"2.1
Let hen) be the unit sample response of an LSI system. Find the frequency response when
(a)
hen)
= 8(n) +
68(n - 1) + 38(n - 2)
(b) hen) = Ur+2u(n - 2).
(a)
This system has a unit sample response that is finite in length. Therefore, the frequency response is a polynomial
in ejw, with the coefficients of the polynomial equal to the values of h(n):
H(ejW) = 1 + 6e-jw + 3e-2jw
This may be shown more formally by writing
00
H(ejW)
00
= n=-oo
L h(n)e-jnw=
L [8(n) + 68(n - 1) + 38(n -
n=-oo
00
Because
L
n=-oo
then
8(n - no)e-jnw = e-jnow
H(ejW) = 1 + 6e-jw + 3e-2jw
2)]e-jnw
(b) For the second system, the frequency response is
00
H(ejW) =
L
h(n)e-jnw=
n=-oo
00
L or+2 e-jnw
n=2
Changing the limits on the sum so that it begins with n = 0, we have
00
H(ejW) =
L
GrHe-j(n+2)W= G)4e-2jWL:o
n=O
Using the geometric series, we find
4 e-2jw
H(ejW)= (13 ) l-l e-'-jw
Ge~jWr
2.4 v Find the magnitude, phase, and group delay of a system that has a unit sample response
hen)
= 8(n) -
a8(n - 1)
where a is real.
The frequency response of this system is
H(ejW) = 1 - ae-jw = 1 - a cosw + ja sinw
Therefore, the magnitude squared is
IH(ejW)12= H(ejW)H*(ejW) = (1 - ae-jW) . (1 - aejW) = 1 + a2 - 2a cos w
The phase, on the other hand, is
a smw
-I H/(ejW) = tan -I
( )cPhw - tan HR(ejW)
1 - a cos w
Finally, the group delay may be found by differentiating the phase (see Prob. 2.19). Alternatively, we may note that
because this system is the inverse of the one considered in Example 2.2.2, the phase and the group delay are simply
the negative of those found in the example. Therefore, we have
a2 - a cosw
Th(W)
=
1 +a2 - 2acosw
.\
~.10
Consider the high-pass filter that has a cutoff frequency Wc = 3Jl'/4 as shown in the following figure:
H(eiw)
1
w
I
-7r
311"
311"
"4
"4
7r
Find the unit sample response, h(n).
(b) A new system is defined so that its unit sample response is I!I(n)
response, HI (elm), of this system.
(a)
=
h(2n). S~tch
the frequency
(a) The unit sample response may be found two different ways. The first is to use the inverse DTFT formula and
perform the integration. The second approach is to use the modulation property and note that if
H/p(ejW)
-'-
n
for Iwl ::: 4"
otherwise
{~
H (eiw) may be written as
H(eiw) = H1p(ei(w-1I"))
Therefore, it follows from the modul,ation property that
hen) = ein1l"h1p(n) = (-l)n h1p(n)
With
h1p(n) = sin(mr
nn /4)
we have
(b)
hen) = (_1)n sin(n7r/4)
nn
The frequency response of the system that has a unit sample response hI (n)
= h(2n)
may be found by evaluating
the discrete-time Fourier transform sum directly:
00
H1(eiw) =
L
n=-oo
00
hl(n)e-inw =
L
n=-oo
h(2n)e-inw=
L
n even
h(n)e-inw/2
However, an easier approach is to note that
hl(n) =h(2n) =(-l)
2nsin(2mr/4)
sin(mr /2)
=
2mr
2mr
which is a low-pass filter with a cutoff frequency of ]'[/2 and a gain of ~. A plot of HI(ejW)is shown in the
following figure:
HI (ejW)
1/2
cv
~
Rn
.. nJl
-J{
1{
Interconnection of Systems
J 2.11
The ideal filters that have frequency responses as shown in the figure below are connected in cascade.
IHI (ejW)1
IH2(ejW)1
1
.
-1T
cv
-1T
/3
1T/3
.
1T
1T
31T/4
1T/4
(a)
1T/4
31T/4
cv
1T
(b)
For an arbitrary input x(n), find the range of frequencies that can be present in the output y(n). Repeat
for the case in which the two systems are connected in parallel.
If thesetwo filtersare connectedin cascade,the frequencyresponseofthe cascadeis
H (ejW)
=
HI (ejW)H2(ejW)
.L
Therefore, any frequencies in the output, y(n), must be passed by both filters. Because the passband for HI (ejW) is
!wl > ]'[/3, and the passband for H2(ejW) is ]'[/4 < Iwl < 3]'[/4, the passband for the cascade (the frequencies for
which both IHI (ejW)1 and IH2(ej(V)1 are equal to 1) is
]'[
- <
3 -
3 ]'[
Iw l < -
-
4 .
With a parallel connection, the overall frequency response is
H (ejW) = HI (ejW) + H2(ejW)
Therefore, the frequencies that are contained in the output are those that are passed by either filter, or
]'[
Iwl> 4
~.a.13
Consider the interconnection of LSI systems shown in the following figure:
c
h2(n)
x(n)
+J
hI (n)
h3(n)
h4(n)
(a) Express the frequency response of the overall system in terms of HI (ejW), H2(ejW), H3(eiw), and
H4(ejW).
(b) Find the frequency response if
hI (n) = 8(n) + 28(n - 2) + 8(n - 4)
h2(n) = h3(n) = (O.2tu(n)
h4(n) = 8(n - 2)
(a)
Because h2(n) is in parallel with the cascade of h3(n) and h4(n), the frequency response of the parallel network is
G(e)W) = H2(e)W) + H3(e)W)H4(e)W)
With h1(n) being in cascade with g(n), the overall frequency response becomes
H(e)W) = H1(e)W)[H2(e)W)+ H3(e)W)H4(e)W)]
(b)
The frequency responses of the systems in this interconnection
are
HI(ejW) = 1 + 2e-j2w + e-j4w = (1 + e-j2w)2
.
"
1
H2(eJW)= H3(eJW)= 1 - O.2e-Jw
.
H4(ejW) = e- j2w
Therefore,
H (ejW) = HI (ejW)[H2(ejW) + H3(ejW)H4(ejW)]
= HI (ejCIJ)H2(ejCIJ)[1
+ H4(ejCIJ)]
(1 + e- j2w)3
1 - O.2e-jw
2.16
A linear shift-invariant system is described by the LCCDE
yen)
= 0.5y(n
- 1) + bx(n)
Find the value of b so that IH(ejW)1 is equal to 1 at (J)= 0, and find the half-power point (i.e., the
frequency at which IH(ejW)12is equal to one-half of its peak value, which occurs at (J)= 0).
The frequency response of the system described by this difference equation is
b
H(eJW)= I - 0.5e-Jw
b2
Because
b2
IH(eJW)12 = (1 - 0.5e-JW)(1 - 0.5eJW) = 1.25 - cosu/)
IH(eJW)1 will be equal to I at w
=
0 if
b2
1.25 - I
This will be true when b
=
I
= ::f: 0.5.
To find the half-power point, we want to find the frequency for which
0.25
IH(eJW)12
= 1.25- cosw
This occurs when
cosw
or cv = 0.23Jl'.
= 0.75
= 0.5
2.20 JFind the DTFT of each of the following sequences:
(a) xl(n)
(b) x2(n)
= Gfu(n + 3)
= an sin(ncvo) u(n)
Gf
(c) x3(n) =
{
n=O,2,4""
"
otherwIse
0
(a) 'For the first sequence, the DTFT may be evaluated directly as follows:
00
00
L ore-jnw = n=-3
L Oe-jWr
X1(ejW) =
n=-3
00
= Oe-jWr3
L
Oe-jwr = 1 -8e3jw
n=O
. 1.e- jw
Z
(b)
The best way to find the DTFT of Xz(n) is to express the sinusoid as a sum of two complex exponentials as follows:
1.
xz(n)
=
"
2j
.
(;
- ane-jnWO]u(n)
[anejnwo
The DTFT of the first term is
1
oo
L
"
1
"
oo
L(
1
n
"
1
anejnWOe-jnw = ae-j (w-wo») = "
2 ] n=O
2] 1 - ae- Jew-wo)
2 ]" n=O
'
'
Similarly, for the second term we have
1
00
1
"
2];- Lane-jnWOe-jnw
n=O
1
= - 2j 1 - ae-j(w+wo)
Therefore,
- 1
X ze
( jW ) --
[
2j
(c)
1
.
1 - ae- jew-wo)
-
1
(a sincvo)e-jw
]
1 - ae- j(w+wo)
1 - (2a cos cvo)e- jw + aZe-Zjw
Finally, for x3(n), we have
00
X3(ejW) =
L
n=-oo
00
X3(ejW) = '"
Therefore,
L
n~
,
00
x3(n)e-jnw=
(1.
)zne-Zjnw =
Z
00
'"
L
n~
L or
e-jnw
n=O,Z,4,...
( 1.e-Zjw
)n =
4
1
"
1- 4
1.e-ZjW