EE-210. Signals and Systems
Homework 1∗
Spring 2010
Exercise Due Date
Week of 1st Feb.
Problems
Q1 Evaluate the following exponentials and express your final answer in rectangular coordinates.
(a) i. e−jπ
solution
e−jπ = ej(−π)
= cos(−π) + j sin(−π)
= cos(π) − j sin(π) = −1 + j · 0 = −1
ii. ej3π/4 + ejπ/4
solution
ej3π/4 + ejπ/4 = (cos(3π/4) + j sin(3π/4)) + (cos(π/4) + j sin(π/4))
1
1
1
1
= (− √ + j √ ) + ( √ + j √ )
2
2
2
2
√
2
= j√ = j 2
2
iii. ejπ/4 − (ejπ/4 )∗
solution
ejπ/4 − (ejπ/4 )∗ = (cos(π/4) + j sin(π/4)) − (cos(π/4) + j sin(π/4))∗
1
1
1
1
= (√ + j √ ) − (√ − j √ )
2
2
2
2
√
2
= j√ = j 2
2
∗ LUMS
School of Science & Engineering, Lahore, Pakistan.
1
(b) If z = x + jy = Aejθ , find x, y, A, and θ, for:
z = (j2ej(3π/2) )2
solution
z = (j2ej(3π/2 )2 = j 2 22 (ej(3π/2) )2
= −1 · 4 · ej(3π/2)·2 = −4ej(3π) = −4(−1) = 4
= 4 + 0 · j = 4ej·0
Q2 Write the following complex signals in polar form, i.e., in the form x(t)= r(t)ejθ(t) , r(t),
θ(t) ∈ R for continuous-time signals, x[n]= r[n]ejθ[n] , r[n], θ[n] ∈ R for discrete-time signals.
(a) x1 (t) = (1 + j)e2 e−j(1+3t)
solution
√
√
π
π
x1 (t) = (1 + j)e2 e−j(1+3t) = 2e2 ej 4 e−j(1+3t) = 2e2 e−j(1− 4 +3t)
√
π
⇒ r1 (t) = 2e2 , θ1 (t) = −1 + − 3t
4
(b) x2 (t) = 2t + jt2 − j
solution
x2 (t) = 2t + jt2 − j = 2t + j(t2 − 1)
p
t2 −1
= 4t2 + (t2 − 1)2 · ej arctan 2t
p
t2 − 1
⇒ r2 (t) = 4t2 + (t2 − 1)2 , θ2 (t) = arctan
2t
(c) x1 [n] = 1/nj , n > 0
solution
x1 [n] = 1/nj = n−j = e−j ln n , n > 0
⇒ r1 [n] = 1, θ1 [n] = − ln n
(d) x2 [n] = j sin(j3n)
solution
x2 [n] = j sin(j3n) = j
⇒ r2 [n] =
ej(j3n) − e−j(j3n)
e−3n − e3n
=
2j
2
e−3n − e3n
, θ2 [n] = 0
2
Q3 Determine if the following systems are
• Memoryless
2
• Time-invariant
• Linear
• Causal
• Stable
(a) y(t) =
d
dt x(t)
solution Time invariant, Linear, Causal
(b) y(t) =
x(t)
1+x(t−1)
solution Time invariant, Causal
(c) y[n] = x[3n − 2]
solution Linear, Stable
(d) y(t) = et x(−t)
solution Linear
Justify your answers!!
Q3 (a) Sketch the signals x[n] = u[n + 3] − u[n] + 0.5n u[n] − 0.5n−4 u[n − 4] , and
y[n] = nu[−n] − δ[n − 1] − nu[n − 3] + (n − 4)u[n − 6].
x[n]
1
0.5
-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10
n
-0.5
-1
y[n]
-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10
-1
-2
-3
-4
-5
3
n
2
x(t)
0
-3
2
-1
3
t
-1
y(t)
2
...
...
0
-3
-1
1
2
3
t
(b) Find expressions for the following signals below:
solution
2
2
(t + 3)u(t + 3) − (t + 3)u(t) − u(t) + u(t − 2) + 2δ(t + 1) − δ(t − 3)
3
3
∞
X
y(t) =
2(t − 3k)u(t − 3k) − 2(t − 3k − 1)u(t − 3k − 1) − 2u(t − 2 − 3k)
x(t) =
k=−∞
Q4 Find the fundamental periods and fundamental frequencies of the following periodic signals.
(a) x[n] = cos(0.01πn)ej0.13πn
solution
7
6
x[n] = cos(0.01πn)ej0.13πn = 0.5ej0.14πn +0.5e−j0.12πn = 0.5ej 100 (2π)n +0.5e−j 100 (2π)n
m2
7
6
1
Thus m
N1 = 100 for the first term and N2 = 100 . The fundamental period common
π
to both terms is N=100, and the fundamental frequency is 50
P∞
(b) x(t) = k=−∞ e−(t−2k) cos(4π(t − 2k))[u(t − 2k) − u((t − 2k) − 1)]]. Sketch this signal
solution This signal is built from a damped cosine term which lasts 1 second(k=0 in the
summation) and is repeated every 2 seconds. Therefore its fundamental period is
T=2 and its fundamental frequency is π radians/s
4
(c) x[n] = e−j
13π
7 n
solution Fundamental period of the signal is N = 14
(d) x(t) = cos(5πt) + sin(12πt)
solution Fundamental period: T = 2
Q5 (a) Show that if x[n] is odd signal, then
+∞
X
x[n] = 0
n=−∞
solution For an odd signal x[n]=-x[n] ⇒ x[0] = 0 and
+∞
X
x[n] = x[0] +
n=−∞
+∞
X
(x[n] − x[n]) = 0
n=1
(b) Show that if x1 [n] is odd and x2 [n] is even, then their product is odd.
solution
x1 [n] = x1 [−n], x2 [n] = −x2 [−n] ⇒ x1 [−n]x2 [−n] = −x1 [n]x2 [n]
(c) Let x[n] be an arbitrary signal with even and odd parts xe [n] = Ex[n], xo [n] = Ox[n]
Show that
+∞
+∞
+∞
X
X
X
x2 [n] =
xe 2 [n] +
xo 2 [n]
k=−∞
k=−∞
k=−∞
solution
+∞
X
x2 [n] =
k=−∞
+∞
X
(xe [n] + xo [n])2 =
k=−∞
=
+∞
X
k=−∞
+∞
X
k=−∞
xe 2 [n] +
+∞
X
xo 2 [n]
k=−∞
5
x2e [n] + 2
+∞
X
k=−∞
xe [n]xo [n] +
+∞
X
k=−∞
x2o [n]