Solutions to Additional Problems

4.16. Find the FT representations for the following periodic signals: Sketch the magnitude and phasespectra.(a) x(t) = 2 cos(t) + sin(2t)

x(t) = ejt + ejt +12j

ej2t 12j

ej2t

o = lcm(, 2) =

x[1] = x[1] = 1

x[2] = x[2] = 12j

X(j) = 2

k=X[k]( ko)

X(j) = 2 [( ) + ( + )] + 1j

[( 2) ( + 2)]

10 5 0 5 10 15 200

1

2

3

4

5

6

7

|X(j

)|

(a) Magnitude and Phase plot

8 6 4 2 0 2 4 6 82

1

0

1

2

arg(

X(j

))

Figure P4.16. (a) Magnitude and Phase plot

(b) x(t) =4

k=0(1)kk+1 cos((2k + 1)t)

x(t) =12

4k=0

(1)kk + 1

[ej(2k+1)t + ej(2k+1)t

]

1

X(j) = 4

k=0

(1)kk + 1

[( (2k + 1)) + ( + (2k + 1))]

30 20 10 0 10 20 300

0.5

1

1.5

2

2.5

3

3.5

|X(j

)|

(b) Magnitude and Phase plot

30 20 10 0 10 20 300

0.5

1

1.5

2

2.5

3

3.5

arg(

X(j

))

Figure P4.16. (b) Magnitude and Phase plot

(c) x(t) as depicted in Fig. P4.16 (a).

x(t) =

{1 |t| 10 otherwise

+

{2 |t| 120 otherwise

X(j) =

k=

[2 sin(k 23 )

k+

4 sin(k 3 )k

]( k 2

3)

2

0 5 10 15 20 250

2

4

6

8

10|X

(j)|

(c) Magnitude and Phase plot

0 5 10 15 20 250

0.5

1

1.5

2

2.5

3

3.5

arg(

X(j

))

Figure P4.16. (c) Magnitude and Phase plot

(d) x(t) as depicted in Fig. P4.16 (b).

T = 4 o =

2

X[k] =14

22

2tej2 ktdt

=

{0 k = 02j cos(k)

k k = 0

X(j) = 2

k=X[k](

2k)

3

0 2 4 6 8 10 12 14 160

0.1

0.2

0.3

0.4

0.5

0.6

0.7|X

(j)|

(d) Magnitude and Phase plot

0 2 4 6 8 10 12 14 162

1

0

1

2

arg(

X(j

))

Figure P4.16. (d) Magnitude and Phase plot

4.17. Find the DTFT representations for the following periodic signals: Sketch the magnitude andphase spectra.(a) x[n] = cos(8 n) + sin(

5 n)

x[n] =12

[ej

8 n + ej

8 n

]+

12j

[ej

5 n ej 5 n

]o = lcm(

8,

5) =

40

X[5] = X[5] = 12

X[8] = X[8] = 12j

X(ej) = 2

k=X[k]( ko)

X(ej) = [(

8) + ( +

8)]

+

j

[(

5) ( +

5)]

4

0.8 0.6 0.4 0.2 0 0.2 0.4 0.6 0.80

0.5

1

1.5

2|X

(ej

)|

(a) Magnitude and Phase plot

0.8 0.6 0.4 0.2 0 0.2 0.4 0.6 0.82

1

0

1

2

arg(

X(e

j)

Figure P4.17. (a) Magnitude and Phase response

(b) x[n] = 1 +

m= cos(4 m)[nm]

N = 8 o =

4

x[n] = 1 +

m=cos(

4m)[nm]

= 1 + cos(

4n)

X[k] =18

3n=4

x[n]ejk4 n

For one period of X[k], k [4, 3]X[4] = 0

X[3] = 1 20.5

8ejk

34

X[2] = 18

ejk24

X[1] = 1 + 20.5

8ejk

4

X[0] =28

X[1] =1 + 20.5

8ejk

4

X[2] =18

ejk24

5

X[3] =1 20.5

8ejk

34

X(ej) = 2

k=X[k]( ko)

= [(1 20.5)

4( +

34

) +14

( +

2) +

(1 + 20.5)4

( +

4) +

14

(2)]

+

[(1 + 20.5)

4(

4) +

14

( 2

) +(1 20.5)

4( 3

4)]

2.5 2 1.5 1 0.5 0 0.5 1 1.5 2 2.50

0.5

1

1.5

2

|X(e

j)|

(b) Magnitude and Phase plot

2.5 2 1.5 1 0.5 0 0.5 1 1.5 2 2.51

0.5

0

0.5

1

arg(

X(e

j))

Figure P4.17. (b) Magnitude and Phase response

(c) x[n] as depicted in Fig. P4.17 (a).

N = 8 o =

4

X[k] =sin(k 58 )8 sin(8 k)

X[0] =58

X(ej) = 2

k=X[k]( k

4)

6

0 2 4 6 8 10 120

1

2

3

4

|X(e

j)|

(c) Magnitude and Phase plot

0 2 4 6 8 10 120

0.5

1

1.5

2

2.5

3

3.5

arg(

X(e

j))

Figure P4.17. (c) Magnitude and Phase response

(d) x[n] as depicted in Fig. P4.17 (b).

N = 7 o =27

X[k] =17

(1 ejk 27 ejk 27

)=

17

(1 2 cos(k 2

7))

X(ej) = 2

k=X[k]( k 2

7)

7

0 2 4 6 8 10 120

0.05

0.1

0.15

0.2

0.25

0.3

0.35|X

(ej

)|

(d) Magnitude and Phase plot

0 2 4 6 8 10 121

0.5

0

0.5

1

arg(

X(e

j))

Figure P4.17. (d) Magnitude and Phase response

(e) x[n] as depicted in Fig. P4.17 (c).

N = 4 o =

2

X[k] =14

(1 + ejk

2 ejk ejk 32

)=

14(1 (1)k) j

2sin(k

2)

X(ej) = 2

k=X[k]( k

2)

8

0 2 4 6 8 10 120

1

2

3

4

5

|X(e

j)|

(e) Magnitude and Phase plot

0 2 4 6 8 10 122

1.5

1

0.5

0

0.5

1

arg(

X(e

j))

Figure P4.17. (e) Magnitude and Phase response

4.18. An LTI system has the impulse response

h(t) = 2sin(2t)

tcos(7t)

Use the FT to determine the system output for the following inputs, x(t).

Let a(t) =sin(2t)

t

FT A(j) =

{1 || < 20 otherwise

h(t) = 2a(t) cos(7t)FT

H(j) = A(j( 7)) + A(j( + 7))

(a) x(t) = cos(2t) + sin(6t)

X(j) = 2

k=X[k]( ko)

X(j) = ( 2) + ( + 2) + j

( 6) j

( + 6)

Y (j) = X(j)H(j)

=

j( 6)

j( + 6)

y(t) = sin(6t)

9

(b) x(t) =

m=(1)m(tm)

X[k] =12(1 ejk) = 1

2(1 (1)k)

=

{0 n even1 n odd

X(j) = 2

l=( l2)

Y (j) = X(j)H(j)

= 24l=3

[( l2) + ( + l2)]

y(t) = 2 cos(6t) + 2 cos(8t)

(c) x(t) as depicted in Fig. P4.18 (a).

T = 1 o = 2

X(j) = 2

k=

(sin(k 4 )

k(1 ejk)

)( k2)

Y (j) = X(j)H(j)

= 2[sin(34 )

3(1 ej3)( 6) + sin(3

4 )

3 (1 ej3)( + 6)

]

=4 sin(34 )

3( 6) + 4 sin(3

4 )

3( + 6)

y(t) =4 sin( 34 )

3cos(6t)

(d) x(t) as depicted in Fig. P4.18 (b).

X[k] = 4 1

8

1816tejk8tdt

=2j cos(k)

k

X(j) = 2

k=X[k]( k8)

Y (j) = 4j( 8) + 4j( + 8)

y[n] = 4

sin(8n)

(e) x(t) as depicted in Fig. P4.18 (c).

T = 1 o = 2

10

X[k] = 1

0

etejk2tdt

=1 e(1+jk2)

1 + jk2

=1 e11 + jk2

X(j) = 2

k=X[k]( k2)

Y (j) = X(j)H(j)

= 2(

1 e11 + j6

( 6) + 1 e1

1 + j8( 8) + 1 e

1

1 j6 ( + 6) +1 e11 j8 ( + 8)

)

Y (j) = (1 e1)[

ej6t

1 + j6+

ej8t

1 + j8+

ej6t

1 j6 +ej8t

1 j8

]

y(t) = 2(1 e1)[Re

{ej6t

1 + j6+

ej8t

1 + j8

}]y(t) = 2(1 e1) [r1 cos(6t + 1) + r2 cos(8t + 2)]

Where

r1 =

1 + (6)2

1 = tan1(6)r2 =

1 + (8)2

1 = tan1(8)

4.19. We may design a dc power supply by cascading a full-wave rectifier and an RC circuit as depictedin Fig. P4.19. The full wave rectifier output is given by

z(t) = |x(t)|

Let H(j) = Y (j)Z(j) be the frequency response of the RC circuit as shown by

H(j) =1

jRC + 1

Suppose the input is x(t) = cos(120t).(a) Find the FT representation for z(t).

o = 240 T =1

120

Z[k] = 120 1

240

1240

12

(ej120t + ej120t

)ejk240tdt

=2(1)k

(1 4k2)

Z(j) = 4

k=

(1)k(1 4k2)( k240)

11

(b) Find the FT representation for y(t).

H(j) =Y (j)Z(j)

In the time domain:

z(t) y(t) = RC ddt

y(t)FT

Z(j) = (1 + jRC)Y (j)

H(j) =1

1 + jRCY (j) = Z(j)H(j)

= 4

k=

(1)k(1 4k2)

(1

1 + jk240RC

)( k240)

(c) Find the range for the time constant RC such that the first harmonic of the ripple in y(t) is less than1% of the average value.The ripple results from the exponential terms. Let = RC.

Use first harmonic only:

Y (j) 4

[() +

13

(( 240)1 + j240RC

+( + 240)1 + j240RC

)]

y(t) =22

+2

32

[ej240t

1 + j240RC+

ej240t

1 j240RC

]

|ripple| = 232

[2

1 + (240)2

]< 0.01

(22

)240 > 66.659

> 0.0884s

4.20. Consider the system depicted in Fig. P4.20 (a). The FT of the input signal is depicted in Fig. 4.20

(b). Let z(t)FT

Z(j) and y(t)FT

Y (j). Sketch Z(j) and Y (j) for the following cases.(a) w(t) = cos(5t) and h(t) = sin(6t)t

Z(j) =12

X(j) W (j)W (j) = (( 5) + ( + 5))

Z(j) =12

(X(j( 5)) + X(j( + 5)))

H(j) =

{1 || < 60 otherwise

C(j) = H(j)Z(j) = Z(j)

Y (j) =12

C(j) [ (( 5) + ( + 5))]

Y (j) =12

[Z(j( 5)) + Z(j( + 5))]

12

=14

[X(j( 10)) + 2X(j) + X(j( + 10))]

5 w

Z(jw)

0.5

Figure P4.20. (a) Sketch of Z(j)

10

Y(jw)

0.5

0.25

w

Figure P4.20. (a) Sketch of Y (j)

(b) w(t) = cos(5t) and h(t) = sin(5t)t

Z(j) =12

[X(j( 5)) + X(j( + 5))]

H(j) =

{1 || < 50 otherwise

C(j) = H(j)Z(j)

Y (j) =12

C(j) [ (( 5) + ( + 5))]

Y (j) =12

[C(j( 5)) + C(j( + 5))]

5 w

Z(jw)

0.5

Figure P4.20. (b) Sketch of Z(j)

13

10

0.25

w

Y(jw)

Figure P4.20. (b) Sketch of Y (j)

(c) w(t) depicted in Fig. P4.20 (c) and h(t) = sin(2t)t cos(5t)T = 2, o = , To = 12

W (j) =

k=

2 sin(k 2 )k

( k)

Z(j) =12

X(j) W (j)

=

k=

sin(k 2 )k

X(j( k))

C(j) = H(j)Z(j)

=7

k=3

sin(k 2 )k

X(j( k)) +7

k=3

sin(k 2 )k

X(j( k))

Y (j) =12

[C(j( 5)) + C(j( + 5))]

w

0.5

37 73

Z(jw)

. . .. . .

Figure P4.20. (c) Sketch of Z(j)

14

2

2150

1

10

8 12

53

57

w

Y(jw)

Figure P4.20. (c) Sketch of Y (j)

4.21. Consider the system depicted in Fig. P4.21. The impulse response h(t) is given by

h(t) =sin(11t)

t

and we have

x(t) =k=1

1k2

cos(k5t)

g(t) =10k=1

cos(k8t)

Use the FT to determine y(t).

y(t) = [(x(t) h(t))(g(t) h(t))] h(t)= [xh(t)gh(t)] h(t)= m(t) h(t)

h(t) =sin(11t)

t

FT H(j) =

{1 || 110 otherwise

x(t) =k=1

1k2

cos(k5t)FT

X(j) = k=1

1k2

[( 5k) + ( + 5k)]

g(t) =10k=1

cos(k8t) = 10k=1

[( 8k) + ( + 8k)]

Xh(j) = X(j)H(j)

= 2

k=1

1k2

[( 5k) + ( + 5k)]

15

Gh(j) = G(j)H(j)

= ( 8) + ( 8)

M(j) =12

Xh(j) Gh(j)

=12

[Xh(j( 8)) + Xh(j( + 8))]

= 2

k=1

1k2

[(( 8 5k) + ( 8 + 5k)) + (( + 8 5k) + ( + 8 + 5k))]

Y (j) = M(j)H(j)

=

2[( + 3) + ( 3)] +

8[( 2) + ( + 2)]

y(t) =12

cos(3t) +18

cos(2t)

4.22. The input to a discrete-time system is given by

x[n] = cos(

8n) + sin(

34

n)

Use the DTFT to find the output of the system, y[n], for the following impulse responses h[n], first notethat

X(ej) = [(

8) + ( +

8)]

+

j

[( 3

4) ( + 3

4)]

(a) h[n] = sin(4 n)

n

H(ej) =

{1 || 40 4 || < , 2 periodic.

Y (ej) = H(ej)X(ej)

= [(

8) + ( +

8)]

y[n] = cos(

8n)

(b) h[n] = (1)n sin(2 n)

n

h[n] = ejnsin(2 n)

n

H(ej) =

{1 | | 40 4 | | < , 2 periodic.

Y (ej) = H(ej)X(ej)

16

=

j

[( 3

4) + ( +

34

)]

y[n] = sin(34

n)

(c) h[n] = cos(2 n)sin( 5 n)

n

h[n] = cos(

2n)

sin(5 n)n

H(ej) =

{12 | 2 | 50 5 | 2 | <

+

{12 | + 2 | 50 5 | + 2 | < , 2 periodic.

Y (ej) = H(ej)X(ej)

= 0

y[n] = 0

4.23. Consider the discrete-time system depicted in Fig. P4.23. Assume h[n] = sin(2 n)

n . Use the DTFTto determine the output, y[n] for the following cases: Also sketch G(ej), the DTFT of g[n].

y[n] = (x[n]w[n]) h[n]= g[n] h[n]

h[n] =sin(2 n)

n

FT H(ej) =

{1 || 20 2 || <

H(ej) is 2 periodic.

(a) x[n] = sin(4 n)

n , w[n] = (1)n

x[n] =sin(4 n)

n

DTFT X(ej) =

{1 || 40 4 || <

w[n] = ejnDTFT W (ej) = 2( )

G(ej) =12

X(ej) W (ej)

=

{1 | | 40 4 | | <

g[n] = ejnsin(4 n)

n

Y (ej) = G(ej)H(ej)

= 0

y[n] = 0

17

G(ej )

34

5

4

1

Figure P4.23. (a) The DTFT of g[n]

(b) x[n] = [n] sin(4 n)

n , w[n] = (1)n

x[n] = [n] sin(4 n)

n

DTFT X(ej) =

{0 || 41 4 || <

w[n] = ejnDTFT W (ej) = 2( )

G(ej) =12

X(ej) W (ej)

=

{0 | | 341 34 | | <

g[n] =sin( 34 n)

n

Y (ej) = G(ej)H(ej)

= H(ej)

y[n] =sin(2 n)

n

G(ej )

34

1

Figure P4.23. (b) The DTFT of g[n]

(c) x[n] = sin(2 n)

n , w[n] = cos(2 n)

18

W (ej) = [(

2) + ( +

2)]

, 2 periodic

G(ej) =12

X(ej) W (ej)

=

{12 | 2 | 20 2 | 2 | <

+

{12 | + 2 | 20 2 | + 2 | <

g[n] =12

sin(2 n)n

(ej

2 n + ej

2 n

)=

sin(2 n)n

cos(

2n)

=sin(n)

2n

=12

(n)

y[n] = g[n] h[n]

=12

h[n]

=sin(n)

2n

G(ej )

. . . . . .0.5

Figure P4.23. (c) The DTFT of g[n]

(d) x[n] = 1 + sin( 16n) + 2 cos(34 n), w[n] = cos(

38 n)

X(ej) = 2() +

j

[[(

16) + ( +

16)]

+ 2[( 3

4) + ( +

34

)]

, 2 periodic.

W (ej) = [( 3

8) + ( +

38

)]

, 2 periodic

G(ej) =12

X(ej) W (ej)

=12

[X(ej(

38 )) + X(ej(+

38 ))

]=

12

[2( 3

8) +

j

(( 7

16) ( 5

16))

+ 2(

( 98

) + ( +38

))]

+12

[2( +

38

) +

j

(( +

516

) ( + 716

))

+ 2(

( 38

) + ( +98

))]

19

g[n] =2

cos(38

n) +12

sin(716

n) 12

sin(516

n) +1

cos(98

n)

Y (ej) = G(ej)H(ej)

y[n] =2

cos(38

n) +12

sin(716

n) 12

sin(516

n)

G(ej )

G(ej )

5

16 167

2167 5

16

2

516

616

716

98

0.5

1

2

Figure P4.23. (d) The DTFT of g[n]

4.24. Determine and sketch the FT representation, X(j), for the following discrete-time signals withthe sampling interval Ts as given:

X(j) =

n=ejTsn

= X(ej)=Ts

(a) x[n] = sin(3 n)

n , Ts = 2

X(ej) =

{1 || 30 3 || <

20

X(j) =

{1 |2| 30 3 |2| <

=

{1 || 60 6 || < 6

X(j) is periodic

X (jw)

1

w6

. . .. . .

Figure P4.24. (a) FT of X(j)

(b) x[n] = sin(3 n)

n , Ts =14

X(ej) =

{1 || 30 3 || <

X(j) =

{1 | 14| 30 3 | 14| <

=

{1 || 430 43 || < 4

X(j) is 8 periodic

X (jw)

43

8

1. . .. . .

w

Figure P4.24. (b) FT of X(j)

21

(c) x[n] = cos(2 n)sin( 4 n)

n , Ts = 2

X(ej) =

{12 | 2 | 40 4 | 2 | <

+

{12 | + 2 | 40 4 | + 2 | <

X(j) =

{12 |2 2 | 30 3 |2 2 | <

+

{12 |2 + 2 | 30 3 |2 + 2 | <

=

{12

8 < 111 , aliasing occurs.

26

X (jw)

10 424

24104. . .

w

7. . .

Figure P4.26. (ii) FT of the sampled signal

(iii) Ts = 15Since Ts > 111 , aliasing occurs.

X (jw)

10 10

. . . . . .

w

Figure P4.26. (iii) FT of the sampled signal

(b) Let x[n] = x(nTs). Sketch the DTFT of x[n], X(ej), for each of the sampling intervals givenin (a).The DTFT simple scales the x axis by the sampling rate.

(i) Ts = 114

X(ej )

1014

1414

3814

3814

1014

1414. . .

14. . .

Figure P4.26. (i) DTFT of x[n]

(ii) Ts = 17

27

X(ej )

107

47

247

247

107

47. . .

7 . . .

Figure P4.26. (ii) DTFT of x[n]

(iii) Ts = 15

10 10

X(ej

)

5 5

. . . . . .

Figure P4.26. (iii) DTFT of x[n]

4.27. Consider subsampling the signal x[n] = sin(6 n)

n so that y[n] = x[qn]. Sketch Y (ej) for the

following choices of q:

X(ej) =

{1 || 60 6 || < 2periodic

q[n] = x[qn]

Y (ej) =1q

q1m=0

X(

ej1q (m2)

)

(a) q = 2

Y (ej) =12

1m=0

X(

ej12 (m2)

)

(b) q = 4

Y (ej) =14

3m=0

X(

ej14 (m2)

)

28

(c) q = 8

Y (ej) =18

7m=0

X(

ej18 (m2)

)

10 8 6 4 2 0 2 4 6 8 10

0

0.5

1Plot of the subsampled signal x[n]

Y(e

j),

q =

2

10 8 6 4 2 0 2 4 6 8 100.1

0

0.1

0.2

0.3

Y(e

j),

q =

4

10 8 6 4 2 0 2 4 6 8 100

0.1

0.2

0.3

0.4

Y(e

j),

q =

8

Figure P4.27. Sketch of Y (ej)

4.28. The discrete-time signal x[n] with DTFT depicted in Fig. P4.28 is subsampled to obtain y[n] =x[qn]. Sketch Y (ej) for the following choices of q:(a) q = 3(b) q = 4(c) q = 8

29

15 10 5 0 5 10 150

0.2

0.4

0.6

0.8Plot of the subsampled signal x[n]

Y(e

j),

q =

3

20 15 10 5 0 5 10 15 200

0.1

0.2

0.3

0.4

Y(e

j),

q =

4

20 15 10 5 0 5 10 15 200

0.05

0.1

0.15

0.2

Y(e

j),

q =

8

. . . . . .

Figure P4.28. Sketch of Y (ej)

4.29. For each of the following signals sampled with sampling interval Ts, determine the bounds on Tsthat gaurantee there will be no aliasing.(a) x(t) = 1t sin 3t + cos(2t)

1t

sin(3t)FT

{

1 || 30 otherwise

cos(2t)FT

( 2) + ( + 2)max = 3

T 0.99

2 sin( Ts2 )

32

(ii) |Hc(j)| 2Ts m

Assuming X(j) = 0 for || > m

(c) Determine the constraints on |Hc(j)| so that the overall magnitude response of this reconstruc-tion system is between 0.99 and 1.01 in the signal passband and less than 104 to the images of the signalspectrum for the following values. Assume x(t) is bandlimited to 12, that is, X(j) = 0 for || > 12.Constraints:

(1) In the pass band:

0.99 < |H1(j)||Hc(j)| < 1.01

0.992

4 sin2( Ts2 )< |Hc(j)| s

>sN

=2

NTsTherefore:

>

4.39. A signal x(t) is sampled at intervals of Ts = 0.1 s. Assume |X(j)| 0 for || > 12 rad/s.Determine the frequency range a < < a over which the DTFS offers a reasonable approximationto X(j), the minimum number of samples required to obtain an effective resolution r = 0.01 rad/s,and the length of the DTFS required so the frequency interval between DTFS coefficients is = 0.001rad/s.

Ts 2. Assume 1 > 21.In this case we can sample x(t) at a rate less than that indicated by the sampling interval and still performperfect reconstruction by using a bandpass reconstruction filter Hr(j). Let x[n] = x(nTs). Determinethe maximum sampling interval Ts such that x(t) can be perfectly reconstructed from x[n]. Sketch thefrequency response of the reconstruction filter required for this case.

We can tolerate aliasing as long as there is no overlap on 1 || 2We require:

s 2 1s 2 1

Implies:

Ts 2

2 1

Hr(j) =

{Ts 1 || 20 otherwise

4.44. Suppose a periodic signal x(t) has FS coefficients

X[k] =

{( 34 )

k, |k| 40, otherwise

The period of this signal is T = 1.(a) Determine the minimum sampling interval for this signal that will prevent aliasing.

X(j) = 24

k=4(34)k( k2)

m = 82Ts

> 2(8)

min Ts =18

(b) The constraints of the sampling theorem can be relaxed somewhat in the case of periodic signals if weallow the reconstructed signal to be a time-scaled version of the original. Suppose we choose a samplinginterval Ts = 2019 and use a reconstruction filter

Hr(j) =

{1, || < 0, otherwise

Show that the reconstructed signal is a time-scaled version of x(t) and identify the scaling factor.

Ts =2019

X(j) =1920

l=

X(j( l1.9))

47

Aliasing produces a frequency scaled replica of X(j) centered at zero. The scaling is by a factor of20 from o = 2 to o = 0.1. Applying the LPF, || < gives x( t20 ), and xreconstructed(t) = 1920x( t20 )

(c) Find the constraints on the sampling interval Ts so that use of Hr(j) in (b) results in the re-construction filter being a time-scaled version of x(t) and determine the relationship between the scalingfactor and Ts.

The choice of Ts is so that no aliasing occurs.

(1)2Ts

< 2

Ts > 1 period of the original signal.

(2)(

2 2Ts

)4