MATH 267
ASSIGNMENT # 6
Due: Feb 29, 2012, in the class (after the midterm break)
You have FIVE problems to hand-in. Hand in written solutions for grading at the BEGINNING of the
lecture on the due date. Illegible, disorganized or partial solutions will receive no credit.
*Staple your HW. You will get F IV E marks OFF if you do not staple your HW! Note that
the instructor will NOT provide stapler.
1. (How to use properties of Fourier transform) In this problem, it is useful to recall that the Fourier
transform of the function
(
1 if − 21 < t < 12 ,
rect(t) =
0 otherwise.
d
is the function sinc( ω2 ), i.e. rect(ω)
= sinc( ω2 ) where the function sinc is defined as
(
sinc(ω) =
1
sin(ω)
ω
if ω = 0,
otherwise.
(See the online notes ”Fourier Transform” page 2, Example 2.)
(a) Find the Fourier transform of
(
e−t
f (t) =
0
if 0 < t < 1,
otherwise.
(Hint: just directly compute the integral for the Fourier transform.)
Solution
Z ∞
b
f (ω) =
f (t)e−iωt dt
−∞
1
Z
e−t e−iωt dt
=
0
1
Z
e−(1+iω)t dt
=
0
h e−(1+iω)t i1
−(1 + iω) 0
i
1 h
=
− e−(1+iω) + 1
1 + iω
=
(b) Without calculating integrals directly, but only using the properties of the Fourier transform and
the result of (a),
(
e−t+4 if 4 < t < 5,
i. find gb1 (ω) where g1 (t) =
0
otherwise.
Solution : (
Note that g1 (t) = f (t − 4) since
(
e−(t−4) if 0 < t − 4 < 1,
e−t+4 if 4 < t < 5,
f (t − 4) =
=
Thus, by the time-shift
0
otherwise
0
otherwise.
property of Fourier series
gb1 (ω) = e−i4ω fb(ω)
i
1 h
= e−i4ω
− e−(1+iω) + 1
1 + iω
1
(
e−t if 4 < t < 5,
0
otherwise.
Solution : Note that g2 (t) = e−4 g1 (t). Thus, (by linearity of Fourier transform), gb2 (ω) =
e−4 gb1 (ω), therefore,
i
1 h
− e−(1+iω) + 1
gb2 (ω) = e−4−i4ω
1 + iω
ii. find gb2 (ω) where g2 (t) =
(Hint: consider the time-shifting property of the Fourier transform.)
(c) Use above (including the results about rect(t)) and linearity and time-shifting property for
Fourier transform, to find Fourier transform of the following functions, i.e. without calculating
integrals directly.

if 0 < t < 1,

1

−2
if 3 < t < 4,
i. h1 (t) =
e−t+4 if 4 < t < 5,



0
otherwise.
Solution Note that h1 (t) = rect(t − 1/2) − 2rect(t − 7/2) + g1 (t).
Let us use the notation F[f (t)](ω) (or F[f ](ω) ) for the Fourier transform of f (t). So, by notation
fb(ω) = F[f (t)])(ω). Thus,
F[h1 (t)](ω) = F[rect(t − 1/2)](ω) − 2F[rect(t − 7/2)](ω) + F[g1 (t)](ω)
by the linearity of the Fourier transform.
Now, use the time-shift property of Fourier transform, so see
7
1
F[h1 (t)](ω) = e−i 2 ω F[rect(t)](ω) − 2e−i 2 ω F[rect(t)](ω) + F[g1 (t)](ω)
i
7
1
1 h
= e−i 2 ω sinc(ω/2) − 2e−i 2 ω sinc(ω/2) + e−i4ω
− e−(1+iω) + 1
1 + iω
h
i
1
7
1
− e−(1+iω) + 1
= e−i 2 ω − 2e−i 2 ω sinc(ω/2) + e−i4ω
1 + iω

−1 if − 1 < t < 0,



3
if 1 < t < 2,
ii. h2 (t) =
−t

e
if 4 < t < 5,



0
otherwise.
Solution Note that h2 (t) = −rect(t + 1/2) + 3rect(t − 3/2) + g2 (t).
Let us use the notation F[f (t)](ω) (or F[f ](ω) ) for the Fourier transform of f (t). So, by notation
fb(ω) = F[f (t)])(ω). Thus,
F[h2 (t)](ω) = −F[rect(t + 1/2)](ω) + 3F[rect(t − 3/2)](ω) + F[g2 (t)](ω)
by the linearity of the Fourier transform.
Now, use the time-shift property of Fourier transform, so see
1
3
F[h1 (t)](ω) = −e+i 2 ω F[rect(t)](ω) + 3e−i 2 ω F[rect(t)](ω) + F[g2 (t)](ω)
i
1
3
1 h
= −e+i 2 ω sinc(ω/2) + 3e−i 2 ω sinc(ω/2) + e−4−i4ω
− e−(1+iω) + 1
1 + iω
i
1 h
+i 12 ω
−i 32 ω
−4−i4ω
= −e
+ 3e
sinc(ω/2) + e
− e−(1+iω) + 1
1 + iω
2. (Time Reversal)
2
(a) Show that if g(t) = f (−t) then gb(ω) = fb(−ω). Also, show that if gb(ω) = fb(−ω), then g(t) = f (−t).
(Hint: Just see what happens to the Fourier transform. You may need to apply change of variables
(substitution) for the integral.)
Solution First, for the first part: Assume g(t) = f (−t). Then, the Fourier transform goes as
Z ∞
g(t)e−iωt dt
gb(ω) =
−∞
Z ∞
=
f (−t)e−iωt dt
−∞
−∞
Z
=
Z∞∞
=
0
f (t0 )eiωt (−dt0 )
( change of variables t0 = −t)
0
f (t0 )e−i(−ω)t dt0
−∞
= fb(−ω)
(by the Fourier transform for f , but with −ω)
For the second part, assume gb(ω) = fb(−ω). Then, the Fourier inversion goes as
Z ∞
1
g(t) =
gb(ω)eiωt dω
2π −∞
Z ∞
1
fb(−ω)eiωt dω
=
2π −∞
Z −∞
0
1
fb(ω 0 )eiω (−t) (−dω 0 ) ( change of variables ω 0 = −ω)
=
2π ∞
Z ∞
0
1
=
fb(ω 0 )eiω (−t) dω 0
2π −∞
= f (−t)
(by the Fourier inversion for f , but with −t)
(b) Show that if f (t) is an even function then fb(ω) is also an even function. Show that if fb(ω) is an even
function, then so is f (t). (Hint: use (a) )
Solution For the first part, assume f (t) is an even function. Then, f (t) = f (−t). Then, by applying
(a) for g(t) = f (t), (then g(t) = f (t) = f (−t)), we see that fb(ω) = gb(ω) = fb(−ω). (Here, (a) is
applied in the second equality.) Thus, fb(ω) = fb(−ω), thus, fb is an even function.
For the second part, assume fb(ω) is an even function. Then, fb(ω) = fb(−ω). Then, by applying (a)
(the second part) for gb(ω) = fb(ω), (then gb(ω) = fb(ω) = fb(−ω)), we see that f (t) = g(t) = f (−t).
(Here, (a) is applied in the second equality.) Thus, f (t) = f (−t), thus, f is an even function.
(c) Show that if f (t) is an odd function then fb(ω) is also an odd function. Show that if fb(ω) is an odd
function, then so is f (t). (Hint: use (a).)
Solution This is similar to (b):
For the first part, assume f (t) is an odd function. Then, −f (t) = f (−t). Then, by applying (a) for
g(t) = −f (t), (then g(t) = −f (t) = f (−t)), we see that −fb(ω) = gb(ω) = fb(−ω). (Here, (a) is applied
in the second equality.) Thus, −fb(ω) = fb(−ω), thus, fb is an odd function.
For the second part, assume fb(ω) is an odd function. Then, −fb(ω) = fb(−ω). Then, by applying
(a) (the second part) for gb(ω) = −fb(ω), (then gb(ω) = −fb(ω) = fb(−ω)), we see that −f (t) = g(t) =
f (−t). (Here, (a) is applied in the second equality.) Thus, −f (t) = f (−t), thus, f is an odd function.
3. This problem is better to be tried after Problems 1 and 2. Sketch the graph of the following functions.
Then find their Fourier transforms using appropriate properties of Fourier transform together with your
answer to Problem 1(a), without directly calculating the integral for the Fourier transform.
(
et if − 1 < t < 0,
(a) f (t) =
0 otherwise.
3
Solution Rename the function in 1 (a) as f1 (t). Then, the function f (t) (in the current problem)
f (t) = f1 (−t).
Then,
fb(ω) = fb1 (−ω) (using 2 (a))
Now, using 1 (a)
fb1 (−ω) =
h
i
1
− e−(1+i(−ω)) + 1
1 + i(−ω)
Therefore,
h
i
1
− e−(1+i(−ω)) + 1
1 + i(−ω)
i
1 h
− e−(1−iω) + 1
=
1 − iω
fb(ω) =
(b)
 t
−1 < t < 0
 e,
e−t ,
0<t<1
g(t) =

0,
otherwise

 −et , −1 < t < 0
e−t ,
0<t<1
h(t) =

0,
otherwise
 −t−1
, −1 < t < 0
 e
e−t ,
0<t<1
s(t) =

0,
otherwise
Solution Let (for notation)
(
f1 (t) =
e−t
0
if 0 < t < 1,
otherwise.
and let
(
f2 (t) =
et
0
if − 1 < t < 0,
otherwise.
Note by (a) and 1 (a), we see that
i
1 h
− e−(1+iω) + 1
1 + iω
i
1 h
fb2 (ω) =
− e−(1−iω) + 1
1 − iω
fb1 (ω) =
Then, using this notation, we see that g(t) = f2 (t) + f1 (t), h(t) = −f2 (t) + f1 (t),
and s(t) = f1 (t + 1) + f1 (t).
Thus, using the linearity of Fourier transform and time-shift property,
gb(ω) = fb2 (ω) + fb1 (ω)
b
h(ω) = −fb2 (ω) + fb1 (ω)
sb(ω) = eiω fb1 (ω) + fb1 (ω)
= (eiω + 1)fb1 (ω)
4
(used time-shift by −1)
Therefore, we get
i
i
1 h
1 h
− e−(1−iω) + 1 +
− e−(1+iω) + 1
1 − iω
1 + iω
h
i
i
1
1 h
−(1−iω)
b
−e
− e−(1+iω) + 1
h(ω) = −
+1 +
1 − iω
1 + iω
h
i
1
− e−(1+iω) + 1
sb(ω) = (eiω + 1)
1 + iω
gb(ω) =
4. (Frequency Shifting)
(a) Show that if g(t) = eiω0 t f (t), then gb(ω) = fb(ω − ω0 ). Also, show that if gb(ω) = fb(ω − ω0 ), then
g(t) = eiω0 t f (t).
Solution First, for the first part: Assume g(t) = eiω0 t f (t). Then, the Fourier transform goes as
Z ∞
gb(ω) =
g(t)e−iωt dt
−∞
Z ∞
eiω0 t f (t)e−iωt dt
=
−∞
Z ∞
=
f (t)e−i(ω−ω0 )t dt
−∞
= fb(ω − ω0 )
(by the Fourier transform for f , but with ω − ω0 )
For the second part, assume gb(ω) = fb(ω − ω0 ). Then, the Fourier inversion goes as
Z ∞
1
g(t) =
gb(ω)eiωt dω
2π −∞
Z ∞
1
fb(ω − ω0 )eiωt dω
=
2π −∞
Z −∞
0
1
=
fb(ω 0 )eiω t+iω0 t dω 0 ( change of variables ω 0 = ω − ω0 )
2π ∞
Z ∞
0
1
= eiω0 t
fb(ω 0 )eiω t dω 0
2π −∞
= eiω0 t f (t)
(by the Fourier inversion for f )
(b) For a function h1 (t), suppose b
h1 (ω) = sinc(ω/2 − 2). Find h1 (t). (Hint: use (a).)
d
Solution Notice that rect(ω)
= sinc(ω/2). Thus,
b
d − 4)
h1 (ω) = sinc((ω − 4)2) = rect(ω
Now, use (a) (the second part) to see that
h1 (t) = ei4t rect(t).
(c) For a function h2 (t), suppose b
h2 (ω) = sinc(ω/2 − 2) + 2 sinc(ω/2 + 2). Find h2 (t). (Hint: use (a) and
the linearity of Fourier transform/ inversion.)
d
Solution Notice that rect(ω)
= sinc(ω/2). Thus,
b
d − 4) + 2rect(ω
d + 4)
h2 (ω) = sinc((ω − 4)2) + 2sinc(ω/2 + 2) = rect(ω
Now, use (a) (the second part) to see that
h2 (t) = ei4t rect(t) + 2e−i4t rect(t)
5
5. (A few warm-up exercises for Fourier inversion) Find the inverse Fourier transform of the following
functions. Namely, for the given Fourier transform function, find the original function.
1
1
− 3+iω
(Hint: Try to use the result of Example 1 in the online notes ”Fourier Transform”
(a) gb(ω) = 2+iω
and the linearity of Fourier transform/inversion.)
Solution Define, for a > 0,
(
e−at if t ≥ 0,
fa (t) =
0
otherwise.
Note using the result of Example 1 in the online notes ”Fourier Transform”, the Fourier transform
1
. So,
(using the notation in the solution of Problem 1 (c) ), F[fa (t)](ω) = a+iω
gb(ω) = F[g(t)](ω) = F[f2 (t)](ω) − F[f3 (t)](ω).
Thus, by the linearity of Fourier transform/inversion, we see that g(t) = f2 (t) − f3 (t). Therefore,
(
e−2t − e−3t if t ≥ 0,
g(t) =
0
otherwise.
(b) fb(ω) = e−i2ω sinc(3ω) (Hint: we are going to discuss the ”scaling property” after the break, but,
here, just use the property that for α > 0 constant, if h(t) = f ( αt ) then b
h(ω) = αfb(αω) and vice
versa: see the online notes ”Fourier Transform” page 4, ”Scaling”. Also, you would need to apply
the ”time shifting” property of Fourier transform. )
d
Solution sinc(3ω) = sinc((6ω)/2) = rect(6ω).
Note that by the scaling property as in the hint,
d
6 rect(6ω)
=b
h(ω) where, h(t) = rect(t/6). Therefore, the Fourier transform (using the notation in
the solution of Problem 1 (c) ) F[ 61 rect(t/6)](ω) = sinc(3ω). Now, by the time-shifting property of
Fourier transform, e−i2ω sinc(3ω) = F[ 16 rect((t−2)/6)](ω). Thus, the original function
f (t) =
1
rect((t−2)/6).
6
*Staple your HW. You will get F IV E marks OFF if you do not staple your HW! Note that
the instructor will NOT provide stapler.
6