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1
EE 221
HW 1
Due: Monday 07-04-2015
Question # 1
a.
b.
c.
2
Question# 2
Determine whether or not each of the following continuous-time signals is periodic. If
periodic, determine i...
3
(d)
𝑇1 =
2𝜋
𝜔
=
2𝜋
2𝜋
= 1 𝑎𝑛𝑑 𝑇2 =
2𝜋
𝜔
=
2𝜋
10𝜋
=
1
5
𝑇1
𝑇2
=
1
1/5
= 5 which is a rational number so 𝑥(𝑡) is periodic....
4
1. Given the following signals: (1) sin 2𝑡 + cos 𝜋𝑡 (2) sin 6𝜋𝑡 + cos 5𝜋𝑡 (3) 𝑒10𝑡
𝑢(𝑡) (4) 𝑒−2𝑡
𝑢(𝑡)
a. Which are perio...
5
𝑃 = lim
𝑇→∞
1
𝑇
∫|𝑥(𝑡)|2
𝑇
2
−𝑇
2
𝑑𝑡 = lim
𝑇→∞
1
𝑇
∫[sin 2𝑡 + cos 𝜋𝑡]2
𝑇
2
−𝑇
2
𝑑𝑡
= lim
𝑇→∞
1
𝑇
∫ (sin2
2𝑡 + cos2
𝜋𝑡 + ...
6
so the energy of the signal is infinite.
As the signal is periodic with period 2Hz, so power can be evaluated as
𝑃 =
1
𝑇...
7
4.
𝐸 = ∫|𝑥(𝑡)|2
∞
−∞
𝑑𝑡 = ∫|𝑒−2𝑡
𝑢(𝑡)|2
∞
0
𝑑𝑡 = ∫ 𝑒−4𝑡
∞
0
𝑑𝑡 = −
𝑒−4𝑡
4
|
0
∞
=
1
4
so the signal has finite energy.
𝑃...
8
(c)
𝑥(𝑡) = 𝑅𝑒 {𝑒
𝑗(6𝜋𝑡−
𝜋
3
)
} + 𝑅𝑒 {𝑒
−𝑗(6𝜋𝑡−
𝜋
3
)
} + 𝑅𝑒 {2𝑒
𝑗(10𝜋𝑡−
𝜋
2
)
} + 𝑅𝑒 {2𝑒
−𝑗(10𝜋𝑡−
𝜋
2
)
}
(d)
,f Hz
Amp...
9
and the third integral can be shown to be
∫ cos (6𝜋𝑡 −
𝜋
3
) sin 10𝜋𝑡
1
0
𝑑𝑡 = 0
So
𝑃 =
1
𝑇𝑂
∫ |𝑥(𝑡)|2
𝑇 𝑂
2
−𝑇 𝑂
2
𝑑𝑡 =...
10
g.
,f Hz
Amplitude
0
( )Phaseshift rad
,f Hz0
(a)Single-sidedspectra
2
4
3 5
3 5
-π/3
-π/2
h.
𝑥(𝑡) = 𝑅𝑒 {𝑒
𝑗(6𝜋𝑡−
𝜋
3
)...
11
∫ cos2
(6𝜋𝑡 −
𝜋
3
)
1
0
𝑑𝑡 =
1
2
∫ [1 − cos (12𝜋𝑡 −
2𝜋
3
)]
1
0
𝑑𝑡 =
1
2
Similarly
∫ (sin2
10𝜋𝑡)
1
0
𝑑𝑡 =
1
2
∫[1 + cos...
12
Question# 6
By definition
And
Both are sketched below.
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Hw1 solution

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Hw1 solution

  1. 1. 1 EE 221 HW 1 Due: Monday 07-04-2015 Question # 1 a. b. c.
  2. 2. 2 Question# 2 Determine whether or not each of the following continuous-time signals is periodic. If periodic, determine its fundamental period. (a) 𝒙(𝒕) = 𝟐 𝐜𝐨𝐬 (𝟑𝒕 + 𝝅 𝟑 ) (b) 𝒙(𝒕) = [𝐜𝐨𝐬 (𝟐𝒕 − 𝝅 𝟑 )] 𝟐 (c) 𝒙(𝒕) = 𝐬𝐢𝐧(𝟒𝝅𝒕)𝒖(𝒕) (d) 𝒙(𝒕) = 𝐜𝐨𝐬(𝟐𝝅𝒕) + 𝐬𝐢𝐧(𝟏𝟎𝝅𝒕) (e) 𝒙(𝒕) = 𝐜𝐨𝐬(𝟐𝒕) 𝐬𝐢𝐧(𝟑𝒕) SOLUTION (a) 𝑇 = 2𝜋 𝜔 = 2𝜋 3 (b) 𝑥(𝑡) = 1 2 [cos (4𝑡 − 2𝜋 3 ) + 1] 𝑇 = 2𝜋 𝜔 = 2𝜋 4 = 𝜋 2 (c) The signal is not periodic therefore fundamental period does not exist.
  3. 3. 3 (d) 𝑇1 = 2𝜋 𝜔 = 2𝜋 2𝜋 = 1 𝑎𝑛𝑑 𝑇2 = 2𝜋 𝜔 = 2𝜋 10𝜋 = 1 5 𝑇1 𝑇2 = 1 1/5 = 5 which is a rational number so 𝑥(𝑡) is periodic. Fundamental period is 𝑇𝑜 = 5𝑇2 = 𝑇1 = 1𝐻𝑧 (e) cos(2𝑡) sin(3𝑡) = 1 2 [sin(5𝑡) − sin(−𝑡)] = 1 2 [sin(5𝑡) + sin(𝑡)] 𝑇1 = 2𝜋 𝜔 = 2𝜋 5 𝑎𝑛𝑑 𝑇2 = 2𝜋 𝜔 = 2𝜋 1 = 2𝜋 𝑇1 𝑇2 = 2𝜋/5 2𝜋 = 1 5 which is a rational number so 𝑥(𝑡) is periodic. Fundamental period is 𝑇𝑜 = 5𝑇1 = 1𝑇2 = 2𝜋 𝐻𝑧
  4. 4. 4 1. Given the following signals: (1) sin 2𝑡 + cos 𝜋𝑡 (2) sin 6𝜋𝑡 + cos 5𝜋𝑡 (3) 𝑒10𝑡 𝑢(𝑡) (4) 𝑒−2𝑡 𝑢(𝑡) a. Which are periodic? Give their periods. b. Which are power signals? Find their average power. c. Which are energy signals? Find their energies. SOLUTION (a) 1. 𝑇1 = 2𝜋 2 = 𝜋 𝑎𝑛𝑑 𝑇2 = 2𝜋 𝜔 = 2𝜋 𝜋 = 2 𝑇1 𝑇2 = 𝜋 2 which is an irrational number so 𝑥(𝑡) is non-periodic. 2. 𝑇1 = 2𝜋 𝜔 = 2𝜋 6𝜋 = 1 3 𝑎𝑛𝑑 𝑇2 = 2𝜋 5𝜋 = 2 5 𝑇1 𝑇2 = 1/3 2/5 = 5 6 which is a rational number so 𝑥(𝑡) is periodic. Fundamental period is 𝑇𝑜 = 5𝑇2 = 6𝑇1 = 2 𝐻𝑧 3. Exponentially increasing function is a non-periodic signal. 4. Exponentially decaying function is a non-periodic signal. (b) & (c) Checking whether the signals are energy or power. 1. 𝐸 = ∫|𝑥(𝑡)|2 ∞ −∞ 𝑑𝑡 = ∫[sin 2𝑡 + cos 𝜋𝑡]2 ∞ −∞ 𝑑𝑡 = ∫ (sin2 2𝑡 + cos2 𝜋𝑡 + 2 sin 2𝑡 cos 𝜋𝑡) ∞ −∞ 𝑑𝑡 Taking each integral at a time we have, ∫ (sin2 2𝑡) ∞ −∞ 𝑑𝑡 = ∞ Similarly ∫ cos2 𝜋𝑡 ∞ −∞ 𝑑𝑡 = ∞ so the energy of the signal is infinite. To see whether it is a power signal we have
  5. 5. 5 𝑃 = lim 𝑇→∞ 1 𝑇 ∫|𝑥(𝑡)|2 𝑇 2 −𝑇 2 𝑑𝑡 = lim 𝑇→∞ 1 𝑇 ∫[sin 2𝑡 + cos 𝜋𝑡]2 𝑇 2 −𝑇 2 𝑑𝑡 = lim 𝑇→∞ 1 𝑇 ∫ (sin2 2𝑡 + cos2 𝜋𝑡 + 2 sin 2𝑡 cos 𝜋𝑡) 𝑇 2 −𝑇 2 𝑑𝑡 So again taking one integral at a time, we have lim 𝑇→∞ 1 𝑇 ∫ (sin2 2𝑡) 𝑇 2 −𝑇 2 𝑑𝑡 = lim 𝑇→∞ 1 𝑇 ∫ 1 2 [1 − cos 4𝑡] 𝑇 2 −𝑇 2 𝑑𝑡 = lim 𝑇→∞ 1 2𝑇 ∫ 𝑑𝑡 𝑇/2 −𝑇/2 − lim 𝑇→∞ 1 2𝑇 ∫ cos 4𝑡 𝑇 2 −𝑇 2 𝑑𝑡 = lim 𝑇→∞ 1 2𝑇 𝑡| − 𝑇 2 𝑇 2 − lim 𝑇→∞ 1 8𝑇 sin 4𝑡| − 𝑇 2 𝑇 2 𝑑𝑡 = 1 2 Similarly lim 𝑇→∞ 1 𝑇 ∫ (cos2 2𝑡) 𝑇 2 −𝑇 2 𝑑𝑡 = lim 𝑇→∞ 1 𝑇 ∫ 1 2 [1 + cos 4𝑡] 𝑇 2 −𝑇 2 𝑑𝑡 = lim 𝑇→∞ 1 2𝑇 ∫ 𝑑𝑡 𝑇/2 −𝑇/2 + lim 𝑇→∞ 1 2𝑇 ∫ cos 4𝑡 𝑇 2 −𝑇 2 𝑑𝑡 = lim 𝑇→∞ 1 2𝑇 𝑡| − 𝑇 2 𝑇 2 + lim 𝑇→∞ 1 8𝑇 sin 4𝑡| − 𝑇 2 𝑇 2 𝑑𝑡 = 1 2 and the third integral can be shown to be lim 𝑇→∞ 2 𝑇 ∫ sin 2𝑡 cos 𝜋𝑡 𝑇 2 −𝑇 2 𝑑𝑡 = 0 So 𝑃 = lim 𝑇→∞ 1 𝑇 ∫|𝑥(𝑡)|2 𝑇 2 −𝑇 2 𝑑𝑡 = 1 < ∞ so the signal is a power signal although it’s not a periodic signal. 2. 𝐸 = ∫|𝑥(𝑡)|2 ∞ −∞ 𝑑𝑡 = ∫[sin 6𝜋𝑡 + cos 5𝜋𝑡]2 ∞ −∞ 𝑑𝑡 = ∫ (sin2 6𝜋𝑡 + cos2 5𝜋𝑡 + 2 sin 6𝜋𝑡 cos 5𝜋𝑡) ∞ −∞ 𝑑𝑡 = ∞
  6. 6. 6 so the energy of the signal is infinite. As the signal is periodic with period 2Hz, so power can be evaluated as 𝑃 = 1 𝑇𝑂 ∫ |𝑥(𝑡)|2 𝑇 𝑂 2 −𝑇 𝑂 2 𝑑𝑡 = 1 2 ∫[sin 6𝜋𝑡 + cos 5𝜋𝑡]2 1 −1 𝑑𝑡 = 1 2 ∫ (sin2 6𝜋𝑡 + cos2 5𝜋𝑡 + 2 sin 6𝜋𝑡 cos 5𝜋𝑡) 1 −1 𝑑𝑡 So again taking one integral at a time, we have ∫ (sin2 6𝜋𝑡) 1 −1 𝑑𝑡 = 1 2 ∫[1 − cos 12𝜋𝑡] 1 −1 𝑑𝑡 = 1 Similarly ∫ (cos2 5𝜋𝑡) 1 −1 𝑑𝑡 = 1 2 ∫[1 + cos 10𝜋𝑡] 1 −1 𝑑𝑡 = 1 and the third integral can be shown to be 2 ∫ sin 6𝜋𝑡 cos 5𝜋𝑡 1 −1 𝑑𝑡 = 0 So 𝑃 = 1 𝑇𝑂 ∫ |𝑥(𝑡)|2 𝑇 𝑂 2 −𝑇 𝑂 2 𝑑𝑡 = 1 2 (1 + 1 + 0) = 1 < ∞ so the signal is a power signal. 3. 𝐸 = ∫|𝑥(𝑡)|2 ∞ −∞ 𝑑𝑡 = ∫|𝑒10𝑡 𝑢(𝑡)|2 ∞ 0 𝑑𝑡 = ∫ 𝑒20𝑡 ∞ 0 𝑑𝑡 = 𝑒20𝑡 20 | 0 ∞ = ∞ so the signal has infinite energy. 𝑃 = lim 𝑇→∞ 1 𝑇 ∫|𝑥(𝑡)|2 𝑇 2 −𝑇 2 𝑑𝑡 = lim 𝑇→∞ 1 𝑇 ∫|𝑒10𝑡 𝑢(𝑡)|2 𝑇 0 𝑑𝑡 = lim 𝑇→∞ 1 20𝑇 𝑒20𝑡|0 𝑇 = 1 20 lim 𝑇→∞ 𝑒20𝑇 − 1 𝑇 = ∞ so the signal is neither energy nor power.
  7. 7. 7 4. 𝐸 = ∫|𝑥(𝑡)|2 ∞ −∞ 𝑑𝑡 = ∫|𝑒−2𝑡 𝑢(𝑡)|2 ∞ 0 𝑑𝑡 = ∫ 𝑒−4𝑡 ∞ 0 𝑑𝑡 = − 𝑒−4𝑡 4 | 0 ∞ = 1 4 so the signal has finite energy. 𝑃 = lim 𝑇→∞ 1 𝑇 ∫|𝑒−2𝑡 𝑢(𝑡)|2 𝑇 0 𝑑𝑡 = lim 𝑇→∞ − 1 4𝑇 𝑒−4𝑡|0 𝑇 = − 1 4 lim 𝑇→∞ 𝑒−4𝑇 − 1 𝑇 = 0 As average power comes out to be 0 so the signal is an energy signal. 2. Given the signal: 𝑥(𝑡) = 2 cos (6𝜋𝑡 − 𝜋 3 ) + 4 sin(10𝜋𝑡) a. Is it periodic? If so, find its period. b. Sketch its single-sided amplitude and phase spectra. c. Write it as the sum of rotating phasors plus their complex conjugates. d. Sketch its two-sided amplitude and phase spectra. e. Show that it is a power signal. SOLUTION (a) Let 𝑥1(𝑡) = 2 cos (𝜔1 𝑡 − 𝜋 3 ) and 𝑥2(𝑡) = 4 cos (𝜔2 𝑡 − 𝜋 2 ) So 𝑇1 = 2𝜋 𝜔1 = 2𝜋 6𝜋 = 1 3 , and 𝑇2 = 2𝜋 10𝜋 = 1 5 For the signal to be periodic, 𝑇1 𝑇2 = 𝑛2 𝑛1 be a rational number. So 𝑇1 𝑇2 = 1 3 1 5 = 5 3 Which is a rational number. So fundamental period of 𝑥(𝑡) is thus given by 𝑇𝑜 = 5𝑇2 = 3𝑇1 = 1 𝐻𝑧 (b) ,f Hz Amplitude 0 ( )Phaseshift rad ,f Hz0 (a)Single-sidedspectra 2 4 3 5 3 5 -π/3 -π/2
  8. 8. 8 (c) 𝑥(𝑡) = 𝑅𝑒 {𝑒 𝑗(6𝜋𝑡− 𝜋 3 ) } + 𝑅𝑒 {𝑒 −𝑗(6𝜋𝑡− 𝜋 3 ) } + 𝑅𝑒 {2𝑒 𝑗(10𝜋𝑡− 𝜋 2 ) } + 𝑅𝑒 {2𝑒 −𝑗(10𝜋𝑡− 𝜋 2 ) } (d) ,f Hz Amplitude 0 ( )Phaseshift rad ,f Hz0 (b)Double-sidedspectra 3 5 1 2 -3-5 12 3 5 -5 -3 -π/3 -π/2 π/2 π/3 (e) 𝐸 = ∫|𝑥(𝑡)|2 ∞ −∞ 𝑑𝑡 = ∫ [2cos (6𝜋𝑡 − 𝜋 3 ) + 4sin 10𝜋𝑡] 2 ∞ −∞ 𝑑𝑡 = ∫ 4cos2 (6𝜋𝑡 − 𝜋 3 ) + 16 sin2 10𝜋𝑡 + 16 cos (6𝜋𝑡 − 𝜋 3 ) sin 10𝜋𝑡) ∞ −∞ 𝑑𝑡 = ∞ The above integral results in infinity (similar to what has been done in previous questions). As the signal is periodic with period 1Hz, so power can be evaluated as 𝑃 = 1 𝑇𝑂 ∫ |𝑥(𝑡)|2 𝑇𝑜 0 𝑑𝑡 = ∫ [2cos (6𝜋𝑡 − 𝜋 3 ) + 4sin 10𝜋𝑡] 2 1 0 𝑑𝑡 = ∫ 4cos2 (6𝜋𝑡 − 𝜋 3 ) + 16 sin2 10𝜋𝑡 + 16 cos (6𝜋𝑡 − 𝜋 3 ) sin 10𝜋𝑡) 1 0 𝑑𝑡 So again taking one integral at a time, we have ∫ cos2 (6𝜋𝑡 − 𝜋 3 ) 1 0 𝑑𝑡 = 1 2 ∫ [1 − cos (12𝜋𝑡 − 2𝜋 3 )] 1 0 𝑑𝑡 = 1 2 Similarly ∫ (sin2 10𝜋𝑡) 1 0 𝑑𝑡 = 1 2 ∫[1 + cos 20𝜋𝑡] 1 0 𝑑𝑡 = 1 2
  9. 9. 9 and the third integral can be shown to be ∫ cos (6𝜋𝑡 − 𝜋 3 ) sin 10𝜋𝑡 1 0 𝑑𝑡 = 0 So 𝑃 = 1 𝑇𝑂 ∫ |𝑥(𝑡)|2 𝑇 𝑂 2 −𝑇 𝑂 2 𝑑𝑡 = 4 ( 1 2 ) + 16 ( 1 2 ) + 0 = 10 < ∞ so the signal is a power signal. Question # 3 Given the following signals: (1) 𝐬𝐢𝐧 𝟐𝒕 + 𝐜𝐨𝐬 𝝅𝒕 (2) 𝐬𝐢𝐧 𝟔𝝅𝒕 + 𝐜𝐨𝐬 𝟓𝝅𝒕 (3) 𝒆 𝟏𝟎𝒕 𝒖(𝒕) (4) 𝒆−𝟐𝒕 𝒖(𝒕) f. Which are periodic? Give their periods. g. Which are power signals? Find their average power. h. Which are energy signals? Find their energies. Question# 4 Given the signal: 𝒙(𝒕) = 𝟐 𝐜𝐨𝐬 (𝟔𝝅𝒕 − 𝝅 𝟑 ) + 𝟒 𝐬𝐢𝐧(𝟏𝟎𝝅𝒕) a. Is it periodic? If so, find its period. b. Sketch its single-sided amplitude and phase spectra. c. Write it as the sum of rotating phasors plus their complex conjugates. d. Sketch its two-sided amplitude and phase spectra. e. Show that it is a power signal. f. Let 𝑥1(𝑡) = 2 cos (𝜔1 𝑡 − 𝜋 3 ) and 𝑥2(𝑡) = 4 cos (𝜔2 𝑡 − 𝜋 2 ) So 𝑇1 = 2𝜋 𝜔1 = 2𝜋 6𝜋 = 1 3 , and 𝑇2 = 2𝜋 10𝜋 = 1 5 For the signal to be periodic, 𝑇1 𝑇2 = 𝑛2 𝑛1 be a rational number. So 𝑇1 𝑇2 = 1 3 1 5 = 5 3 Which is a rational number. So fundamental period of 𝑥(𝑡) is thus given by 𝑇𝑜 = 5𝑇2 = 3𝑇1 = 1 𝐻𝑧
  10. 10. 10 g. ,f Hz Amplitude 0 ( )Phaseshift rad ,f Hz0 (a)Single-sidedspectra 2 4 3 5 3 5 -π/3 -π/2 h. 𝑥(𝑡) = 𝑅𝑒 {𝑒 𝑗(6𝜋𝑡− 𝜋 3 ) } + 𝑅𝑒 {𝑒 −𝑗(6𝜋𝑡− 𝜋 3 ) } + 𝑅𝑒 {2𝑒 𝑗(10𝜋𝑡− 𝜋 2 ) } + 𝑅𝑒 {2𝑒 −𝑗(10𝜋𝑡− 𝜋 2 ) } i. ,f Hz Amplitude 0 ( )Phaseshift rad ,f Hz0 (b)Double-sidedspectra 3 5 1 2 -3-5 12 3 5 -5 -3 -π/3 -π/2 π/2 π/3 j. 𝐸 = ∫|𝑥(𝑡)|2 ∞ −∞ 𝑑𝑡 = ∫ [2cos (6𝜋𝑡 − 𝜋 3 ) + 4sin 10𝜋𝑡] 2 ∞ −∞ 𝑑𝑡 = ∫ 4cos2 (6𝜋𝑡 − 𝜋 3 ) + 16 sin2 10𝜋𝑡 + 16 cos (6𝜋𝑡 − 𝜋 3 ) sin 10𝜋𝑡) ∞ −∞ 𝑑𝑡 = ∞ The above integral results in infinity (similar to what has been done in previous questions). As the signal is periodic with period 1Hz, so power can be evaluated as 𝑃 = 1 𝑇𝑂 ∫ |𝑥(𝑡)|2 𝑇𝑜 0 𝑑𝑡 = ∫ [2cos (6𝜋𝑡 − 𝜋 3 ) + 4sin 10𝜋𝑡] 2 1 0 𝑑𝑡 = ∫ 4cos2 (6𝜋𝑡 − 𝜋 3 ) + 16 sin2 10𝜋𝑡 + 16 cos (6𝜋𝑡 − 𝜋 3 ) sin 10𝜋𝑡) 1 0 𝑑𝑡 So again taking one integral at a time, we have
  11. 11. 11 ∫ cos2 (6𝜋𝑡 − 𝜋 3 ) 1 0 𝑑𝑡 = 1 2 ∫ [1 − cos (12𝜋𝑡 − 2𝜋 3 )] 1 0 𝑑𝑡 = 1 2 Similarly ∫ (sin2 10𝜋𝑡) 1 0 𝑑𝑡 = 1 2 ∫[1 + cos 20𝜋𝑡] 1 0 𝑑𝑡 = 1 2 and the third integral can be shown to be ∫ cos (6𝜋𝑡 − 𝜋 3 ) sin 10𝜋𝑡 1 0 𝑑𝑡 = 0 So 𝑃 = 1 𝑇𝑂 ∫ |𝑥(𝑡)|2 𝑇 𝑂 2 −𝑇 𝑂 2 𝑑𝑡 = 4 ( 1 2 ) + 16 ( 1 2 ) + 0 = 10 < ∞ so the signal is a power signal. Question# 5 Show that complex exponential function 𝒙(𝒕) = 𝒆𝒋𝒘 𝟎 𝒕 is periodic and find its fundamental period.
  12. 12. 12 Question# 6 By definition And Both are sketched below.

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