Then
Taking the inverse Fourier transform of Y(R) and using Eq.
(6.1351, we get
6.33. Consider a discrete-time LTI system with impulse response
Find the output y[n] if the input x[n] is a periodic sequence with
fundamental period
No = 5 as shown in Fig. 6

sin [( R - R,) ~ / 2 ]
sin[(R - R,)/2]
(b) Note from Eq. (6.98) that
we obtain
6.52. Show that if x[n] is real, then its DFT X [ k ] satisfies the
relation
where * denotes the complex conjugate.
From Eq. (6.92)
Hence, if x[n] is real, then x * [ n ] = x[n

Noting that (1 - e-jn) = 0 for R = 0, X ( R ) must be of the form
where A is a constant. To determine A we proceed as follows.
From Eq. (1.5) the even
component of u[nl is given by
u,[n] = $ + f 6 [ n ]
Then the odd component of u[n] is given by
uo[n] = u

samples of x[n], respectively.
( a ) Show that
N
f [ n ] = 4.1 = 0 outside 0 s n 5 - - 1
2
CHAP. 61 FOURIER ANALYSIS OF DISCRETE-TIME
SIGNALS AND SYSTEMS 35 1
( b ) Show that the N-point DFT X [ k ] of x[n] can be expressed
as
(N/2)- 1 N
where F [ k ] = C

n=O
where tn = n At and T, = NAt. Setting w = w, in the above
expression, we have
N- 1
X(wk) = At x(t,) e-'"'kln (6.237)
n=O
Next, since the highest frequency of x(t) is w, the inverse
Fourier transform of ~ ( wis) g iven
by [Eq. (5.3211
Dividing the freq

Equation (6.49) is the discrete-time counterpart of Eq. (6.47). It
states again the inverse
relationship between time and frequency. That is, as the signal
spreads in time (m > I), its
Fourier transform is compressed (Prob. 6.22). Note that X(rnR)
is peri

( a ) x(t) = e-"'u(t), a > 0
sin at
( b ) x(t) = rr l
(a) From Eq. (5.155)
1
~ ( t=) e - "u( t ) - X( W ) = a +jo
From Eq. (1.14)
Now, by Eq. (5.180)
FOURIER ANALYSIS OF TIME SIGNALS AND SYSTEMS
[CHAP. 5
from which we get
Thus.
( b ) From Eq. (5.137)
sin

From Fig. 6-7 we see that x [ n ] is the periodic extension of
(0,1,2,3) with fundamental
period No = 4. Thus,
By Eq. (6.8) the discrete-time Fourier coefficients c, are
Note that c, = c,-, = cT [Eq. (6.17)].
FOURIER ANALYSIS OF DISCRETE-TIME SlGNALS AND

which are sketched in Fig. 6-24. We see that the system is a
discrete-time high-pass FIR filter.
Fig. 6-23
6.39. The system function H(z) of a causal discrete-time LTI
system is given by
where a is real and la1 < 1. Find the value of b so that the
frequen

CHAP. 51 FOURIER ANALYSIS OF TIME SIGNALS AND
SYSTEMS
From Eq. (5.155)
Then
The normalized energy of x ( t ) is
Using Parseval's identity (5.641, the normalized energy of y ( t )
is
=1 -/w, -d o 1 - w, 1 1 tan-' - = - E = T O 4 + 0 2 27 2 2 " 8
from which

Ix(-fl)I= Ix(R)I + a ) = - 9 ( ~ ) (6.646)
From Eqs. ( 6 . 6 3 ~()6~.6 36), and (6.64~w) e see that if x[n] is
real and even, then X(R) is
real and even, while if x[n] is real and odd, X(R) is imaginary
and odd.
0. Parseval's Relations:
Equation (6.66 ) i

Fig. 5-38
5.68. Find the inverse Fourier transform of
Hint: Differentiate Eq. (5.155) N times with respect to (a).
CHAP. 51 FOURIER ANALYSIS OF TIME SIGNALS AND
SYSTEMS
5.69. Find the inverse Fourier transform of
1
X(w)=
2 - w2 + j3w
Hint: Note that
2 - w

Thus, the z-transform and the Fourier transform of 6[n] are the
same. Note that 6[n] is absolutely
summable and that the ROC of the z-transform of 6[nl contains
the unit circle.
EXAMPLE 6.2. Consider the causal exponential sequence
x [ n ] = anu[n] a real

(a) From Fig. 6-20 we have
y[n] - ay[n - 11 =x[n] (6.147)
Taking the Fourier transform of Eq. (6.147) and by Eq. (6.771,
we have
( b ) Using Eq. (6.371, we obtain
h[n] = anu[n]
(c) From Eq. (6.148)
and
which is sketched in Fig. 6-21 for a = 0.9 and a = 0.

Q ) = 1 - A ) - ~ Z ~ (+O ()2 1 - A) - ' ~ x ( z ) (7.38)
Hence, taking the inverse unilateral z-transform of Eq. (7.38),
we get
q [ n ]= ~ ; ' ( ( Z I- A) - ' z )~( o )+ ~ ~ ' (-( A~) - '1b ~ ( z ) )
(7.39)
Substituting Eq. (7.39) into Eq. (7.2261, we ge

5. Time Reversal:
6. Duality:
7. Circular Convolution:
where
The convolution sum in Eq. (6.108) is known as the circular
conuolution of x,[n] and
4nI.
8. Multiplication:
where
9. Additional Properties:
When x[n] is real, let
where x,[n] and xo[n] are the

Equations (6.27) and (6.28) are the discrete-time counterparts of
Eqs. (5.31) and (5.32).
C. Fourier Spectra:
The Fourier transform X(R) of x[n] is, in general, complex and
can be expressed as
As in continuous time, the Fourier transform X(R) of a
nonperi

From Eq. (6.126) we have
1 N0-I No- 1
ck = - C x , [ n ] x 2 [ n ]e -jknon = C d d k - m
NO n=o m =O
Setting k = 0 in the above expression, we get
6.10. (a) Verify Parseval's identity [Eq. (6.19)] for the discrete
Fourier series, that is,
(6) Using x[n] i

The method of evaluating X [ k I based on Eqs. (6.217~a) nd
(6.217b) is known as the
decimation-in-time fast Fourier transform (FFT) algorithm. Note
that since N/2 is even,
using the same procedure, F[kl and G [ k ] can be found by first
determining the
(

The Fourier coefficients c, are often referred to as the spectral
coefficients of x[n].
C. Convergence of Discrete Fourier Series:
Since the discrete Fourier series is a finite series, in contrast to
the continuous-time
case, there are no convergence issu

396 STATE SPACE ANALYSIS (CHAP. 7
and by Eq. (7.29) we obtain
7.22. Repeat Prob. 7.20 using the spectral decomposition
method.
Since all eigenvalues of A are distinct, by Eq. (7.33) we have
Then, by Eq. (7.34) we obtain
7.23. Repeat Prob. 7.20 using the z

Using the Laplace transform, we can calculate eA'. Comparing
Eqs. (7.63) and (7.49), we
see that
E. Stability:
From Eqs. (7.63) and (7.68) or (7.70), we see that if all
eigenvalues A, of the system
matrix A have negative real parts, that is,
Recfw_A,) < 0

twice the highest frequency present in x ( t ) and w, = wJ2, then
y(t) = x(t). If this condition
on the bandwidth of x ( t ) is not satisfied, then y ( t ) z x ( t ) .
However, if w, = 0,/2, then
y(mT,) = x(mT5) for any integer value of m.
CHAP. 51 FOURIE

(6.198)
otherwise
the mean-square error e2 defined by
is minimized.
By definition (6.27)
Let
ffi m
H ( R ) = C h[n]e-J'" and H , ( R ) = z ho[n]e-JRn
n = -m n - - m
where e[n] = h[n] - h,[n]. By Parseval's theorem (6.66) we have
The last two terms in Eq.

y [n] = y,[n] + jy,[n] = H(R) eJRn
~ , [ n ]= Recfw_y[n])= R ~ ( H ( Re)J Rn)
y,[n] = I ~ cfw_[Ynl ) = Imcfw_H(R)
Fig. 6-4 System responses to elnn, cos Rn, and sin Rn.
CHAP. 61 FOURIER ANALYSIS OF DISCRETE-TIME
SIGNALS AND SYSTEMS 303
When a sinusoid cos

6.7. Let x[n] be a real periodic sequence with fundamental
period N, and Fourier
coefficients ck =ak + jb, where a, and b, are both real.
(a) Show that a - , = a k and b-,= -bk.
(b) Show that c, is real if No is even.
(c) Show that x [ n ] can also be exp

Thus, the frequency response Hd(fl) of the discrete-time system
is
1-1
Hd ( f l )= Hd ( z ) l , , , , ~=~ 1 - e-nTs e - ~ n 1 - e - 2 n ~ ,- in
(6.179)
Note that if the system function of a continuous-time LTI system
is given by
then the impulse-invarianc

cC sin wM(t - kT,)
41)= C x(kTs)
& - -m - kT-)
where T, = rr/w,.
Let
From Eq. ( 5.183 we have
Xi
T,X,(w) = C X(o - ko,)
k = -m
Then, under the following two conditions,
7T
(1) X(o)=O,IwI>o, and (2) T,=WM
we see from Eq. (5.1185 that
CHAP. 51 FOURIER ANALY

q[2] = Aq[l] + bx[l] = Acfw_~q[0+] bx[O] ] + bx[l]
= A ~ ~ [ O+] A bx[O] + bx[l]
By continuing this process, we obtain
If the initial state is q[n,] and x[n] is defined for n 2 no, then,
proceeding in a similar
manner, we obtain
The matrix An is the n-fol