1
EG1110 Signals and Systems
Exercise 1: Simple signal problems
Some useful identities
Complex number in Cartesian form z = x + jy
Complex number in Polar form z = rejθ where
r = |z| =
q
x2 + y 2
θ = arg z = tan−1
y
x
Euler’s Identity
ejθ = cos θ + j sin θ
Problems
1. Using Euler’s identity prove that
1
cos θ = (ejθ + e−jθ )
2
[Hint: derive an expression for e−jθ ]
2. Using Euler’s identity prove that
sin(θ + φ) = sin θ cos φ + cos θ sin φ
Note that Euler’s Identity can be used to prove many trigonometric formulae.
3. Show that the phase shifted sine wave
r cos(ωt + φ)
can be written as the sum of scaled sine and cosine waves of the same frequency
a cos(ωt) + b sin(ωt)
Give expressions for r and θ. [Hint: use the trigonometric identity cos(x + y) = cos(x) cos(y) −
sin(x) sin(y) (prove if you have time)]
4. Show that
d jωt
(e ) = jωejωt
dt
[Hint: use Euler’s identity]
5. A decaying exponential sinusoid is given by
x(t) = ke−at cos(ωt),
a>0
Sketch this signal, labelling points at which x(t) = 0 and points at which x(t) = ±ke −at .
2
Solutions
1. We have the Euler Identity:
ejθ = cos θ + j sin θ
(1)
e−jθ = cos θ − j sin θ
(2)
From this it follows that
So adding the two above equations we have
ejθ + e−jθ = 2 cos θ
from which one can find cos θ.
2. Similar to above, subtracting equation (2) from equation (1) gives:
ejθ − e−jθ = 2j sin θ
and so
sin θ =
1 jθ
(e − e−jθ )
2j
Therefore, replacing θ by θ + φ we have
sin(θ + φ) =
=
1
j(θ+φ) − e−j(θ+φ) )
2j (e
1
jθ jφ
−jθ e−jφ )
2j (e e − e
Using Euler’s identity on each e±jx then gives:
sin(θ + φ) =
1
2j
=
1
2j
=
1
2j
[(cos θ + j sin θ)(cos φ + j sin φ)
−(cos θ − j sin θ)(cos φ − j sin φ)]
cos θ cos φ + j cos θ sin φ + j sin θ cos φ − sin θ sin φ
−cos θ cos φ + j sin θ cos φ + j cos θ sin φ + sin θ sin φ
[2j sin θ cos φ + 2j cos θ sin φ]
= sin θ cos φ + cos θ sin φ
Q.E.D
3. Using the trigonometric identity:
r cos(ωt + φ) = r cos(φ) cos(ωt) − r sin(φ) sin(ωt)
and comparing the right hand side to the expression given in the question:
a cos(ωt) + b sin(ωt)
it follows that
a = r cos(φ)
b
= −r sin(φ)
3
From this we have that
•
a2 + b2 = r2 cos2 (φ) + r 2 sin2 (φ)
= r2 [cos2 (φ) + sin2 (φ)]
= r2
Thus r =
|
{z
}
=1
√
a2 + b2
• Performing the division
b
−r sin(φ)
=
r cos(φ)
a
Thus implies − tan(φ) = b/a and hence φ = tan−1 (−b/a).
4.
d
jωt )
dt (e
=
d
dt [cos(ωt)
+ j sin(ωt)]
= −ω sin(ωt) + jω cos(ωt)
= ω[j cos(ωt) − sin(ωt)]
= jω[cos(ωt) + j sin(ωt)]
= jωejωt
5. x(t) = ke−at cos(ωt).
Thus we have zeros when cos(ωt) = 0, i.e. at points:
t=
π 3π 5π
,
,
...
2ω 2ω 2ω
Similarly x(t) = ±ke−at when cos(ωt) = 1, i.e. when
t = 0,
π 2π
, ,...
ω ω
At t = 0, x(t) = k. Thus we have a waveform as shown below (plotted with ω = 5 and a = 0.5 and
k = 1)
1
0.8
2π/ω
0.6
4π/ω
Amplitude [units]
0.4
6π/ω
0.2
0
−0.2
5π/ω
−0.4
3π/ω
−0.6
π/ω
−0.8
−1
0
1
2
3
4
5
Time [sec]
6
7
8
9
10