10.6 PRACTICE PROBLEM SOLUTIONS

y( D)

( D + 10) ( D + 5)

=

( D + 10)

-----------

= ---------------------------------------

--------------------

x(

D)

(

D + 5)

2

( D + 5)

6dB

0dB

0.8

1.6

90

0.8

1.6

-90

x( t)

= 20 cos ( 9t + 0.3)

y

=

( D + 10)

=

9j + 10

5 – 9j

=

50 + 81 – 45j

= 1.236 – 0.425j

-

--------------------

----------------

-------------

-------------------------------

x

( D + 5)

9j + 5

5 – 9j

25 + 81

y

=

1.236

2

+ 0.425

2

atan

–0.425

= 1.307 –0.3312rad

-

----------------

x

1.236

y

x- = 2.33dB –9.49°

y( t)

= 20( 1.307) sin ( 9t + 0.3 + ( –0.3312) )

y( t)

= 26.1 sin ( 9t – 0.031)

Aside: This can also be done entirely with phasors in cartesian notation

y

=

( D + 10)

=

9j + 10

5 – 9j

=

50 + 81 – 45j

1.236 – 0.425j

-

--------------------

----------------

-------------

------------------------------- =

x

( D + 5)

9j + 5

5 – 9j

25 + 81

x

=

20( cos ( 0.3rad)

+ j sin ( 0.3rad) )

=

19.1 + 5.91j

y

–0.031

---------------------------- =

19.1 + 5.91j

y( t)

= 26.11 sin ( 9t – 0.031)

( D + 10)

=

9j + 10

=

13.45 0.733

=

13.45

0.733 – 1.064 = 1.31

--(--D------+-----5---)--

----------------

10.30-------------------1.064------------

10.30------------

9j + 5

= 26.2 –0.031

= 26.2 sin ( 9t – 0.031)

for the numerator, (zero)

1000

rad

= 159Hz

--------

s

D2 + 2ω

nζ D +

ω n2 = D2 + 5D + 100

ω

n2 = 100

ω

n

=

rad

fn

= 1.59Hz

10--------

s

2ω

nζ

= 5

ζ

5

=

-------------

= 0.25

(underdamped)

2

( 10)

fd

= fn

1 – ζ

2

= 1.54Hz

G( 0)

=

0 + 1000

= 10 = 20dB

-------------------------------------

02 + 5( 0) + 100

Gain

(dB)

20

0

159

f(Hz)

0.1

1

1.54

10

100

1000

-40dB/dec

-60

-20dB/dec

-80

phase

(deg)

0

0.154

15.4,15.9

1590 f(Hz)

0.1

1

1.54

10

100

159

1000

-90

-180

b) ω

= 1

D =

1j

x

0.0667

–3

m

---

----------------------------

---

F

=

( 1j) 2 + 33.3

= 2.07

× 10

0rad N

F( t) = 10 cos ( t + 1) N

Fω(

)

= ( 10 1rad) N

x

–3

m

-------------------------------

---

( 10 1rad) N = 2.07

× 10 0rad N

x( ω

)

= 2.07 × 10

–3

m

( 10 1rad) N

---

0rad N

x( ω

)

= ( 10) 2.07 ×

10–3 ( 0rad + 1rad) m

x( ω

)

= 0.0207 1radm

x( t)

= 0.0207 cos ( t + 1) m

d)

for the denominator, (poles)

D2 + 2ω

nζ D + ω

n2 =

D2 + 33.3

ω

rad

fn

=

0.918Hz

n = 5.77--------

s

ζ

= 0

(undamped)

fd

= fn

1 – ζ 2

= 0.918Hz

G( 0) =

0.0667

for the initial gain

---------------

= 2 = –54dB

33.3

Gain

∞

(dB)

-54

0

f(Hz)

0.1

1

10

100

-94

-40dB/dec

-134

-20dB/dec

phase

(deg)

f(Hz)

0

0.1

1

10

100

-90

-180

e)

x( t)

= ( – 0.020 cos ( 5.77t)

+ 0.020) m

f)

x( t)

= ( – 0.2162 cos ( 5.77t) + 0.2162) m

a)

·

v

y =

·

v = – v – 7y + F

b)

yi + 1

=

yi

+ h

0 1

yi

+

0

vi + 1

vi

–7 –1

vi

F

given

y0

=

1

F = 10

v0

0

using h=0.001s

y1

=

1

+ 0.001

0

1

1

+

0

v1

0

–7 –1

0

10

1.428m

etc.. until

y100

=

m

v100

---

0.000 s

c)

y( t) = ( – 0.437e–0.5t cos ( 2.598t – 0.19) + 1.429) m

d)

y

=

1

=

1

=

1

---

--------------------------

-----

----------

-------------------

2--)----+-----j--(---ω----)-

F

D2 + D + 7 ( jω ) 2

+ jω + 7 ( 7 – ω

y

=

1

--

-------------------------

---------

----

-

F

2

2

2

( 7 – ω )

+

ω

θ

0

= ( 0 – angle( ω

2

2

= ----------------------------------------------

, 7 – ω ) )

= –angle( ω , 7 – ω )

angle( ω

, 7 – ω 2)

0.01Hz

1Hz

100Hz

1Hz

100Hz

a)

x

=

1

---

-------------

F

5 + D

b)

x

=

1

1

=

1 0

=

1

ω

---

-------------

= --------------

-----------------------------------------------

--------------------- – atan

---

F 5 + D 5 + jω

2

+ ω

2

atan

ω

5

2

+ ω

2

5

5

---

5

c)

gain(dB)

1

initial gain

=

20log

=

–14dB

-----------

-14dB

5

5 + 0