Exponential and Sinusoidal Signals

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1
CLASS 2
(Sections 1.3)
Exponential and Sinusoidal Signals
βˆ™
They arise frequently in applications, and many other signals can be constructed from them.
Continuous-time complex exponential and sinusoidal signals:
π‘₯(𝑑) = πΆπ‘’π‘Žπ‘‘
where 𝐢 and π‘Ž are in general complex numbers.
Real exponential signals: 𝐢 and π‘Ž are reals.
at
0
βˆ™
C>0 and a<0.
Ce
Ce
at
C>0 and a>0.
C
0
t
0
C
0
t
The case π‘Ž > 0 represents exponential growth. Some signals in unstable systems exhibit exponential
growth.
βˆ™
The case π‘Ž < 0 represents exponential decay. Some signals in stable systems exhibit exponential
decay.
2
Periodic complex exponential:
𝑒𝑗𝑀0 𝑑
where 𝑗 =
√
−1, 𝑀0 βˆ•= 0 is real, and 𝑑 is the time.
Euler’s formula: 𝑒𝑗𝑀0 𝑑 = cos(𝑀0 𝑑) +𝑗 sin(𝑀0 𝑑) . Note that
| {z }
| {z }
Re{𝑒𝑗𝑀0 𝑑 }
Im{𝑒𝑗𝑀0 𝑑 }
1
jw t
sin(w t)
e
Im
0
0
0
cos(w0t)
−1
−1
βˆ™
βˆ£π‘’π‘—π‘€0 𝑑 ∣ = 1 and ∠𝑒𝑗𝑀0 𝑑 = 𝑀0 𝑑.
βˆ™
𝑒𝑗2πœ‹π‘˜ = 1, for π‘˜ = 0, ±1, ±2, . . . .
0
Re
1
Since
(
)
2πœ‹
𝑗𝑀0 𝑑+ βˆ£π‘€
∣
𝑒
0
𝑀
𝑗2πœ‹ βˆ£π‘€0 ∣
= 𝑒𝑗𝑀0 𝑑 𝑒
0
𝑗𝑀0 𝑑
0)
= 𝑒𝑗𝑀0 𝑑 𝑒|𝑗2πœ‹sign(𝑀
{z } = 𝑒
=1
we have
𝑒𝑗𝑀0 𝑑 is periodic with fundamental period
2πœ‹
.
βˆ£π‘€0 ∣
Note that
βˆ™
𝑒𝑗𝑀0 𝑑 and 𝑒−𝑗𝑀0 𝑑 have the same fundamental period.
βˆ™
Energy in 𝑒𝑗𝑀0 𝑑 :
βˆ™
Average Power in 𝑒𝑗𝑀0 𝑑 : lim𝑇 →∞
βˆ™
{π‘’π‘—π‘˜π‘€0 𝑑 }π‘˜=0,±1,... , are all periodic with period
∫∞
−∞
βˆ£π‘’π‘—π‘€0 𝑑 βˆ£π‘‘π‘‘ =
∫∞
−∞
1
2𝑇
1.𝑑𝑑 = ∞
∫𝑇
−𝑇
βˆ£π‘’π‘—π‘€0 𝑑 βˆ£π‘‘π‘‘ = lim𝑇 →∞
2πœ‹
.
βˆ£π‘€0 ∣
1
2𝑇
∫𝑇
−𝑇
1.𝑑𝑑 = 1.
They are called a harmonically related set of
complex exponentials with π‘’π‘—π‘˜π‘€0 𝑑 being the π‘˜−th harmonic.
3
Sinusoidal signals:
𝐴 cos(𝑀0 𝑑 + πœ™)
and
𝐴 sin(𝑀0 𝑑 + πœ™).
where 𝐴 is real, 𝑀0 is real, πœ™ is real, and 𝑑 is the time. (Graph one of the signals!)
βˆ™
βˆ™
They arise in systems that conserve energy such as an ideal LC circuit or an ideal mass-spring system.
– Periodic with the same fundamental period 𝑇0 = 2πœ‹/βˆ£π‘€0 ∣
– βˆ£π‘€0 ∣ is the fundamental frequency
– 𝑓0 := 1/𝑇0 = βˆ£π‘€0 ∣/(2πœ‹) is the number of cycles per unit time (large 𝑓0 means more oscillatory)
– ∣𝐴∣ is the amplitude
– βˆ£πœ™βˆ£ is the size of the phase shift.
βˆ™
Since
𝑒𝑗(𝑀0 𝑑+πœ™) = cos(𝑀0 𝑑 + πœ™) + 𝑗 sin(𝑀0 𝑑 + πœ™)
we can write
𝐴 cos(𝑀0 𝑑 + πœ™) = 𝐴Re(𝑒𝑗(𝑀0 𝑑+πœ™) )
𝐴 sin(𝑀0 𝑑 + πœ™) = 𝐴Im(𝑒𝑗(𝑀0 𝑑+πœ™) ).
βˆ™
Recall, for any complex number 𝑧,
𝑧 = Re(𝑧) + 𝑗Im(𝑧)
𝑧 ∗ = Re(𝑧) − 𝑗Im(𝑧)
therefore
Re(𝑧) =
𝑧 + 𝑧∗
2
Im(𝑧) =
𝑧 − 𝑧∗
.
2𝑗
Hence, we can also write
)
𝐴 ( 𝑗(𝑀0 𝑑+πœ™) ( 𝑗(𝑀0 𝑑+πœ™) )∗ ) 𝐴 ( 𝑗(𝑀0 𝑑+πœ™)
𝑒
+ 𝑒
=
𝑒
+ 𝑒−𝑗(𝑀0 𝑑+πœ™)
2
2
𝐴
𝐴
= π‘’π‘—πœ™ 𝑒𝑗𝑀0 𝑑 + 𝑒−π‘—πœ™ 𝑒−𝑗𝑀0 𝑑
2
2
(
)
𝐴 𝑗(𝑀0 𝑑+πœ™) ( 𝑗(𝑀0 𝑑+πœ™) )∗ ) 𝐴 −π‘—πœ‹/2 ( 𝑗(𝑀0 𝑑+πœ™)
𝐴 sin(𝑀0 𝑑 + πœ™) =
𝑒
− 𝑒
= 𝑒
𝑒
− 𝑒−𝑗(𝑀0 𝑑+πœ™)
2𝑗
2
𝐴
𝐴
= 𝑒𝑗(πœ™−πœ‹/2) 𝑒𝑗𝑀0 𝑑 − 𝑒−𝑗(πœ™+πœ‹/2) 𝑒−𝑗𝑀0 𝑑 .
2
2
𝐴 cos(𝑀0 𝑑 + πœ™) =
4
General complex exponential signals:
πΆπ‘’π‘Žπ‘‘
where 𝐢 and π‘Ž are complex numbers.
If
𝐢 = βˆ£πΆβˆ£π‘’π‘—πœƒ
and
π‘Ž = π‘Ÿ + 𝑗𝑀0
then
πΆπ‘’π‘Žπ‘‘ = βˆ£πΆβˆ£π‘’π‘—πœƒ 𝑒(π‘Ÿ+𝑗𝑀0 )𝑑 = βˆ£πΆβˆ£π‘’π‘Ÿπ‘‘ 𝑒𝑗(𝑀0 𝑑+πœƒ) = βˆ£πΆβˆ£π‘’π‘Ÿπ‘‘ cos(𝑀0 𝑑 + πœƒ) +𝑗 βˆ£πΆβˆ£π‘’π‘Ÿπ‘‘ sin(𝑀0 𝑑 + πœƒ) .
|
{z
}
|
{z
}
Re(πΆπ‘’π‘Žπ‘‘ )
Im(πΆπ‘’π‘Žπ‘‘ )
r>0.
r<0.
|C|ert
at
Re(Ce )
at
Re(Ce )
|C|ert
0
0
−|C|ert
0
−|C|e
0
t
rt
t
βˆ™
If π‘Ÿ = 0, the real and imaginary part are sinusoidals.
βˆ™
If π‘Ÿ > 0, the real and imaginary part are sinusoidals multiplied by a growing exponential.
Such signals arise in unstable systems.
βˆ™
If π‘Ÿ < 0, the real and imaginary part are sinusoidals multiplied by a decaying exponential.
Such signals arise in stable systems, for example, in RLC circuits, or in mass-spring-friction system,
where the energy is dissipated due to the resistors, friction, etc.
5
Discrete-time complex exponential and sinusoidal signals:
π‘₯[𝑛] = 𝐢𝑒𝛽𝑛
where 𝐢 and 𝛽 are complex numbers.
Analogous to the continuous-time case with the following differences: (𝑀0 is real below)
βˆ™
𝑒𝑗𝑀0 𝑑 = 𝑒𝑗𝑀1 𝑑 are different signals if 𝑀0 βˆ•= 𝑀1 , whereas
𝑒𝑗𝑀0 𝑛 = 𝑒𝑗𝑀1 𝑛
if
𝑀0 − 𝑀1 = 2π‘˜πœ‹, for some π‘˜ ∈ {0, ±1, . . . }.
(Explain this on the unit circle!)
Therefore, it is sufficient to consider only the case 𝑀0 ∈ [0, 2πœ‹) or 𝑀0 ∈ [−πœ‹, πœ‹).
βˆ™
As 𝑀0 increases 𝑒𝑗𝑀0 𝑛 oscillates at higher frequencies, whereas this is not the case for 𝑒𝑗𝑀0 𝑛 .
In the figure below, the frequency of oscillations increases as 𝑀0 changes from 0 to πœ‹ then it decreases
as 𝑀0 changes from πœ‹ to 2πœ‹.
βˆ™
𝑒𝑗𝑀0 𝑑 is periodic with fundamental period 2πœ‹/βˆ£π‘€0 ∣, whereas
𝑒𝑗𝑀0 𝑛 is periodic
⇔ 𝑒𝑗𝑀0 𝑛 = 𝑒𝑗𝑀0 (𝑛+𝑀 ) for some integer 𝑀 > 0, for all 𝑛
⇔ 𝑒𝑗𝑀0 𝑀 = 1 for some integer 𝑀 > 0
⇔ 𝑀0 𝑀 = 2πœ‹π‘š for some integers π‘š, 𝑀 > 0
⇔
βˆ™
If
𝑀0
2πœ‹
=
is 𝑀 =
π‘š
𝑀
𝑀0
is rational.
2πœ‹
for some integers π‘š and 𝑀 which have no common factors, then the fundamental period
2π‘šπœ‹
𝑀0
because
𝑒𝑗𝑀0 (𝑛+𝑁 ) = 𝑒𝑗𝑀0 𝑛 𝑒𝑗
The same observations hold for discrete-time sinusoids.
2πœ‹π‘š
𝑁
𝑀
.
6
w =1π/8
w =0π/8
0
n=4
n=5
n=3
n=6
n=2
n=7
n=1
n=8
n=0
n=9
n=15
n=10 n=14
n=11
n=13
n=12
n=3
Im
n=0
Im
Im
w =2π/8
0
0
n=4
n=6
n=7
Re
Re
w =4π/8
w =8π/8
w =12π/8
0
0
n=2
n=0
n=1
n=0
Im
n=3
Im
Im
n=0
n=5
n=1
n=3
n=2
n=0
n=1
Re
Re
w0=14π/8
w0=15π/8
w0=16π/8
n=4
n=7
n=0
n=3
n=2
Re
n=1
Im
n=6
n=12
n=11
n=13
n=10 n=14
n=9
n=15
n=8
n=0
n=7
n=1
n=6
n=2
n=5
n=3
n=4
Im
Re
n=5
Im
n=1
Re
0
Fig. 1.
n=2
Re
To determine the fundamental period, count the number of steps to get back to 1!
n=0
Re
7
Examples:
1) Is π‘₯[𝑛] = 𝑒𝑗𝑛2πœ‹/3 + 𝑒𝑗𝑛3πœ‹/4 periodic? If it is periodic, what’s its fundamental period?
For 𝑒𝑗𝑛2πœ‹/3 , 𝑀0 /(2πœ‹) = 1/3, so 𝑒𝑗𝑛2πœ‹/3 is periodic with fundamental period 3.
For 𝑒𝑗𝑛3πœ‹/4 , 𝑀0 /(2πœ‹) = 3/8, so 𝑒𝑗𝑛3πœ‹/4 is periodic with fundamental period 8.
π‘₯[𝑛] is periodic with fundamental period 24 = π‘™π‘π‘š(3, 8).
2) Is π‘₯[𝑛] = sin(3𝑛/4) periodic? If it is periodic, what’s its fundamental period?
Since
𝑀0
2πœ‹
=
3
8πœ‹
is irrational, π‘₯[𝑛] is not periodic; see the figure where π‘₯[𝑛] = 0 only at 𝑛 = 0.
sin(3t/4) and sin(3n/4)
1
0
−1
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
t
3) Is π‘₯[𝑛] = sin(8πœ‹π‘›/31) periodic? If it is periodic, what’s its fundamental period?
Since 𝑀0 /(2πœ‹) = 4/31, π‘₯[𝑛] is periodic with fundamental period 31; see the figure where π‘₯[0] =
π‘₯[31] = 0. Note that the continuous-time signal sin(8πœ‹π‘‘/31) has fundamental period 31/4, hence it
is 0 at 𝑑 = 31/4. But π‘₯[𝑛] has no 31/4−th sample and it misses 0 between π‘₯[7] and π‘₯[8].
sin(8π t/31) and sin(8π n/31)
1
0
−1
0 1 2 3 4 5 6 7 8 9 1011121314151617181920212223242526272829303132333435
t
8
Harmonically related discrete-time periodic exponentials:
πœ™π‘˜ [𝑛] = {π‘’π‘—π‘˜(2πœ‹/𝑁 )𝑛 }π‘˜=0,±1,... , are all periodic with period 𝑁 .
However, unlike the continuous-time signals, these signals are not all distinct because
πœ™π‘˜+𝑁 [𝑛] = 𝑒𝑗(π‘˜+𝑁 )(2πœ‹/𝑁 )𝑛 = π‘’π‘—π‘˜(2πœ‹/𝑁 )𝑛 𝑒𝑗2πœ‹π‘› = πœ™π‘˜ [𝑛].
This implies that there are only 𝑁 distinct signals in this set, for example,
πœ™0 [𝑛] = 1
πœ™1 [𝑛] = 𝑒𝑗2πœ‹π‘›/𝑁
πœ™2 [𝑛] = 𝑒𝑗4πœ‹π‘›/𝑁
..
.
πœ™1 [𝑛] = 𝑒𝑗2(𝑁 −1)πœ‹/𝑁 .
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