + α r × 2 j sin(Ω0 − θ ) z 1 − 2α cos(Ω0 ) z −1 + α 2 z −2

−2

ROC : z > α ROC : z > α ,

Solutions

r sin θ + α r sin(Ω 0 − θ ) z −1 r × α sin(Ω0 k + θ )u[ k ] ←⎯→ 1 − 2α cos Ω0 z −1 + α 2 z −2 k

or,

ROC : z > α

Z

.

A + Bz −1 = 1 + 2γ z −1 + α 2 z −2

where

A = r sin θ , B = α r sin(Ω0 − θ ), γ = −α cos Ω0 . Alternatively, the parameters

expressed in terms of

A, B, γ , α

A2α 2 + B 2 − 2 ABγ

r=

2

α −γ

2

,

13

r , Ω0 , θ

can be

as follows (as given in Entry 12).

( ) , and θ = tan −γ

Ω0 = cos −1

α

−1

(

A α 2 −γ 2 B − Aγ

)

.

Proof for the expressions of r , Ω0 ,θ in Problem 13.5 A2α 2 + B 2 − 2 ABγ

α 2 −γ 2

r 2α 2 sin 2 θ + r 2α 2 sin 2 ( Ω0 −θ ) + 2 r 2α 2 sin θ cos( Ω0 )sin( Ω0 −θ )

=

α 2 −α 2 cos 2 ( Ω0 ) sin 2 θ + sin 2 ( Ω0 −θ ) + 2sin θ cos( Ω0 )sin( Ω0 −θ )

=r

1− cos 2 ( Ω0 ) = sin 2 ( Ω0 )

=r

because sin 2 θ + sin 2 (Ω 0 − θ ) + 2sin θ cos(Ω0 ) sin(Ω 0 − θ ) = sin 2 θ + [sin(Ω 0 ) cos θ − cos(Ω0 ) sin θ ] + 2sin θ cos(Ω0 ) [sin(Ω 0 ) cos θ − cos(Ω 0 ) sin θ ] 2

= sin 2 θ + sin 2 (Ω0 ) cos 2 θ + cos 2 (Ω 0 ) sin 2 θ − 2sin(Ω 0 ) cos(Ω 0 ) cos θ sin θ + 2sin(Ω0 ) cos(Ω0 ) sin θ cos θ − 2 cos 2 (Ω0 ) sin 2 θ = sin 2 θ + sin 2 (Ω0 ) cos 2 θ − cos 2 (Ω 0 ) sin 2 θ = sin 2 (Ω 0 ) cos 2 θ + sin 2 θ (1 − cos 2 (Ω0 ) ) = sin 2 (Ω 0 ) cos 2 θ + sin 2 θ sin 2 (Ω0 ) = sin 2 (Ω 0 )

cos −1 tan −1

(

( ) = cos ( −γ

−1

α

A α 2 −γ 2 B − Aγ

α cos Ω0 α

) = tan ( −1

(b) By comparing X ( z ) =

)=Ω

0

r sin θ α 2 −α 2 cos 2 Ω0 α r sin( Ω0 −θ ) +α r sin θ cos Ω0

) = tan

(

α r sin Ω0 sin θ α r sin Ω0 cosθ

) = tan

−1

( tan θ ) = θ

1 with the above transform pair, we get A = 1, B = 0, γ= −0.5, and α = 1 − z −1 + z −2

1. Substituting the values to compute α, θ, and Ω0, gives

r=

−1

A2α 2 + B 2 − 2 ABγ α2 −γ2

Ω 0 = cos −1

( ) = cos −γ α

= −1

=

1 1− 0.25

(0.5) =

π 3

,

2 3

,

14

Chapter 13

θ = tan −1

and

(

A α 2 −γ 2 B − Aγ

) = tan

−1

(

0.75 0.5

) = tan ( 3 ) = −1

π

3

Hence, the inverse z-transform of X(z) is given by

x[k ] = rα k sin(Ω0 k + θ ) u[k ] =

2 3

sin ( π3k + π3 ) u[k ] .

Alternative Solution of Part (b)

X ( z) =

{

=

2 3

=

2 3

=

1 3

= =

2 3

}

sin ( π3k + π3 ) u[k ] =

2 3

{sin (

πk 3

+ π3 ) u[k ]}

{sin ( ) cos ( ) u[k ] + cos ( ) sin ( ) u[k ]} πk

π

3

{

1 2

πk

3

sin ( π3k ) u[k ] +

π

3

3 2

3

}

cos ( π3k ) u[k ]

{sin ( ) u[k ]} + {cos ( ) u[k ]} πk

πk

3

3

−1

1 3

sin( π3 ) z 1 − cos( π3 ) z −1 + 1 − 2 cos( π3 ) z −1 + z −2 1 − 2 cos( π3 ) z −1 + z −2 1 3

3 2

z −1

1 − z −1 + z −2 1 = −1 1 − z + z −2

ROC: z > 1

1 − 12 z −1 + 1 − z −1 + z −2

ROC: z > 1

▌ Program 13.5. MATLAB Program to calculate and verify solutions in Problem 13.5. k=[0:10] ; x = (2/sqrt(3))*sin(pi*k/3 + pi/3) % x = [1 1 0 -1 -1 0 1 1 0 -1 -1] stem(k, x, 'filled'), grid xlabel('k') % Label of X-axis ylabel('x[k]') % Label of Y-axis print -dtiff plot.tiff % Save the figure as a TIFF file %MATLAB Verification 1 % X ( z) = 1 − z −1 + z −2 sys = filt([1],[1 -1 1]) output = impulse(sys,10) % output= y1= [1 1 0 -1 -1 0 1 1 0 -1 -1]

Problem 13.6

(a) (i) From pair 9 in Table 13.1, we know

{sin(Ω0 k )u[k ]} =

z sin(Ω 0 ) z − 2 z cos(Ω 0 ) + 1 2

ROC: z > 1 .

Solutions

d ⎡ {sin(Ω 0 k )u[k ]}⎤⎦ dz ⎣ ⎤ z sin(Ω 0 ) d ⎡ = −z ⎢ 2 ⎥ dz ⎣ z − 2 z cos(Ω 0 ) + 1 ⎦

{k sin(Ω0 k )u[k ]} = − z

15

[ using frequency differentiation property] ROC: z > 1

= −z

( z 2 − 2 z cos(Ω 0 ) + 1) sin(Ω 0 ) − z sin(Ω 0 ) × (2 z − 2 cos(Ω 0 )) ( z 2 − 2 z cos(Ω 0 ) + 1)2

ROC: z > 1

= −z

z sin(Ω 0 )( z 2 − 1) − z 2 sin(Ω 0 ) + sin(Ω 0 ) = ( z 2 − 2 z cos(Ω 0 ) + 1)2 ( z 2 − 2 z cos(Ω 0 ) + 1)2

ROC: z > 1

(ii) Substituting r = α = 1 , in Entry 12 of Table 13.1, we get the following pair.

{sin(Ω0 k + θ )u[k ]} =

z [ z sin θ + sin(Ω0 − θ ) ] , ROC: z > 1 z 2 − 2 z cos Ω0 + 1

{k sin(Ω0 k + θ )u[k ]} d ⎡ {sin(Ω 0 k + θ )u[k ]}⎤⎦ [ using frequency differentiation property] dz ⎣ d ⎡ z [ z sin θ + sin(Ω 0 − θ ) ] ⎤ = −z ⎢ ⎥ dz ⎣ z 2 − 2 z cos Ω 0 + 1 ⎦ ( z 2 − 2 z cos(Ω 0 ) + 1) [ 2 z sin θ + sin(Ω 0 − θ ) ] − z { z sin θ + sin(Ω 0 − θ )} × (2 z − 2 cos(Ω 0 )) = −z ( z 2 − 2 z cos(Ω 0 ) + 1) 2 = −z

= =

ROC: z > 1 ROC: z > 1

−z ⎡ − {2 cos(Ω 0 ) sin θ + sin(Ω 0 − θ )} z 2 + 2 z sin θ + sin(Ω 0 − θ ) ⎤⎦ ( z − 2 z cos(Ω 0 ) + 1) 2 ⎣ 2

z ⎡⎣ z 2 sin(Ω 0 + θ ) − 2 z sin θ − sin(Ω 0 − θ ) ⎤⎦

ROC: z > 1

( z 2 − 2 z cos(Ω 0 ) + 1) 2

(b) Substituting Ω0 = π3 , θ =

π

6

, in pair a(ii) derived above, we get the following pair

z ⎡⎣ z 2 sin( π2 ) − 2 z sin( π6 ) − sin( π6 ) ⎤⎦ π π k k u k sin( ) [ ] + = { } 3 6 ( z 2 − 2 z cos( π3 ) + 1) 2 z ⎡ z 2 − z − 0.5⎤⎦ = ⎣ 2 ( z − z + 1) 2

ROC: z > 1 ROC: z > 1

On the other hand, from pair 9 in Table 13.1, we know the following z-transform pair.

{sin( π3 k )u[k ]} =

3 z sin( π3 ) 2 z = z 2 − 2 z cos( π3 ) + 1 z 2 − z + 1

The given z-transform pair can now be proved as follows.

ROC: z > 1 .

16

Chapter 13

=

{⎡⎣

2 3

2 3

{sin( π3 k )u[k ]} −

sin( π3 k ) −

3 2

z = × 2 − z − z +1 2 3

k 3

1 3

}

sin( π3 k + π6 ) ⎤⎦ u[k ]

×

{k sin( π3 k + π6 )u[k ]}

1 3

z ⎡⎣ z 2 − z − 0.5⎤⎦

ROC: z > 1

( z 2 − z + 1) 2

=

z 3

⎡ 1 z 2 − z − 0.5 ⎤ − = ⎢ 2 2 2⎥ ⎣ z − z + 1 ( z − z + 1) ⎦

=

z 3

3 ⎡ ⎤ 1.5 2 z = ⎢ ( z 2 − z + 1) 2 ⎥ ( z 2 − z + 1) 2 ⎣ ⎦

z 3

⎡ z 2 − z + 1 − ( z 2 − z − 0.5) ⎤ ⎢ ⎥ ( z 2 − z + 1) 2 ⎣ ⎦

ROC: z > 1 ROC: z > 1 ▌

Alternative (Direct) Solution of Part (b): Direct calculation of y[ k ] using partial fraction method: First express,

1 as follows. ( z − z + 1) 2 2

⎡ 1 =⎢ 2 2 ( z − z + 1) ⎣ ( z − 0.5 − j

1 3 z − 0.5 + j )( 2

1⎡ 1 =− ⎢ 3 ⎣ z − 0.5 − j ⎡ 1 1 =− ⎢ 3 ⎢ z − 0.5 − j ⎣⎢

(

⎡ 1 1 =− ⎢ 3 ⎢ z − 0.5 − j ⎣⎢

(

3 2

1 − z − 0.5 + j

3 2

3 2

+

2

⎤ ⎡ 1 ⎪⎧ 1 =⎢j 3⎨ 3 ⎥ − −j z ) 0.5 ⎪ ⎢⎣ ⎩ 2 ⎦ ⎤ 3 ⎥ 2 ⎦

3 2

1 − z − 0.5 + j

⎪⎫⎤ ⎥ 3 ⎬ 2 ⎪ ⎭⎦⎥

2

1

2

2

2

) ( z − 0.5 + j ) ( z − 0.5 − j )( z − 0.5 + j ) 2

+

3 2

1

) ( z − 0.5 + j ) 2

2

3 2

3 2

⎤ 2 ⎥ z 2 − z + 1⎥ ⎦⎥

The function y[ k ] can then be calculated as follows.

3 2

⎤ ⎥ ⎥ ⎦⎥

Solutions

y[k ] =

−1

17

3 ⎪⎧ ⎪⎫ 2 z ⎨ 2 2⎬ ⎩⎪ ( z − z + 1) ⎭⎪

⎧ ⎡ ⎤⎫ z z 2z ⎥⎪ ⎪ 1 ⎢ = + − 2 ⎨− ⎬ 2 2 ⎢ z − z + 1⎥ ⎪ z − 0.5 + j 23 ⎪ 2 3 ⎢ z − 0.5 − j 23 ⎥⎦ ⎭ ⎣ ⎩ ⎡ ⎧ ⎫ ⎧ ⎫⎤ 3 z z 1 ⎢ −1 ⎪ 2 ⎪ ⎪⎥ 1 −1 ⎪ 2 z =− + + × ⎨ ⎬ ⎨ ⎬ 2 2 ⎥ 2 2 3⎢ 3 3 z − z +1 ⎪ z − 0.5 − j 23 ⎪ ⎪ z − 0.5 + j 23 ⎪⎥ ⎢⎣ ⎩ ⎭ ⎩ ⎭⎦ − − 1 1 k k 1 ⎡ ⎤ u[k ] + 2 sin( π k )u[k ] =− + k 0.5 − j 23 k 0.5 + j 23 3 ⎢ ⎥⎦ 3 2 3⎣ 2 k ⎡ j3π ( k −1) − j3π ( k −1) ⎤ e u[k ] + sin( π3 k )u[k ] =− +e ⎢ ⎥ ⎦ 3 2 3⎣ 2 k =− × 2 cos ( π3 (k − 1) ) u[k ] + sin( π3 k )u[k ] 3 2 3 −1

(

) (

(

)

(

(

)

(

)

= 23 sin( π3 k )u[k ] − = ⎡⎣ 23 sin( π3 k ) −

k 3

k 3

)

)

cos ( ( π3 k + π6 ) − π2 ) u[k ]

sin( π3 k + π6 ) ⎤⎦ u[k ] ▌

Program 13.6. MATLAB Program (for verification) k=0:9 ; y1=(2/3)*sin(pi*k/3).*(k>=0); y2=-(k/sqrt(3)).*sin(pi*k/3+pi/6).*(k>=0); y= y1+y2 % y = [0 0 0 0.8660 1.7321 0.8660 -1.7321

-3.4641

-1.7321

2.5981]

sys = filt([0 0 0 sqrt(3)/2],[1 -2 3 -2 1]) h = impulse(sys,10) plot(k,y,k,h) % h = [0 0 0 0.8660 1.7321 0.8660 -1.7321 -3.4641

-1.7321

2.5981]’

%MATLAB Verification: %

(1 − z

3 2 −1

z −3 + z −2 )

2

=

z −3 1 − 2 z −1 + 3 z −2 − 2 z −3 + z −4 3 2

Problem 13.7

Example 13.3(v) showed that

k = 0,1 ⎧1, ⎪ Z x5 [k ] = ⎨2, k = 2,5 ←⎯ → X ( z ) = 1 + z −1 + 2 z −2 + 2 z −5 ⎪0, otherwise. ⎩ Since g[k] = x5[k − 10], G(z) = z−10X5 (z), which yields

ROC: entire z-plane except ( z ≠ 0 ) .

18

Chapter 13

G ( z ) = z −10 + z −11 + 2 z −12 + 2 z −15

ROC: entire z-plane except (z ≠ 0).

Problem 13.8

For a causal sequence x[k], the z-transform is given by X ( z ) = x[0] + x[1]z −1 + x[2] z −2 + x[3] z −3 +

+ x[k ] z − k +

Applying the limit, z → ∞, the above equation reduces to

x[0] = lim X ( z ).

z →∞

Problem 13.9

Using the time shifting property,

( x[k + 1] − x[k ]) = ( x[k + 1]) − ( x[k ]) = ( z − 1) X ( z ) . N

( x[k + 1] − x[k ]) = Nlim ∑ ( x[k + 1] − x[k ]) z − n →∞

However, from definition,

k =0

Therefore, ( z − 1) X ( z ) = lim

N →∞

N

∑ ( x[k + 1] − x[k ]) z

−n

.

k =0

Applying the limit, z → 1, the above equation reduces to N

lim( z − 1) X ( z ) = lim ∑ ( x[k + 1] − x[k ]) z →1

N →∞

k =0

= lim ⎡⎣( x[1] − x[0]) + ( x[2] − x[1]) + ...... + ( x[ N + 1] − x[ N ]) ⎤⎦ . N →∞ = lim x[ N + 1] N →∞

= lim x[k ] k →∞

which is the final value theorem.

Problem 13.10

(i) x[k] = (5/6)k u[k − 6]. We know that Z (5 6) k u[k ] ←⎯ →

1 1 − (5 / 6) z −1

ROC : z > 5 6

Using the time shifting property, Z (5 6) k −6 u[k − 6] ←⎯ →

or, (ii) x[k] = k(2/9)k u[k]

z −6 1 − (5 / 6) z −1

z −6 (5 6) u[ k − 6] ←⎯→ (5 6) 1 − (5 / 6) z −1 k

Z

6

ROC : z > 5 6 ROC : z > 5 6 .

Solutions We know that Z (2 9) k u[k ] ←⎯ →

1 1 − (2 / 9) z −1

ROC : z > 2 9

Using the frequency differentiation property, Z k (2 9) k u[k ] ←⎯ → −z

or,

Z k (2 9) k u[k ] ←⎯ → −z

1

ROC : z > 2 9

(−1)(−2 / 9)(−1)( z −2 ) ROC : z > 2 9 .

(1 − (2 / 9) z )

−1 2

Z k (2 9) k u[k ] ←⎯ →

or,

d 1 dz 1 − (2 / 9) z −1

(2 / 9) z −1

ROC : z > 2 9 .

(1 − (2 / 9) z )

−1 2

(iii) x[k] = ramp(k) = ku[k]. k

k u [ k ] = ∑ u [ m] − u [ k ] .

We know that

m=0

Calculating the z-transform of both sides and applying the time-accumulation property, we get Z k u[k ]←⎯→

z 1 1 , − −1 z −1 1− z 1 − z −1

Z k u[k ]←⎯→

which reduces to

z −1

(1 − z )

−1 2

,

which can be expressed in the alternate form Z k u[k ]←⎯→ ,

z

(z − 1)2

.

(iv) x[k] = ek sin(k)u[k]. We know that Z e(1+ j ) k u[k ] ←⎯ →

and

Z e(1− j ) k u[k ] ←⎯ →

1 1− e

(1+ j ) −1

1− e

(1− j ) −1

ROC : z > e

z

1

ROC : z > e

z

Adding the two transform pairs, Z e(1+ j ) k u[k ] + e(1− j ) k u[k ] ←⎯ →

or,

Z 2 j sin(k )ek u[k ] ←⎯ →

1 1− e

(1+ j ) −1

z

1 1− e

(1− j ) −1

1 − e(1− j ) z −1 − 1 + e(1+ j ) z −1

(1 − e

(1+ j ) −1

z

) (1 − e

(1− j ) −1

z

z

)

ROC : z > e ROC : z > e .

19

20

Chapter 13

e sin(1) z −1 sin( k )e u[k ] ←⎯→ 1 − 2e cos(1) z −1 + e 2 z −2 k

or,

Z

ROC : z > e

Problem 13.11

The four possible regions of convergences are: 1.

|z| < 0.5.

2.

0.5 < |z| < 0.75

3.

0.75 < |z| < 1.25

4.

|z| > 1.25

Since a stable system must have the unit circle within its ROC, we choose 0.75 < |z| < 1.25 as the ROC. This ROC corresponds to a double sided sequence, and hence the system is not causal. To determine the impulse response, we perform the partial fraction expansion

H ( z) z ( z − 1) −4 / 3 3/ 2 25 / 3 = ≡ + + z ( z − 0.5)( z − 0.75)( z − 1.25) z − 0.5 z − 0.75 z − 1.25 H ( z) =

or,

−4 / 3 3/ 2 5/ 6 + + . −1 −1 1 − 0.5 z 1 − 0.75 z 1 − 1.25 z −1 ROC :| z | > 0.5

ROC :| z | > 0.75

ROC :| z| 0.2 .

The z-transfer function of the inverse LTID system is given by

H inv ( z ) = 1 − 0.2 z −1

ROC: Entire z - plane except z = 0.

Calculating the inverse z-transform, we obtain

hinv [k ] = δ [k ] − 0.2δ [k − 1]. (ii) (a) Input: x1[k] = u[k]. Using the transform pair Z u[k ] ←⎯ →

the z-transform of the output is given by

1 1 − z −1

ROC: z > 1 ,

Solutions

Y1 ( z ) = X 1 ( z ) H ( z ) =

1 −1

(1 − 0.2 z )(1 − z −1 )

ROC: z > 1 .

Taking the partial fraction expansion, we obtain

Y1 ( z ) 1.25 0.25 = − z z − 1 z − 0.2 ROC :| z| >1

ROC :| z | > 0.2

Calculating the inverse z-transform, we obtain

y1[k ] = (1.25 − 0.25 × 0.2k )u[k ]. (b) Input: x2[k] = 5δ[k − 4] − 2δ[k + 4]. Using the time domain convolution method, the output is obtained as

y2 [k ] = 5− k u[k ] ∗ (5δ [k − 4] − 2δ [k + 4]) = 5 × 5− ( k −4) u[k − 4] − 2 × 5− ( k + 4) u[k + 4]. (c) Input: x3 [k ] = e k + 2u[ − k + 2] = e5 ( e) k −3 u[ − k + 2] . From Example 13.2, we know the transform pair Z −e k u[ −k − 1] ←⎯ →

1 1 − ez −1

ROC: z < e .

Using the time shifting property, we obtain Z −e k −3u[ −( k − 3) − 1] = − e k −3u[ − k + 2] ←⎯ →

In other words,

X 3( z) =

{e (e) 5

k −3

u[− k + 2]} = − e5

z −3 1 − ez −1

z −3 1 − ez −1

ROC: z < e .

ROC: z < e .

The z-transform of the output is then given by

Y3 ( z ) = X 3 ( Z ) H ( z ) = − e5 or,

z −3 (1 − 0.2 z −1 )(1 − ez −1 )

Y3 ( z ) 1 = − e5 2 z z ( z − 0.2)( z − e)

ROC:0.2 < z < e

ROC: 0.2 < z < e

Calculating the partial fraction expansion, we obtain

⎡ ⎤ ⎢ Y3 ( z ) 9.8737 1.8394 9.9274 0.0537 ⎥ = −e5 ⎢ + − + ⎥ ROC:0.2 < z < e , z z2 ( z − 0.2) ( z − e) ⎥ ⎢ z ⎢⎣ ROC:|z|>0 ROC:|z|>0 ROC:|z|>0.2 ROC:|z| 0 ⎢ ROC:|z|>0 ⎥ ROC:|z|>0.2 ROC:|z|< e ⎦ ⎣⎢

21

22

Chapter 13

Calculating the inverse z-transform, we obtain

y3[k ] = − e5 ⎡⎣9.8737δ [k ] + 1.8394δ [k − 1] − 9.9274(1/ 5) k u[k ] − 0.0537u[ − k − 1]⎤⎦ .

Problem 13.13

(i) Taking the z-transform of the input and output, we get

X ( z) = =

1 z −1 − 1 − (1 / 3) z −1 1 − (1 / 4) z −1 1 − (5 / 4) z −1 + (1 / 3) z −2

(1 − (1/ 3) z ) (1 − (1 / 4) z )

Y ( z) =

and

ROC : z > (1 / 3)

−1

1 1 − (1 / 4) z −1

−1

ROC : z > (1 / 4).

The transfer function of the system is given by H ( z) =

Y ( z) 1 − (1 / 3) z −1 = X ( z ) 1 − (5 / 4) z −1 + (1 / 3) z −2

ROC : z > 0.8694

(ii) By expressing the transfer function as

H ( z) z − (1 / 3) = ( z − 0.8644)( z − 0.3856) z

ROC : z > 0.8694 .

and taking the partial fraction expansion, we get H ( z) 1.1093 0.1093 =− − z ( z − 0.8644) ( z − 0.3856) ROC:| z| >0.8644

ROC:| z| >0.3856

Taking the inverse z-transform, we get h[k ] = (1.1093 × 0.8644 k − 0.1093 × 0.3856 k )U [k ]. (iii) By expressing the transfer function as H ( z) =

Y ( z) 1 − (1 / 3) z −1 = X ( z ) 1 − (5 / 4) z −1 + (1 / 3) z −2

and cross multiplying, we get Y ( z ) − (5 / 4) z −1Y ( z ) + (1 / 3) z −2Y ( z ) = X ( z ) − (1 / 3) z −1 X ( z ) Taking the inverse z-transform, we get y[k ] − (5 / 4) y[ k − 1] + (1 / 3) y[k − 2] = x[k ] − (1 / 3) x[k − 1] with initial conditions: y[−2] = y[−1] = 0.

Problem 13.14

(i) We express the transfer function as

Solutions

H ( z) z2 = z ( z − 0.3)( z − 0.5)( z − 0.7)

23

ROC: z > 0.7

where the ROC is selected to obtain a causal LTID system. Taking the partial fraction expansion, we get

H ( z ) 1.125 6.25 6.125 = − + z z − 0.3 z − 0.5 z − 0.7 ROC :| z| > 0.3

ROC :| z| > 0.5

ROC :| z | > 0.7

Taking the inverse z-transform, we get

h[k ] = (1.125 × 0.3k − 6.25 × 0.5k + 6.125 × 0.7 k ) u[k ]. (ii) By expressing the transfer function as

H ( z) =

Y ( z) 1 = −1 X ( z ) 1 − 1.5 z + 0.71z −2 − 0.105 z −3

and cross multiplying, we obtain

Y ( z ) − 1.5 z −1Y ( z ) + 0.71z −2Y ( z ) − 0.105 z −3Y ( z ) = X ( z ) Taking the inverse z-transform, we obtain

y[k ] − 1.5 y[k − 1] + 0.71y[k − 2] − 0.105 y[k − 3] = x[k ] with initial conditions: y[−3] = y[−2] = y[−1] = 0. (iii) Using the convolution property, the output for the unit step function is given by

Y ( z) = H ( z) X ( z) =

z4 . ( z − 1)( z − 0.3)( z − 0.5)( z − 0.7)

Taking the partial fraction expansion, we obtain

Y ( z ) 200 / 21 27 / 56 25 / 4 343 / 24 = − + − z z −1 z − 0.3 z − 0.5 z − 0.7 ROC :| z| >1

ROC :| z| > 0.3

ROC :| z | > 0.5

ROC :| z | > 0.7

Taking the inverse z-transform, we get

y[k ] = (9.524 − 0.4821× 0.3k + 6.25 × 0.5k − 14.2917 × 0.7 k )u[k ]. (iv) Using linear convolution, the output is given by

y[k ] = u[k ] ∗ (1.125 × 0.3k − 6.25 × 0.5k + 6.125 × 0.7 k )u[k ] . Note that u[k ] ∗ α k u[k ] =

∑ u[k − n]α

n =−∞

k

n

u[n] = ∑ α n = n =0

α k +1 − 1 . α −1

Hence, the output is given by

⎡ 0.3k +1 − 1 0.5k +1 − 1 0.7 k +1 − 1 ⎤ − 6.25 × + 6.125 × y[k ] = ⎢1.125 × ⎥ u[k ] −0.7 −0.5 −0.3 ⎦ ⎣

24

Chapter 13 which reduces to

y[k ] = (9.524 − 0.4821× 0.3k + 6.25 × 5k − 14.2917 × 7 k )u[k ].

Problem 13.15

(i)

Using Eq. (13.35), the transfer function of the system (assuming a causal system) is calculated as follows: H (z) =

Y (z)

X ( z)

=

1 − z −2 1 − z −2 = −1 − 2 1 + z + 14 z 1 + 0.5 z −1

(

)

ROC: z > 0.5 .

2

(ii) In order to calculate the impulse response, we represent the H ( z ) as follows. H (z) =

1 − z −2

(

1 + 0.5 z −1

=

1

) ( 2

1 + 0.5 z −1

+

− z −2

) ( 2

1 + 0.5 z −1

)

= (−2 z ).

2

−0.5 z −1

(

1 − (−0.5) z −1

)

2

+ 2 z −1

−0.5 z −1

(

1 − ( −0.5) z −1

)

2

.

From Entry 7 of Table 13.1, we get the following z-transform pair. Z p[k ] = k (−0.5) k u[k ] ←⎯ → P( z ) =

−0.5 z −1

(1 − (−0.5) z )

−1 2

ROC: z > 0.5

Therefore, applying the time shifting property, we get

h[k ] = −2 p[k + 1] + 2 p[k − 1] = −2(k + 1)(−0.5) k +1 u[k + 1] + 2(k − 1)(−0.5)k −1 u[k − 1] = (k + 1)(−0.5) k u[k + 1] − 4(k − 1)(−0.5) k u[k − 1] = 0, at k =−1

= (k + 1)(−0.5) k u[k ] − 4(k − 1)(−0.5) k u[k − 1] =−4 at k = 0

= −4δ [k ] + (k + 1)(−0.5) u[k ] − 4(k − 1)(−0.5) k u[k ] k

= −4δ [k ] − (3k − 5)(−0.5) k u[k ] The impulse response function is shown in the Fig. S13.15(a). (iii) X ( z ) =

1 . Using the convolution property, the output for the unit step function is given by 1 − 0.5 z −1

Y ( z ) = X ( z ) H ( z) =

1 − z −2

(1 − 0.5z )(1 + 0.5z )

where ⎡ 1 − z −2 k1 = ⎢ ⎢ 1 + 0.5 z −1 ⎣

(

)

⎤ 3 ⎥ =− 2 ⎥ 4 ⎦ z −1 = 2

−1

−1

2

k3 k1 k2 + + 1 − 0.5 z −1 1 + 0.5 z −1 1 + 0.5 z −1

(

)

2

Solutions

⎡ ⎧ 1 − u 2 ⎫⎤ ⎡ (1 − 0.5u )(−2u ) − (1 − u 2 )(−0.5) ⎤ −19 =⎢ = k2 = ⎢ dud ⎨ ⎬⎥ ⎥ 2 (1 − 0.5u ) 8 ⎦ u =−2 ⎣ ⎩1 − 0.5u ⎭⎦ u =−2 ⎣ ⎡ 1 − z −2 ⎤ 3 k3 = ⎢ =− −1 ⎥ 2 ⎣1 − 0.5 z ⎦ z −1 =−2

(a)

(b)

Fig. S13.15. (a) Impulse response function and (b) output response in Problem 13.15. Substituting the above values, we obtain Y ( z) = =

−3 / 4 13 / 4 −3 / 2 + + 1 − 0.5 z −1 1 + 0.5 z −1 1 + 0.5 z −1

(

)

2

−3 / 4 13 / 4 −0.5 z −1 + + 3 z 1 − 0.5 z −1 1 + 0.5 z −1 1 − (−0.5) z −1

Calculating the inverse z-transform, we get

(

)

2

25

26

Chapter 13 y[k ] = − 34 (0.5) k u[k ] + 134 (−0.5) k u[ k ] + 3( k + 1)(−0.5) k +1 u[k + 1] = − 34 (0.5) k u[k ] + 134 (−0.5) k u[k ] − 23 (k + 1)(−0.5) k u[k + 1] = 0, at k =−1

= − (0.5) u[k ] + (−0.5) u[k ] − (k + 1)(−0.5) k u[k ] 3 4

3 2

k

} = − {3(0.5) + (6k − 7)(−0.5) } u[k ] ⎧⎪− {3(0.5) + (6k − 7)(0.5) } u[ k ] =⎨ ⎪⎩ − {3(0.5) − (6k − 7)(0.5) } u[k ] =−

{

13 4

k

3 4

(0.5) k + ( 32 k − 74 )( −0.5) k u[ k ]

1 4

k

k

1 4

k

k

k = even

1 4

k

k

k = odd

⎧⎪− 14 {3 + (6k − 7)} (0.5) k u[ k ] k = even =⎨ k 1 ⎪⎩ − 4 {3 − (6k − 7)} (0.5) u[k ] k = odd ⎧⎪ − 14 (6k − 4)(0.5) k u[ k ] k = even =⎨ k ⎪⎩− 14 (10 − 6k )(0.5) u[k ] k = odd ⎧⎪−(3k − 2)(0.5) k +1 u[k ] k = even =⎨ k +1 ⎪⎩ (3k − 5)(0.5) u[k ] k = odd The output response is shown in the Fig. S13.15(b). k (iv) The impulse response of the system, h[ k ] = −4δ [k ] − (3k − 5)(−0.5) u[k ] , and the input signal,

x[k ] = ( 12 ) k u[k ] = (0.5) k u[k ] . The output signal can be calculated by applying linear convolution as follows.

{

}

y[k ] = h[k ] ∗ x[ k ] = −4δ [k ] − (3k − 5)(−0.5) k u[k ] ∗ x[ k ]

{

}

= −4δ [k ] ∗ x[k ] − (3k − 5)(−0.5)k u[k ] ∗ x[k ] = −4 x[k ] −

∑ {(3m − 5)(−0.5)

m

m =−∞ ∞

{

}

u[k ] ⋅ x[k − m]

}

= −4 x[k ] − ∑ (3m − 5)(−0.5) m ⋅ (0.5) k − m u[k − m] m =0

k

= −4(0.5) u[k ] − (0.5) u[k ]∑ (3m − 5)(−0.5) m (0.5) − m k

k

m =0

k ⎡ ⎤ = (0.5) k u[k ] ⎢ −4 − ∑ (3m − 5)(−1) m ⎥ m =0 ⎣ ⎦

Solutions

k

m Note: ∑ (3m − 5)(−1) = 3 m =0

27

k ⎧⎪ 12 (3k − 10) k = even m m − − − = m ( 1) 5 ( 1) . ⎨ 3 ∑ ∑ m =0 m =0 ⎪⎩ − 2 (k + 1) k = odd k

k = even ⎪⎧ k / 2 =⎨ ⎩⎪− ( k +1) / 2 k = odd

⎪⎧1 k = even =⎨ ⎩⎪0 k = odd

Therefore, k ⎡ ⎤ y[k ] = (0.5) u[k ] ⎢ −4 − ∑ (3m − 5)(−1) m ⎥ m =0 ⎣ ⎦ ⎧⎪−4 − 12 (3k − 10) k = even k = (0.5) u[k ] × ⎨ k = odd ⎪⎩ −4 + 32 (k + 1) ⎧⎪− 12 (3k − 2) k = even = (0.5) k u[k ] × ⎨ ⎪⎩ 12 (3k − 5) k = odd ⎧⎪−(3k − 2)(0.5) k +1 u[k ] k = even =⎨ k +1 ⎪⎩ (3k − 5)(0.5) u[k ] k = odd k

It is observed that the output obtained above by applying time-domain convolution is identical to that obtained in step (iii) using the z-transform approach.

28

Chapter 13

Program 13.15. MATLAB Program %Plotting the impulse response k=[0:10] ; h = -(3*k-5).*((-0.5).^k) ; h(1)=h(1)-4 ; stem(k, h, 'filled'), grid xlabel('k') % Label of X-axis ylabel('h[k]') % Label of Y-axis print -dtiff plot.tiff % Save the figure as a TIFF file %MATLAB Verification sys = filt([1 0 -1],[1 1 0.25]) h1 = impulse(sys,10) %i_response = [1.0 -1.0 -0.25 0.50 -0.4375 0.3125 -0.2031 0.1250 -0.0742 0.0430] % Part (iii) – Output response k=[0:10] ; y1 = (0.5).^k ; y2 = (-0.5).^k; y = -0.25*(3*y1+(6*k-7).*y2) % y= [1.0000 -0.5000 -0.5000 0.2500 -0.3125 0.1563 0.0625 -0.0430 0.0215] stem(k, y, 'filled'), grid xlabel('k') % Label of X-axis ylabel('y[k]') % Label of Y-axis print -dtiff plot.tiff % Save the figure as a TIFF file %MATLAB Verification % 1 − z −2 1 − z −2 Y (z) =

(1 − 0.5z )(1 + 0.5z ) −1

−1

2

=

-0.1250

1 + 0.5 z −1 − 0.25 z −2 − 0.125 z −3

sys = filt([1 0 -1],[1 0.5 -0.25 -0.125]) output = impulse(sys,10) % output= [1.0000 -0.5000 -0.5000 0.2500 0.1250 0.0625 -0.0430 0.0215]

-0.3125

0.1563

-

Solutions

Problem 13.16

(i) x[k ] = u[k + 2] − u[k − 3] and h[k ] = u[k − 5] − u[k − 6] . The z-transform of the input and impulse response is given by

X ( z) =

z 2 − z −3 z −5 − z −6 = z −6 ROC : z > 1 . ROC : z > 1 and H ( z ) = z −1 z −1

The output is then given by

Y ( z) =

z −4 − z −9 ROC : z > 1 . z −1

Calculating the inverse z-transform, we obtain the output

y[k ] = u[k − 4] − u[k − 9] . h[k ] = 3− k u[k − 4] = 3−4 × 3− ( k −4) u[k − 4] .

(ii) x[k ] = u[k ] − u[k − 9] and

The z-transform of the input and impulse response is given by

X ( z) =

z − z −8 3−4 z −4 ROC : z > 1 and H ( z ) = ROC : z > (1/ 3) . z −1 z − 13

The output is then given by

Y ( z) =

3−4 ( z −3 − z −12 )

3⎡ 1 1 ⎤ −4 −3 −12 3 ROC : z > 1 = z − z × − ( ) ⎢ ( z − 1)( z − 13 ) 2 ⎣ z − 1 z − 13 ⎥⎦ Y ( z) =

or,

⎡ z 1 −4 z ⎤ z − z −13 ) ⎢ ROC : z > 1 − ( 1⎥ 54 ⎣ z −1 z − 3 ⎦

⎡ z z ⎤ , the output is obtained as − 1⎥ ⎣ z −1 z − 3 ⎦

Noting that ⎡⎣1 − 3− k ⎤⎦ u[k ] ↔ ⎢

y[ k ] =

1 1 ⎡⎣1 − 3− ( k −4) ⎤⎦ u[k − 4] − ⎡⎣1 − 3− ( k −13) ⎤⎦ u[k − 13] . 54 54

(iii) x[k ] = 2 − k u[k ] = 0.5k u[k ] and

h[k ] = k (u[k ] − u[k − 4])

The z-transform of the input signal is given by

X ( z) =

z ROC : z > 0.5 . z − 0.5

The z-transform of the impulse response is given by ∞

3

k =0

k =0

H ( z ) = ∑ k (u[k ] − u[k − 4]) z − k = ∑ kz − k = z −1 + 2 z −2 + 3z −3 The z-transform Y ( z ) of the output response is then given by

ROC : z > 0

29

30

Chapter 13

Y ( z ) = ( z −1 + 2 z −2 + 3z −3 ) X(z)

ROC : z > 0.5

Calculating the inverse z-transform, we obtain the output

y[k ] = x[k − 1] + 2 x[k − 2] + 3x[k − 3] = 0.5k −1 u[k − 1] + 2 × 0.5k −2 u[k − 2] + 3 × 0.5k −3 u[k − 3]. (iv) x[k ] = u[k ] and h[k ] = 4

−k

.

Z α |k | ←⎯→

Recall that

(α − 1 / α ) z ( z − α)( z − 1 / α)

ROC : α < z < (1 / α) .

The z-transform of the input and impulse response is given by X ( z) =

z − 3.75 z ROC : z > 1 and H ( z ) = ROC : 0.25 < z < 4 . z −1 ( z − 0.25)( z − 4)

The output is then given by Y ( z) =

− 3.75 z 2 ROC : 1 < z < 4 . ( z − 1)( z − 0.25)( z − 4)

By partial fraction expansion, we obtain Y ( z ) 1.6667 0.3333 1.3333 = − − z z −1 z − 0.25 z−4 ROC:| z| >1

ROC:| z| >0.25

ROC:| z|< 4

which results in the output

y[k ] = 1.6667u[k ] − 0.3333 × 0.25k u[k ] + 1.3333 × 4k u[ − k − 1] . (v) x[k ] = 2 − k u[k ] and h[k ] = 2 k u[−k − 1] . The z-transform of the input and impulse response is given by

X ( z) =

−z z ROC : z > 0.5 and H ( z ) = ROC : z < 2 . z − 0 .5 z−2

The output is then given by Y ( z) =

− z2 ROC : 0.5 < z < 2 . ( z − 0.5)( z − 2)

By partial fraction expansion, we get Y ( z ) 0.3333 1.3333 = − z z − 0.5 z−2 ROC:| z| >0.5

ROC:| z|< 2

which results in the output

y[k ] = 0.3333 × 0.5k u[k ] + 1.3333 × 2k u[ − k − 1] .

Solutions

31

Problem 13.17

(i) Calculating the z-transform of the input and output, we obtain

X ( z) =

1 1 2 − (7 /12) z −1 + = 1 − (1/ 4) z −1 1 − (1/ 3) z −1 (1 − (1/ 3) z −1 ) (1 − (1/ 4) z −1 )

ROC: z > (1/ 3)

2 4 2 + (1/ 2) z −1 Y ( z) = − =− ROC: z > (3/ 4) 1 − (1/ 4) z −1 1 − (3/ 4) z −1 (1 − (1/ 3) z −1 )(1 − (3/ 4) z −1 ) The transfer function of the system is given by

⎡⎣ 2 + (1/ 2) z −1 ⎤⎦ Y ( z) H ( z) = =− X ( z) ⎡⎣ 2 − (7 /12) z −1 ⎤⎦ ⎡⎣ 2 + (1/ 2) z −1 ⎤⎦ =− ⎡⎣ 2 − (7 /12) z −1 ⎤⎦

⎡⎣1 − (1/ 3) z −1 ⎤⎦ ⎡⎣1 − (1/ 3) z −1 ⎤⎦

⎡⎣1 − (1/ 4) z −1 ⎤⎦ ⎡⎣1 − (3 / 4) z −1 ⎤⎦

ROC: z > (3 / 4)

⎡⎣1 − (1/ 4) z −1 ⎤⎦ ⎡⎣1 − (3 / 4) z −1 ⎤⎦

1 − (1/16) z −1 =− 1 − (25 / 24) z −1 + (21/ 96) z −2

ROC: z > (3 / 4)

(ii) Calculating the partial fraction expansion of H(z), we obtain H ( z) z − (1 / 16) 1.5 0.5 =− 2 ≡− + . z ( z − 3 / 4) ( z − 7 / 12) z − (25 / 24) z + ( 21 / 96) ROC:| z| >( 3 / 4 )

ROC:| z| > ( 7 / 12 )

Calculating the inverse z-transform, we obtain

h[k ] = ( −1.5 × 0.75k − 0.5 × 0.2917k )u[k ]. (iii) By expressing the transfer function as H ( z) =

Y ( z) 1 − (1 / 16) z −1 =− X ( z) 1 − ( 25 / 24) z −1 + (21 / 96) z −2

and cross multiplying, we obtain Y ( z ) − ( 25 / 24) z −1Y ( z ) + (21 / 96) z −2Y ( z ) = − X ( z ) + (1 / 16) z −1 X ( z ) Calculating the inverse z-transform, we obtain y[k ] − (25 / 24) y[k − 1] + (21 / 96) y[k − 2] = − x[k ] + (1 / 16) x[k − 1] with initial conditions: y[−2] = y[−1] = 0.

Problem 13.18

The impulse response of the LTIC system is given by ∞

h(t ) = ∑ 0.3kT δ (t − kT ) k =0

32

Chapter 13

The impulse response includes a series of causal CT impulse functions with decaying amplitude. Note that the input also is a train of causal impulse functions with decaying magnitude. It appears that both the input and the system are CT representation of discrete signal and system, respectively. Therefore, it will be easier to calculate the output in the DT domain. To solve the problem in the discrete domain, we use the impulse transformation. The equivalent DT input is given by f [k ] = 0.2 k u[k ] , with F ( z ) =

z . z − 0.2

The transfer function of the equivalent DT system is given by

H ( z) =

z z − 0.3

ROC : z > 0.3 .

The z-transform of the DT output is therefore given by Y ( z) = H ( z)F ( z) =

z2 . ( z − 0.3)( z − 0.2)

Using partial fraction expansion, the output is expressed as

Y ( z) z 3 2 = = − , ( z − 0.3)( z − 0.2) z − 0.3 z − 0.2 z y[k ] = 3 × 0.3k u[k ] − 2 × 0.3k u[k ] = ( 3 × 0.3k − 2 × 0.3k ) u[k ] .

or,

(

)

The equivalent CT output is given by, y (t ) = ∑ 3 × 0.3kT − 2 × 0.2kT δ (t − kT ) . k =0

The above answer can also be obtained by using the Laplace transform in the CT domain.

Problem 13.19

To determine the stability, we will assume that the systems are physically realizable, i.e., causal. (i)

H ( z) =

z−2 ( z − 0.6 + j 0.8)( z 2 + 0.25)

The pole-zero plot is shown in Fig. S13.21(i). There is one zero, at z = 2, and three poles at z = 0.6 – j0.8, ±j0.5. Two poles at z = ±0.5 are inside the unit circle ( z = 0.5 ), and the pole at z = 0.6 − j 0.8 is on the unit circle ( z = 1 ). Therefore, the system is a marginally stable system. (ii)

H ( z) =

( z − 2)( z − 1) ( z − 2)( z − 1) z −1 = = 2 2 ( z − 2.5 z + 1)( z + 0.25) ( z − 2)( z − 0.5)( z + 0.25) ( z − 0.5)( z 2 + 0.25) 2

The pole-zero plot is shown in Fig. S13.21(ii). There is one zero at z = 1 and three poles at z = 0.5, ±j0.5. As all poles are inside the unit circle ( z = 0.5 ), the system is absolutely stable. (iii)

H ( z) =

z − 0.2 ( z + 0.1)( z 2 + 4)

Solutions

33

The pole-zero plot is shown in Fig. S13.21(iii). There is one zero at z = 0.2, and three poles at z = 0.1, ±j2. The pole at z = 0.1 is inside the unit circle. However, the two poles at z = ± j 2 are outside the unit circle ( z = 2 ).Therefore, the system is an unstable system. For system (iii), by selecting the ROC, 0.1 < |z| < 2, a stable implementation of H(z) can be obtained (as the ROC includes the unit circle). However, such an implementation will not be causal (physically realizable). A stable and causal implementation is not possible for this transfer function. (iv)

H ( z ) = z −1 − 2 z −2 + z − 3 =

z 2 − 2 z + 1 ( z − 1) 2 = z3 z3

The pole-zero plot is shown in Fig. S13.21(iv). There are two zeros at z = 1, and three poles at z = 0. As all three poles are inside the unit circle ( z = 0 ), the system is absolutely stable. (v)

H ( z) = =

( z 2 + 2.5 z + 0.9 + j 0.15) z ( z 2 + 2.5 z + 0.9 + j 0.15) z = 2 z + (1.8 + j 0.3) z + (0.6 + j 0.6) z − 0.2 + j 0.3 ( z + 1) 2 ( z − 0.2 + j 0.3) 3

z ( z + 0.4309 + j 0.0916)( z + 2.0691 − j 0.0916) ( z + 1) 2 ( z − 0.2 + j 0.3)

The pole-zero plot is shown in Fig. S13.21(v). There is a double pole on the unit circle. Therefore, the system is unstable. Note that if it was a single pole on the unit circle, the system would have been marginally stable. (vi)

H ( z) = =

z 3 − 1.2 z 2 + 2.5 z + 0.8 z 6 + 0.3z 5 + 0.23z 4 + 0.209 z 3 + 0.1066 z 2 − 0.04162 z − 0.07134 ( z + 0.2753)( z − 0.7376 − j1.5369)( z − 0.7376 + j1.5369) ( z − 0.5)( z + 0.6)( z − 0.3 − j 0.7)( z − 0.3 + j 0.7)( z + 0.4 − j 0.5)( z + 0.4 + j 0.5)

The pole-zero plot is shown in Fig. S13.21(vi). All six poles are inside the unit circle. Therefore, the system is stable. ▌

Problem 13.20

The frequency response of the system is given by H (Ω) = H ( z ) z = exp( jΩ ) =

e jΩ 1 . = jΩ e + 0.1 1 + 0.1e − jΩ

The frequency response is plotted in Figure S13.20 using the following MATLAB code. A blown-up is also included to calculate the amplitude and phase gain at Ω = π/10.

34

Chapter 13

omega = [-pi:pi/20:pi] ; H = 1./(1+0.1*exp(-j*omega)) ; subplot(2,1,1), plot(omega, abs(H)), grid xlabel('Omega') % Label of X-axis ylabel('|H(Omega)|') % Label of Y-axis % axis([-3.2 3.2 0.9 1.2]) print -dtiff plot.tiff % Save figure as a TIFF file % subplot(2,1,2), plot(omega, angle(H)), grid xlabel('Omega') % Label of X-axis ylabel('