Slide Tín hiệu và Hệ thống – Lesson 5 System exercises – Hoàng Gia Hưng – UET

Last lesson review ❑ Time domain representation of DT LTI systems ➢ Difference equation: characteristic polynomial, eigenvalues, characteristic modes, zero input response, zero state res[r]

ELT2035 Signals & Systems

Hoang Gia Hung Faculty of Electronics and Telecommunications University of Engineering and Technology, VNU Hanoi

Lesson 5: System exercises

Last lesson review

❑ Time domain representation of DT LTI systems

➢ Difference equation: characteristic polynomial, eigenvalues,

characteristic modes, zero input response, zero state response, natural response, forced response.

➢ Impulse response: convolution sum and properties

➢ Relationship between impulse response with LTI system properties

➢ Block diagram representation: system reduction

➢ State space representation: state variable, state space, state equations and output equation

❑ Today lesson: system exercises.

Exercise 1

a Consider a CT system whose input x(t) and output y(t) are related

by 𝑦 𝑡 = ׬𝜏=0𝑡+1𝑥 𝜏 𝑑𝜏 for t>0 Is the system memoryless? stable?

causal?

b Consider a DT system whose input and output are related by

𝑦 𝑛 = 3𝑥 𝑛 − 2 − 0.5𝑥 𝑛 + 𝑥 𝑛 + 1 Is the system memoryless? stable? causal?

c Consider 𝑦 𝑡 = cos 𝜔𝑐𝑡 𝑥(𝑡) Is the system memoryless? linear? time-invariant?

d Consider 𝑦 𝑛 = 2𝑛 + 1 𝑥 𝑛 Is the system memoryless? linear? time-invariant?

SOLUTION

a Dynamic, stable, noncausal

b Dynamic, stable, noncausal

c Memoryless, linear, time varying

d Memoryless, linear, time varying

Exercise 2

a Compute the impulse response of a system described by 𝑦 𝑛 =

2𝑥 𝑛 − 1 − 4𝑥[𝑛 − 3]

b Find the impulse response of a system specified by the equation

𝐷2 + 4𝐷 + 3 𝑦 𝑡 = 𝐷 + 5 𝑥(𝑡)

SOLUTION

a ℎ 𝑛 = 2𝛿 𝑛 − 1 − 4𝛿 𝑛 − 3 ⟹ ℎ[𝑛] = ቐ

2, 𝑛 = 1

−4, 𝑛 = 3

0, 𝑛 ∉ 1,3

b Characteristic equation: 𝜆2 + 4𝜆 + 3 = 0 ⟹ 𝜆1 = −1, 𝜆2 = −3

𝑦𝑛 𝑡 = 𝑐1𝑒−𝑡 + 𝑐2𝑒−3𝑡 Set 𝑦𝑛 0 = 0 and ሶ𝑦𝑛 0 = 1, we obtain 𝑐1 = 1

2, 𝑐2 = − 1

2 ⟹ 𝑦𝑛 𝑡 =

1

2 𝑒−𝑡 − 𝑒−3𝑡 Since bn=0, thus

ℎ 𝑡 = 𝑃 𝐷 𝑦𝑛 𝑡 𝑢 𝑡 = 𝐷 + 5 𝑦𝑛 𝑡 𝑢(𝑡)

Hence ℎ 𝑡 = 2𝑒−𝑡 − 𝑒−3𝑡 𝑢(𝑡)

Exercise 3

If 𝑓 𝑡 ∗ 𝑔 𝑡 = 𝑐(𝑡), then show that 𝑓 𝑎𝑡 ∗ 𝑔 𝑎𝑡 = 1

𝑎 𝑐(𝑎𝑡) with 𝑎 ≠ 0

SOLUTION

By definition:

𝑓 𝑎𝑡 ∗ 𝑔 𝑎𝑡 = ׬−∞∞ 𝑓 𝑎𝜏 𝑔 𝑎 𝑡 − 𝜏 𝑑𝜏 = ׬−∞∞ 𝑓 𝑎𝜏 𝑔(𝑎𝑡 − 𝑎𝜏)𝑑𝜏 (1) Perform the variable change: 𝑥 = 𝑎𝜏 ⟹ 𝑑𝑥 = 𝑎𝑑𝜏, (1) becomes

𝑓 𝑎𝑡 ∗ 𝑔(𝑎𝑡) =

1

𝑎න−∞

∞

𝑓(𝑥)𝑔(𝑎𝑡 − 𝑥)𝑑𝑥 , 𝑎 > 0 1

𝑎න∞

−∞

𝑓 𝑥 𝑔 𝑎𝑡 − 𝑥 𝑑𝑥 , 𝑎 < 0

Hence 𝑓 𝑎𝑡 ∗ 𝑔(𝑎𝑡) = 1

𝑎 𝑐(𝑎𝑡), 𝑎 ≠ 0

Note: This is the time-scaling property of convolution: if both signals are

time-scaled by a, their convolution is also time-scaled by a (and multiplied

by 1

𝑎 )

Exercise 4

Find [sin 𝑡 𝑢(𝑡)] ∗ 𝑢(𝑡)

SOLUTION

By definition:

[sin 𝑡 𝑢(𝑡)] ∗ 𝑢(𝑡) = ׬−∞𝑡 sin 𝜏 𝑢 𝜏 𝑢 𝑡 − 𝜏 𝑑𝜏 (1) Since 𝑢 𝜏 = 0 ∀𝜏 < 0, (1) becomes

[sin 𝑡 𝑢(𝑡)] ∗ 𝑢(𝑡) = ׬0𝑡sin 𝜏 𝑢 𝜏 𝑢 𝑡 − 𝜏 𝑑𝜏 𝑢(𝑡) (2) Because 𝑢 𝜏 𝑢 𝑡 − 𝜏 = 1 ∀𝜏 ∈ [0, 𝑡], (2) becomes

[sin 𝑡 𝑢(𝑡)] ∗ 𝑢(𝑡) = ׬0𝑡sin 𝜏 𝑑𝜏 𝑢(𝑡) = 1 − cos 𝑡 𝑢(𝑡)

Exercise 5

Find 𝑓 𝑘 ∗ 𝑔[𝑘], with 𝑓 𝑘 , 𝑔[𝑘] are depicted below

SOLUTION

Exercise 6

SOLUTION

Choose the state variables as: 𝑞1(𝑡)

𝑞2(𝑡) =

𝑦 (𝑡)

𝑑

𝑑𝑡𝑦 (𝑡) , we have:

൞

𝑑

𝑑𝑡𝑞1 𝑡 = 𝑞2 𝑡

𝑑

𝑑𝑡𝑞2 𝑡 = −26𝑞1 𝑡 − 9𝑞2 𝑡 + 24𝑥 𝑡

Hence, the state space representation of the system is:

ሶ𝒒 𝑡 = 0 1

−26 −9 𝒒 𝑡 +

0

24 𝑥 𝑡

𝑦 𝑡 = 1 0 𝒒 𝑡 + 0 𝑥(𝑡)

Find a state-space representation for the system:

𝑑2

𝑑𝑡2𝑦 𝑡 + 9 𝑑

𝑑𝑡𝑦 𝑡 + 26𝑦(𝑡) = 24𝑥(𝑡)

Exercise 7

SOLUTION

Exercise 8

SOLUTION

➢ The description with state variable 𝑞1 𝑡 , 𝑞2 𝑡 is 𝑨 = 𝛼1 0

0 𝛼2 , 𝒃 =

𝑏1

𝑏2 , 𝒄 =

𝑐1 𝑐2 , 𝐷 = [0].

➢ Similarity transformation 𝑻 = 1 −1

2 0 ⟹ 𝑻

−1 = 0 1 2Τ

−1 1 2 Τ .

➢ The new state variable description is: 𝑨′ = 𝑻𝑨𝑻−1 = 𝛼2

𝛼1−𝛼2 2

0 𝛼1 , 𝒃

′ = 𝑻𝒃 =

𝑏1 − 𝑏2

2𝑏1 , 𝒄′ = 𝒄𝑻−1 = −𝑐2

𝑐1+𝑐2

2 , 𝐷′ = 𝐷 = [0]

Consider the system in the given

block diagram If we define new

states as 𝑞1′ 𝑡 = 𝑞1 𝑡 − 𝑞2 𝑡 and

𝑞2′ 𝑡 = 2𝑞1 𝑡 , find the new state

variable description and draw its

corresponding block diagram

Exercise 8 (cont.)

SOLUTION (cont.)

➢ The corresponding block diagram is

𝛼2

𝛼1 0.5(𝛼1 − 𝛼2)