Lecture03 Fourier Representations of Signals

Fourier Representations of Signals & LTI Systems

CHAPTER

X [k ]

Even function

arg ^ X [ k ]`

X[ ]k { Magn Magnit itud udee spec spectr trum um of

[x ]n

arg ^ X[ k]` { Phase spectrum of


CHAPTER

Example 3.3 Computation of DTFS by Inspection by Inspection Determine the DTFS coefficients of x x [n] = cos cos (n /3 + ), using the method of inspection. 1. Period: N = N = 6 o = 2 /6 = /3 2. Using Euler’s formula, x formula, x [n] can be expressed as

x[ n]

Odd function

j(

e

x[ n] 

S 3

nI )

e

 j(

S 3

nI )

2



1 2

S

 j n

e jI e

3



1 2

S j n

ejI e 3

(3.13)

3. Compare Eq. (3.13) with the DTFS of Eq. (3.10) with summing from k = k = 2 to k = k = 3:

x> n@ 

o

= /3, written by

3

¦ X[ k] e

jk jkS n/ 3

k  2

 X[ 2] e j2S n/ 3  X[1] e jS n/ 3  X [0]  X [1] e jS n/ 3  X [2] e j2S n/ 3  X [3] e jS n Equating terms in Eq.(3.13) with those in Eq. (3.14) having equal frequencies, k /3, gives

Figure 3.6 (p. 204) 204) 204) Magnitude and phase of the DTFS coefficients for the signal in Fig. 3.5. Signals and Systems_Simon Haykin & Barry Van Veen


CHAPTER

x> n@

DTFS ;

S 3

m  o

-e  jI / 2, k  1 ° X> k@  ® e jI / 2, k 1 ° 0, o t h e r w i s e o n  2 d k d 3 ¯

4. Magnitude spectrum and phase spectrum of X of X [k ]: ]: Fig. 3.8. 3.8. Figure 3.8 (p. 206) 20 6) 206) Magnitude and phase of DTFS coefficients for Example 3.3.

13

(3.14)

Signals and Systems_Simon Haykin & Barry Van Veen

14


CHAPTER

Example 3.4 DTFS Representation of An Impulse Train Find the DTFS coefficients of the N -periodic -periodic impulse train

x> n@ 

f

¦ G > n lN@

l f

as shown in Fig. 3.9. 3.9. It is convenient to evaluate Eq.(3.11) over the i nterval n = 0 to n = N 1 to obtain

X> k@ 

1 N

N 1

¦ G > n@ e n 0

 jk2nS / N



1 N


CHAPTER


CHAPTER

i The Inverse DTFS The similarity between Eqs. (3.10) and (3.11) ind icates that the same mathematical methods can be used to find the time-domain signal corresponding to a set of DTFS coefficients. Example 3.5 The Inverse DTFS Use Eq. (3.10) to determine the time-domain signal x [n] from the DTFS coefficients depicted in Fig. 3.10. 3.10. 1. Period of DTFS coefficients = 9

o

x> n@ 

S

= 2 /9

2. It is convenient to evaluate Eq.(3.11) over the interval k = k =

4 to k = k = 4 to obtain

4

¦ X > k @e

S/3

jk2S n/ 9

k  4

 e j2S / 3e j6 S n/ 9  2e jS / 3 e j4 S n/ 9  1  2 e jS / 3 e j4S n / 9  e j2S / 3 e j6S n / 9  2 cos  6S n / 9  2S /3 / 3  4 cos  4S n / 9  S / 3  1

Example 3.6 DTFS Representation of A Square Wave Find the DTFS coefficients for the N -periodic -periodic square wave given by Signals and Systems_Simon Haykin & Barry Van Veen

-1, x>n@  ® ¯0,


X[ k] 

 3. For k For k = = 0,

e

:ojk

N ,

 e

1. Period of DTFS coefficients = N

o

= 2 / /N N

2. It is convenient to evaluate Eq.(3.11) over the interval n =

¦

 jk:0 ( m M)

 jk:0 n

x[ n] e

M to M to n = N

M

1.

X[ k] 

1

N

e

2 M

¦e

jk:0 M

e

Change of variable on the index of summation: m = n + M

(3.15)  jk:0 m

m 0

2N , …, we have

1 X> k@ 

3. For k For k 0, obtain

X[ k] 

2 M

¦e N

:ojk

(3.15)

N  M 1

1

m0

That is, each period contains 2M 2M + + 1 consecutive ones and the remaining N (2M (2M +1) +1) values are zero, as depicted in Fig. 3.11. 3.11. Note that this definition requires that N > N > 2M 2M +1. +1. Figure 3.11 209) 3.11 (p. 209) Square wave for Example 3.6.

18


CHAPTER

 M d n d M M  n  N  M

1

S

17


CHAPTER

 S/3

Figure 3.10 3.1 0 (p. 208) 20 8) 3.10 208) Magnitude and phase of DTFS coefficients for Example 3.5.

N ,

1

2 M

2 M  1

m0

N

¦ 1 N

r N, r 2 N,



2N , …, we may sum the geometric series in Eq. (3.15) to

§ 1  e jk: M( 2 M1 ) · ¨ ¸, N © 1  e  jk : ¹   ·§ § jk: 0 M

k 0,

,

0

0

kz 0, 

r N, r 2 N, ..... . 

·

§



(3.16) 





·


CHAPTER

X >k @ 

4. Substituting

o

1 sin k : 0 2M  1 / 2  sin k : 0 / 2 

N

,

k

 0, r N , r 2 N ,

CHAPTER




The numerator and denominator of above Eq. are divided by 2 j .

= 2 / /N N , yields

- 1 sin  kS  2 M  1 / N  , k z 0, r N , r 2 N , ° X > k @  ® N sin  kS / N  ° k  0, r N , r 2 N ,  2 M  1 / N , ¯ 1 sin  kS  2 M  1 / N  2 M  1  lim





k

o 0 , r N , r 2N ,



sin  kS / N

N

N

L’Hôpital’s Rule

For this reason, the expression for X X [k ] is commonly written as

X >k @ 

1 sin k S 2M  1 / N  N

Figure 3.12 211 3.12 (p. 211) 211)) The DTFS coefficients for the square wave shown in Fig. 3.11, assuming a period N N = = 50: (a) M M = = 4. (b) M M = = 12.

The value of X X [k ] for k for k = = 0, N , 2N , …, is obtained from the limit as k 0.

sin k S / N 

Fig. 3.12. Signals and Systems_Simon Haykin & Barry Van Veen

21



CHAPTER

CHAPTER

h Symmetry property of DTFS coefficient: X coefficient: X [k ] = X = X [ k ]. ].

x>n@ 

N / 2

¦ X >k @e

B > k @  ®

¯ 2 X> k@ ,

k 0, k 1, 2,

x[ n] 



N / 2 1

¦  X >m@e

jk : 0 n

 X > m@e  jk : n  0

m 1

3. Use X Use X [m] = X = X [ m] and the identity N



N / 2 1

¦

m 1

o

= 2 to obtain

§ e 2 X > m@ ¨ ©

jm:0 n

 e 2

· ¸ ¹

jm:0 n

e j n = cos( n) since sin( n) = 0 for integer n integer n.

N/ 2  1

N / 2

¦ B[ k] cos( k: k  0

k   N / 2 1

x>n@  X >0@  X > N / 2@e jk : 0 n / 2

N/ 2 ,



and write the DTFS for the square wave in terms of a series of harmonically related cosines as

jk : 0 n

x> n@  X >0@  X > N / 2@ e jS n


- X> k@ ,

1. DTFS of Eq. (3.10) can be written as a series involving harmonically related cosines. 2. Assume that N N is is even and let k range k range from ( N /2 ) + 1 to N /2. Eq. (3.10) becomes

22

0

n)

(3.17)

h A similar expression may be derived for N for N odd. odd. Example 3.7 Building a Square Wave From DTFS Coefficients The contribution of each term to the square wave may be illustrated by defining the partial-sum approximation to x to x [n] in Eq. (3.17) as

x J [ n] 



J

¦ B[ k] cos( k: n) k  0

0

(3.18)


CHAPTER

Example 3.9 Direct Calculation of FS Coefficients Determine the FS coefficients for the signal x (t ) depicted in Fig. 3.16. 3.16.

Figure 3.17 (p. 217) Magnitude and phase spectra for Example 3.9.

1. The period of x x (t ) is T = T = 2, so o = 2 /2 = . 2. One period of x x (t ): x ): x (t ) = e 0 t 2. 3. FS of x ): x (t ):

X> k@ 

1

2³

2

0

e2 t e jkS t dt dt

X> k@ 

1

2 ³

2

0

2t ,

  2  jkS  t

e

1   2 e 2  2  S  jk

Figure 3.16 3.16 (p. 216) 216)

dt Time-domain signal for Example 3.9. 2

jkS  t 0



1 4

1  2S jk

e4 e jk 2S  

Fig. 3.17.

h The Magnitude of X X [k ] the magnitude spectrum of x x (t ); ); the phase of X X [k ]

1  e 4 4

2S jk e

jk 2

=1

the phase spectrum of x x (t ). ). Signals and Systems_Simon Haykin & Barry Van Veen

33


CHAPTER

Example 3.10 FS Coefficients for An Impulse Train Determine the FS coefficients for the signal x (t ) defined by x t 

l f

1. Fundamental period of x x (t ) is T = T = 4, each period contains an impulse. 2. By integrating over a period that is symmetric about the origin 2 < t 2, to obtain X obtain X [k ]: ]:

1

4 ³

2

2


34


CHAPTER

Example 3.11 Calculation of FS Coefficients By Inspection Determine the FS representation of the signal

xt   3 cosS t / 2  S / 4 

f

¦ G  t 4 l

X> k@ 


CHAPTER

1 dt 4

 jkS / 2  t

G  t e

3. The magnitude spectrum is constant and the phase spectrum is zero. i Inspection method for finding X finding X [k ]: ]: Whenever x x (t ) is expressed in terms of sinusoid, it is easier to obtain X obtain X [k ] by inspection. The method of inspection is based on expanding all real sinusoid s in terms of complex sinusoids and

1. Fundamental period of x x (t ) is T = T = 4, and as f

x( t) 

¦

X[ k] ejkS t / 2

(3.21)

jS t/ 2 S / 4 

jS t/ 2  S / 4 

o

= 2 /4 = /2. Eq. (3.19) is written

k f

x t  3

e

 e 2



3 2

ejS / 4 ejS t / 2 

3 2

e jS / 4 e jS t / 2

2. Equating each term in this expression to the terms in Eq. (3.21) gives the FS coefficients: -3

°2 e ° °3

 jS / 4

, k  1

Magnitude and


CHAPTER


CHAPTER

f

f

 ¦ 1 / 2k e jk S / 20 e jk S t  ¦ 1 / 2l e  jl S / 20 e  jl S t

S/4

k  0

l 1

xt  

xt  

h Time-domain representation obtained from FS coefficients

x t 

f

¦ 1 / 2

e

Sjk/ 20

e

 ¦ 1 / 2 

Sjk t

k 0

 k

e Sjk/ 20 e Sjk

t

k  1



CHAPTER

1. Fundamental frequency:

o

= 2 /T /T .

2. X 2. X [k ]: ]: by integrating over T /2

X> k@ 

  

1 T

³

T /2

 Zjk0 t

T / 2

1 Tjk Z 0

dt

x t e

3 5  4 cosS t  S / 20 

= 2 /T /T = = . From Eq. (3.19), we obtain

o

f

k

1 T

³

T 0


2 sin k Z 0T 0 

k o0

jk0 t  Z

e

dt

§ e ¨ TkZ 0 © 2

, k z 0

jkZ 0 t

T 0 jkZ0 T 0

2sin  kZ 0T 0  Tk Z 0

For k For k = = 0, we have

 e

· ¸ , k z 0 ¹

jkZ 0 T0

Tk Z 0 X >k @ 



2j , k z 0

2T 0

T

2 sin k Z 0T 0 

Tk Z 0

4. X 4. X [k ] is real valued. Using

X [ k ] 

2 sin(k 2S T0 / T )

k 2S

o

sin(S u ) S u

(3.24)

= 2 /T /T gives gives X X [k ] as a function of the ratio T o /T /T :

(3.23)

5. Fig. 3.22 (a)-(c) depict X depict X [k ], ], 1/64, respectively. h Defin Definition ition of Sinc functi function: on:

sinc( sinc(u ) 

38


CHAPTER

lim

t T /2 T 0

37

T 0

e

1

Figure 3.21 221) 3.21 (p. 221) Square wave for Example 3.13.

Assume that the fundamental period is T = T = 2.

1. Fundamental frequency:

1 1  1 / 2e  j S t S / 20 

Example 3.13 FS for A Square Wave Determine the FS representation of the square wave depicted in Fig. 3.21. 3.21.

Example 3.12 Inverse FS Find the time-domain signal x signal x (t ) corresponding to the FS coefficients k



1  1 / 2e j S t S / 20 

2. Putting the fraction over a common denominator results in

 S/4

Figure 3.18 (p. 219) 219) Magnitude and phase spectra for Example 3.11.

X >k @  1 / 2 e jk S / 20

1

50

k 50, for T for T o /T /T =1/4, =1/4, T o /T /T =1/16, =1/16, and T o /T /T =

Fig. 3.23.

CHAPTER


Figure 3.22a&b (p. 222) The FS coefficients, X [k ], ], – 50 d k d 50, for three square waves. (see Fig. 3.21.) (a) T o/T = 1/4 . (b) T o/T T = = 1/16. (c) T o/T = 1/64.

Figure 3.22c (p. 222)


CHAPTER


CHAPTER


41



CHAPTER

3) The smaller ripples outside the mainlobe are termed sidelobe sidelobe.. 4) The FS coefficients in Eq. (3.23) are expressed as X> k@ 

2T0

T

§ 2T · sinc ¨ k 0 ¸ © T ¹

i Fourier series pair

Trigonometric FS

1. Trigonometric FS of real-valued signal x signal x (t ): ):

x( t)  B[0] 

f

¦ B[ k] cos( kZ t)  A[ k] sin( kZ t) 0

k 1

FS coefficients:

B[0]  B[ k] 

1

T 2 T

³

T

0

(3.25)

B[0] = X = X [0] [0] represents the timeaveraged value of the signal.

x( t) dt

³ x( t) cos( kZ t) dt T

0

0

0

(3.26)

42

CHAPTER


CHAPTER

Example 3.14 Square Wave Partial-Sum Approximation

2. Relation between exponential FS and trigonometric FS coefficients:

B[ k]  X [ k]  X [ k] A[ k]  j( X[ k]  X[  k] )

for k for k 0

(3.27)

Let the partial-sum approximation to the FS in Eq. (3.29), be given by

Euler’s Formula



x J t  

Ex. Find trigonometric FS coefficients of the square wave studied in Example 3.13. 1. Substituting Eq. (3.20) into Eq. (3.27), gives

B[0]  2 T0 / T 2 sin(k 2S T0 / T ) [B ]k , k S A[ k ]  0

(3.28)

k  0

0

Because x Because x (t ) is an even function

(3.29)


CHAPTER


Figure 3.25a (p. 225) 225) Individual terms (left panel) in the FS expansion of a square wave and the corresponding partial-sum approximations J (t ) (right panel). The square wave has period T T = = 1 and T s/T = ¼. The J = 0 term is 0(t ) = ½ and is not not shown. (a) J J = = 1.

0

This approximation involves the exponential FS coefficients coefficients with indices J k J . Consider a square wave with T = T = 1 and T o /T /T = = ¼. Depict one period of the J th th term in this this sum, and and find

J 1, 3, 3, 7, 7, 29 29, an and 99 99.

1. The individual terms and p artial-sum approximation are depicted in Fig. 3.25. 3.25. 2. Each partial-sum approximation passes through the average value (1/2) of the discontinuity, the approximation exhibits ripple. 3. This ripple near discontinuities in partial-sum FS approximations is termed the Gibbs phenomenon. 4. As J J increase, increase, the ripple in the partial-sum approximations becomes more and more concentrated near the discontinuities.

f

¦ B[ k] cos( kZ t)

k  0

x J  t for

2. Trigonometric FS expression of x x (t ): ):

x( t) 

J

¦ B>k @cosk Z t 



kz 0


45


CHAPTER


46

CHAPTER



CHAPTER

1. If the input to an LTI system LTI system is expressed as a weighted sum of sinusoid, then the output is also a weighted sum of sinusoids. 2. Input: 4. Frequency response of the system H ( j j ):

xt  

yt  

H  jZ  

f

¦ H  jk Z  X >k @e

jk Z 0t

0

k  f

6. Substituting for H for H ( jk jk o) with RC = RC = 0.1 s and

Y >k @ 

Figure 3.21 221 3.21 (p. 221) 221)) Square wave for Example 3.13.

FS ;Z m o Y> k@  H  jkZ 0  X> k@ 0

5. Frequency response of the RC circuit: RC circuit:

3. Output:

Example 3.15 RC Circuit: Calculating The Output By Means of FS Let us find the FS representation for the output y (t ) of the RC circuit RC circuit depicted in Fig. 3.2 in response to the square-wave input depicted in Fig. 3.21, 3.21, assuming that T o /T /T = = ¼, T = T = 1 s, and RC = RC = 0.1 s.

10

sin k S / 2 

j 2S k  10

k S

1 / RC jZ  1 / RC o

= 2 , and T o /T /T = = ¼, gives

7. We determine y (t ) using the approximation Figure 3.2 3.2 (p. 197) 197) 197) RC circuit RC circuit for Example 3.1. Signals and Systems_Simon Haykin & Barry Van Veen


Figure 3.26 (p. 228 228)) The FS coefficients Y[k Y[ k ], ], –25 d k d 25, for the RC RC circuit circuit output in response to a square-wave input. (a) Magnitude spectrum. (b) Phase spectrum. c) One period of the input signal x (t ) dashed line) and output signal y (t ) (solid line). The output signal y (t ) is computed from the partial-sum

y t

jk Z 0t

k  f

Figure 3.25e 3.25e (p. 226) (e) J J = = 99.

CHAPTER

f

¦ X >k @e

y( t) |

100

¦

Y[ k] ejkZ 0t

k  100

49

(3.30)

The magnitude and phase spectra for the range 25 k 25: Fig. 3.26 (a) and (b). Waveform of y of y (t ): ): Fig. 3.26(c).


CHAPTER


Figure 3.26 (p. 228 228)) The FS coefficients Y[k Y[ k ], ], –25 d k d 25, for the RC RC circuit circuit output in response to a square-wave input. (a) Magnitude spectrum. (b) Phase spectrum. c) One period of the input signal x (t ) dashed line) and output signal y (t ) (solid line). The output signal y (t ) is computed from the partial-sum

50


CHAPTER

h If x x [n] is not absolutely absolutely summable, but does satisfy (i.e., if x x [n] has finite energy),

f

¦ x>n@

2

4. Magnitude and phase spectra:

It can be shown that the sum in Eq. (3.32) converges in a mean-square error sense, but does not converge pointwise.

f

nf

Example 3.17 DTFT of An Exponential Sequence Find the DTFT of the sequence x sequence x [n] = 1. DTFT of x x [n]:

X e j:  

f

¦D u>n@e n

 j:n

n  f

nu[n].

This sum diverges for D

f

¦ D e 

 j: n

f

 ¦ D n e  j:n

n 0

1 1  D e

 j:

, D

X e j:  

1

 D sin :  § D sin : · arg^ X e j: `   arctan¨ ¸ © 1  D cos : ¹ 2

2

Fig. 3.29 for = 0.5 and

X e:j  

Euler’s Formula

1  D cos :  jD sin :


Figure 3.29 (p.232) The DTFT of an exponential signal x [n] = (D)nu [n]. (a) Magnitude spectrum for D = 0.5. (b) Phase spectrum for D = 0.5. (c) Magnitude spectrum for D = 0.9. (d) Phase spectrum for D = 0.9.

1  D cos :

2

1/ 2



Even function

1

D

2

 1  2D cos : 

1/ 2

Odd function = 0.9.

d M ! M

as depicted in Fig. 3.30 (a). Find the DTFT of x x [n]. 1. DTFT of x x [n]:

(3.33)


CHAPTER

1

-1, n x > n@  ® ¯0, n

3. If is real valued, Eq. (3.33) becomes

1



Example 3.18 DTFT of A Rectangular Pulse Let

n 0



X e j: 

t1

2. For < 1, we have

X( e j: ) 


CHAPTER

M

¦1e

 :j n

.

n  M

57


CHAPTER

Figure 3.30 (p. 233) Example 3.18. (a) Rectangular pulse in the time domain. (b) DTFT in the frequency domain.

58


p


CHAPTER

X ej:  

2M

2 M

¦e

 ej:M ¦ e j: m

 j: ( m M )

m0

m 0

Change of variable m = n + M

- j:M 1  e j:2 ( M 1) , : z 0, r 2S , r 4S , °e ® 1  e j: °¯ 2 M  1, :=0, r 2S , r 4S , 2. The expression for X X (e j ) when : z 0, r 2S , r 4S ,

3. Graph of X X (e j ): Fig. 3.30(b). Example 3.19 Inverse DTFT of A Rectangular Pulse Find the inverse DTFT of





X( e :j )  e :j M

 j: 2 M1 / 2

e

 e :

:j / 2

j

2 M1 / 2

e :

j /2

Figure 3.31 3 .31 (p. 234) 23 4)) 3.31 234 Example 3.19. (a) Rectangular pulse in the frequency domain. (b) Inverse DTFT in the time domain.

: d W , W  n  S

-1, X  e j:   ® ¯0,



e


CHAPTER

which is depicted in Fig. 3.31 (a).

 e j: 2 M1 / 2 

 e :j / 2 

j 2 M1 / 2

 e :j 2 M1 / 2  e j: / 2  e j: / 2 sin  :  2 M  1 / 2  L’Hôptital’s L’Hôpt ital’s Rule lim  2 M  1; : o 0,r 2S , r 4S , , sin  : 2  With understanding that X that X (e j ) for sin  :  2 M  1 / 2  : z 0, r 2S , r 4S , is obtained as X (e j: )  sin  : 2  limit.



 e :



1. Note that X that X (e j ) is specified only for W 2. Inverse DTFT x DTFT x [n]: 1

x> n@ 




1



e j:n

W W

, nz0

2S nj 1  sin Wn  , n z 0. S n

3. For n For n = 0, the integrand is unity and we have x have x [0] [0] = W / . Using L’Hôpital’s rule, we easily show that

lim n o0

1

nS

sin Wn  

W S

,

and thus we usually write

>x @n

1 S n

2S W

61


CHAPTER

sin  Wn

as the inverse DTFT of X X (e j ), with the understanding that the value at n = 0 is obtained as limit.

³ e

j :n

<

.

p

d:



CHAPTER

Example 3.20 DTFT of The Unit Impulse Find the DTFT of x x [n] = [n]. 1. DTFT of x x [n]:

X e :j  

f

¦ G > n@ e

 :j n

1

n f

G > n @

DTFT m m o 1.

2. This DTFT pair is depicted in Fig. 3. 32. 32.

Figure 3.32 3.3 2 (p. 235) 23 5) 3.32 235) Example 3.20. (a) Unit impulse in the time domain. (b) DTFT of unit impulse in the frequency domain.

62

CHAPTER


CHAPTER

Figure 3.47 (p. 252) The spectrum of the raisedraised-cosine pulse consists of a sum of three frequency--shifted frequency sinc functions.

Figure 3.48 (p. 252) Spectrum of the raisedraisedcosine pulse in dB, normalized by T 0.


CHAPTER



3.8 Properties of Fourier of Fourier Representation 1. Four Fourier representations: Table 3.2. 3.2.

89


CHAPTER

90


2. Signals that are periodic in time have discrete frequency-domain representations, while nonperiodic time signals have continuous frequencydomain representations: Table 3.3. 3.3.

3.9 Linearity and Symmetry Properties Symmetry Properties 1. Linearity property for four Fourier representations:

z t  ax t  by by t

FT m o

z t  ax t  by t

m o FS ;Z o

Z Z j   aX Z j   bY Z j  Z> k@  aX> k@  bY b Y> k@

CHAPTER


Example 3.30 Linearity in The FS Suppose z (t ) is the periodic signal depicted in Fig. 3.49(a). Use the linearity and the results of Example of Example 3.13 to determine the FS coefficients Z [k ]. ]. 1. Signal z (t ): ):

z t 

3 2

x t 

1 2


CHAPTER

x t

FS ;2S m o X> k@  1  kS   sin  kS 4 

y t

FS ;2S m o Y> k@  1  kS   sin  kS 2 

From Example 3.13.

Linearity

3. FS of z of z (t ): ):

z t

FS ;2S m o Z> k@   3  2 kS   sin  kS 2   1  2 kS   sin  kS 2 

3.9.1 Symmetry Properties: Real and Imaginary Signals

h Symmetry property for real-valued signal x (t ): ): 1. FT of x x (t ): ):

X* ( jZ )  ª

¬« ³

y  t

where x(t) and y(t): Fig. 3.49(b) and (c). 2. X 2. X [k ] and Y [k ]: ]:

f

f

x( t) e

j t Z

dtº

*

f

 x* ( t) e Z j t dt ¼» ³ f

(3.37)

2. Since x(t) is real valued, x valued, x (t ) = x = x (t ). ). Eq. (3.37) becomes

X*  jZ  

f

³

f

x t e j ( 

Z ) t

dt

X* ( jZ )  X( jZ )

X ( j j ) is complex-conjugate symmetric (3.38)

Re^ X( jZ )`  Re^ X( jZ )` and Im^ X( jZ )`   Im^ X(  jZ )` Signals and Systems_Simon Haykin & Barry Van Veen

CHAPTER

93


i The conjugate symmetry property for DTFS:


CHAPTER

X [ k]  X [ N  k ] X[ k]  X[ N  k]

h The symmetry conditions in all four Fourier representations of real-valued signals are indicated in Table 3.4. 3.4.


h Complex-conjugate symmetry in FT:

*

Because the DTFS coefficients are N N periodic, periodic, and thus

94

A simple characterization of the output of an LTI system with a realvalued impulse response when the input is a real-valued sinusoid.

i Continuous-time case: 1. Input signal of LTI system: Euler’s formula

x t  Acos Z t  I  jZ  t I

 x t   A/ 2  e

  A/ 2  e

jZ tI 

2. Real-valued impulse response of LTI system is denoted by h(t ). ). 3. Output signal of LTI system:

y( t)  H( jZ ) ( A/ 2) e

j(Z tI arg^ H( jZ )`)

 H( jZ ) ( A/ 2) e j(Z tI arg^ H( jZ )`)

Exploiting the symmetry conditions:

H( jZ )

 H( jZ )

Applying Eq. (3.2) and linearity

CHAPTER



CHAPTER

The Differentiation property for a periodic signal 1. The FS representation of a periodic signal x signal x (t ): ):

v

x  t 

f

Differentiating both

¦ X > k@ e

jk jkZ 0 t

k f

sides with respect to t

f d xt   X >k @ jk Z 0 e jk Z 0t dt k  f

¦

2. Differentiation property of a periodic signal: d dt

FS ; Z m m o

x t

jkZ 0 X> k@

0

Differentiation of x x (t ) in Time-Domain 

Figure 3.58 3 .58 (p. 273) 2 73)) 3.58 273 Magnitude of frequency response in dB for MEMS accelerometer for Zn = 10,000 rad/s, Q = 2/5, Q = 1, and Q = 200. Signals and Systems_Simon Haykin & Barry Van Veen

CHAPTER

ª d x( t) º  ( jkZ ) u X[ k]  0 0 k 0 «¬ dt »¼ k  0

Example 3.39 Differentiation Property Use the differentiation property to find the FS representation of the triangular wave depicted in Fig. 3. 59 (a). (a). Fig. 3. 59 (b) z( t) 

1. Define a waveform:

121


( jk jk 0) X [k ] in Frequency-Domain

d

dt

y( t)


CHAPTER

122


3. The differentiation property implies that Y [ y] 

Z[ k]  jkZ 0 Y[ y]

1 Z0 jk Z

Except for k for k = =0 Z [k ]

The quantity Y [0] [0] is the average value of y of y (t ) and is determined by inspection Fig. 3.59 (b) to be T /2. Therefore, Figure 3.59 3 .59 (p. 274) 27 4)) 3.59 274 Signals for Example 3.39. (a) Triangular wave y (t ). ). (b) The derivative of y of y (t ) is the square wave z (t ). ). 2. Comparing z (t ) in Fig. 3. 59 (b) and the sqare sqare wave x(t) x(t) with T 0 /T /T = = ¼ in Example 3.13, 3.13, we have z( t)  4 x( t)  2

Z[ k]  4 X[ k]  2G [ k]

°

0,

k  0

y t

m  o FS ; Z 0

k  0 - T / 2, °° k S · § Y> k@  ® 2T sin ¨ ¸ © 2 ¹ , k z 0 ° 2 2 ¯° jk S

3.11.2 Differentiation in Frequency 1. FT of signal x signal x (t ): ):

X  jZ  

f

³ xt e f

 jZ t

dt

Differentiating both sides with respect to

d X  jZ   d Z

f

³  jtxt e f

 jZ t

dt


CHAPTER


CHAPTER

M

3. In Example 3.25, we obtained

X  jZ  

2 Z

¦ b e 

2. Frequency response:

Z  jZ   e

sin Z T 0 

2

 jZ T 1

Z

 

H e j:

sin Z T 0 

k  0 N

k  0

M

k

Taking Takin g DTFTof DTFTof both

x( t) DTFT m o e

z> n  k@

sides of this equation N

M

:jk

Z e : 

k  0

j:

k

 j:

k  0

M

k

z t  

j:

k

j:

z t  

 j: k

k  0 N

k  0

k

The frequency-shift properties of Fourier representations are summarized in Table 3.8. 3.8. Table 3.8 Frequency-Shift Properties of Fourier Representations FT m o

X  j Z  J  

F S0 ; Z

x > k  k0 @

DTFT e j*n x[ n] m m o

j :  * º X ªe 

e

e

jk0 0Z

jk 0 0:

x t t

m o

¬

¼

n F ;: S0 x[ n ] m  o X > k  k0 @

Example 3.42 Finding an FT by Using Using the Frequency-Shift Frequency-Shift Property Use the frequency-shift property to determine the FT of the complex sinusoidal pulse: j10 t

-

1

f

Z  jZ e 2S ³

d Z

1 2S

f

³ X  jK e 

X  j Z  J e 2S ³

jZ t

f

=

j K  J t

f

f

1



jZ t

f

d Z

into above Eq. gives

d K

133


e jJ t x  t 

FT Z( jZ Z ) mo

z( t)

and

 e jJ t

Frequency-shifting of X X ( j j ) by (e t ) x (t ) in Time-Domain

 jZ k

k


CHAPTER

X( jZ Z )

Substituting variables

¦ b e  Y e   X e  ¦ a e  j:

FT mo

2. By the definition of the inverse FT, we have

j

¦ a e  Y e   ¦ b e  X e  k

 j:

The problem is to express the inverse FT of Z of Z ( j j ) = X = X ( j j ( )) in terms of x x (t ). ).

1. Suppose that:

k

k  0

 jZ k

k

3.12.2 Frequency - -Shift Shift Property

¦ a y>n  k @  ¦ b x>n  k @ k  0

(3.55)

¦ a e 

Frequency response of a discrete-time system 1. Difference equation: N

 j: k

k

1 2S

f

³ X  jK e f

jK t

jJ t d K  e jJ xt 

in Frequency-Domain [i.e. X [i.e. X ( j j (

))]] ))


CHAPTER

-1, xt   ® ¯0,

134

Fourier Representations of Signals & LTI Systems t  S t ! S

2. Use the results of Example of Example 3.25, 3.25, we have

x t

FT m o

X  jZ  

2 Z

sin ZS  FT m o

x t

Frequency-shift property

X Z j 

2 Z

sin  ZS 

3. FT of z of z (t ): ):

e j10t x  t 

FT m o X  j Z  10 

FT m o

z  t 

Example 3.43 Using multiple Properties to Find an FT Find the FT of the signal



d

^

3

 





`

2 Z  10

sin  Z  10  S 


CHAPTER

H  jZ  

2. Partial-fraction expansion:

1

 jZ 2  25000 jZ   100002



C 1 jZ  20000

C 2



20,000 and d 2 =

5,0000.

 C2

  jZ  5000  

1

jZ  5,000 jZ 20,000

1 / 15000 jZ  5000

 e 20000t  u  t 

= 10,000 10,000 rads/s rads/s and Q = Q = 1

1

 jZ   10000  jZ   10000 2 2

2. Partial-fraction expansion: The roots of the denominator polynomial are

1

 jZ   25000  jZ   10000 

2

H  jZ  

jZ  5,000

 1/15,000 141




h  t    j 10000 10000 3

  e

5000t

e j 5000

3t

 e5000t e j 5000

n

H  jZ  

1

 jZ 2  50 jZ   10000 2

2. Impulse response:

ht   1 /10000e 25t sin10000t ut 

 5000  j 5000 3

3t




CHAPTER


= 10,000 10,000 rads/s rads/s and Q = Q = 200


d2

 j / 10000 3  j / 10000 3   jZ  5000  j 5000 3 jZ  5000  j 5000 3

 1 /  5000 3  e 5000t sin 5000 3t  u  t  Case (c): (c) :

d1  5000  j 5000 3


jZ  5,000


CHAPTER

n

H  jZ  

 1/15000

2

1 jZ  20,000




2 2  jZ   25000  jZ   10000 jZ 20,000

1

h  t   1/15000   e 5000t Case (b): (b):

Coefficients C 1 and C 2:

  jZ  20,000 

 1 / 15000 jZ  20000


jZ  5000

The roots of the denominator polynomial are d 1 =

C1


CHAPTER

The roots of the denominator polynomial are

 25  j10,000 ,000 d 2  25  j10,000 ,000 d1

Impulse response for (a) Q Q = = 2/5, (b) Q Q = = 1, and (c) Q Q = = 200: Fig. 3.64 (a) ~ (c). (c). 1. For both Q = Q = 2/5 and Q = Q = 1, the impulse response is approximately zero for t > t > 1 ms.

Figure 3.64 3 .64 (p. 289) 28 9)) 3.64 289 Impulse response of MEMS accelerometer.

142


CHAPTER

3.13.2 Inverse Discrete - Time Time Fourier Transform

n DTFT m o  d k  u > n @ m

1. Suppose X Suppose X (e j ) is given by a ratio of polynomial in e j , i.e.

X e j:  

E M e  j:M D N e

 j: N



 D N 1e



 E 1e  j:  E 0   D 1e  j:  1

 j:  N 1

Normalized to 1



D N e

 D N 1e

 j:  N 1





 D 1e

 j:

 1   1  d k e





 j:



C k k


145


CHAPTER

6

6

e  j:

1



 e  j 2 :

1

6

C 1 1  j: e 2

C 2  1 1  e  j: 3

1

6

1

e  j:

1

 e  j 2 : 6

1

 v 0 6

6

1/2 and d 2 = 1/3.

5  e  j:  5 § 1 · 6 C1  ¨ 1  e  j: ¸ © 2 ¹ 1  1 e : j 1 e 6 6

e  j:  2

5 6

 e  j:  5

 §¨ 1  e  j: ·¸ © 3 ¹ 1  1 e  : j 1 e 1

 2 :j

6

6

 2 :j e  j: 3


146


CHAPTER

Non-periodic continuous-time signals 1. Non-periodic signals: x signals: x (t ), ), z (t ), ), and y (t ) = x = x (t )z (t ). ).

Find the FT of y of y (t ). ).

2. FT of x x (t ) and z (t ): ):

Coefficients C 1 and C 2

C2

v2

6

1

3.14 Multiplication Property

5

 e  j:  5 1

1. Characteristic polynomial:



 e  j:  5

3. Partial-Fraction Expansion:

Since

1

j:

2. The roots of above polynomial: d 1 =

j

k 1

5

X e

Find d k , the roots of this polynomial

 D N 1v  D N  0

  ¦  :  1  d e

X e j:

k

Find the inverse DTFT of

3. Partial-fraction expansion: Assuming that M < M < N and N and all the d k are distinct, we may express X express X (e j )as N

n

k

Example 3.45 Inverse by Partial-Fraction Expansion

k 1

Replace e j with the generic variable v :

v N  D 1v N 1  D 2 v N  2

N

¦ C d  u>n@ k 1

N

Expansions for repeated roots are treated in Appendix B.

1 1  d k e  j:

The linearity property implies that

x>n@ 

2. Factor the denominator polynomial as  j: N


CHAPTER

5  e j :  5 6 1 1  e : j 3 e  j :  2

f

1 2S

4

³ X  jve y t  

y t 

1

:j e j: 3

3. FT of (t ): ):

jvt

dv

f

1

and f

f

2S 2 ³f ³ f

Change variable:

5 6

 e  j:  5 1 1  e 2

xt  

1 2S

=

³

f

f

v

z t  

1 2S

f

³ Z  jK e f

jK t

d K

X  jv Z  jK e j K  v t d K dv Inner Part: Z ( j j ) X ( j j )

ª 1 f X jv Z j Z  v dvº e jZ t dZ  » «¬ 2S ³ f     ¼


CHAPTER

v

CHAPTER


,u

v

,u

H t (e j ) is the area under F ( ) between = /2 and = /2.

Figure 3.66 3 .66 (p. 294) 29 4)) 3.66 294 The effect of truncating the impulse response of a discrete-time system. (a) Frequency response of ideal system. (b) F :(T) for : near zero.

Figure 3.66 294 3.66 (p. 294) 294)) The effect of truncating the impulse response of a discrete-time system. (c) F :(T) for : slightly greater than S/2.(d) Frequency response of system with truncated impulse response.



CHAPTER

3. Let

153

DTFT m m o H t (e : )

ht [n ]

j

Using the multiplication property in Eq. (3.57), we have

Ht  e j:   4. Since

and

1

S

³ H  e  W  e 

j T  : 

j T

2S S

 dT

-1, T  S / 2 H >e jT @  ® 0 , S / 2  T  S ¯ sin   :  T   2 M  1 / 2  j  :T  W e   sin   :  T  / 2 

 

H t e j:

1

S / 2

³

F : T d T 2S S / 2


CHAPTER

154


Multiplication of periodic time-domain signals corresponds to convolution of the Fourier representations. Same fundamental 2S / T y (t )  x (t ) z (t ) mF;T o Y [k ]  X [k ] * Z[k ] (3.58) periods of x of x (t ) and z (t ) where

On the basis of Example 3.18

X [ k ]  Z [k ] 

Non-periodic convolution of the FS coefficients

f

¦ X [ m] Z [ k  m ]

m f

x (t )

z (t ) FS (Least common multiple)

X [k ]

Z [k ]

Example 3.47 Radar Range Measurement: Spectrum of RF Pulse Train Discussion: refer to p. 295 in textbook.

The RF pulse train used to measure the range and introduced in Section 1.10 may be defined as the product of a square wave p wave p((t ) and a sine wave s(t ), ), as shown in Fig. 3.68. 3.68. Assume that s(t ) = sin (1000 t /T /T ). ). Find the FS coefficients of x (t ). ).

CHAPTER



CHAPTER

Fundamental frequency

2. Fundamental period: For p( p(t ): ): 0 = 2 /T. /T.

s(t ) = sin (1000 t /T / T ) = sin(500

t ). ). 0t

3. FS coefficients of s of s(t ): ):

- 1 /(2 j ), k  500 °   500 °0, otherwise ¯ S[k ]  1 /( /( 2 j )G [k  500]  1 /( /(2 j )G [k  5 00]

S [k ]  ® 1 /( 2 j ), k

4. FS coefficients of p( p(t ): ):

P[ k]  e jkT oZ o

1) Result of Example 3.13 2) Time Time--shift property

sin(kZ oT o ) k S

5. FS coefficients of x x (t ): ):

[ ]k X Figure 3.68 297 3.68 (p. 297) 297)) The RF pulse is expressed as the product of a periodic square wave and a sine wave. Signals and Systems_Simon Haykin & Barry Van Veen

CHAPTER


1 2

 j( k500 ) ToZo

e

sin(( k  500)ZoTo ) ( k  500)S



1 2

 (j k500 ) ToZ o s in(( k

e

6. Fig. 3.69 depicts the magnitude spectrum for 0

157

 500)Z oTo )

(k  500)S

k 1000.


CHAPTER

158


Multiplication property for discrete-time periodic signals:

y[ n]  x[ n] z[ n] Figure 3.69 (p. 298) FS magnitude spectrum for 0 d k d 1000. The result is depicted as a continuous curve, due to the difficulty of displaying 1000 stems

where

TF;T2S / T mD  o Y [ k ]  X [ k ]  Z[ k]

X [ k ]  Z[ k ] 

N 1

¦ X [ m] Z[k  m]

m 0

(3.59)

Periodic convolution of DTFS coefficients

Fundamental period = N. The multiplication properties of all four Fourier representations are summarized in Table 3.9. 3.9. Table 3.9 Multiplication Properties of Fourier Representations

1

x( t) z( t)

FT m o

x( t) z( t)

m  o FS ; Z o

2S

.

X ( jZ )  Z( jZ ) X [ k ]  Z[ k ]


CHAPTER

CHAPTER


3.15 Scaling Property

v

1. Let z (t ) = x = x (at ). ). 2. FT of z of z (t ) :

³

Z( jZ ) 

f

f

z( t) e

dt 

Zj t

f

³

f

x( at) e

j t Z

Changing variable: W = at

dt

- (1 / a ) f x (W )e  j (Z / a )W dW , a ! 0 ³ f ° Z ( jZ )  ® f °(1 / a ) ³ x (W )e j (Z / a )W dW , a  0 f ¯ Z( jZ )  (1 / a ) z( t)  x( at)

f

³

f

x(W ) e j ( Z / a )W dW ,

FT m o (1 / a ) X( jZ / a).

Scaling in Time-Domain

(3.60)

Inverse Scaling in Frequency-Domain

Figure 3.70 3.70 (p. 300) 300) The FT scaling property. The figure assumes that 0 < a < 1.

Signal expansion or compression! Refer to Fig. 3.70. 3.70. Signals and Systems_Simon Haykin & Barry Van Veen

161


CHAPTER

Example 3.48 Scaling a Rectangular Pulse Let the rectangular pulse Use the FT of x x (t ) and the scaling property to find the FT of the scaled rectangular pulse

°-1, x(t )  ® ¯°0,

t  1

t ! 1

°-1, y (t )  ® ¯°0,

t  2 t ! 2

1. Substituting T 0 = 1 into the result of Example of Example 3. 25 gives

X ( jZ ) 

2 Z

sin(Z ) Scaling property and a = 1/2

2. Note that y (t ) = x = x (t /2).

§ 2 · sin(2Z )  2 sin( 2Z ) ¸ Z © 2Z ¹

Y ( jZ )  2 X ( j 2Z )  2 ¨

Fig. 3.71

Substituting T 0 = 2 into the result of Example of Example 3.25 can also give the answer!


CHAPTER


Figure 3.71 (p. 301) Application of the FT scaling property in Example 3.48. (a) Original time signal. (b) Original FT. (c) Scaled time signal y (t ) = x (t /2). /2). (d) Scaled FT Y ( j Z) = 2 X ( j 2Z).

162


CHAPTER

CHAPTER


1/ 2

ª T t 2dt º ³ T » Td  « T « » «¬ ³ T dt »¼ o

 [(1 /(2To ))(1 / 3) 3)t 3

o

o

T o 1/ 2  To

]

v

 Td / 3

; p

o

2. The uncertainty principle given by Eq. (3.65) states that

Bw

t 1 /( /( 2T d )

t 3 / (2 (2T o )

Bw

3.18 Duality

Difference in the factor 2 S and the sign change in the complex sinusoid.

3.18.1 The Duality Property of the FT 1. FT pair: f

1

x( t) 

³

X ( jZ) e jZ t dZ

f

2S

and

Rectangular in time-domain l sinc function in frequency-domain

X( jZ ) 

f

³

f

x( t) e jZ t dt

2. General equation:

y( v) 

f

1 2S

Choose

³

f

z(K ) e jvK dv

= t and t and

=

Figure 3.73 3.73 (p. 307) 307) Duality of rectangular rectangular pulses and sinc functions.

(3.66)

the Eq. (3.66) implies that Signals and Systems_Simon Haykin & Barry Van Veen


CHAPTER

y( t) 

173

f

1

2S ³

f

FT mo

1

f

2S ³

f

(z )t

z(Z )

(3.67) =

and

= t , then

z( t) e jZ t dt

FT m o 2S y(Z )

(3.68) Fig. 3.74

4. If we are given by an FT pair

(z )t

FT m o 2S y(Z )

Fourier Representations of Signals & LTI Systems v

3. Interchange the role of time and frequency by letting Eq. (3.66) implies that

y(Z ) 

CHAPTER

z(Z ) e jZ t dZ

y( )t


(3.69)

Example 3.52 Applying 3.52 Applying Duality

f( t)

FT m o

Z ) F( jZ

(3.70)

; p

174

CHAPTER


>> N=50; >> M=4; >> x=[ones(1,M+1),zeros(1,Nx=[ones(1,M+1),zeros(1,N-2*M-1),ones(1,M)]; 2*M-1),ones(1,M)]; >> X=fft(x)/N; >> k=[0:N-1]; k=[0:N-1]; %frequency index 0.2 >> stem(k,real(fftshift(X)))

CHAPTER


3. Effective duration and bandwidth for the discrete-time periodic signals: 1

Fig. 3. 12 (a)

2 ª ( N 1) /2/2 2 2º « ¦ n x[n] » n  ( N 1) /2 /2 » T d  « ( N 1) /2 /2 « 2 » « ¦ x[ n] » ¬ n  ( N 1) /2/2 ¼

1

(3.75)

2 ª ( N 1) /2/2 2 2 º « ¦ k X [k ] » k  ( N 1) /2 /2 » Bw  « ( N 1) /2 /2 « 2 » ¦ X [k ] »¼ « ¬ n ( N 1) /2/2

(3.76)

0.15

MATLAB function for computing the product T d Bw :

0.1

0.05

0

-0.05

0

10

20

30

40


CHAPTER

50

1) The length of input vector x = odd and centers on middle 2) .* = element-by-element product 3) * = inner product 4) ' = complex-conjugate transpose

197


4. Use the function TdBw to evaluate the time-bandwidth product for two rectangular, raised cosine, and Gaussian pulse train as follo ws: >> x=ones(1,101); % 101 point rectangular pulse >> TdBw(x) ans = 788.0303 >> x=ones(1,301); % 301 point rectangular pulse >> TdBw(x) ans = 1.3604e+003 >> x=0.5*ones(1,101)+cos(2*pi*[-50:50]/101); x=0.5*ones(1,101)+cos(2*pi*[-50:50]/101); % 101 point rectangular pulse >> TdBw(x) ans = 277.7327 >> x=0.5*ones(1,301)+cos(2*pi*[-150:150]/301); x=0.5*ones(1,301)+cos(2*pi*[-150:150]/301); % 301 point rectangular pulse

>> function TBP=TdBw(x) >> %Compute the Time-Bandwidth product using the DTFS >> %One period must be less than 1025 points >> N=1025; >> M=(N-max(size(x)))/2; >> xc=[zeros(1,M),x,zeros(1,M)]; >> % center pulse within a period >> n=[-(N-1)/2:(N-1)/2]; >> n2=n.*n; >> Td=sqrt((xc.*xc)*n2'/(xc*xc')); >> X=fftshift(fft(xc)/N); % evaluate DTFS and center >> Bw=sqrt(real((X.*conj(X))*n2'/(X*X'))); >> TBP=Td*Bw; Signals and Systems_Simon Haykin & Barry Van Veen

CHAPTER

198


>> n=[-500:500]; >> x=exp(-0.001*(n.*n)); % Narrow Gaussian pulse >> TdBw(x) ans = 81.5669 >> x=exp(-0.0001*(n.*n)); % Broad Gaussian pulse >> TdBw(x) ans = 81.5669

h Note that the Gaussian pulse t rains have the smallest time-bandwidth product. h Time-bandwidth product is identical for both the narrow and broad Gaussian pulse trains .

Lecture03 Fourier Representations of Signals

Recommend Documents