ECE 320 – Network Analysis II. Sprint 2018.
Instructor: Dr. Aly Farag
Test 2: Closed Book and No Calculator.
Test is four problems, equal weight. Answer on the given problems sheets – nothing else would be looked at nor graded.
Problem 1: Consider a continuous function 𝑥(𝑡), defined for 𝑡 ≥ 0. The Laplace Transform (LT) for 𝑥(𝑡) is defined as:
∞
𝑋(𝑠) = ∫0 𝑥(𝑡)𝑒 −𝑠𝑡 𝑑𝑡. Derive the following properties:
a)
LT(𝛿(𝑡)) = 1, the 𝛿(𝑡) is the Dirac-delta function … 5 points
∞
∞
∞
LT(𝛿(𝑡)) = ∫ 𝛿(𝑡)𝑒 −𝑠𝑡 𝑑𝑡 ≡ ∫ 𝛿(𝑡 − 0)𝑥(𝑡) 𝑑𝑡 ≡ ∫ 𝛿(𝑡 − 0)𝑥(𝑡) 𝑑𝑡 = 𝑒 −𝑠𝑡 |𝑡=0
0
0−
−∞
= 1; by the sifting property of the Delta function.
b) LT(𝑢(𝑡)) = 1/𝑠, where 𝑢(𝑡) is the unit-step function … 5 points
∞
1
1
LT(𝑢(𝑡)) = ∫ 𝑒 −𝑠𝑡 𝑑𝑡 = − 𝑒 −𝑠𝑡 |
=
𝑠
𝑠
𝑡=0
0
LT(𝑠𝑖𝑛(𝜔𝑡)𝑢(𝑡)) = 𝜔/(𝑠 2 + 𝜔2 ) … 5 points
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c)
By Euler Formula: 𝑠𝑖𝑛(𝜔𝑡) = 1/2𝑗(𝑒 𝑗𝜔𝑡 − 𝑒 −𝑗𝜔𝑡 )
∞
LT(𝑠𝑖𝑛(𝜔𝑡)) = ∫
0
2𝑗(𝑒 𝑗𝜔𝑡
1
1 ∞
𝑑𝑡 = ∫ (𝑒 −(𝑠−𝑗𝜔)𝑡 − 𝑒 −(𝑠+𝑗𝜔)𝑡 )𝑑𝑡
−𝑗𝜔𝑡
−𝑠𝑡
)𝑒
−𝑒
2𝑗 0
1
1
−(𝑠+𝑗𝜔)𝑡 𝑡=∞
= −2𝑗(𝑠−𝑗𝜔) 𝑒 −(𝑠−𝑗𝜔)𝑡 |𝑡=∞
|𝑡=0
𝑡=0 + 2𝑗(𝑠+𝑗𝜔) 𝑒
=
1
1
𝜔
−
= 2
2𝑗(𝑠 − 𝑗𝜔) 2𝑗(𝑠 + 𝑗𝜔) (𝑠 + 𝜔 2 )
d) LT(𝑥(𝑡 − 𝜏)𝑢(𝑡 − 𝜏)) = 𝑒 −𝑠𝜏 𝑋(𝑠), 𝜏 > 0 … 5 points
∞
∞
LT(𝑥(𝑡 − 𝜏)𝑢(𝑡 − 𝜏)) = ∫ 𝑥(𝑡 − 𝜏)𝑢(𝑡 − 𝜏)𝑒 −𝑠𝑡 𝑑𝑡 ≡ ∫ 𝑥(𝑡 − 𝜏)𝑢(𝑡 − 𝜏)𝑒 −𝑠𝑡 𝑑𝑡
∞
∞
Let 𝛾 = 𝑡 − 𝜏, hence, 𝑑𝑡 = 𝑑𝛾; 𝑡|𝜏 , 𝛾|0
0
𝜏
, therefore,
∞
∞
LT(𝑥(𝑡 − 𝜏)𝑢(𝑡 − 𝜏)) = ∫ 𝑥(𝛾)𝑒 −𝑠(𝛾+𝜏) 𝑑𝛾 = 𝑒 −𝑠𝜏 ∫ 𝑥(𝛾)𝑒 −𝑠𝛾 𝑑𝛾 = 𝑒 −𝑠𝜏 𝑋(𝑠).
0
e)
LT(𝑡𝑥(𝑡)) = −
𝑋(𝑠) =
𝑑
𝑑𝑠
0
𝑋(𝑠)… 5 points
∞
∫0 𝑥(𝑡)𝑒 −𝑠𝑡
𝑑𝑡; differentiating the two sides with respect to 𝑠; get
Hence,
Th
∞
𝑑𝑋(𝑠)
𝑑 ∞
𝑑
(𝑥(𝑡)𝑒 −𝑠𝑡 ) 𝑑𝑡; as the limits are deterministic … we can also use Libniz Rule.
=
∫ 𝑥(𝑡)𝑒 −𝑠𝑡 𝑑𝑡 ≡ ∫
𝑑𝑠
𝑑𝑠 0
0 𝑑𝑠
𝑑𝑋(𝑠)
𝑑𝑠
∞ 𝑑
= ∫0
𝑑𝑠
∞
(𝑥(𝑡)𝑒 −𝑠𝑡 ) 𝑑𝑡 = ∫0 −𝑡𝑥(𝑡)𝑒 −𝑠𝑡 𝑑𝑡.
sh
Therefore, by the definition of LT, it is evident that LT(𝑡𝑥(𝑡))
=−
𝑑
𝑑𝑠
𝑋(𝑠).
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Problem 2: Evaluate the Laplace transform (LT) of the following functions, using properties of the LT in problem 1.
a.
𝑥(𝑡) = (𝑡 − 1) sin(𝜔𝑡)𝑢(𝑡) … 12 points
From Problem 1(c): LT(𝑠𝑖𝑛(𝜔𝑡)𝑢(𝑡)) = 𝜔/(𝑠 2 + 𝜔2 )
From Problem 1(e): LT(𝑡𝑥(𝑡))
=−
𝑑
𝑑𝑠
𝑋(𝑠)
Therefore, 𝐿𝑇(𝑥(𝑡) = (𝑡 − 1) sin(𝜔𝑡)𝑢(𝑡)) = 𝐿𝑇(𝑡𝑠𝑖𝑛(𝜔𝑡)𝑢(𝑡) − 𝐿𝑇(sin(𝜔𝑡) 𝑢(𝑡)
𝑋(𝑠) = −
𝑑
𝜔
𝜔
𝜔(2𝑠)
𝜔
𝜔(2𝑠) + 𝜔𝑠 2 + 𝜔3
+
=
+
=
(𝑠 2 + 𝜔 2 )2
𝑑𝑠 (𝑠 2 + 𝜔 2 ) (𝑠 2 + 𝜔 2 ) (𝑠 2 + 𝜔 2 )2 (𝑠 2 + 𝜔 2 )
b.
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Note: If
𝑥(𝑡) = 𝑡 (𝑢(𝑡) − 𝑢(𝑡 − 5))… 13 points
From Problem 1(b): LT(𝑢(𝑡)) = 1/𝑠
From Problem 1(d): LT(𝑥(𝑡 − 𝜏)𝑢(𝑡 − 𝜏)) = 𝑒 −𝑠𝜏 𝑋(𝑠), 𝜏 > 0.
From Problem 1(e): LT(𝑡𝑥(𝑡))
𝑑
𝑑𝑠
𝑋(𝑠)
𝑑 1
𝑒 −5𝑠
1
−5𝑠𝑒 −5𝑠 −𝑒 −5𝑠
1−𝑒 −5𝑠 (1+5𝑠)
( −
)= 2+
=
𝑑𝑠 𝑠
𝑠
𝑠
𝑠2
𝑠2
sh
Th
Hence, 𝑋(𝑠) = −
=−
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Problem 3: The switch in the circuit shown has been is position a for a long time. At 𝑡 = 0, it moves instantaneously to position b.
(a) Derive the output voltage in the Laplace domain Vo (s). 15 points
(b) Find the output voltage in the Time domain vo (t). 10 points
180
; ii) If you stumbled in part (a), move on to part (b) and get the inverse LT.
𝑠 2 +5𝑠+4
Th
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Hint: i) The form will be Vo (s) =
sh
Note: you can also apply KVL as usual on three meshes; you’ll get the same final results.
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Problem 4: Consider the following simple RL circuit. Assume unit values for the inductance and the resistances, and the
input current 𝑖𝑖 (𝑡) = cos 𝑡 . Suppose the switch closes at 𝑡 = 0. Assume the inductor has no initial charge.
a) Evaluate the response 𝑣0 (𝑡) using the Laplace Transform Method. Note: LT(cos ω 𝑡) = 𝑠/(𝑠 2 + ω2 ) . 13 points
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At 𝑡 > 0, the equivalent circuit is:
KVL: 𝑣(𝑡) = 𝑖(𝑡) × (𝑅1 + 𝑅2 ) + 𝐿
𝑑𝑖(𝑡)
𝑑𝑡
With unit values for 𝑅1 , 𝑅2 and 𝐿; 𝑣(𝑡) =
𝑉0 (𝑠) = 𝐼(𝑠)𝑍3 =
𝑑𝑖(𝑡)
+
𝑑𝑡
2𝑖(𝑡); 𝑖(0) = 0 𝐴. The Laplace circuit will be:
̅̅̅2
𝑉(𝑠)
𝑠
𝑘1
𝑘2
𝑘
𝑍3 =
=
+
+
(𝑍1 + 𝑍2 +𝑍3 )
(𝑠 + 2)(𝑠 2 + 1) (𝑠 + 2) (𝑠 + 𝑗) (𝑠 − 𝑗)
0.4
= − (𝑠+2) +
0.5
√5(𝑠+𝑗)
𝑒 𝑗𝑡𝑎𝑛
The Inverse Laplace Transform is: 𝑣0 (𝑡) = (−0.4 𝑒 −2𝑡 +
1
√5
+
0.5
√5(𝑠−𝑗)
0.5 𝑗26.56° −𝑗𝑡
𝑒
𝑒
√5
𝑒 −𝑗𝑡𝑎𝑛
+
−1 (1)
2
1
; Note: 𝑡𝑎𝑛−1 (2) = 26.56°
0.5 −𝑗26.56° 𝑗𝑡
𝑒
𝑒 ) 𝑢(𝑡)
√5
cos(𝑡 + 90° − 26.56°) 𝑢(𝑡) Volts.
sh
Th
Which can be simplified to: 𝑣0 (𝑡) = −0.4 𝑒 −2𝑡 +
−1 (1)
2
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b) Repeat (a) using the Phasors method. 12 points.
KVL of the Phasor Circuit:
𝑉 = 𝐼(𝑍1 + 𝑍2 +𝑍3 ) = 𝐼(2 + 𝑗);
𝑉0 = 𝐼 × 𝑍3 = 𝑗𝐼;
𝑍3
1 +𝑍2 +𝑍3 )
Therefore, 𝑣0 (𝑡) =
1
√5
=
𝑗
2+𝑗
=
(1+2𝑗)
5
=
−1 (2)
𝑒𝑗𝑡𝑎𝑛
√5
=
1
√5
𝑒 𝑗63.434° ≡
1
√5
𝑒 𝑗(90°−26.56°)
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Hence, 𝑉0 = 𝑉 (𝑍
cos(𝑡 + 90° − 26.56°) 𝑢(𝑡) Volts.
sh
Th
Note: The Phasor solution provides only the steady state solution.
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