EE3210 Signal and System
Final Exam
Problem 1. Questions Answer
a) (i)
b) (ii)
c) (iv)
d) (i)
Problem 2.
Question Answer
a) (iii) b) (iv) c) (iv)
Problem 3. (a)
𝑑𝑦(𝑡) 𝑑𝑡
∞
+ 10𝑦(𝑡) = ∫ −∞
𝑥(𝜏)𝑧(𝑡 − 𝜏)𝑑𝜏 − 𝑥(𝑡)
𝑤h𝑒𝑟𝑒 𝑧(𝑡) = 𝑒−𝑡𝑢(𝑡) + 𝛿(𝑡)
𝑑𝑦(𝑡) + 10𝑦(𝑡) = 𝑥(𝑡)𝑧(𝑡) − 𝑥(𝑡) 𝑑𝑡
(𝑗2𝜋𝑓 + 10)𝒴(𝑓) = 𝒳(𝑓)[𝑧(𝑓) − 1]
H(𝑓)=𝒴(𝑓)= 𝑧(𝑓)−1 = 𝒳(𝑓) 𝑗2𝜋𝑓 + 10
1
(𝑗2𝜋𝑓 + 10)(𝑗2𝜋𝑓 + 1)
= 𝑐1 + 𝑐2
(b)
H(𝑓)=
1
(𝑗2𝜋𝑓 + 10)(𝑗2𝜋𝑓 + 1)
𝑗2𝜋𝑓 + 10
𝑗2𝜋𝑓 + 1
𝑐1 = (𝑗2𝜋𝑓 + 10)H(𝑓)|𝑗2𝜋𝑓=−10 = − 1 9
𝑐2 = (𝑗2𝜋𝑓 + 1)H(𝑓)|𝑗2𝜋𝑓=−1 = 1 9
1 1𝐹−1 1−𝑡−10𝑡 H(𝑓)=9(𝑗2𝜋𝑓+1)−9(𝑗2𝜋𝑓+10)⇒ h(𝑡)=9(𝑒 −𝑒 )𝑢[𝑡]
Problem 4. .
(a) The LT of the differential equation is (𝑠2 + 6𝑠 + 8)𝒴(𝑠) = 𝑠𝒳(𝑠)
H(𝑠)=𝒴(𝑠)= 𝑠 = c1 + 𝑐2 𝒳(𝑠) 𝑠2 +6𝑠+8 𝑠+4 𝑠+2
By using the LT table,
𝑐1 =(𝑠+4)H(𝑠)|𝑠=−4 =−4=2 −2
𝑐2 =(𝑠+2)H(𝑠)|𝑠=−2 =−2=−1 2
h(𝑡) = 2𝑒−4𝑡𝑢(𝑡) − 𝑒−2𝑡𝑢(𝑡)
𝐿𝐼 1 𝒴(𝑠)=H(𝑠)∙𝒳(𝑠) 𝑥(𝑡)=𝑢(𝑡)⇒𝒳(𝑠)=𝑠
𝒴(𝑠)= 1 = c1 + 𝑐2 (𝑠+4)(𝑠+2) 𝑠+4 𝑠+2
𝑐1 = (𝑠 + 4)H(𝑧)|𝑠=−4 = −1/2 𝑐2 = (𝑠 + 2)H(𝑧)|𝑠=−2 = 1/2
The step response of the given equation.
𝑦(𝑡) = − 1 𝑒−4𝑡𝑢(𝑡) + 1 𝑒−2𝑡𝑢(𝑡) 22
𝑅𝑂𝐶 𝑅𝑒(𝑠) > −2
(b)𝑦(𝑡)=𝑒𝑡[4+4∫𝑡𝑒−𝜏𝑦(𝜏)𝑑𝜏] 𝑡≥0 0
𝑑𝑦(𝑡) 𝑡
𝑑𝑡 = 𝑒𝑡 [4 + 4 ∫ 𝑒−𝜏𝑦(𝜏)𝑑𝜏] + 4𝑒𝑡(𝑒−𝑡𝑦(𝑡)) = 5𝑦(𝑡)
0
By obtaining the initial condition as 𝑦(0−) = 4
𝑠𝑌 (𝑠) − 𝑦(0−) = 4𝑌 (𝑠) 𝐼𝐼
𝑠𝑌 (𝑠) − 4 = 5𝑌 (𝑠) 𝐼𝐼
4 𝑠−5
After applying the inverse LT, 𝑦(𝑡) = 4𝑒5𝑡𝑢(𝑡).
𝑌 (𝑠) = 𝐼
Problem 5.
(a) Apply the time-shifting property,
(1 − 3 𝑍−1 + 1 𝑍−2) 𝒴(𝑧) = 𝒳(𝑧) 48
𝒴(𝑧) 1 𝑍2 1 H(𝑧) = 𝒳(𝑧) = 3 1 = 1 1 𝑅𝑂𝐶, |𝑍| > 2
1−4𝑍−1 +8𝑍−2 (𝑍−4)(𝑍−2) Let’s represent H(𝑧) in the following form,
H(𝑧)=
Based on the Z-transform table,
1 = 𝑐1 + 𝑐2
1−3𝑍−1 +1𝑍−2 1−1𝑍−1 1−1𝑍−1 4842
𝑐1 =(1−1𝑍−1)H(𝑧)| −1 =−1
4 𝑍=1 4
𝑐2 =(1−1𝑍−1)H(𝑧)| −1
=2
2 𝑍=1 2
1𝑛1𝑛1𝑛1𝑛 h[𝑛]=2∙(2) 𝑢[𝑛]−(4) 𝑢[𝑛]=[2∙(2) −(4) ]𝑢[𝑛]
(b) Since the |𝑍| = 1/2 , which contain the |𝑍| = 1, so, it is a stable system.
In addition, we can find that the 𝑅𝑂𝐶 > 1/2, which means that it is a casual system.
(c) Since 𝑢[𝑛]↔ 1 , |𝑍|>1 1−𝑍−1
𝒴(𝑧)=
𝒴(𝑧) = H(𝑧)𝒳(𝑧) 𝑤h𝑒𝑟𝑒 𝒳(𝑧) = 𝑍(𝑢[𝑛])
1 = 𝑐1 + 𝑐2 + 𝑐3
(1−1𝑍−1)(1−1𝑍−1)(1−𝑍−1) 1−1𝑍−1 1−1𝑍−1 1−𝑍−1 42 42
𝑐1 =(1−1𝑍−1)𝒴(𝑧)| −1 =1 4 𝑍=13
418
= (1 − 1 𝑍−1) 𝒴(𝑧)| = −2 𝒴(𝑧) = 3 + −2 + 3
2 𝑍−1=1 1−1𝑍−1 1−1𝑍−1 1−𝑍−1
𝑐 2
Using the z-transform table,
11𝑛1𝑛8 𝑦[𝑛]=[3(4) −2(2) +3]𝑢[𝑛]
242
𝑐3 = (1 − 𝑍−1)𝒴(𝑧)|𝑍−1=1 = 8 3