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2.161 Signal Processing: Continuous and Discrete
Fall 2008
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MASSACHUSETTS INSTITUTE OF TECHNOLOGY
DEPARTMENT OF MECHANICAL ENGINEERING
2.161 Signal Processing – Continuous and Discrete
Fall Term 2008
Problem Set 1 Solution: Convolution and Fourier Transforms
Problem 1:
Use the convolution definition y (t ) = f $ h =
"
! f (% )h(t # % )d%
#"
(a) f (t ) = " (t + 1.5) + " (t ) + " (t ! 0.75)
#1% | t |, % 1 $ t $ 1
h(t ) = "
!0, otherwise
y (t ) =
"
! (% ($ + 1.5) + % ($ ) + % ($ # 0.75))h(t # $ )d$
#"
"
"
"
#"
#"
#"
= ! % ($ + 1.5)h(t # $ )d$ + ! % ($ )h(t # $ )d$ + ! % ($ # 0.75)h(t # $ )d$ ,
and using the sifting property of the impulse function,
y (t ) = h(t + 1.5) + h(t ) + h(t ! 0.75)
1.4
1.2
1
0.8
0.6
0.4
0.2
0
-3
-2.5
-2
-1.5
(b) h(t) is the as same used in part (a)
#1, % 1.5 $ t $ 0.75
,
f (t ) = "
!0, otherwise
-1
-0.5
0
0.5
1
1.5
2
2.5
3
then y (t ) = f $ h =
#
!
f (% )h(t " % )d% =
"#
0.75
! h(t " % )d%
"1.5
Four basic cases can be observed while varying t (sliding the triangle waveform).,
Case
Range
Equation
Picture
1
Triangle
outside
t≤-2.5
0
0.5
1.75≤t
0
-4
t +1
Less
than
half
triangle
inside
" (1 ! (# ! t ))d# = # + #t !
-2.5≤t≤ -1.5
!1.5
0.75
" (1 + (# ! t ))d# = # ! #t +
0.75≤t≤1.75
t !1
2 t +1
#
2
t
More
than
half
triangle
inside
#2
(
1
+
(
#
!
t
))
d
#
+
0
.
5
=
#
!
#
t
+
"
2
!1.5
-1.5≤t≤ -0.5
#
"t (1 ! (# ! t ))d# + 0.5 = # + #t + 2
-0.25≤t≤0.75
t !1
=
(t ! 1.75)
2
+ 0.5 = 1 !
!1.5
t
-2
-1
0
1
-3
-2
-1
0
1
1
0.5
t
2 0.75
0.75
(t + 2.5) 2
2
!1.5
2 0.75
#
2
=
-3
2
0
-4
(t + 0.5) 2
2
1
0.5
(t + 0.25) 2
+ 0.5 = 1 !
2
0
-3
-2
-1
0
1
1
Whole
triangle
inside
-0.5≤t≤
0.25
1
0.5
0
-2
The result of the convolution, y(t), is plotted in the following figure
1
0.75
0.5
0.25
0
-3
-2.5
-2
-1.5
-1
-0.5
0
0.5
1
1.5
2
-1
0
1
(c)
T
$
!1, | t |%
f (t ) = #
2
!"0, otherwise
then,
$t +T / 2
! ( d* = t + T , & T % t % 0
! &T 2
T
! T2
)
2
!
y (t ) = f ' f = ( f (* ) f (t & * )d* = ( f (t & * )d* = # ( d* = t & T , 0 % t % T
&T
&)
! t &T 2
2
!
0, otherwise
!
!
"
T
0
F ( j! ) =
&
' f (t )e
"&
" j !t
-T
T /2
dt =
'e
"T / 2
0
T
T /2
" j !t
e " j !t %
e " j!T / 2 " e j!T sin(!T / 2)
dt =
=
=
#
" j! $ "T / 2
" j!
! /2
when T=1, h(t ) = f (t ) ! f (t ) and from the convolution theorem
2
& sin(' / 2) #
& sin( x) #
H ( j' ) = F ( j' ) F ( j' ) = $
! =$
!
% ' /2 "
% x "
2
Problem 2
2
f1 ( x) = e ! ax , f 2 ( x) = e ! bx
"
!
y ( x) = f1 $ f 2 =
2
#"
"
= !e
"
f1 (% ) f 2 ( x # % )d% =
!e
#a% 2
#"
)
#( a +b )'
'
(
* 2 # 2abx+b* + abx+b &$$
2
%
#"
"
d* = ! e
!
2
*
bx & abx 2
)
#( a +b )' #
$ #
a +b % a +b
(
d* = e
#
abx 2 "
a +b
!e
#"
ab
a +b
"
x2
!&
$e
d&
=
a+b
2
!"
*
bx &
)
#'' a + b #
$$
a +b %
(
#"
using the variable substitution: " = a + b# !
# y (t ) = e
2
e #b ( x #% ) d%
bx
a+b
, d# =
d"
a+b
ab
% ! a +b x 2
e
, which is a Gaussian function.
a+b
Problem 3
f (t ) = " (t + 0.5) + " (t ) + " (t ! 0.5)
F ( jw) =
$
%
f (t )e
#
j! t
#$
F ( j! ) = e
$
dt = % " (t + 0.5)e
#
j! t
#$
!
j
2
+1+ e
!
#j
2
$
dt + % " (t )e
#
#$
j! t
$
dt + % " (t # 0.5)e
#
j! t
#$
!
= 1 + 2 cos( )
2
3
j
](
)F
!
[
e
R
2
1
0
-1
-10
-8
-6
-4
-2
0
2
4
6
8
10
2
4
6
8
10
!
1
0.5
j
](
)F
0
!
[
m
I -0.5
-1
-10
-8
-6
-4
-2
0
dt
2
d*
Problem 4
These solutions are all based on the elementary properties of the Fourier transform (see the class
handout).
*
*
1
1
1 (1
% 1
j"t
(a) x (0) =
X
(
j
"
)
e
d
"
|
=
X ( j" )d" =
& X 0 .2W + X 0 .W # = X 0W
t =0
)
)
2! +*
2! +*
2! ' 2
$ !
(b) Using the symmetry properties, we note that X ( j! ) is real, therefore x ( !t ) = x (t ), that is
they are complex conjugates.
(c) This one is a little tricky! We use the property that
!
# x(t )dt = X ( j$ ) |
$ =0
"!
BUT note that there is a singularity at ! = 0 . The question is: what is the value of X ( j 0) ?
The problem statement specifies that X ( j! ) = X 0 for 0 " ! <W, so you can argue that
"
! x(t )dt = X
0
.
#"
On the other hand if you approximate the step discontinuity with a smooth function (say erf())
around ! = 0 , you can argue that the value of X ( j 0) = 0.75 X 0 , or
"
! x(t )dt = 0.75 X
0
.
#"
So the answer is dependent on your assumption about the discontinuity!
(d) From Parseval’s theorem
$
$
1
1 1 2
3 2
2
|
x
(
t
)
|
dt
=
| X ( j" ) |2 d" =
( X 0 .2W + X 02 .W ) =
X 0W .
#%$
#
2! %$
2! 4
4!
Problem 5
# 1, | $ |< $ c
H ( j$ ) = "
!0, otherwise
If and impulse is passed through the filter, we obtain the impulse response h(t ) = F "1 {H ( j! }
1
h(t ) =
2"
$
% H ( j! ) e
#$
j !t
1
d! =
2"
!c
%e
#! c
j !t
d! =
! sin ! c t ! c
1
( e j! c t # e # j! c t ) = c
=
sinc(! c t )
2"jt
" !ct
"
w/pi
0
0
t
The filter is acausal.
Problem 6
h(t ) = 5e
!3t
"
Let’s compute the Fourier transform of h(t) H ( j$ ) = ! 5e #3t e # j$t dt =
0
5
j$ + 3
Note: lower limit in integral is 0 because a real filter is a causal system.
a) The transfer function can be found by taking the Laplace transform, which can be viewed as
a Fourier transform where jω is replaced by s=σ+jω .
H (s) =
5
s+3
b) The frequency response is given by H(jω) computed previously.
c) We find the cut-off frequency by solving:
| H ( j# c ) | 1
H ( j 0)
5
5
= !| H ( j# c ) =
!
= ! # c2 = 36 " 9 = 27 ! # c = 27 rad / s
| H ( j 0) |
2
2
9 + # c2 6