MATH 267
Due: March 7, 2012, in the class
ASSIGNMENT # 7
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.
Note: throughout the assignment, the function u(t) denotes the unit step function:
1, t ≥ 0
u(t) =
0, t < 0
Also, in the following problems, feel free to use properties of Fourier Transform / Fourier Inversion and standard
examples, e.g. Fourier transforms of e−at u(t) (for a > 0) and rect(t).
1. (Scaling, time-shift, duality, differentiation)
(a) Find Fourier transform of


t + 1, −1 ≤ t ≤ −1/2;
f (t) = −t,
−1/2 ≤ t ≤ 0;


0,
otherwise.
(Hint: This is similar to one of class examples about differentiation rule for Fourier transform.)
Solution : Note that
d
f (t) = rect(2(t + 3/4)) − rect(2(t + 1/4)).
dt
[For this, do first the scaling of the class example (scale by 1/2) and do the appropriate time-shift
(by -1/4). ]
Therefore, the Fourier transform
F[
d
f (t)](ω) = F[rect(2(t + 3/4)) − rect(2(t + 1/4))](ω)
dt
= F[rect(2(t + 3/4))](ω) − F[rect(2(t + 1/4))](ω)
iω3/4
=e
F[rect(2t)](ω) − e
iω/4
(by linearity of F.T.)
F[rect(2t)](ω)
(by time-shift property: practically it can be better to do this step first before handling scaling.)
= [eiω3/4 − eiω/4 ]F[rect(2t)](ω)
1
= [eiω3/4 − eiω/4 ] F[rect(t)](ω/2)
(by scaling property)
2
1
= [eiω3/4 − eiω/4 ]sinc(ω/4) (see ω/4 in sinc instead of ω/2!)
2
eiω/2 iω/4
=
[e
− e−iω/4 ]sinc(ω/4)
2
= ieiω/2 sin(ω/4)sinc(ω/4)
d
But, on the other hand F[ dt
f (t)](ω) = iωF[f (t)](ω) by the differentiation rule.
Therefore, for ω 6= 0, we see
1 iω/2
ie
sin(ω/4)sinc(ω/4)
iω
eiω/2
=
sinc(ω/4)sinc(ω/4)
4
eiω/2
=
[sinc(ω/4)]2
4
F[f (t)](ω) =
1
For ω = 0, we can directly compute the integral
Z
Z ∞
f (t)e−i0t dt =
F[f (t)](0) =
ei0/2
2
4 [sinc(0/4)]
f (t)dt = 1/4
−∞
−∞
(Note that when ω = 0,
∞
= 1/4.) Therefore, we have
F[f (t)](ω) =
eiω/2
[sinc(ω/4)]2 .
4
(b) Find Fourier transform of


t + 2, −2 ≤ t ≤ −1;
f (t) = −t,
−1 ≤ t ≤ 0;


0,
otherwise.
(Hint: Use (a) and scaling property of Fourier transform.)
Solution
Let f1 (t) denote the function f (t) in part (a). Now for f (t) in this part (b), we see that
f (t) = 2f1 (t/2).
Therefore,
F[f (t)](ω) = 2×2F[f (1 (t)](2ω)
So, we have
iω
e
fb(ω) = 4
[sinc(ω/2)]2
4
= eiω [sinc(ω/2)]2
Remark: In fact, it can be easier to do this part (b) first and to use this to do part (a). The
function in part (b) is nothing but a time-shift of the class example and the function in part (a) is
the scaled function of the function in part (b) by scale factor 1/2.
(c) Let f (t) = e−|t| .
i. Find fb(ω). (Hint: this is a class example. You can use the result for e−t u(t) and apply properties
of Fourier transform: here time-reversal property is relevant.)
Solution Let
f0 (t) = e−t u(t)
Note that F[f0 (t)](ω) =
can write
1
iω+1
( this is one of the standard example given in the class). Now, we
f (t) = f0 (t) + f0 (−t).
Therefore,
F[f (t)](ω) = F[f0 (t)](ω) + F(f0 (−t)](ω)
= F[f0 (t)](ω) + F(f0 (t)](−ω)
(used time-reversal property F[g(−t)](ω) = F[g(t)](−ω).)
2
Therefore,
1
1
+
iω + 1 −iω + 1
2
= 2
ω +1
F[f (t)](ω) =
ii. Use part (i) and the duality property to find the Fourier transform gb(ω) of the function
g(t) =
1 1
π 1 + t2
Solution See http://www.iam.ubc.ca/˜sospedra/a7MATH267-sol.pdf, #1 (b).
2. (Differentiation in frequency)
(a) Prove the following:
d b
if g(t) = tf (t) then gb(ω) = i dω
f (ω)
(Hint: differentiate the definition (I mean, the integral) of fb(ω) with respect to ω: i.e.
Z ∞
d b
d
f (ω) =
f (t) e−itω dt. )
dω
dω
−∞
Solution See http://www.iam.ubc.ca/˜sospedra/a7MATH267-sol.pdf, #2 (a).
(b) Use (a) to show
if gb(ω) =
d b
dω f (ω),
then g(t) = −itf (t)
(Here, use the fact that if fb1 (ω) = fb2 (ω) then f1 (t) = f2 (t). In other words, the Fourier transform
fb(ω) determine the original function f (t).)
Solution See http://www.iam.ubc.ca/˜sospedra/a7MATH267-sol.pdf, #2 (a).
(c) Using the frequency differentiation property in part (a), compute the Fourier transform of:
(i) f (t) = t rect(t)
Solution See http://www.iam.ubc.ca/˜sospedra/a7MATH267-sol.pdf, #2 (b).
(ii) g(t) = t2 e−3t u(t) (Hint: you can apply the frequency differentiation property twice.)
Solution See http://www.iam.ubc.ca/˜sospedra/a7MATH267-sol.pdf, #2 (b).
(d) [Fourier inversion] For a real nonzero constant a, find the function g(t) if
gb(ω) =
1
(iω + a)2
(Hint: You can use (b). Can you express gb(ω) as a ω-derivative of certain function? )
Solution Observe that
i
1
d h 1 i
=
dω iω + a
(iω + a)2
Also, note that
F[e−at u(t)](ω) =
1
iω + a
Therefore, by part (a) (or (b))
g(t) = te−at u(t).
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3. (RLC circuit) Consider the ODE for RLC circuit:
LCy 00 (t) + RCy 0 (t) + y(t) = x(t)
(a) Let R = 4, L = 3, C = 1 and x
b(ω) = 1. Find y(t) using Fourier transform method.
Solution The left-hand side is 3y 00 (t) + 4y 0 (t) + y(t). Thus, the Fourier transform gives
−3ω 2 yb(ω) + 4iωb
y (ω) + yb(ω) = x
b(ω).
Therefore,
1
x
b(ω)
−3ω 2 + 4iω + 1
1
(since we assumed w(ω)
b
= 1. )
=
2
−3ω + 4iω + 1
yb(ω) =
Note that −3ω 2 + 4iω + 1 = (3iω + 1)(iω + 1) Now, by partial fraction,
1
1
=
−3ω 2 + 4iω + 1
(3iω + 1)(iω + 1)
A
B
=
+
.
3iω + 1 iω + 1
Here, A and B are determined by
A(iω + 1) + B(3iω + 1) = 1
(A + 3B)iω + A + B = 1
Comparing the real and imaginary parts, we get,
A + 3B = 0
A+B =1
Therefore, B = − 21 , A = 32 . Thus,
yb(ω) =
1
3
1
=
−
−3ω 2 + 4iω + 1
2(3iω + 1) 2(iω + 1)
Now, for the Fourier inversion y(t) = F −1 [b
y (ω)](t),
i
1
3
−
(t)
2(3iω + 1) 2(iω + 1)
h 1 i
h 1 i
3
1
= F −1
(t)− F −1
(t)
2
3iω + 1
2
iω + 1
h
i
h 1 i
1
1
1
= F −1
(t)− F −1
(t)
2
iω + 1/3
2
iω + 1
1
1
= e−t/3 u(t) − e−t u(t).
2
2
y(t) = F −1
h
(b) Let R = 2, L = 1, C = 1 and x
b(ω) = 1. Find y(t) using Fourier transform method. (Hint: You may
want to use Problem 2 (d). )
Solution The left-hand side is y 00 (t) + 2y 0 (t) + y(t). Thus, the Fourier transform gives
−ω 2 yb(ω) + 2iωb
y (ω) + yb(ω) = x
b(ω).
Therefore,
1
x
b(ω)
−ω 2 + 2iω + 1
1
=
(since we assumed w(ω)
b
= 1. )
−ω 2 + 2iω + 1
1
=
(iω + 1)2
yb(ω) =
4
Now use the result of Problem 2 (d), to get
y(t) = te−t u(t).
(c) Let R = 4, L = 3, C = 1 and x(t) = u(t)e−2t . (Note u(t) is defined in the beginning of the HW).
Find y(t) using Fourier transform method.
Solution The left-hand side is 3y 00 (t) + 4y 0 (t) + y(t). Thus, the Fourier transform gives
−3ω 2 yb(ω) + 4iωb
y (ω) + yb(ω) = x
b(ω).
Therefore,
yb(ω) =
−3ω 2
1
x
b(ω)
+ 4iω + 1
Now, from our class example,
1
iω + 2
x
b(ω) = F[e−2t u(t)](ω) =
Therefore,
1
1
−3ω 2 + 4iω + 1 iω + 2
1
=
(3iω + 1)(iω + 1)(iω + 2)
yb(ω) =
(Note that −3ω 2 + 4iω + 1 = (3iω + 1)(iω + 2) )
Now, by partial fraction,
yb(ω) =
A
B
C
+
+
.
3iω + 1 iω + 1 iω + 2
Here, A, B and C are determined by
A(iω + 1)(iω + 2) + B(3iω + 1)(iω + 2) + C(3iω + 1)(iω + 1) = 1
The left hand side is simplified by
A(iω + 1)(iω + 2) + B(3iω + 1)(iω + 2) + C(3iω + 1)(iω + 1)
= A(−ω 2 + 3iω + 2) + B(−3ω 2 + 7iω + 2) + C(−3ω 2 + 4iω + 1)
= −(A + 3B + 3C)ω 2 + (3A + 7B + 4C)iω + 2A + 2B + C
Comparing the last line with 1 (since they should be the same as functions of ω), we have,
A + 3B + 3C = 0
3A + 7B + 4C = 0
2A + 2B + C = 1
Therefore,
B = −1/2,
A = 9/10,
C = 1/5
Therefore,
yb(ω) =
1
1 1
1 1
9
−
+
.
10 3iω + 1 2 iω + 1 5 iω + 2
Therefore,
5
h
i
1 1 1
1
9 −1 1 (t) − F −1
(t) + F −1
(t)
y(t) = F −1 yb(ω) (t) =
F
10
3iω + 1
2
iω + 1
5
iω + 2
1 1 9 −1 1
1
1
=
F
(t) − F −1
(t) + F −1
(t)
30
iω + 1/3
2
iω + 1
5
iω + 2
3 −t/3
1
1
=
e
u(t) − e−t u(t) + e−2t u(t)
(by using the standard example e−at u(t))
10
2
5
i
h3
1
1
= u(t)
e−t/3 − e−t + e−2t
10
2
5
4. (Fourier Inversion) In the following use properties of Fourier transform/inversion.
(a) fb(ω) = sinc(ω + 1).
d
Solution Recall rect(ω)
= sinc(ω/2). Also, recall the frequency-shift property and the scaling
property.
First, by the frequency-shift, for the inverse Fourier transform F −1 ,
Note that sincω =
1
2
F −1 [sinc(ω + 1)](t) = e−it F −1 [sinc(ω)](t).
2sinc(2ω/2) , thus, by linearity and scaling property,
1
2sinc(2ω/2) ](t)
2
1 −1
= F [ 2sinc(2ω/2) ](t)
2
1
= F −1 [sinc(ω/2)](t/2)
2
1
= rect(t/2)
2
F −1 [sinc(ω)](t) = F −1 [
(by linearity)
(by scailing)
Therefore,
e−it
1
rect(t/2)
f (t) = F −1 [sinc(ω + 1)](t) = e−it rect(t/2) =
2
2
i.e.
(
f (t) = F
−1
[sinc(ω + 1)](t) =
e−it
2 ,
0,
−1 < t < 1;
otherwise.
d
b
(b) Suppose that a function δ(t) satisfies δ(ω)
= 1 and suppose dt
u(t) = δ(t), where u(t) is the unit step
function defined in the beginning of the assignment. Express the inverse Fourier transform of
b
h(ω) = sin(ω)
using δ(t) as well as its time-shift and scaling. (Hint: Find the relation to sinc function. Then, try
to use properties of Fourier transform. You may have to express a rectangular function using the
function u(t). For example, rect(t) = u(t + 1/2) − u(t − 1/2). We will learn on Monday, what such
function δ(t) is.)
Solution
There are two methods.
First method (harder)
Realize that b
h(ω) = ω sinc(ω). By the differentiation rule for the Fourier transform, if we let
F[k(t)](ω) = sinc(ω), then,
iω sinc(ω) = F[
6
d
k(t)](ω)
dt
Remark: In other words,
iω sinc(ω) = F
ii
hdh
F −1 [sinc(ω)](t) (t).
dt
The above shows that
h(t) =
1 d
k(t).
i dt
for F[k(t)](ω) = sinc(ω).
d
= sinc(ω/2). Therefore, by
Now, for k(t), notice that sin ω = 12 2 sinc(2ω/2). Also, note that rect(ω)
scaling property,
k(t) =
1
rect(t/2).
2
Here, it is convenient to notice that rect(t/2) = u(t + 1) − u(t − 1). (See the remark below.) Thus,
i
1h
k(t) = u(t + 1) − u(t − 1) .
2
Therefore,
i
1 d
1hd
d
k(t) =
u(t + 1) − u(t − 1)
i dt
2i dt
dt
i
1h
δ(t + 1) − δ(t − 1)
=
2i
h(t) =
Remark: This problem canh be harder if iyou doh not realize rect(t/2)
= u(t + 1) − u(t − 1). In that
i
d
d
case, you have to handle dt u(t/2 + 1/2) − dt u(t/2 − 1/2) . We can still do this. For example,
+ 1/2) = 12 δ(t/2 + 1/2) by the chain rule. Here, δ(t/2 + 1/2) = δ( 21 (t + 1)) is the time-shift
of δ(t/2) by −1. We now have to note that δ(t/2) = 2δ(t). This can be verified by using the scailing
property of the Fourier transform,
d
dt u(t/2
F[δ(t/2)](ω) = 2F[δ(t)](2ω) = 2
(because F[δ(t)](ω) = 1)
By the inverse Fourier transform, this shows δ(t/2) = 2δ(t).
h
i
h
i
d
d
Therefore, dt
u(t/2 + 1/2) = δ(t + 1), similarly dt
u(t/2 − 1/2) = δ(t − 1). Using this we get the
same result as in the given solution. //
1 iω
[e − e−iω ]. Then, for the inverse Fourier
Second method (easier) Notice that sin(ω) = 2i
−1
transform F ,
h1
i
F −1 [sin ω](t) = F −1
[eiω − e−iω ]
2i
i
1 h −1 iω
=
F [e ] − F −1 [e−iω ]
2i
Now, notice that by the time-shift property,
F −1 [eiω ] = F −1 [1](t + 1)
F −1 [e−iω ] = F −1 [1](t − 1)
Since F −1 [1](t) = δ(t), we have
F −1 [eiω ] = δ(t + 1)
F −1 [e−iω ] = δ(t − 1)
Back to the original problem, we see
h(t) = F −1 [sin ω](t) =
7
1
[δ(t + 1) − δ(t − 1)]
2i
(c) Use the same function δ(t) as in part (b) to express the inverse Fourier transform of
gb(ω) = sin(5ω + π/6).
(Hint: Use (b).)
Solution
We can use the result of part (b). Note that sin(5ω + π/6) = 15 [5 sin(5(ω + π/30))] (this latter
expression is more convenient for using the frequency shifting and scaling property). For the inverse
Fourier transform F −1 , we see
F −1 [sin(5ω + π/6)](t)
1
= F −1 [5 sin(5(ω + π/30))](t)
5
1
= e−itπ/30 F −1 [5 sin(5ω)](t)
(frequency scaling)
5
1
= e−itπ/30 F −1 [sin(ω)](t/5)
(by scaling)
5
1
1
( by using part (b))
= e−itπ/30 [δ(t/5 + 1) − δ(t/5 − 1)]
5
2i
1 −itπ/30
=
e
[δ t/5 + 1 − δ t/5 − 1 ]
10i
5. (Convolution) We will work out a couple of examples in class on Monday.
Consider the functions

 −2, −2 ≤ t < 0
1, 0 ≤ t < 4
f (t) =

0, otherwise
1, −1 ≤ t < 0
g(t) =
0, otherwise
and h(t) = (f ∗ g)(t)
(a) Find h(t) and draw an accurate graph of this function on the interval −4 ≤ t ≤ 5. Hint: You should
obtain a collection of straight line segments.
Solution See http://www.iam.ubc.ca/˜sospedra/a7MATH267-sol.pdf, #3 (a).
\
(b) Find b
h(ω). Hint: Use the convolution property, (f
∗ g)(ω) = fb(ω)b
g (ω).
Solution See http://www.iam.ubc.ca/˜sospedra/a7MATH267-sol.pdf, #3 (b).
(c) Compute the integral
1
2π
Z
∞
ω 3ω
fb(ω)sinc
ei 2 dω
2
−∞
(Hint: First, realize sinc(ω/2) as Fourier transform of a function. Then, see what this integral
means: it is the inverse Fourier transform of a certain function. At what t? Then, use a property of
convolution. )
Solution See http://www.iam.ubc.ca/˜sospedra/a7MATH267-sol.pdf, #3 (c).
6. (NOT TO HAND IN)
Recall that the convolution of two functions f and g is the function f ∗ g defined by
Z ∞
(f ∗ g) (t) =
f (s)g(t − s)ds
−∞
\
and that (f
∗ g)(ω) = fb(ω)b
g (ω). Justify the following properties of the convolution:
(a) f ∗ g = g ∗ f
Solution See http://www.iam.ubc.ca/˜sospedra/a7MATH267-sol.pdf, #6.
8
(b) For constants A1 and A2 , we have (A1 f1 + A2 f2 ) ∗ g = A1 (f1 ∗ g) + A2 (f2 ∗ g)
Solution See http://www.iam.ubc.ca/˜sospedra/a7MATH267-sol.pdf, #6.
(c) (f1 ∗ f2 ) ∗ f3 = f1 ∗ (f2 ∗ f3 )
Solution See http://www.iam.ubc.ca/˜sospedra/a7MATH267-sol.pdf, #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.
9