MATH 267
Due: March 21, 2012, in the class
ASSIGNMENT # 8
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
1. [NOT TO HAND-IN] [Scaling property of the delta function]
(a) For α > 0, show that
Z
∞
δ(αt)dt =
−∞
1
α
(Hint: This is a one line proof. Use change of variables.)
(b) For α < 0, show that
Z ∞
1
δ(αt)dt = −
α
−∞
(Hint: This is a one line proof. Use change of variables.)
(c) From parts (a) and (b), we conclude that, for α 6= 0,
Z ∞
1
.
δ(αt)dt =
|a|
−∞
Solution
See Problem 2 in http://www.iam.ubc.ca/˜sospedra/a8MATH267-sol.pdf
2. [Delta function and integral]
R∞
δ(t − π)eit/3 dt.
√
R∞
Solution Since eit/3 is continuous, −∞ δ(t − π)eit/3 dt = eiπ/3 = 1/2 + i 3/2.
R∞
d
(b) Find the value of the integral −∞ sin(t + π2 ) dt
u(t)dt.
R∞
R∞
d
d
u(t)dt = −∞ sin(t + π2 )δ(t)dt =
Solution Notice that dt u(t) = δ(t). Therefore, −∞ sin(t + π2 ) dt
sin(0 + π/2) = 1.
h
i
R∞
(c) Find the value of the integral −∞ δ(t − 1) u(1 − 2t ) + u(t − 1) + 2 dt. (Hint: Here applies the class
R∞
example about −∞ δ(t)u(t)dt.)
Solution We have to be carefulR because the step functions u(1 − 2t ), u(t − 1) are not continuous at
∞
t = 2, t = 1, respectively. Note −∞ δ(t − 1)u(1 − 2t )dt since u(1 − t/2) is continuous at t = 1 (t = 1
is the center of the delta function δ(t − 1). But, since u(t − 1) is discontinuous
at the center t = 1 of
R∞
the delta function u(t − 1), we have to do as in the class example, to get −∞ δ(t − 1)u(t − 1)dt = 1/2
(Here 1/2 is nothing but the average of the left and right limit of u(t − 1) at t = 1.)
(a) Find the value of the integral
Z
−∞
∞
h
i
t
δ(t − 1) u(1 − ) + u(t − 1) + 2 dt
2
−∞
Z ∞
Z ∞
Z ∞
t
=
δ(t − 1)u(1 − )dt +
δ(t − 1)u(t − 1)dt +
δ(t − 1)2dt
2
−∞
−∞
−∞
1
= u(1 − 1/2) + + 2. = 1 + 1/2 + 2 = 3 + 1/2
2
1
(d) Find the value of the integral
ih
i
R∞ h
u(t
+
1)
+
rect(t)
δ(1
−
2t)
+
2δ(t
+
1)
dt. (Hint: The result of
−∞
Problem 1 is relevant here.)
Solution We can follow as in (c). Since rect(t) is not continuous at t = 1/2, −1/2, we need a little
care. First,,
Z ∞
rect(t)δ(t + 1)dt = rect−1 = 0.
−∞
since rect(t) is continuous at t = −1.
R∞
Now, to handle with −∞ rect(t)δ(1 − 2t)dt, note δ(1 − 2t) (or equivalently δ(2t − 1) (since δ(t) is an
even function) and it has the center at t = 1/2. We also use change of variables as in Problem 1:
Namely,
Z ∞
Z ∞
Z
1 ∞
rect(t)δ(2t − 1)dt =
rect(t)δ(1 − 2t)dt =
rect(s/2)δ(s − 1)ds
2 −∞
−∞
−∞
1 1
= ×
2 2
(The value 1/2 comes from the average of the left and right limit of rects/2 at s = 1. )
Using above and proceeding as in part (c), we see
Z ∞h
ih
i
u(t + 1) + rect(t) δ(1 − 2t) + 2δ(t + 1) dt
−∞
Z ∞
Z ∞
Z
=
u(t + 1)δ(1 − 2t)dt +
2u(t + 1)δ(t + 1)dt +
−∞
−∞
∞
−∞
Z
∞
rect(t)δ(1 − 2t)dt +
2rect(t)δ(t + 1)dt
−∞
= u(1/2) + 2 × 1/2 + 1/2 × 1/2+) = 1 + 1 + 1/4 = 2 + 1/4
(e) Find the value of the integral
Solution
Notice that
d
dt u(t)
Z
∞
R∞ h
3δ(t − 2) +
−∞
= δ(t), therefore,
d
dt u(t
d
dt u(t
ih
i
+ 1) u(t + 1) − u( 2t − 1) dt.
+ 1) = δ(t + 1). We proceed similarly as part (d) to see
ih
i
d
t
u(t + 1) u(t + 1) − u( − 1) dt
dt
2
−∞
Z ∞h
i
ih
t
=
3δ(t − 2) + δ(t + 1) u(t + 1) − u( − 1) dt
2
−∞
Z ∞
Z ∞
t
=
3δ(t − 2)u(t + 1)dt +
−3δ(t − 2)u( − 1)dt
2
−∞
−∞
Z ∞
Z ∞
t
+
δ(t + 1)u(t + 1)dt +
−δ(t + 1)u( − 1)dt
2
−∞
−∞
1
1
1
1
= 3u(2 + 1) − 3 × × 1 + × 1 − u(−1/2 − 1) = 3 − 3 × × 1 + × 1 − 0
2
2
2
2
=2
h
3δ(t − 2) +
3. [Delta function and Fourier transform]
(a) Let f (t) = δ(t − 1) + 3 + δ(1 − 2t) + ei2t . Find fb(ω).
Solution
F[f (t)](ω) = F[δ(t − 1)](ω) + F[3[(ω) + F[δ(2(1/2 − t))](ω) + F[ei2t ](ω)
1
= e−iω + 3 × 2πδ(ω) + e−iω/2 + 2πδ(ω − 2)
2
1 −iω/2
−iω
=e
+ 6πδ(ω) + e
+ 2πδ(ω − 2)
2
2
(b) Let gb(ω) = δ(2ω − 1) + 1 + δ(2 − 2ω) + eiω . Find g(t).
Solution
F −1 [g(t)](ω) = F −1 [δ(2(ω − 1/2))](t) + F −1 [1[(t) + F −1 [δ(2(1 − ω))](t) + F[eiω ](t)
1
1
= eit/2 F −1 [δ((ω))](t/2) + δ(t) + eit F −1 [δ(ω)](t/2) + δ(t + 1)
2
2
1 it/2 1
1 it 1
= e
+ δ(t) + e
+ δ(t + 1)
2
2π
2 2π
1
1
= eit/2
+ δ(t) + eit
+ δ(t + 1)
4π
4π
(c) Let h(t) = u(t + 1) + u(2t + 1). Find the values b
h(π) and b
h(π/2). (Hint: Here, the class example
d
u(t) = δ(t), so
about u
b(ω) applies.) Solution Note that dt
d
h(t) = δ(t + 1) + 2δ(2t + 1)
dt
= δ(t + 1) + δ(2(t + 1/2)).
Now,
F[
d
h(t)](ω) = iωF[h(t)](ω)
dt
on one hand, and on the other hand, we have
F[
1
d
h(t)](ω) = F[δ(t + 1) + δ(2(t + 1/2))](ω) = eiω + eiω/2
dt
2
(here, of course, we used the time-shift and scaling property).
So, we get
1
iωF[h(t)](ω) = eiω + eiω/2
2
In particular,
1
iπb
h(π) = eiπ + eiπ/2 = −1 + i/2
2
√
√
1 iπ/4
2
2
iπ/2
b
iπ/2h(π/2) = e
+ e
=i+
+i
2
4
4
Thus,
−1 + i/2
b
h(π) =
iπ
√
√
i(2 + 2/2) + 2/2
b
.
h(π/2) =
iπ
4. [Discrete Fourier transform]
(In the following, we do not distinguish between “length= N discrete-time signals” and “N -periodic
discrete-time signals”.)
For the given discrete-time signal x, find x
b, i.e. its discrete Fourier transform (or in other words, discrete
Fourier series).
(a) x = [0, 1, 0, 0].
Solution Note e−i2π/4k = e−ikπ/2 = 1, −i, −1, i for k = 0, 1, 2, 3, respectively. Therefore,
1
1
1
i
(0 + 1 + 0 + 0) =
x
b[1] = (0 + (−i) + 0 + 0) = −
4
4
4
4
1
1
1
i
x
b[1] = (0 + (−i)2 + 0 + 0) = −
x
b[3] = (0 + (−i)3 + 0 + 0) = .
4
4
4
4
x
b[0] =
3
Thus,
1
i
1 i
x
b = [ ,− ,− , ]
4 4 4 4
Note that this is nothing but x
b[k] = eikπ/2 /4 for k = 0, 1, 2, 3.
(b) x = [1, 1, 1, 1].
Solution Note e−iπ/4k = e−ikπ/2 = 1, −i, −1, i for k = 0, 1, 2, 3, respectively. Therefore,
1
1
(1 + 1 + 1 + 1) = 1
x
b[1] = (1 + (−i) − 1 + i) = 0
4
4
1
1
x
b[1] = (1 + (−i)2 + (−1)2 + (i)2 ) = 0
x
b[3] = (1 + (−i)3 + (−1)3 + i3 ) = 0.
4
4
x
b[0] =
(You can also use ‘orthogonality’ to see the above immediately.)
Thus,
x
b = [1, 0, 0, 0]
N entries
z
}|
{
Remark Let us consider more general case: x = [1, 1, · · · , 1]. By definition,
x
b[k] =
Clearly x
b[0] =
x
b[k] =
1
N [N ]
i
(N −1)k
k
2k
1h
1 + e−2πi N + e−2πi N + · · · + e−2πi N
N
= 1. For 1 ≤ k ≤ N − 1, let w = wN = e
2πi
N
to get
i
1h
1 1 − w−N k
1 + w−k + w−2k + · · · + w−(N −1)k =
=0
N
N 1 − w−k
since w−kN = e−2πki = 1
Hence x
b = [1, 0, · · · , 0] (N entries).
(c) x = [1, −1, 1, −1].
Solution Note e−i2π/4k = e−ikπ/2 = 1, −i, −1, i for k = 0, 1, 2, 3, respectively. Therefore,
1
1
(1 − 1 + 1 − 1) = 0
x
b[1] = (1 − (−i) − 1 − i) = 0
4
4
1
1
2
2
2
x
b[1] = (1 − (−i) + (−1) − i ) = 1
x
b[3] = (1 − (−i)3 + (−1)3 − i3 ) = 0.
4
4
x
b[0] =
(You can also use ‘orthogonality’ to see the above immediately, since x[n] = (−1)n for n = 0, 1, 2, 3.)
Thus,
x
b = [0, 0, 1, 0]
(d) x = [1, 2, 3, 4].
Solution
Note ei2π/4k = eikπ/2 = 1, i, −1, −i for k = 0, 1, 2, 3, respectively. Therefore,
1
1
−1 + i
(1 − 2 + 3 − 4) = −1/2
x
b[1] = (1 + 2(−i) − 3 + 4i) =
4
4
2
1
1
1
−1 − i
x
b[1] = (1 + 2(−i)2 + 3(−1)2 + 4i2 ) = −
x
b[3] = (1 + 2(−i)3 + 3(−1)3 + 4i3 ) =
.
4
2
4
2
x
b[0] =
Thus,
1 −1 + i 1 −1 − i
x
b = [− ,
,− ,
]
2
2
2
2
4
(e) x = [1, 31 , 312 , · · · , 3110 , 3111 ].
Solution
Let us consider more general case: x = [1, r, r2 , · · · , rN −1 ]. (In the problem, N = 12, r = 1/3. )
Let w = wN = e
2πi
N
Then,
x
b[0] =
i
1 1 − rN
1h
1 + r + r2 + · · · + r(N −1) =
N
N 1−r
(note that we could use the geometric sum for the case r 6= 1) and
x
b[k] =
2
(N −1) i
1h
1 − (rw−k )N
1 − rN
1 + rw−k + rw−k + · · · + rw−k
=
=
N
N (1 − rw−k )
N (1 − rw−k )
for 1 ≤ k ≤ N − 1. (Here, we used that wN = 1. )
Back to our specific case, we see that
x
b[k] =
1 − ( 31 )12
12(1 − 13 e−iπk/6 )
for k = 0, 1, 2, · · · , 11.
5. [Inverse discrete Fourier transform]
(a) x
b = [0, 0, 3, 0].
Solution Note eiπ/4k = eikπ/2 = 1, i, −1, −i for k = 0, 1, 2, 3, respectively. Therefore,
x[1] = 0 + 0 − 3 − 0 = −3
x[0] = 0 + 0 + 3 + 0 = 3
2
x[1] = 0 + 0 + 3(−1) + 0 = 3
x[3] = 0 + 0 + 3(−1)3 + 0 = −3.
Thus,
x = [3, −3, 3, −3]
(b) x
b = [1, 1, 1, 1].
Solution
Note eiπ/4k = eikπ/2 = 1, i, −1, −i for k = 0, 1, 2, 3, respectively. Therefore,
x[0] = 1 + 1 + 1 + 1 = 4
2
2
x[1] = 1 + i − 1 − i = 0
2
x[1] = 1 + i + (−1) + (−i) = 0
x[3] = 1 + i3 + (−1)3 + (−i)3 ) = 0.
(You can also use ‘orthogonality’ to see the above immediately.)
Thus,
x = [4, 0, 0, 0]
N entries
z
}|
{
Remark Let us consider more general case: x
b = [1, 1, · · · , 1]. By definition,
n
2n
x[n] = 1 + e2πi N + e2πi N + · · · + e2πi
Clearly x[0] = N . For 1 ≤ n ≤ N − 1, let w = wN = e
x[n] = 1 + wn + w2n + · · · + w(N −1)n =
Hence x = [N, 0, · · · , 0], (N entries).
5
2πi
N
(N −1)n
N
to get
1 − wN n
=0
1 − wn
since wnN = e2πni = 1
(c) x
b = [1, 41 , 412 , · · · , 4110 , 4111 ]
Solution
Let us consider more general case: x
b = [1, r, r2 , · · · , rN −1 ]. (In the problem, N = 12, r = 1/4. )
2πi
Let w = wN = e N . Then,
1 − rN
1−r
(note that we could use the geometric sum for the case r 6= 1) and
2
(N −1) i 1 − (rwn )N
1 − rN
x[n] = 1 + rwn + rwn + · · · + rwn
=
=
(1 − rwn )
1 − rwn
x[0] = 1 + r + r2 + · · · + r(N −1) =
for 1 ≤ n ≤ N − 1. (Here, we used that wN = 1. )
Back to our specific case, we see that
x[n] =
1 − ( 41 )12
1 − 14 eiπn/6
for k = 0, 1, 2, · · · , 11.
6. [Discrete complex exponentials]
(a) Compute
10 h
2π
i
X
2π
2π
2π
2π
e−i 11 2n + ei 11 3n ei 11 2n + e−i 11 3n + ei 11 8n
n=0
P1
Solution Note that in the following sum n=0 0 three are N = 11 terms. This matches well with
2π
e0i 11 kn to apply orthogonality of discrete complex exponentials.
10 h
X
2π
2π
e−i 11 2n + ei 11 3n
2π
2π
2π
ei 11 2n + e−i 11 3n + ei 11 8n
i
n=0
=
10
X
2π
2π
e−i 11 2n ei 11 2n +
2π
2π
e−i 11 2n e−i 11 3n +
n=0
n=0
+
10
X
10
X
2π
2π
ei 11 3n ei 11 2n
n=0
10
X
2π
2π
e−i 11 2n ei 11 8n
n=0
10
X
2π
2π
ei 11 3n e−i 11 3n +
n=0
10
X
2π
2π
ei 11 3n ei 11 8n
n=0
= 11 + 0 + 0 + 0 + 11 + 11
In the last line we have used the orthogonality of discrete complex exponentials. (For the last term,
notice that 3 + 8 = 11 an integer multiple of 11.) So, the answer is
33
(b) Compute
10 h
2π
i
X
2π
2π
2π
2π
e−i 10 2n + ei 10 3n ei 10 2n + e−i 10 3n + ei 10 7n
n=0
P10
2π
Solution Notice that the sum n=0 has 11 terms, but, we have complex exponentials e−i 10 kn with
N = 10. Because of this we decompose the sum as
10 h
X
2π
2π
e−i 10 2n + ei 10 3n
2π
2π
2π
ei 10 2n + e−i 10 3n + ei 10 7n
i
n=0
=
9 h
X
2π
2π
e−i 10 2n + ei 10 3n
2π
2π
2π
ei 10 2n + e−i 10 3n + ei 10 7n
i
n=0
2π
2π
2π
2π
2π
+ e−i 10 2×10 + ei 10 3×10 ei 10 2×10 + e−i 10 3×10 + ei 10 7×10
6
For the summation
P9
n=0
part, we use the orthogonality of discrete complex exponentials, and see
9 h
2π
i
X
2π
2π
2π
2π
e−i 10 2n + ei 10 3n ei 10 2n + e−i 10 3n + ei 10 7n
n=0
=
9
X
2π
2π
e−i 10 2n ei 10 2n +
n=0
+
9
X
2π
2π
e−i 10 2n e−i 10 3n +
n=0
9
X
2π
2π
ei 10 3n ei 10 2n +
n=0
9
X
2π
2π
2π
2π
e−i 10 2n ei 10 7n
n=0
9
X
2π
2π
ei 10 3n e−i 10 3n +
n=0
9
X
ei 10 3n ei 10 7n
n=0
= 10 + 0 + 0 + 0 + 10 + 10 = 30
For the remaining part,
2π
2π
2π
2π
2π
e−i 10 2×10 + ei 10 3×10 ei 10 2×10 + e−i 10 3×10 + ei 10 7×10
(since ei2πk = 1 for integer k).
= (1 + 1)(1 + 1 + 1)
= 6.
Therefore the final answer is 36.
7. [NOT TO HAND-IN] Two N -periodic (or length= N ) signals x, y and their discrete Fourier transforms
x
b, yb are given. Prove:
np
f [n] = e2πi N x[n]
f [n] =
1
N
N
−1
X
x[m] y[n − m]
=⇒
fb[k] = x
b[k − p]
(frequency shifting)
=⇒
fb[k] = x
b[k] yb[k]
(convolution)
=⇒
fb[k] =
m=0
f [n] = x[n] y[n]
N
−1
X
x
b[m] yb[k − m]
(multiplication)
m=0
(Notice some typos are corrected in red color above!!!)
Solution
Use the definition of the discrete fourier transforms for the three properties.
kp
• • Frequency Shifting: Given that f [n] = e2πi N x[n], substitution yields
N −1
N −1
N −1
(k−p)n
kn
kn
1 X
1 X 2πi kp
1 X
fb[k] =
f [n]e−2πi N =
e N x[n]e−2πi N =
x[n]e−2πi N = x
b[k − p]
N n=0
N n=0
N n=0
• • Convolution: Substituting
f [n] =
N −1
1 X
x[m] y[n − m]
N m=0
into
N −1
kn
1 X
fb[k] =
f [n]e−2πi N
N n=0
gives
N −1 N −1
N −1 N −1
kn
kn
1 XX
1 X X
fb[k] = 2
x[m] y[n − m] e−2πi N = 2
x[m] y[n − m]e−2πi N
N n=0 m=0
N m=0 n=0
Make a change of variables replacing the summation variable n with r = n − m:
N −1 N −1−m
N −1
N −1−m
k(r+m)
kr
1 X X
1 X
1 X
−2πi N
−2πi km
b
N
f [k] = 2
x[m] y[r]e
=
x[m]e
y[r]e−2πi N
N m=0 r=−m
N m=0
N r=−m
The summand of the r sum is periodic of period N . That is, replacing r by r + N has no effect on
the summand. So all domains
of summation consisting
PN −1−m
PN −1 of a single full period give the same sum and
we may replace the sum r=−m by the sum r=0 , which gives fb[k] = x
b[k] yb[k].
7
• • Multiplication: Substituting
x[n] =
N
−1
X
x
b[m]e2πi
mn
N
into
m=0
N −1
N −1
kn
kn
1 X
1 X
fb[k] =
f [n]e−2πi N =
x[n]y[n]e−2πi N
N n=0
N n=0
gives
NX
N −1
−1
kn
1 X
2πi mn
b
N
e−2πi N
f [k] =
y[n]
x
b[m]e
N n=0
m=0
=
N −1 N −1
(k−m)n
1 X X
y[n] x
b[m]e−2πi N
N n=0 m=0
=
N −1 N −1
(k−m)n
1 X X
y[n] x
b[m]e−2πi N
N m=0 n=0
=
N
−1
X
m=0
=
N
−1
X
x
b[m]
N −1
(k−m)n
1 X
y[n]e−2πi N
N n=0
x
b[m] yb[k − m]
m=0
*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.
8