Engineering Chapter 11 Homework Clearly the characteristic polynomial of the system is

c) Taking the Laplace transform with zero initial conditions, we obtain

14. If an LTIC system has the transfer function ˆ

H(s) = ˆ

Y(s)

ˆ

F(s)=s+1

(s+2)2determine a

linear ODE that describes the relationship between the system input f(t)and the

output y(t).

Solution:

From the system transfer function, we have

15. Determine the characteristic polynomial P(s), characteristic poles, characteristic

modes, and the zero-input solution for each of the LTIC systems described below.

a) d2y

dt2+ 2dy

dt −8y(t) = 6f(t),y(0−) = 0, y′(0−) = 1

b) d3y

dt3+ 2d2y

dt2−dy

dt −2y(t) = f(t),y(0−) = 1, y′(0−) = 1, y′′(0−) = 0

c) d2y

dt2+ 2dy

dt +y(t) = 2f(t),y(0−) = 1, y′(0−) = 1.

Solution:

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a) Clearly the characteristic polynomial of the system is

b) The characteristic polynomial is

20

Applying the initial conditions:

c) The characteristic polynomial is

16.

21

a) Take the Laplace transform of the following ODE to determine ˆ

Y(s)assuming

f(t) = u(t),y(0−) = 1, and y′(0−) = 0. Determine y(t)for t > 0by taking

the inverse Laplace transform of ˆ

Y(s).

d2y

dt2+ 5dy

dt + 4y(t) = df

dt + 2f(t).

b) Repeat (a) for y(0−) = 0 and

3dy

dt + 6y(t) = δ(t).

c) Repeat (a) for y(0−) = 0 and

dy

dt −y(t) = e−tu(t).

Solution:

a) Taking the Laplace transform gives

b) Taking the Laplace transform gives

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c) Taking the Laplace transform, we obtain

17. The transfer function of a particular LTIC system is ˆ

H(s) = ˆ

Y(s)

ˆ

F(s)=4

s−2. Is the

system asymptotically stable? Explain. Is the system BIBO stable? Explain.

Solution:

18. What are the resonance frequencies in a system with the transfer function

ˆ

H(s) = s

(s+ 1)(s2+ 4)(s2+ 25)?

Is the system marginally stable? BIBO-stable? Explain.

Solution:

19. Determine the zero-state response y(t) = h(t)∗f(t)of the marginally stable system

ˆ

H(s) = 1

(s2+ 4)(s2+ 9)

to an input f(t) = cos(2t)u(t).

Solution:

23

Then the output is

20. a) Determine the transfer function and characteristic modes of the circuit shown

below assuming that Ca=Cb= 1/2F, R= 1.5 Ω, and L= 1/2H.

+

–

f(t)

y(t)

Ca

Cb

L

R

i(t)

v(t)

b) Given that v(0−) = 1 V and i(0−) = 0.5A,and using the element values given

in part (a), determine y(t)for t > 0in the circuit:

f(t) = 0

y(t)

Ca

Cb

L

R

i(t)

v(t)

Solution:

a) The equivalent circuit in the s-domain is

24

Simplifying, we obtain

b) We already have the zero-state transfer function. Now we want to calculate

the zero-input response, which is a linear combination of the characteristic

21. Consider the following circuit, which is in DC steady-state until the switch is opened

at t= 0,

+

–

t= 0

4V 1F 2H

2Ω 2Ω

v(t)

i(t)

+

–

25

a) Determine i(0−)and v(0−).

b) Determine the characteristic modes of the circuit to the right of the switch for

t > 0.

c) Determine i(t)for t > 0.

Solution:

a) Before the switch opens, we have

2Ω 2Ω

i(t)

b) After the switch closes, we have

2Ω

ˆ

I(s)

c) The zero-input response is a linear combination of the characteristic modes,

then

26

Applying initial conditions, we have

22. In the circuit:

+

–

t= 0

1H

f(t)y(t)

+

–

1Ω

1F

i(t)

a) Determine the transfer function ˆ

H(s) = ˆ

Y(s)

ˆ

F(s)for t > 0.

b) Determine the zero-state response for t > 0if f(t) = e3t.

c) Determine the zero-input response for t > 0if y(0−) = 1 V and i(0−) = 0.

Solution:

a) After the switch closes, we have the following equivalent circuit:

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b) We are given the causal input

Hence, we have

ˆ

3

13

13 s+1

13

3

13

13 s+1

2+5/2

13

Finally, taking the inverse Laplace transform, we obtain the zero-state re-

sponse

s+s

28

Modifying before taking the inverse Laplace transform,

Alternatively, we could form the zero-input response as a linear combination

of the characteristic modes. Since, for applying the initial conditions, we need

i(t), we can write

Then from the voltage-current relation of the inductor, we have

23. Consider the circuit:

+

–1H

f(t)

2Ω 1F

t= 0

2Ω

y(t)

+

–

i(t)

a) Show that the transfer function of the circuit for t > 0is ˆ

H(s) = ˆ

Y(s)

ˆ

F(s)=

s

4s2+5s+2 .

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b) What are the characteristic modes of the circuit?

c) Determine y(t)for t > 0if f(t) = 1 V, y(0−) = 1 V, and i(0−) = 0.

Solution:

a) The s-domain equivalent circuit is

c) The s-domain equivalent including initial-value sources is

+

–s

ˆ

F(s) = 1

s

2Ω 1

s

2Ω ˆ

Y(s)

+

–Cv(0−) = 1A

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24. The system shown below can be implemented as a cascade of two 1st-order systems

ˆ

H1(s)and ˆ

H2(s). Identify the possible forms of ˆ

H1(s)and ˆ

H2(s).

Σ

ˆ

F(s)ˆ

Y(s)

1

s+ 2

2

s

s+ 1

Solution:

From the graph, we have

25. Determine the impulse response h(t)of the system shown below. Also determine

whether the system is BIBO stable.

31

Σ

ˆ

F(s)ˆ

Y(s)

s

s+ 1

1

s+ 1

Σ

Solution:

Labeling the output of the summations

s

s+ 1

We can write the output ˆ

Q(s)of the lower-left summation as

ˆ

Q(s) = ˆ

F(s)1 + s

s+ 1=2s+ 1

s+ 1 ˆ

F(s).

32

26. Determine the transfer function ˆ

H(s)of the system shown below. Also determine

whether the system is BIBO stable.

Σ

ˆ

F(s)ˆ

Y(s)

1

s+ 1

2

s

s+ 1

Solution:

Let the output of the sum be

27. Determine the transfer function ˆ

H(s)of the system shown below and determine

for which Kthe system is BIBO stable if

a) ˆ

Hf(s) = 1

s+K.

b) ˆ

Hf(s) = s+K.

Σ

Y(s)

1

s+ 1

Hf(s)

F(s)

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Solution:

We can write the output as

ˆ

Y(s) = 1

s+ 1 ˆ

F(s) + ˆ

Hf(s)ˆ

Y(s)

ˆ

Y(s)h1−ˆ

Hf(s)i=1

s+ 1 ˆ

F(s).

Consequently, the general transfer function can be written as

b) For ˆ

Hf(s) = s+K, we have the following transfer function

ˆ

H(s) = 1

(s+ 1) (1 −s−K).

28. Consider a system with transfer function ˆ

H(s) = 2

s+1−j3.Draw a block diagram

that implements this transfer function. Individual blocks in the diagram may de-

note addition of real-valued signals, ampliﬁcation by real values, and integration of

real-valued signals. Your diagram should contain no complex numbers. Hints: 1)

The transfer function ˆ

H(s) = a

s+bcorresponds in the time domain to the equation

dy

dt +by(t) = af (t)or, equivalently, y(t) = −bRy(τ)dτ +aRf(τ)dτ. So, it is easy

to draw a block diagram of this system. 2) In this homework exercise, you may

assume that f(t)is real-valued, but y(t)is complex-valued, which means that y(t)

34

is a pair of real-valued signals. Your diagram will need to show yR(t)and yI(t),

the real and imaginary parts of y(t),separately. 3) Multiplication of a complex sig-

nal by a complex number involves four real multiplications. Addition of complex

signals requires two real additions.

Solution:

From the transfer function we can obtain the following relation

ˆ

Y(s) [s+ 1 −j3] = 2 ˆ

F(s),

With this two expressions, we can build the following block diagram

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following diagram:

ΣyR(t)

fR(t)1

s

2

36