MASSACHUSETTS INSTITUTE OF TECHNOLOGY
Department of Electrical Engineering and Computer Science
6.003,Signals and Systems|Fall 2003
Problem Set 5 Solutions
Issued,October 22,2003
Exercise for home study:
O&W 4.47
This problem examines the Fourier transform of a continuous-time LTI system with a
real,causal impulse response,h(t).
(a) To prove that H(j!) is completely speci ed by<efH(j!)gfor a real and causal h(t),
we explore the even part of a function,he(t).
By de nition,he(t) = 12h(t) + 12h( t),Since h(t) = 0 for t < 0,
h(t) =
8<
:
2he(t) for t > 0
he(t) for t = 0
0 for t < 0
(1)
Therefore,if we know He(j!),then we can nd both he(t) and h(t),If h(t) was not
causal,we couldn’t determine h(t) from he(t) alone,We would need ho(t),the odd
part of h(t) also,What is He(j!)? If h(t) is real then He(j!) =<efH(j!)g,This is
shown below:
He(j!) =
Z 1
1
he(t)e j!tdt =
Z 1
1
1
2h(t)e
j!tdt +
Z 1
1
1
2h( t)e
j!tdt
=
Z 1
1
1
2h(t)e
j!tdt +
Z 1
1
1
2h(t)e
j!tdt =
Z 1
1
h(t) cos(!t)dt =<efH(j!)g:
(b) Given we know <efH(j!)gis cos !,we need to nd h(t).
<efH(j!)g= He(j!) = cos ! = 0:5ej! + 0:5e j!
.
he(t) =F 1fcos(!)g=F 1f0:5ej!g+F 1f0:5e j!g= he1(t) + he2(t)
For he1(t) we use the time shift property that for any to,e j!toX(j!) F ! x(t to).
Thus for he1(t),to = 1,We have
he1(t) =F 1f0:5e j! 1 1g= 0:5 (t + 1):
1
We nd he2(t) using the same method:
he2(t) =F 1f0:5e j! 1 1g= 0:5 (t 1):
We combine the two signals to get he(t) = 0:5 (t + 1) + 0:5 (t 1),Finally,we know
because h(t) is causal,our nal answer is h(t) = 2he(t) = (t 1):
(c) We need to show that a real and causal h(t) can be recovered from ho(t) everywhere
except at t = 0,By de nition,ho(t) = 12h(t) 12h( t),Because h(t) is causal,
ho(t) = 12h(t) for t > 0 and ho(t) = 12h( t) for t < 0,This means that if we know
ho(t),we know h(t) = 2ho(t) when t > 0 and h(t) = 0 when t < 0,However,at t = 0 we
have a problem,ho(0) = 0 no matter what h(0) is,For example,if h(t) = (t)+ (t 1),
then ho(t) = 12 (t 1) + 12 ( t 1),The delta function at t = 0 was lost when we
looked at the odd part of h(t).
If we do not have a singularity at t = 0,but instead has some arbitrary nite value at
t = 0,then the imaginary part of H(j!) can be used to specify H(j!),If we have
h(t) =
1 + u(t) for t = 0
u(t) for t6= 0 (2)
Then H(j!) = R 1 1 u(t)e j!tdt,The nite value of 1 at t = 0 has no area so it doesn’t
show up under the integral.
Ho(j!) = 12 R 1 1(h(t) h( t))e j!tdt = 12(R 1 1 h(t)e j!tdt R 1 1 h(t)ej!tdt) = j R 1 1 h(t) sin !t:
This shows that Ho(j!) = =mfH(j!)g,This also shows that Ho(j!) = 12H(j!)
1
2H( j!),Thus,H(j!) can be recovered from Ho(j!),Also the imaginary part canbe used to nd h
o(t) which can be used to nd h(t) everywhere except at t = 0:
Problems to be turned in:
Problem 1 Consider the signal x(t) with spectrum depicted in Figure p4.28 (a) of O&W.
Sketch the spectrum of
y(t) = x(t) [cos(t=2) + cos(3t=2)]:
Solution:
To draw Y (j!),the spectrum of y(t),we use the linearity and the multiplication property.
Thus,Y (j!) = 12 [X(j!) Ffcos t2g] + 12 [X(j!) Ffcos 3t2g].
Ffcos t2g= [ (! 0:5) + (! + 0:5)]:
Ffcos 3t2g= [ (! 1:5) + (! + 1:5)]:
Thus,
Y (j!) = 12 X(j!) (! 0:5) + 12 X(j!) (! + 0:5)
2
+ 12 X(j!) (! 1:5) + 12 X(j!) (! + 1:5):
= 12 (X(j(! 0:5)) + X(j(! + 0:5)) + X(j(! 1:5)) + X(j(! + 1:5)))
So,X(j!) is convolved with 4 shifted impulse functions,Convolving a signal with a shifted
impulse function causes the signal to be shifted and replicated about the location on the
x-axis where the impulse function is located,Thus,centered at t = 1:5; 0:5; 0:5,and 1:5,
we replicate X(j!),We also need to scale these 4 replications by a factor of 12 due to the
multiplication of the constants,12, This can be seen in the gure below:
X(j!)
! 2:5 1:5 1:5 2:5
0:5
Problem 2 Consider the system depicted below:
x(t)
p(t)
a(t)
H(j!)
b(t)
q(t)
c(t)
where x(t) = sin 4 t t,p(t) = cos 2 t,q(t) = sin 2 t t,and the frequency response of H(j!) is
given by
H(j!)
!
1
2 2
3
(a) Let A(j!) be the Fourier transform of a(t),Sketch and clearly label A(j!).
(b) Let B(j!) be the Fourier transform of b(t),Sketch and clearly label B(j!).
(c) Let C(j!) be the Fourier transform of c(t),Sketch and clearly label C(j!).
(d) Compute the output c(t).
Solution:
(a) To nd A(j!),we use the multiplication property,Since a(t) = x(t) p(t),then
A(j!) = 12 [X(j!) P(j!)],We need to nd X(j!) and P(j!),To nd X(j!) from
x(t),we recognize x(t) as being in O & W’s Table 4.2 Basic Fourier Transform Pairs.
It is a sinc function with W = 4, Thus,
X(j!) =
1 forj!j< 4
0 forj!j> 4 (3)
X(j!)
!
4 4 0
1
Because p(t) = cos 2 t,P(j!) = [ (! 2 ) + (! + 2 )],Since P(j!) is two impulse
functions,the convolution of X(j!) with P(j!) results in the superposition of two
copies of X(j!),one centered at ! = 2 and the other centered at ! = 2, The
resulting A(j!) is shown below:
A(j!)
!
6 6 2 2 0
1
0:5
4
(b) To nd B(j!),we use the convolution property,Thus,B(j!) = A(j!)H(j!),A(j!)
is a low pass lter and H(j!) is a high pass lter,Multiplying the two together cre-
ates a bandpass lter,A(j!) cuts o all frequencies for j!j > 6, H(j!) cuts o all
frequencies forj!j< 2, The resulting signal,B(j!) is shown below:
B(j!)
!
0:5
1:0
6 6 2 2 0
(c) To nd C(j!),we need to convolve B(j!) with Q(j!).
Q(j!) =
1 forj!j< 2
0 forj!j> 2 (4)
Q(j!) is shown below:
Q(j!)
!
2 2 0
1
Thus,C(j!) can be drawn as shown below:
5
C(j!)
!
8 8 4 4 0
1
(d) To compute c(t),we multiply b(t) with q(t),B(j!) is the sum of two ideal frequency-
shifted unity-gain lters,Filters in the frequency domain become sinc functions in the
time domain,In addition,a frequency shift of !o corresponds to multiplying by e j!ot
in the time domain,Hence,
b(t) = 12e j4 t sin 2 t t + 12ej4 t sin 2 t t = cos 4 tsin 2 t t,
Therefore
c(t) = b(t)q(t) = cos 4 tsin
2 2 t
2t2,
Problem 3 O&W 4.44,In addition to parts (a) and (b),answer the following,
(c) Find the di erential equation relating the input and output of this system.
Solution:
(a) We are given the following equation that relates the output y(t) of a causal LTI system
to the input x(t).
dy(t)
dt + 10y(t) =
Z 1
1
x( )z(t )d x(t) (5)
where
z(t) = e tu(t) + 3 (t):
We want to nd the frequency response H(j!) = Y (j!)=X(j!) of this system,To do
this we take the transform of each term in Equation 5,Because of linearity,to nd the
Fourier transform of the entire equation,we take the Fourier transform of each term in
the equation,For the rst term,dy(t)dt,we use the di erentiation property to nd that
F
dy(t)
dt
F !j!Y (j!):
6
For the second term,we have 10Y (j!),For the next term,the integral,we recognize
this integral as being the convolution of x(t) and z(t),Thus,the convolution property
tells us that
Ffx(t) z(t)g F !X(j!)Z(j!):
Finally,x(t) becomes X(j!),The new equation in the frequency domain is
j!Y (j!) + 10Y (j!) = X(j!)Z(j!) X(j!):
Algebraic manipulations are used to separate out Y (j!) on the left side of the equation
and separate out X(j!) on the right side of the equation,We then nd
H(j!) = Y (j!)X(j!) = Z(j!) 110 + j! (6)
We insert the function Z(j!),By linearity,Ffz(t)g = Ffe tu(t)g+Ff3 (t)g,The
Fourier transform for each of the terms can be found in O & W’s Table 4.2.
Z(j!) = 11 + j! + 3,= 4 + 3j!1 + j!
This can be plugged into Equation( 6) to give
H(j!) = Y (j!)X(j!) = 3 + 2j!(10 + j!)(1 + j!) (7)
(b) The impulse response can be found by doing a partial fraction expansion of H(j!) as
found in Equation( 7).
H(j!) = 3 + 2j!(10 + j!)(1 + j!) = A10 + j! + B1 + j! = 17=910 + j! + 1=91 + j!:
The inverse Fourier transform of the last two terms can be determined from Table 4.2
of O & W to give:
h(t) =
1
9e
t + 17
9 e
10t
u(t):
(c) To nd the di erential equation relating the input to the output we go back to Equa-
tion 7 and rewrite it as
H(j!) = Y (j!)X(j!) = 3 + 2j!(j!)2 + 11j! + 10:
After cross-multiplying we get
7
10Y (j!) + 11j!Y (j!) + (j!)2Y (j!) = 3X(j!) + 2j!X(j!):
Using the di erentiation property,we do an inverse Fourier transform to get the dif-
ferential equation
10y(t) + 11dy(t)dt + 11d
2y(t)
dt2 = 3x(t) + 2
dx
dt,
Problem 4 O&W 5.21 (c),(g)
Solution:
(c) We need to compute the Fourier transform of x[n] = 13jnju[ n 2],To do this we use
the analysis equation as shown below:
X(ej!) =
2X
n= 1
1
3
n
e j!n =
1X
n=2
1
3e
j!
n
= 11 1
3e
j! 1
1
3e
j! =
1
9e
j2!
1 13ej!
(g) We need to compute the Fourier transform of x[n] = sin( 2 n) + cos(n),To do this we
use Table 5.2 in O & W to nd the Fourier transform of each term and then sum the
two transforms for x[n].
F
n
sin
2n
o
= j
1X
l= 1
( (! 2 2 l) (! + 2 2 l))
Ffcos(n)g=
1X
l= 1
( (! 1 2 l) + (! + 1 2 l))
X(ej!) = j
1X
l= 1
( (! 2 2 l) (!+ 2 2 l))+
1X
l= 1
( (! 1 2 l)+ (!+1 2 l))
Problem 5 The following are Fourier transforms of discrete-time signals,Determine the
signal corresponding to each transform.
(a) X(ej!) = 4ej4! ej! + 6 + 8e j3! 16e j11!
(b)
X(ej!) =
(
1; 0 j!j< 4 ; 2 <j!j
0; 4 <j!j< 2
(c) X(ej!) = 1 + 3e
j3!
1 + 14e j!
8
Solutions:
(a) When the Fourier transform is a sum of exponentials,often the easiest way to nd x[n]
is to use the analysis equation
X(ej!) =
1X
n= 1
x[n]e j!n
to match each term to what the particular value of n is,For example,in this problem
we can see that the rst term,4ej4! can only result from the n = 4 term,Thus,we
can rewrite the equation as
X(ej!) = x[ 4]e j!( 4) + x[ 1]e j!( 1) + x[0]e j!0 + x[3]e j!(3) + x[11]e j!(11):
We match terms to nd
x[ 4] = 4; x[ 1] = 1; x[0] = 6; x[3] = 8; x[11] = 16:
x[n] = 0 for all other n.
(b) X(ej!) for this problem looks like:
X(ej!)
!
2
2
4
40
1
The Fourier transform for this signal as the sum of 3 ideal low pass lters with two
of them frequency-shifted to ! = 3 4 and ! = 3 4, Table 5.1 in O & W shows that a
frequency-shift corresponds to multiplication by an exponential in the time domain:
X(ej(! !o)) F !ej!onx[n]:
Thus,in the time domain,we can write this as the sum of three sinc functions each
with W = 4 and with two of them multiplied by the appropriate exponential,This
gives
x[n] = e j 3 4 n sin
4n
n +
sin 4 n
n + e
j 3 4 n sin
4n
n =
2 cos(3 4 n) + 1
sin(
4 n)
n,
9
(c) This Fourier transform looks similar to the Fourier transform for anu[n] in Table 5.2
of O & W,We will manipulate it to look more like that,First,we separate it into two
terms so that
X(ej!) = 11 + 1
4e
j! +
3e j3!
1 + 14e j! = X1(e
j!) + X2(ej!):
By linearity we can solve for the time signal for each transform and then x[n] =
x1[n]+x2[n],X1(ej!) matches to the transform pair mentioned above in Table 5.2 with
a = 14,Thus,
x1[n] =
14
n
u[n]:
X2(ej!) matches to the transform pair in Table 5.2 also except for the numerator term
of 3e j3!,This numerator term corresponds to a scaling of 3 and a time shift of 3.
Thus,
x2[n] = 3
14
n 3
u[n 3]:
Summing the two terms,we get
x[n] =
14
n
u[n] + 3
14
n 3
u[n 3]:
Problem 6 Let X(ej!) denote the Fourier transform of the signal x[n] depicted below.
x[n]
n 5 4 3
2 1 0
1 2 3
4 5 6 7 8
2
1
2 2
1
2
(a) Find X(1) = X(ej0).
(b) Find such that ej !X(ej!) is real.
(c) Evaluate R X(ej!)d!.
(d) Find X(ej ).
10
(e) Determine and sketch the signal whose Fourier transform is<efX(ej!)g.
(f) Evaluate each of the following integrals:
(f.1)
Z
jX(ej!)j2d!
(f.2)
Z
dX(ej!)
d!
2
d!
Solutions:
(a) We use the analysis equation:
X(1) = X(ej0) =
1X
n= 1
x[n]ej 0 n = 2:
(b) If x[n] is real and even,X(ej!) is real and even,Since this signal is real,we just need
to make it even,If we time shift x[n] 2 steps to the left,it will be real and even,A
left time shift of 2 corresponds to multiplying the frequency spectrum by ej!2,= 2
will make the frequency spectrum real (and even).
(c) We use the synthesis equation:
Z
X(ej!)d! = 2
1
2
Z
X(ej!)ej! 0d!
= 2 x[0] = 2,
(d) We use the analysis equation:
X(ej ) =
1X
n= 1
x[n]e j n =
1X
n= 1
x[n]( 1)n = 10
(e) <efX(ej!)g,which is the real (and even) part of X(ej!),is the Fourier transform of
the even part of the time signal,Thus,we need to sketch xe[n] = 0:5x[n] + 0:5x[ n].
This is shown below:
11
xe[n]
n
6 5 4 4 5 6 7
1
0:5
1
3
1
2
1
1
1
0
1
1
1
2
1
3
0:5
1
(f) To evaluate the following integrals we use Parseval’s relations.
(f.1) With Parseval’s relation,we know that
1X
n= 1
jx[n]j2 = 12
Z
jX(ej!)j2d!:
Since we know x[n],we can use it in Parseval’s relation to nd the quantity we
are interested in,Thus,
1X
n= 1
jx[n]j2 = 18
and
Z
jX(ej!)j2d! = 18 2 = 36,
(f.2) To use Parseval’s relation again,we need to de ne a new variable,Y (ej!) =
j dX(ej!)d!, Then according to Table 5.1 of O&W,
F 1
jdX(e
j!)
d!
=F 1fY (ej!)g= y[n] = nx[n]:
Thus,
Z
dX(ej!)
d!
2
d! =
Z
jdX(ej!)
d!
2
d! = 2
1
2
Z
jY (ej!)j2d!
= 2
1X
n= 1
jy[n]j2 = 2
1X
n= 1
jnx[n]j2 = 2 96 = 432,
12
Problem 7 Answer the questions asked in O&W 5.24 for the following two signals.
x1[n]
n 5 4
3 2 1
0 1 2 3 4
5 6 7
1
1
8
x2[n]
n 5 4 3 1 0 1 3 4 5 6 7 8
1
2
3
2
2
3
1
Solution:
(a)(a.1) We want to know if<efX1(ej!)g= 0,This is true only if x1[n] is purely real and
odd or purely imaginary and even,Looking at the signal,x1[n] we see that it is
real and odd,or xe[n] = 0:5x[n] + 0:5x[ n] = 0,Yes,<efX1(ej!)g= 0:
(a.2) Because this signal is real and odd,it is purely imaginary,=mfX1(ej!)g6= 0:
(a.3) The real part of a Fourier transform corresponds to the part of the time signal
which is real and even and the part of the time signal which is odd and imaginary.
By multiplying X1(ej!) by ej !,we can shift the signal in time,Thus,we can
shift by the correct number of units to make the signal even,Looking at the graph
of x1[n] reveals that a time shift of = 2 will cause the signal to be even and
thus,the resulting Fourier transform,ej 2!X1(ej!) will be real.
13
(a.4) We can use the synthesis equation as follows:
Z
X(ej!)d! = 2
1
2
Z
X(ej!)ej! 0d!
= 2 x[0] = 0:
Yes,R X(ej!)d! = 0.
(a.5) By de nition,the Fourier transform of any discrete sequence is periodic,so X(ej!)
is periodic.
(a.6) Since x1[n] is periodic,
X(ej!) =
1X
k= 1
2 ak
! 2 N k
where ak = 1N Pn=<N> x[n]e j 2 N kn,Plugging ! = 0 into the equation above shows
that
X(ej0) =
1X
k= 1
2 ak
2 N k
:
From this equation,we can see that all terms will be zero except possibly 2 a0 (0).
If a0 = 0 then X(ej0) = 0,Using the formula for ak above we see that,
a0 = 18
X
n=<N>
x[n] = 0:
We could also note that a0 is the DC average of the signal and from the graph we
can see that the DC average is zero,So,yes,X(ej0) = 0:
(b)(b.1) To determine if<efX(ej!)g= 0,we need to see if x2[n] is real or imaginary and
even or odd,From the graph,we can see that it is real and even,Thus,X(ej!)
is real and even,So<efX(ej!)g6= 0.
(b.2) This means that=mfX(ej!)g= 0.
(b.3) Since X(ej!) is already real,= 0.
(b.4) Using the same strategy employed for x1[n],we see that x2[0]6= 0 so R X(ej!)d!6=
0.
(b.5) By de nition,all X(ej!) are periodic so X2(ej!) is periodic.
(b.6) X2(ej0) is the DC component of the signal,We can see that the DC component
is equal to zero so this condition is met,That is,
X2(ej0) =
1X
n= 1
x[n] = 0:
14
Problem 8 Consider the same question as asked in O&W 5.27 (a) but with X(ej!) as
depicted below
X(ej!)
!
1
2 4 2 4
and with
p[n] = cos n cos( n=2):
Solution:
To sketch W(ej!),we use the multiplication property,
p[n]x[n] = w[n] F !W(ej!) = 12 P(ej!) X(ej!):
This allows us to use circular convolution to solve the problem,X(ej!) is as shown above
and
P(ej!) =
1X
l= 1
( (! 2 l)+ (! + 2 l))
1X
l= 1
( (! 2 2 l)+ (! + 2 2 l)):
The area under the impulses at ! = 2 l is equal to 2 not just because the impulses
from the term P1l= 1 (! 2 l) overlap with the impulses from the term P1l= 1 (! +
2 l).
These impulses and the impulses at ! = 2 2 l cause replicas of X(ej!) that are centered
at these !’s,The resulting W(ej!) is shown below:
15
W(ej!)
!
3 2 3 2
5 4 5 4
3 4 3 4
2 2
4 4
1
0:5
Problem 9 Answer the same questions as asked in O&W 5.30 (b) with x[n] as given in
that problem and for each of the following LTI unit sample responses:
(a) h[n] = sin n=16 n sin( n=12) n
(b) h[n] = sin( n=8) sin( n=2) 2n2
Solution:
We are given that x[n] = sin( n8 ) 2 cos( n4 ):
(a) To determine y[n],we will use the convolution property,That is,we will compute
Y (ej!) = X(ej!)H(ej!) and then we can use the synthesis equation to determine y[n]
from Y (ej!),To nd H(ej!) for the given h[n],we use Table 5.1 and Table 5.2 in
O&W and nd H(ej!) = H1(ej!) H2(ej!) where
H1(ej!) =
1 forj!j<
160 for
16 <j!j<
(8)
and
H2(ej!) =
1 forj!j<
120 for
12 <j!j<
(9)
Subtracting H2(ej!) from H1(ej!) yields
H(ej!) =
1 for 3
48 <j!j<
4
480 for 3
48 >j!j>
4
48
(10)
H(ej!) is a bandpass lter whose gain is 1.
16
X(ej!) is found from Table 5.2:
X(ej!) = j
1X
l= 1
( (! 8 2 l) (!+ 8 2 l)) 2
1X
l= 1
( (! 4 2 l)+ (!+ 4 2 l)):
It is clear that the bandpass lter doesn’t allow this x[n] through it and hence Y (ej!) =
0 and y[n] = 0.
(b) Again we will use the convolution property to nd Y (ej!) = X(ej!)H(ej!) and then
nd y[n] from Y (ej!),X(ej!) is the same as the previous problem,However,
h[n] = sin(
n
8 ) sin
n
2
2n2 = h1[n]h2[n]:
We can use the multiplication property to nd H(ej!),Here,
H(ej!) = 12 H1(ej!) H2(ej!):
h1[n] and h2[n] are sinc functions,thus,
H1(ej!) =
1 forj!j<
80 for
8 <j!j<
(11)
H2(ej!) =
1 forj!j<
20 for
2 <j!j<
(12)
The convolution of these two signals is to be calculated.
H(ej!) is shown below:
H(ej!)
! 5
8
5
8
3
8
3
8
1=8
0
17
Multiplying X(ej!) with H(ej!) gives the following spectrum for Y (ej!):
Y (ej!)
!
8j
4
8j
4
8
8
4 4
The above gure shows that y[n] is just a scaled version of x[n],Thus,
y[n] = 18 sin
n
8
14 cos
n
4
:
Problem 10 Consider a system consisting of the cascade of two LTI systems as depicted
below
x[n] System 1
z[n]
System 2 y[n]
System 1 is LTI and has a unit-sample response of
h[n] =
1
4
n
u[n]:
System 2 is LTI,and we know that if the input is
x[n] = [n] + 12 [n 1]
the output is
y[n] = 10 [n] [n 1]:
In the following parts,please show your work.
18
(a) What is the frequency response X(ej!) of the overall system?
(b) Find the di erence equation for the overall system.
(c) Find the impulse response of the overall system.
Solution:
(a) For this cascade of two LTI systems,
H(ej!) = H1(ej!)H2(ej!):
Taking a Fourier transform of h1[n] = (14)nu[n] gives
H1(ej!) = 11 0:25e j!,
For H2(ej!),we take Fourier transforms of z[n] and y[n] as given above and then
H2(ej!) = Y (ej!)Z(ej!),This gives
H2(ej!) = 10 e
j!
1 + 0:5e j!,
Multiplying the two terms gives the overall frequency response as
H(ej!) = 11 0:25e j! 10 e
j!
1 + 0:5e j!,
(b) To nd the di erence equation for the whole system,we need to multiply Y (ej!) by
the denominator of the above equation and multiply Z(ej!) by the numerator of the
above equation,That is
Y (ej!)(1 + 0:25e j! 0:125e j2!) = X(ej!)(10 e j!):
Then,
Y (ej!) = 0:25e j!Y (ej!) + 0:125e j2!Y (ej!) + 10X(ej!) 1e j!X(ej!):
Because of linearity,we take the inverse Fourier transform of each term,noting that
multiplication of e j!no in the frequency domain corresponds to a shift of no in the
time domain,Thus,the di erence equation is
y[n] = 0:25y[n 1] + 0:125y[n 2] + 10x[n] x[n 1]:
19
(c) To nd the impulse response,h[n],we rewrite the equation for H(ej!) found in (a) as
the sum of two rst order systems and do term by term matching to nd what the
numerator values should be for each term,That is,
H(ej!) = 10 e
j!
(1 0:25e j!)(1 + 0:5e j!) =
A
(1 0:25e j!) +
B
(1 + 0:5e j!)
= A(1 + 0:5e
j!)
(1 0:25e j!)(1 + 0:5e j!) +
B(1 0:25e j!)
(1 0:25e j!)(1 + 0:5e j!)
= A + B + (0:5A 0:25B)e
j!)
(1 0:25e j!)(1 + 0:5e j!),
Matching terms gives us A = 2 and B = 8,Then the inverse transform for each term
can be determined from Table 5.2:
H(ej!) = 2(1 0:25e j!) + 8(1 + 0:5e j!)
which gives,
h[n] = 2(0:25)nu[n] + 8( 0:5)nu[n]:
20
Department of Electrical Engineering and Computer Science
6.003,Signals and Systems|Fall 2003
Problem Set 5 Solutions
Issued,October 22,2003
Exercise for home study:
O&W 4.47
This problem examines the Fourier transform of a continuous-time LTI system with a
real,causal impulse response,h(t).
(a) To prove that H(j!) is completely speci ed by<efH(j!)gfor a real and causal h(t),
we explore the even part of a function,he(t).
By de nition,he(t) = 12h(t) + 12h( t),Since h(t) = 0 for t < 0,
h(t) =
8<
:
2he(t) for t > 0
he(t) for t = 0
0 for t < 0
(1)
Therefore,if we know He(j!),then we can nd both he(t) and h(t),If h(t) was not
causal,we couldn’t determine h(t) from he(t) alone,We would need ho(t),the odd
part of h(t) also,What is He(j!)? If h(t) is real then He(j!) =<efH(j!)g,This is
shown below:
He(j!) =
Z 1
1
he(t)e j!tdt =
Z 1
1
1
2h(t)e
j!tdt +
Z 1
1
1
2h( t)e
j!tdt
=
Z 1
1
1
2h(t)e
j!tdt +
Z 1
1
1
2h(t)e
j!tdt =
Z 1
1
h(t) cos(!t)dt =<efH(j!)g:
(b) Given we know <efH(j!)gis cos !,we need to nd h(t).
<efH(j!)g= He(j!) = cos ! = 0:5ej! + 0:5e j!
.
he(t) =F 1fcos(!)g=F 1f0:5ej!g+F 1f0:5e j!g= he1(t) + he2(t)
For he1(t) we use the time shift property that for any to,e j!toX(j!) F ! x(t to).
Thus for he1(t),to = 1,We have
he1(t) =F 1f0:5e j! 1 1g= 0:5 (t + 1):
1
We nd he2(t) using the same method:
he2(t) =F 1f0:5e j! 1 1g= 0:5 (t 1):
We combine the two signals to get he(t) = 0:5 (t + 1) + 0:5 (t 1),Finally,we know
because h(t) is causal,our nal answer is h(t) = 2he(t) = (t 1):
(c) We need to show that a real and causal h(t) can be recovered from ho(t) everywhere
except at t = 0,By de nition,ho(t) = 12h(t) 12h( t),Because h(t) is causal,
ho(t) = 12h(t) for t > 0 and ho(t) = 12h( t) for t < 0,This means that if we know
ho(t),we know h(t) = 2ho(t) when t > 0 and h(t) = 0 when t < 0,However,at t = 0 we
have a problem,ho(0) = 0 no matter what h(0) is,For example,if h(t) = (t)+ (t 1),
then ho(t) = 12 (t 1) + 12 ( t 1),The delta function at t = 0 was lost when we
looked at the odd part of h(t).
If we do not have a singularity at t = 0,but instead has some arbitrary nite value at
t = 0,then the imaginary part of H(j!) can be used to specify H(j!),If we have
h(t) =
1 + u(t) for t = 0
u(t) for t6= 0 (2)
Then H(j!) = R 1 1 u(t)e j!tdt,The nite value of 1 at t = 0 has no area so it doesn’t
show up under the integral.
Ho(j!) = 12 R 1 1(h(t) h( t))e j!tdt = 12(R 1 1 h(t)e j!tdt R 1 1 h(t)ej!tdt) = j R 1 1 h(t) sin !t:
This shows that Ho(j!) = =mfH(j!)g,This also shows that Ho(j!) = 12H(j!)
1
2H( j!),Thus,H(j!) can be recovered from Ho(j!),Also the imaginary part canbe used to nd h
o(t) which can be used to nd h(t) everywhere except at t = 0:
Problems to be turned in:
Problem 1 Consider the signal x(t) with spectrum depicted in Figure p4.28 (a) of O&W.
Sketch the spectrum of
y(t) = x(t) [cos(t=2) + cos(3t=2)]:
Solution:
To draw Y (j!),the spectrum of y(t),we use the linearity and the multiplication property.
Thus,Y (j!) = 12 [X(j!) Ffcos t2g] + 12 [X(j!) Ffcos 3t2g].
Ffcos t2g= [ (! 0:5) + (! + 0:5)]:
Ffcos 3t2g= [ (! 1:5) + (! + 1:5)]:
Thus,
Y (j!) = 12 X(j!) (! 0:5) + 12 X(j!) (! + 0:5)
2
+ 12 X(j!) (! 1:5) + 12 X(j!) (! + 1:5):
= 12 (X(j(! 0:5)) + X(j(! + 0:5)) + X(j(! 1:5)) + X(j(! + 1:5)))
So,X(j!) is convolved with 4 shifted impulse functions,Convolving a signal with a shifted
impulse function causes the signal to be shifted and replicated about the location on the
x-axis where the impulse function is located,Thus,centered at t = 1:5; 0:5; 0:5,and 1:5,
we replicate X(j!),We also need to scale these 4 replications by a factor of 12 due to the
multiplication of the constants,12, This can be seen in the gure below:
X(j!)
! 2:5 1:5 1:5 2:5
0:5
Problem 2 Consider the system depicted below:
x(t)
p(t)
a(t)
H(j!)
b(t)
q(t)
c(t)
where x(t) = sin 4 t t,p(t) = cos 2 t,q(t) = sin 2 t t,and the frequency response of H(j!) is
given by
H(j!)
!
1
2 2
3
(a) Let A(j!) be the Fourier transform of a(t),Sketch and clearly label A(j!).
(b) Let B(j!) be the Fourier transform of b(t),Sketch and clearly label B(j!).
(c) Let C(j!) be the Fourier transform of c(t),Sketch and clearly label C(j!).
(d) Compute the output c(t).
Solution:
(a) To nd A(j!),we use the multiplication property,Since a(t) = x(t) p(t),then
A(j!) = 12 [X(j!) P(j!)],We need to nd X(j!) and P(j!),To nd X(j!) from
x(t),we recognize x(t) as being in O & W’s Table 4.2 Basic Fourier Transform Pairs.
It is a sinc function with W = 4, Thus,
X(j!) =
1 forj!j< 4
0 forj!j> 4 (3)
X(j!)
!
4 4 0
1
Because p(t) = cos 2 t,P(j!) = [ (! 2 ) + (! + 2 )],Since P(j!) is two impulse
functions,the convolution of X(j!) with P(j!) results in the superposition of two
copies of X(j!),one centered at ! = 2 and the other centered at ! = 2, The
resulting A(j!) is shown below:
A(j!)
!
6 6 2 2 0
1
0:5
4
(b) To nd B(j!),we use the convolution property,Thus,B(j!) = A(j!)H(j!),A(j!)
is a low pass lter and H(j!) is a high pass lter,Multiplying the two together cre-
ates a bandpass lter,A(j!) cuts o all frequencies for j!j > 6, H(j!) cuts o all
frequencies forj!j< 2, The resulting signal,B(j!) is shown below:
B(j!)
!
0:5
1:0
6 6 2 2 0
(c) To nd C(j!),we need to convolve B(j!) with Q(j!).
Q(j!) =
1 forj!j< 2
0 forj!j> 2 (4)
Q(j!) is shown below:
Q(j!)
!
2 2 0
1
Thus,C(j!) can be drawn as shown below:
5
C(j!)
!
8 8 4 4 0
1
(d) To compute c(t),we multiply b(t) with q(t),B(j!) is the sum of two ideal frequency-
shifted unity-gain lters,Filters in the frequency domain become sinc functions in the
time domain,In addition,a frequency shift of !o corresponds to multiplying by e j!ot
in the time domain,Hence,
b(t) = 12e j4 t sin 2 t t + 12ej4 t sin 2 t t = cos 4 tsin 2 t t,
Therefore
c(t) = b(t)q(t) = cos 4 tsin
2 2 t
2t2,
Problem 3 O&W 4.44,In addition to parts (a) and (b),answer the following,
(c) Find the di erential equation relating the input and output of this system.
Solution:
(a) We are given the following equation that relates the output y(t) of a causal LTI system
to the input x(t).
dy(t)
dt + 10y(t) =
Z 1
1
x( )z(t )d x(t) (5)
where
z(t) = e tu(t) + 3 (t):
We want to nd the frequency response H(j!) = Y (j!)=X(j!) of this system,To do
this we take the transform of each term in Equation 5,Because of linearity,to nd the
Fourier transform of the entire equation,we take the Fourier transform of each term in
the equation,For the rst term,dy(t)dt,we use the di erentiation property to nd that
F
dy(t)
dt
F !j!Y (j!):
6
For the second term,we have 10Y (j!),For the next term,the integral,we recognize
this integral as being the convolution of x(t) and z(t),Thus,the convolution property
tells us that
Ffx(t) z(t)g F !X(j!)Z(j!):
Finally,x(t) becomes X(j!),The new equation in the frequency domain is
j!Y (j!) + 10Y (j!) = X(j!)Z(j!) X(j!):
Algebraic manipulations are used to separate out Y (j!) on the left side of the equation
and separate out X(j!) on the right side of the equation,We then nd
H(j!) = Y (j!)X(j!) = Z(j!) 110 + j! (6)
We insert the function Z(j!),By linearity,Ffz(t)g = Ffe tu(t)g+Ff3 (t)g,The
Fourier transform for each of the terms can be found in O & W’s Table 4.2.
Z(j!) = 11 + j! + 3,= 4 + 3j!1 + j!
This can be plugged into Equation( 6) to give
H(j!) = Y (j!)X(j!) = 3 + 2j!(10 + j!)(1 + j!) (7)
(b) The impulse response can be found by doing a partial fraction expansion of H(j!) as
found in Equation( 7).
H(j!) = 3 + 2j!(10 + j!)(1 + j!) = A10 + j! + B1 + j! = 17=910 + j! + 1=91 + j!:
The inverse Fourier transform of the last two terms can be determined from Table 4.2
of O & W to give:
h(t) =
1
9e
t + 17
9 e
10t
u(t):
(c) To nd the di erential equation relating the input to the output we go back to Equa-
tion 7 and rewrite it as
H(j!) = Y (j!)X(j!) = 3 + 2j!(j!)2 + 11j! + 10:
After cross-multiplying we get
7
10Y (j!) + 11j!Y (j!) + (j!)2Y (j!) = 3X(j!) + 2j!X(j!):
Using the di erentiation property,we do an inverse Fourier transform to get the dif-
ferential equation
10y(t) + 11dy(t)dt + 11d
2y(t)
dt2 = 3x(t) + 2
dx
dt,
Problem 4 O&W 5.21 (c),(g)
Solution:
(c) We need to compute the Fourier transform of x[n] = 13jnju[ n 2],To do this we use
the analysis equation as shown below:
X(ej!) =
2X
n= 1
1
3
n
e j!n =
1X
n=2
1
3e
j!
n
= 11 1
3e
j! 1
1
3e
j! =
1
9e
j2!
1 13ej!
(g) We need to compute the Fourier transform of x[n] = sin( 2 n) + cos(n),To do this we
use Table 5.2 in O & W to nd the Fourier transform of each term and then sum the
two transforms for x[n].
F
n
sin
2n
o
= j
1X
l= 1
( (! 2 2 l) (! + 2 2 l))
Ffcos(n)g=
1X
l= 1
( (! 1 2 l) + (! + 1 2 l))
X(ej!) = j
1X
l= 1
( (! 2 2 l) (!+ 2 2 l))+
1X
l= 1
( (! 1 2 l)+ (!+1 2 l))
Problem 5 The following are Fourier transforms of discrete-time signals,Determine the
signal corresponding to each transform.
(a) X(ej!) = 4ej4! ej! + 6 + 8e j3! 16e j11!
(b)
X(ej!) =
(
1; 0 j!j< 4 ; 2 <j!j
0; 4 <j!j< 2
(c) X(ej!) = 1 + 3e
j3!
1 + 14e j!
8
Solutions:
(a) When the Fourier transform is a sum of exponentials,often the easiest way to nd x[n]
is to use the analysis equation
X(ej!) =
1X
n= 1
x[n]e j!n
to match each term to what the particular value of n is,For example,in this problem
we can see that the rst term,4ej4! can only result from the n = 4 term,Thus,we
can rewrite the equation as
X(ej!) = x[ 4]e j!( 4) + x[ 1]e j!( 1) + x[0]e j!0 + x[3]e j!(3) + x[11]e j!(11):
We match terms to nd
x[ 4] = 4; x[ 1] = 1; x[0] = 6; x[3] = 8; x[11] = 16:
x[n] = 0 for all other n.
(b) X(ej!) for this problem looks like:
X(ej!)
!
2
2
4
40
1
The Fourier transform for this signal as the sum of 3 ideal low pass lters with two
of them frequency-shifted to ! = 3 4 and ! = 3 4, Table 5.1 in O & W shows that a
frequency-shift corresponds to multiplication by an exponential in the time domain:
X(ej(! !o)) F !ej!onx[n]:
Thus,in the time domain,we can write this as the sum of three sinc functions each
with W = 4 and with two of them multiplied by the appropriate exponential,This
gives
x[n] = e j 3 4 n sin
4n
n +
sin 4 n
n + e
j 3 4 n sin
4n
n =
2 cos(3 4 n) + 1
sin(
4 n)
n,
9
(c) This Fourier transform looks similar to the Fourier transform for anu[n] in Table 5.2
of O & W,We will manipulate it to look more like that,First,we separate it into two
terms so that
X(ej!) = 11 + 1
4e
j! +
3e j3!
1 + 14e j! = X1(e
j!) + X2(ej!):
By linearity we can solve for the time signal for each transform and then x[n] =
x1[n]+x2[n],X1(ej!) matches to the transform pair mentioned above in Table 5.2 with
a = 14,Thus,
x1[n] =
14
n
u[n]:
X2(ej!) matches to the transform pair in Table 5.2 also except for the numerator term
of 3e j3!,This numerator term corresponds to a scaling of 3 and a time shift of 3.
Thus,
x2[n] = 3
14
n 3
u[n 3]:
Summing the two terms,we get
x[n] =
14
n
u[n] + 3
14
n 3
u[n 3]:
Problem 6 Let X(ej!) denote the Fourier transform of the signal x[n] depicted below.
x[n]
n 5 4 3
2 1 0
1 2 3
4 5 6 7 8
2
1
2 2
1
2
(a) Find X(1) = X(ej0).
(b) Find such that ej !X(ej!) is real.
(c) Evaluate R X(ej!)d!.
(d) Find X(ej ).
10
(e) Determine and sketch the signal whose Fourier transform is<efX(ej!)g.
(f) Evaluate each of the following integrals:
(f.1)
Z
jX(ej!)j2d!
(f.2)
Z
dX(ej!)
d!
2
d!
Solutions:
(a) We use the analysis equation:
X(1) = X(ej0) =
1X
n= 1
x[n]ej 0 n = 2:
(b) If x[n] is real and even,X(ej!) is real and even,Since this signal is real,we just need
to make it even,If we time shift x[n] 2 steps to the left,it will be real and even,A
left time shift of 2 corresponds to multiplying the frequency spectrum by ej!2,= 2
will make the frequency spectrum real (and even).
(c) We use the synthesis equation:
Z
X(ej!)d! = 2
1
2
Z
X(ej!)ej! 0d!
= 2 x[0] = 2,
(d) We use the analysis equation:
X(ej ) =
1X
n= 1
x[n]e j n =
1X
n= 1
x[n]( 1)n = 10
(e) <efX(ej!)g,which is the real (and even) part of X(ej!),is the Fourier transform of
the even part of the time signal,Thus,we need to sketch xe[n] = 0:5x[n] + 0:5x[ n].
This is shown below:
11
xe[n]
n
6 5 4 4 5 6 7
1
0:5
1
3
1
2
1
1
1
0
1
1
1
2
1
3
0:5
1
(f) To evaluate the following integrals we use Parseval’s relations.
(f.1) With Parseval’s relation,we know that
1X
n= 1
jx[n]j2 = 12
Z
jX(ej!)j2d!:
Since we know x[n],we can use it in Parseval’s relation to nd the quantity we
are interested in,Thus,
1X
n= 1
jx[n]j2 = 18
and
Z
jX(ej!)j2d! = 18 2 = 36,
(f.2) To use Parseval’s relation again,we need to de ne a new variable,Y (ej!) =
j dX(ej!)d!, Then according to Table 5.1 of O&W,
F 1
jdX(e
j!)
d!
=F 1fY (ej!)g= y[n] = nx[n]:
Thus,
Z
dX(ej!)
d!
2
d! =
Z
jdX(ej!)
d!
2
d! = 2
1
2
Z
jY (ej!)j2d!
= 2
1X
n= 1
jy[n]j2 = 2
1X
n= 1
jnx[n]j2 = 2 96 = 432,
12
Problem 7 Answer the questions asked in O&W 5.24 for the following two signals.
x1[n]
n 5 4
3 2 1
0 1 2 3 4
5 6 7
1
1
8
x2[n]
n 5 4 3 1 0 1 3 4 5 6 7 8
1
2
3
2
2
3
1
Solution:
(a)(a.1) We want to know if<efX1(ej!)g= 0,This is true only if x1[n] is purely real and
odd or purely imaginary and even,Looking at the signal,x1[n] we see that it is
real and odd,or xe[n] = 0:5x[n] + 0:5x[ n] = 0,Yes,<efX1(ej!)g= 0:
(a.2) Because this signal is real and odd,it is purely imaginary,=mfX1(ej!)g6= 0:
(a.3) The real part of a Fourier transform corresponds to the part of the time signal
which is real and even and the part of the time signal which is odd and imaginary.
By multiplying X1(ej!) by ej !,we can shift the signal in time,Thus,we can
shift by the correct number of units to make the signal even,Looking at the graph
of x1[n] reveals that a time shift of = 2 will cause the signal to be even and
thus,the resulting Fourier transform,ej 2!X1(ej!) will be real.
13
(a.4) We can use the synthesis equation as follows:
Z
X(ej!)d! = 2
1
2
Z
X(ej!)ej! 0d!
= 2 x[0] = 0:
Yes,R X(ej!)d! = 0.
(a.5) By de nition,the Fourier transform of any discrete sequence is periodic,so X(ej!)
is periodic.
(a.6) Since x1[n] is periodic,
X(ej!) =
1X
k= 1
2 ak
! 2 N k
where ak = 1N Pn=<N> x[n]e j 2 N kn,Plugging ! = 0 into the equation above shows
that
X(ej0) =
1X
k= 1
2 ak
2 N k
:
From this equation,we can see that all terms will be zero except possibly 2 a0 (0).
If a0 = 0 then X(ej0) = 0,Using the formula for ak above we see that,
a0 = 18
X
n=<N>
x[n] = 0:
We could also note that a0 is the DC average of the signal and from the graph we
can see that the DC average is zero,So,yes,X(ej0) = 0:
(b)(b.1) To determine if<efX(ej!)g= 0,we need to see if x2[n] is real or imaginary and
even or odd,From the graph,we can see that it is real and even,Thus,X(ej!)
is real and even,So<efX(ej!)g6= 0.
(b.2) This means that=mfX(ej!)g= 0.
(b.3) Since X(ej!) is already real,= 0.
(b.4) Using the same strategy employed for x1[n],we see that x2[0]6= 0 so R X(ej!)d!6=
0.
(b.5) By de nition,all X(ej!) are periodic so X2(ej!) is periodic.
(b.6) X2(ej0) is the DC component of the signal,We can see that the DC component
is equal to zero so this condition is met,That is,
X2(ej0) =
1X
n= 1
x[n] = 0:
14
Problem 8 Consider the same question as asked in O&W 5.27 (a) but with X(ej!) as
depicted below
X(ej!)
!
1
2 4 2 4
and with
p[n] = cos n cos( n=2):
Solution:
To sketch W(ej!),we use the multiplication property,
p[n]x[n] = w[n] F !W(ej!) = 12 P(ej!) X(ej!):
This allows us to use circular convolution to solve the problem,X(ej!) is as shown above
and
P(ej!) =
1X
l= 1
( (! 2 l)+ (! + 2 l))
1X
l= 1
( (! 2 2 l)+ (! + 2 2 l)):
The area under the impulses at ! = 2 l is equal to 2 not just because the impulses
from the term P1l= 1 (! 2 l) overlap with the impulses from the term P1l= 1 (! +
2 l).
These impulses and the impulses at ! = 2 2 l cause replicas of X(ej!) that are centered
at these !’s,The resulting W(ej!) is shown below:
15
W(ej!)
!
3 2 3 2
5 4 5 4
3 4 3 4
2 2
4 4
1
0:5
Problem 9 Answer the same questions as asked in O&W 5.30 (b) with x[n] as given in
that problem and for each of the following LTI unit sample responses:
(a) h[n] = sin n=16 n sin( n=12) n
(b) h[n] = sin( n=8) sin( n=2) 2n2
Solution:
We are given that x[n] = sin( n8 ) 2 cos( n4 ):
(a) To determine y[n],we will use the convolution property,That is,we will compute
Y (ej!) = X(ej!)H(ej!) and then we can use the synthesis equation to determine y[n]
from Y (ej!),To nd H(ej!) for the given h[n],we use Table 5.1 and Table 5.2 in
O&W and nd H(ej!) = H1(ej!) H2(ej!) where
H1(ej!) =
1 forj!j<
160 for
16 <j!j<
(8)
and
H2(ej!) =
1 forj!j<
120 for
12 <j!j<
(9)
Subtracting H2(ej!) from H1(ej!) yields
H(ej!) =
1 for 3
48 <j!j<
4
480 for 3
48 >j!j>
4
48
(10)
H(ej!) is a bandpass lter whose gain is 1.
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X(ej!) is found from Table 5.2:
X(ej!) = j
1X
l= 1
( (! 8 2 l) (!+ 8 2 l)) 2
1X
l= 1
( (! 4 2 l)+ (!+ 4 2 l)):
It is clear that the bandpass lter doesn’t allow this x[n] through it and hence Y (ej!) =
0 and y[n] = 0.
(b) Again we will use the convolution property to nd Y (ej!) = X(ej!)H(ej!) and then
nd y[n] from Y (ej!),X(ej!) is the same as the previous problem,However,
h[n] = sin(
n
8 ) sin
n
2
2n2 = h1[n]h2[n]:
We can use the multiplication property to nd H(ej!),Here,
H(ej!) = 12 H1(ej!) H2(ej!):
h1[n] and h2[n] are sinc functions,thus,
H1(ej!) =
1 forj!j<
80 for
8 <j!j<
(11)
H2(ej!) =
1 forj!j<
20 for
2 <j!j<
(12)
The convolution of these two signals is to be calculated.
H(ej!) is shown below:
H(ej!)
! 5
8
5
8
3
8
3
8
1=8
0
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Multiplying X(ej!) with H(ej!) gives the following spectrum for Y (ej!):
Y (ej!)
!
8j
4
8j
4
8
8
4 4
The above gure shows that y[n] is just a scaled version of x[n],Thus,
y[n] = 18 sin
n
8
14 cos
n
4
:
Problem 10 Consider a system consisting of the cascade of two LTI systems as depicted
below
x[n] System 1
z[n]
System 2 y[n]
System 1 is LTI and has a unit-sample response of
h[n] =
1
4
n
u[n]:
System 2 is LTI,and we know that if the input is
x[n] = [n] + 12 [n 1]
the output is
y[n] = 10 [n] [n 1]:
In the following parts,please show your work.
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(a) What is the frequency response X(ej!) of the overall system?
(b) Find the di erence equation for the overall system.
(c) Find the impulse response of the overall system.
Solution:
(a) For this cascade of two LTI systems,
H(ej!) = H1(ej!)H2(ej!):
Taking a Fourier transform of h1[n] = (14)nu[n] gives
H1(ej!) = 11 0:25e j!,
For H2(ej!),we take Fourier transforms of z[n] and y[n] as given above and then
H2(ej!) = Y (ej!)Z(ej!),This gives
H2(ej!) = 10 e
j!
1 + 0:5e j!,
Multiplying the two terms gives the overall frequency response as
H(ej!) = 11 0:25e j! 10 e
j!
1 + 0:5e j!,
(b) To nd the di erence equation for the whole system,we need to multiply Y (ej!) by
the denominator of the above equation and multiply Z(ej!) by the numerator of the
above equation,That is
Y (ej!)(1 + 0:25e j! 0:125e j2!) = X(ej!)(10 e j!):
Then,
Y (ej!) = 0:25e j!Y (ej!) + 0:125e j2!Y (ej!) + 10X(ej!) 1e j!X(ej!):
Because of linearity,we take the inverse Fourier transform of each term,noting that
multiplication of e j!no in the frequency domain corresponds to a shift of no in the
time domain,Thus,the di erence equation is
y[n] = 0:25y[n 1] + 0:125y[n 2] + 10x[n] x[n 1]:
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(c) To nd the impulse response,h[n],we rewrite the equation for H(ej!) found in (a) as
the sum of two rst order systems and do term by term matching to nd what the
numerator values should be for each term,That is,
H(ej!) = 10 e
j!
(1 0:25e j!)(1 + 0:5e j!) =
A
(1 0:25e j!) +
B
(1 + 0:5e j!)
= A(1 + 0:5e
j!)
(1 0:25e j!)(1 + 0:5e j!) +
B(1 0:25e j!)
(1 0:25e j!)(1 + 0:5e j!)
= A + B + (0:5A 0:25B)e
j!)
(1 0:25e j!)(1 + 0:5e j!),
Matching terms gives us A = 2 and B = 8,Then the inverse transform for each term
can be determined from Table 5.2:
H(ej!) = 2(1 0:25e j!) + 8(1 + 0:5e j!)
which gives,
h[n] = 2(0:25)nu[n] + 8( 0:5)nu[n]:
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