[1] (17분) Given 𝑥𝑥 𝑡𝑡 = 2∏ 𝑡𝑡 sin 𝜋𝜋𝑡𝑡, answer the following questions.
(∏ 𝑡𝑡 : The rectangular pulse function with the duration of 1 sec.)
(주의: 주요 지점의 수치들이 기재되지 않으면 감점함. )
(a) (10점) Sketch 𝑥𝑥 𝑡𝑡
(b) (10점) Sketch 𝑥𝑥 −2𝑡𝑡 +
1
2
(c) (10점) Evaluate 𝑥𝑥 𝑛𝑛 = 𝑥𝑥(𝑡𝑡)|𝑡𝑡=𝑛𝑛/8 . [수치로 나열하세요.]
(d) (10점) Evaluate the energy of 𝑥𝑥 𝑡𝑡 . [과정 2단계 이상 기재.]
2
(e) (10점) ∫−2(1 + 𝑥𝑥(𝑡𝑡))𝛿𝛿 𝑡𝑡 +
𝜋𝜋
2
𝑑𝑑𝑑𝑑. [과정 2단계 이상 기재.]
[1] (17분) Given 𝑥𝑥 𝑡𝑡 = 2∏ 𝑡𝑡 sin 𝜋𝜋𝑡𝑡, answer the following questions.
(∏ 𝑡𝑡 : The rectangular pulse function with the duration of 1 sec.)
(주의: 주요 지점의 수치들이 기재되지 않으면 감점함. )
(a) (10점) Sketch 𝑥𝑥 𝑡𝑡
𝑥𝑥 𝑡𝑡
Sol)
+2
+2
+2
+2
+2
𝑡𝑡
[1] (17분) Given 𝑥𝑥 𝑡𝑡 = 2∏ 𝑡𝑡 sin 𝜋𝜋𝑡𝑡, answer the following questions.
(∏ 𝑡𝑡 : The rectangular pulse function with the duration of 1 sec.)
(주의: 주요 지점의 수치들이 기재되지 않으면 감점함. )
(b) (10점) Sketch 𝑦𝑦(𝑡𝑡) = 𝑥𝑥 −2𝑡𝑡 +
Sol)
𝑦𝑦(𝑡𝑡)
1
2
+2
+2
+2
𝑡𝑡
+2
+2
[1] (17분) Given 𝑥𝑥 𝑡𝑡 = 2∏ 𝑡𝑡 sin 𝜋𝜋𝑡𝑡, answer the following questions.
(∏ 𝑡𝑡 : The rectangular pulse function with the duration of 1 sec.)
(주의: 주요 지점의 수치들이 기재되지 않으면 감점함. )
(c) (10점) Evaluate 𝑥𝑥 𝑛𝑛 = 𝑥𝑥(𝑡𝑡)|𝑡𝑡=𝑛𝑛/8 . [수치로 나열하세요.]
Sol)
𝑛𝑛
𝑥𝑥 𝑛𝑛
+1
+1
+1
+1
+1
+1
+1
+1
+1
+1
-5
-4
-3
-2
-1
0
1
2
3
4
5
0
-2
-sin 3𝜋𝜋
8
- 2
-sin 𝜋𝜋8
0
sin 𝜋𝜋8
2
sin 3𝜋𝜋
8
0
0
(d) (10점) Evaluate the energy of 𝑥𝑥 𝑡𝑡 . [과정 2단계 이상 기재.]
Sol)
∞
𝐸𝐸∞ = � |𝑥𝑥 𝑡𝑡
−∞
|2 𝑑𝑑𝑑𝑑
1/2
=�
4sin2 (𝜋𝜋𝑡𝑡) 𝑑𝑑𝑑𝑑
−1/2
+5
1/2
= 2�
[1 − cos 2𝜋𝜋𝑡𝑡] 𝑑𝑑𝑑𝑑 = 2
−1/2
+3
+2
[1] (17분) Given 𝑥𝑥 𝑡𝑡 = 2∏ 𝑡𝑡 sin 𝜋𝜋𝑡𝑡, answer the following questions.
(∏ 𝑡𝑡 : The rectangular pulse function with the duration of 1 sec.)
(주의: 주요 지점의 수치들이 기재되지 않으면 감점함. )
2
(e) (10점) ∫−2(1 + 𝑥𝑥(𝑡𝑡))𝛿𝛿 𝑡𝑡 +
Sol)
𝜋𝜋
2
𝑑𝑑𝑑𝑑. [과정 2단계 이상 기재.]
𝜋𝜋
2
The argument 𝑡𝑡 + = 0. ⇒ 𝑡𝑡 = −
𝜋𝜋
2
+4
𝜋𝜋
2
The integration interval includes the time instant 𝑡𝑡 = − .
1 + 𝑥𝑥 𝑡𝑡 |𝑡𝑡=−𝜋𝜋 = 1 + 𝑥𝑥 −
2
+2
𝜋𝜋
2
+2
= 1 (∵ 𝑥𝑥 − 𝜋𝜋⁄2 = 0 from (a))
+2
[2] (17분) Given a system such as
𝑑𝑑 2 𝑦𝑦(𝑡𝑡)
+
𝑑𝑑𝑑𝑑 2
3
𝑑𝑑𝑑𝑑(𝑡𝑡)
+
𝑑𝑑𝑑𝑑
2𝑦𝑦 𝑡𝑡 = 𝑥𝑥 2 (𝑡𝑡).
Determine if it is linear, time-invariant, causal, and/or memoryless. Give appropriate
proofs. [Note: 𝑦𝑦 𝑡𝑡 = 𝑓𝑓 𝑥𝑥 𝑡𝑡 , 𝑓𝑓 � : system function]
(주의: 적절한 증명 또는 근거가 없는 경우, 감점함. )
- Linearity (26점)
- Time Invariance (18점)
- Causality (3점)
- Memoryless (3점)
𝑑𝑑 2 𝑦𝑦(𝑡𝑡)
+
𝑑𝑑𝑑𝑑 2
[2] (17분) Given a system such as
3
𝑑𝑑𝑑𝑑(𝑡𝑡)
+
𝑑𝑑𝑑𝑑
2𝑦𝑦 𝑡𝑡 = 𝑥𝑥 2 (𝑡𝑡). Determine if it
is linear, time-invariant, causal, and/or memoryless. Give appropriate proofs.
[Note: 𝑦𝑦 𝑡𝑡 = 𝑓𝑓 𝑥𝑥 𝑡𝑡 , 𝑓𝑓 � : system function]
(주의: 적절한 증명 또는 근거가 없는 경우, 감점함. )
Sol)
- Linearity (26점)
𝑑𝑑2 𝑦𝑦1 (𝑡𝑡)
𝑑𝑑𝑦𝑦 (𝑡𝑡)
+3 1
2
𝑑𝑑𝑑𝑑
𝑑𝑑𝑑𝑑
𝑑𝑑2 𝑦𝑦1 (𝑡𝑡)
𝑎𝑎1
+
𝑑𝑑𝑑𝑑 2
𝑑𝑑2 𝑦𝑦1 (𝑡𝑡)
𝑎𝑎1
+
𝑑𝑑𝑑𝑑 2
⇎
+2
+ 2𝑦𝑦1 (𝑡𝑡) =
𝑑𝑑𝑦𝑦 (𝑡𝑡)
3 1 +
𝑑𝑑𝑑𝑑
𝑑𝑑𝑦𝑦 (𝑡𝑡)
3 1 +
𝑑𝑑𝑑𝑑
𝑑𝑑2 𝑎𝑎1 𝑦𝑦1 𝑡𝑡 +𝑎𝑎2 𝑦𝑦2 𝑡𝑡
𝑑𝑑𝑑𝑑 2
+2
+2
+
𝑥𝑥12 (𝑡𝑡)
+2
2𝑦𝑦1 (𝑡𝑡) =
2𝑦𝑦1 (𝑡𝑡) +
→
𝑎𝑎1 𝑥𝑥12 (𝑡𝑡),
𝑑𝑑𝑦𝑦 (𝑡𝑡)
+3 2
𝑑𝑑𝑑𝑑
+3
+2
+2
+ 2𝑦𝑦2 (𝑡𝑡) = 𝑥𝑥22 (𝑡𝑡) → 𝑦𝑦2 (𝑡𝑡)
𝑑𝑑𝑦𝑦 (𝑡𝑡)
+3 2 +
𝑑𝑑𝑑𝑑
+2
=
+ 2 𝑎𝑎1 𝑦𝑦1 𝑡𝑡 + 𝑎𝑎2 𝑦𝑦2 𝑡𝑡
𝑎𝑎1 𝑥𝑥1 𝑡𝑡 + 𝑎𝑎2 𝑥𝑥2 (𝑡𝑡) ⇏ 𝑎𝑎1 𝑦𝑦1 𝑡𝑡 + 𝑎𝑎2 𝑦𝑦2 𝑡𝑡
∴ Non-linear (or signal multiplication term.)
+2
2𝑦𝑦2 (𝑡𝑡) = 𝑎𝑎2 𝑥𝑥22 (𝑡𝑡)
+2
+ 2𝑦𝑦2 (𝑡𝑡) =
+ 2 𝑎𝑎1 𝑦𝑦1 𝑡𝑡 + 𝑎𝑎2 𝑦𝑦2 𝑡𝑡
𝑑𝑑 𝑎𝑎 𝑦𝑦 𝑡𝑡 +𝑎𝑎2 𝑦𝑦2 𝑡𝑡
+3 1 1
𝑑𝑑𝑑𝑑
+2
𝑑𝑑2 𝑦𝑦2 (𝑡𝑡)
𝑎𝑎2
𝑑𝑑𝑑𝑑 2
𝑑𝑑2 𝑦𝑦2 (𝑡𝑡)
𝑎𝑎2
𝑑𝑑𝑑𝑑 2
𝑑𝑑 𝑎𝑎 𝑦𝑦 𝑡𝑡 +𝑎𝑎2 𝑦𝑦2 𝑡𝑡
3 1 1
𝑑𝑑𝑑𝑑
𝑑𝑑2 𝑎𝑎1 𝑦𝑦1 𝑡𝑡 +𝑎𝑎2 𝑦𝑦2 𝑡𝑡
𝑑𝑑𝑑𝑑 2
𝑑𝑑2 𝑦𝑦2 (𝑡𝑡)
𝑑𝑑𝑦𝑦 (𝑡𝑡)
𝑦𝑦1 (𝑡𝑡),
+3 2
2
𝑑𝑑𝑑𝑑
𝑑𝑑𝑑𝑑
𝑎𝑎1 𝑥𝑥12 (𝑡𝑡)+𝑎𝑎2 𝑥𝑥22 (𝑡𝑡)
+1
𝑎𝑎1 𝑥𝑥12 (𝑡𝑡)+𝑎𝑎2 𝑥𝑥22 (𝑡𝑡)
+2
= [𝑎𝑎1 𝑥𝑥1 𝑡𝑡 + 𝑎𝑎2 𝑥𝑥2 𝑡𝑡 ]2
[2] (17분) Given a system such as
𝑑𝑑 2 𝑦𝑦(𝑡𝑡)
+
𝑑𝑑𝑑𝑑 2
3
𝑑𝑑𝑑𝑑(𝑡𝑡)
+
𝑑𝑑𝑑𝑑
2𝑦𝑦 𝑡𝑡 = 𝑥𝑥 2 (𝑡𝑡). Determine if it
is linear, time-invariant, causal, and/or memoryless. Give appropriate proofs.
[Note: 𝑦𝑦 𝑡𝑡 = 𝑓𝑓 𝑥𝑥 𝑡𝑡 , 𝑓𝑓 � : system function]
- Time Invariance (18점)
+2
+2
+2
𝑥𝑥(𝑡𝑡) → 𝑦𝑦(𝑡𝑡)
𝑥𝑥 𝑡𝑡 − 𝑡𝑡0 → 𝑦𝑦 𝑡𝑡 − 𝑡𝑡0 ?
𝑑𝑑2 𝑦𝑦(𝑡𝑡−𝑡𝑡0 )
+
𝑑𝑑𝑑𝑑 2
𝑑𝑑2 𝑦𝑦𝑡𝑡0 (𝑡𝑡)
𝑑𝑑𝑑𝑑 2
+2Let τ
+2
+3
3
𝑑𝑑𝑑𝑑(𝑡𝑡−𝑡𝑡0 )
+
𝑑𝑑𝑑𝑑
𝑑𝑑𝑦𝑦𝑡𝑡0 (𝑡𝑡)
𝑑𝑑𝑑𝑑
2𝑦𝑦(𝑡𝑡 − 𝑡𝑡0 ) = 𝑥𝑥 2 𝑡𝑡 − 𝑡𝑡0 ?
+ 2𝑦𝑦𝑡𝑡0 (𝑡𝑡) = 𝑥𝑥 2 𝑡𝑡 − 𝑡𝑡0
= 𝑡𝑡 − 𝑡𝑡0 & 𝑑𝑑𝑑𝑑 = 𝑑𝑑𝑑𝑑 & 𝑑𝑑𝑑𝑑 2 = 𝑑𝑑𝑑𝑑 2 .
𝑑𝑑2 𝑦𝑦𝑡𝑡0 (𝜏𝜏+𝑡𝑡0 )
𝑑𝑑𝑑𝑑2
+3
𝑑𝑑𝑦𝑦𝑡𝑡0 (𝜏𝜏+𝑡𝑡0 )
𝑑𝑑𝑑𝑑
+ 2𝑦𝑦𝑡𝑡0 (𝜏𝜏 + 𝑡𝑡0 ) = 𝑥𝑥 2 𝜏𝜏
+2
+2 𝑦𝑦𝑡𝑡0 𝜏𝜏 + 𝑡𝑡0 = 𝑦𝑦 𝜏𝜏 → 𝑦𝑦𝑡𝑡0 𝑡𝑡 = 𝑦𝑦 𝑡𝑡 − 𝑡𝑡0
+2
𝑥𝑥 𝑡𝑡 − 𝑡𝑡0 → 𝑦𝑦 𝑡𝑡 − 𝑡𝑡0
+2 ∴ Time-invariant (or
All coefficients are constants.)
두 가지는 같은 개념이므로 둘 중
하나만 인정함.
[2] (17분) Given a system such as
𝑑𝑑 2 𝑦𝑦(𝑡𝑡)
+
𝑑𝑑𝑑𝑑 2
3
𝑑𝑑𝑑𝑑(𝑡𝑡)
+
𝑑𝑑𝑑𝑑
2𝑦𝑦 𝑡𝑡 = 𝑥𝑥 2 (𝑡𝑡). Determine if it
is linear, time-invariant, causal, and/or memoryless. Give appropriate proofs.
[Note: 𝑦𝑦 𝑡𝑡 = 𝑓𝑓 𝑥𝑥 𝑡𝑡 , 𝑓𝑓 � : system function]
- Causality (3점)
The output is determined by only the current input. ∴ Causal
+2
- Memoryless (3점)
𝑑𝑑2 𝑦𝑦(𝑡𝑡) 𝑑𝑑𝑑𝑑(𝑡𝑡)
,
𝑑𝑑𝑑𝑑 2
𝑑𝑑𝑑𝑑
terms ∴ With memory
+2
+1
+1
[3] (9분) Given 𝑧𝑧1 = −2𝑒𝑒 −𝑗𝑗𝜋𝜋/3 and 𝑧𝑧2 = 2𝑒𝑒 −𝑗𝑗𝜋𝜋/4 , evaluate the following
operations and represent them in Euler’s form (i.e., 𝑟𝑟𝑒𝑒 𝑗𝑗𝜃𝜃 , with 𝑟𝑟 > 0 and
−π ≤ 𝜃𝜃 < 𝜋𝜋). (𝑧𝑧�1 : conjugate of 𝑧𝑧1 )
(주의: 최소한의 풀이과정이 기재되지 않으면 정답으로 인정하지 않음.
Rectangular forms of 𝑧𝑧1 , 𝑧𝑧2 , 𝑧𝑧�1 , and 𝑧𝑧�1 + 𝑧𝑧2 , and at least two steps (최소 2 단계) for
𝑧𝑧
evaluating 𝑧𝑧2 must be given.)
1
(a) (20점) 𝑧𝑧�1 + 𝑧𝑧2
(b) (10점)
𝑧𝑧2
𝑧𝑧1
[3] (9분) Given 𝑧𝑧1 = −2𝑒𝑒 −𝑗𝑗𝜋𝜋/3 and 𝑧𝑧2 = 2𝑒𝑒 −𝑗𝑗𝜋𝜋/4 , evaluate the following
operations and represent them in Euler’s form (i.e., 𝑟𝑟𝑒𝑒 𝑗𝑗𝜃𝜃 , with 𝑟𝑟 > 0 and
−π ≤ 𝜃𝜃 < 𝜋𝜋). (𝑧𝑧�1 : conjugate of 𝑧𝑧1 )
(a) (20점) 𝑧𝑧�1 + 𝑧𝑧2 (𝑧𝑧�1 : conjugate of 𝑧𝑧1 )
Sol)
+4
𝑧𝑧1 = −2 cos −
+4 𝑧𝑧2
+2 𝑧𝑧�1
= 2 cos −
= −1 − 𝑗𝑗 3
𝜋𝜋
3
𝜋𝜋
4
+ 𝑗𝑗 sin −
+ 𝑗𝑗 sin −
𝜋𝜋
3
𝜋𝜋
4
= −2 cos
= 2 cos
+4
𝜋𝜋
3
𝜋𝜋
4
− 𝑗𝑗 sin
− 𝑗𝑗 sin
𝜋𝜋
3
𝜋𝜋
4
= −1 + 𝑗𝑗 3
= 1 − 𝑗𝑗
+3
𝑧𝑧�1 + 𝑧𝑧2 = −1 − 𝑗𝑗 3 + 1 − 𝑗𝑗 = −𝑗𝑗 1 + 3 = 1 + 3 𝑒𝑒
+3
𝜋𝜋
−𝑗𝑗
2
[3] (9분) Given 𝑧𝑧1 = −2𝑒𝑒 −𝑗𝑗𝜋𝜋/3 and 𝑧𝑧2 = 2𝑒𝑒 −𝑗𝑗𝜋𝜋/4 , evaluate the following
operations and represent them in Euler’s form (i.e., 𝑟𝑟𝑒𝑒 𝑗𝑗𝜃𝜃 , with 𝑟𝑟 > 0 and
−π ≤ 𝜃𝜃 < 𝜋𝜋). (𝑧𝑧�1 : conjugate of 𝑧𝑧1 )
(b) (10점)
Sol)
𝑧𝑧2
𝑧𝑧1
+4
𝑧𝑧2
𝑧𝑧1
=
2𝑒𝑒
𝜋𝜋
−𝑗𝑗 4
−2𝑒𝑒
𝜋𝜋
−𝑗𝑗 3
=−
1
𝑒𝑒
2
𝑗𝑗𝜋𝜋 𝑗𝑗𝜋𝜋
−4 3
𝑒𝑒
=−
1
𝑒𝑒
2
𝜋𝜋
𝑗𝑗 12
=
+3
+3
1 −𝑗𝑗 11𝜋𝜋
𝑒𝑒 12
2