37151512 network analysis by van valkenburg solution chap 6

8/6/2019 37151512 Network Analysis by Van Valkenburg Solution CHAP 6

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Muhammad Irfan Yousuf (Peon of Holy Prophet (P.B.U.H)) 2000-E-41 118

cos ω t:

ω A + [1/CR 1]B = 0 ⇒A = -B/ω CR 1

sin ω t:

-ω B + [1/CR 1]A = I0/C

-ω B + [1/CR 1][-B/ω CR 1] = I0/C

B = I0/C[-ω - 1/C2R 12]

A = -[I0/C[-ω - 1/C2R 12]]/ω CR 1

Complete solution

V(t) = Ke-(1/CR1)t + [-[I0/C[-ω - 1/C2R 12]]/ω CR 1]sin ω t + [I0/C[-ω - 1/C2R 1

2]]cos ω t

At t = 0+

V(0+) = Ke-(1/CR1)0+ + [-[I0/C[-ω - 1/C2R 12]]/ω CR 1]sin ω (0+) + [I0/C[-ω - 1/C2R 1

2]]cos

ω (0+)

I0sin ω t [R 1 + R 2] = K(1) + [I0/C[-ω - 1/C2R 12]]

I0sin ω t [R 1 + R 2] = K + [I0/C[-ω - 1/C2R 12]]

K = I0sin ω t [R 1 + R 2] - [I0/C[-ω - 1/C2R 12]]

V(t) = [I0sin ω t [R 1 + R 2] - [I0/C[-ω - 1/C2R 12]]]e-(1/CR1)t + [-[I0/C[-ω -

1/C2R 12]]/ω CR 1]sin ω t + [I0/C[-ω - 1/C2R 1

2]]cos ω t

Q#6.29: Consider a series RLC network which is excited by a voltage source.

1. Determine the characteristic equation.

2. Locus of the roots of the equation.

3. Plot the roots of the equation.

Solution:

R L

C

V(t)

i(t)

For t ≥ 0

According to KVL

di 1

L + ∫ idt + Ri = V(t)

dt C

Differentiating with respect to ‘t’

d2i i di




L + + R = 0

dt2 C dt

Dividing both sides by ‘L’

d2i i Rdi

+ + = 0 (i)

dt

2

LC LdtThe characteristic equation can be found by substituting the trial solution i = e st or

by the equivalent of substituting s2 for (d2i/dt2), and s for (di/dt); thus

1 R

s2 + + s = 0

LC L

2)

ζ = 0 jω jω n

ζ = 1

ζ→∞ σ

-jω n

ζ = 0

1 R

s2 + + s = 0

LC L

Characteristic equation:

as2 + bs + c = 0

Here

a 1

bR

L

c1

LC

-b ± √b2 – 4ac

s1, s2 =

2a

ζ →∞




R R 2 1

- ± - 4(1)

L L LC

s1, s2 =

2(1)

R R 2 1

- ± - 4(1)

L L LC

s1, s2 =

2 2

R R 2 1

- ± - 4(1)

L L LC

s1, s2 =

2 4

R R 2 1

= - ± - 4(1)

2L 2L 4LC

R R 2 1

= - ± -

2L 2L LC radical term (ii)

Hint: 4 = 2

To convert equation (i) to a standard form, we define the value of resistance that causes

the radical (pertaining to the root) term in the above equation as the critical resistance, R cr .

This value is found by solving the equation

2

R 1

- = 0

2L LC




R = R cr

2

R cr 1

- = 0

2L LC

2

R cr 1

=

2L LC

Taking square root of both the sides

2

R cr 1

=

2L LC

R cr 1

=

2L LC

Using cross multiplication

L

R cr = 2C

Hint: 1 = 1

R

ζ =

R cr

R C

ζ =

2 L




1

ω n =

LC

R

2ζ ω n =

L

1

ω n2 =

LC

Substituting the corresponding values in equation (i) we get

s2 + 2ζ ω ns + ω n2 = 0

roots of the characteristic equation are

Characteristic equation:

as2 + bs + c = 0

Here

a 1

b 2ζ ω n

c ω n2

-b ± √b2 – 4ac

s1, s2 =

2a

-2ζ ω n ± √(2ζ ω n)2 – 4(1)(ω n

2)

s1, s2 =

2(1)

-2ζ ω n √4ζ 2ω n2 – 4ω n

2

s1, s2 = ±2 2

Simplifying we get

s1, s2 = -ζ ω n ± ω n√ζ 2 – 1

when ζ = 0

s1, s2 = -(0)ω n ± ω n√(0)2 – 1




s1, s2 = ± ω n√ –1

s1, s2 = ± jω n

Hint: √ –1 = j

3)

R 500 ΩL 1 H

C 1 × 10-6 F

Substituting the corresponding values in equation (ii)

500 500 2 1

= - ± -

2(1) 2(1) (1)(10-6) (ii)

= -250 ± √62500 - 1000000

= -250 ± √-937500

= -250 ± √937500√-1

s1, s2 = -250 ± j968.246

R 1000 ΩL 1 H

C 1 × 10-6 F


1000 1000 2 1= - ± -

2(1) 2(1) (1)(10-6) (ii)

= -500 ± √250000 - 1000000

= -500 ± √-750000

= -500 ± √750000√-1




s1, s2 = -500 ± j 866.025

R 3000 Ω

L 1 HC 1 × 10-6 F


3000 3000 2 1

= - ± -

2(1) 2(1) (1)(10-6) (ii)

= -1500 ± √2250000 - 1000000= -1500 ± √ 1250000

= -1500 ± 1118.034

= (-1500 + 1118.034), (-1500 - 1118.034)

s1, s2 = -381.966, -2618.034

R 5000 ΩL 1 H

C 1 × 10-6 F


5000 5000 2 1

= - ± -

2(1) 2(1) (1)(10-6) (ii)

= -2500 ± √6250000 - 1000000

= -2500 ± √ 5250000

= -2500 ± 2291.288

= (-2500 + 2291.288), (-2500 - 2291.288)

s1, s2 = -208.712, -4791.288




Q#6.31: Analyze the network given in the figure on the loop basis, and determine the

characteristic equation for the currents in the network as a function of k 1. Find

the values of k 1 for which the roots of the characteristic equation are on the

imaginary axis of the s plane. Find the range of values of k 1 for which the roots

of the characteristic equation have positive real parts.

Solution: 1 H

i2

1 ΩK 1i1

1 Ω

1 Ω 1 ΩV1(t)

i1 i3

1 F

Loop i1:

For t ≥ 0

According to KVL

V1(t) = (i1)(1 Ω ) + (i1 – i2)(1 Ω ) + (i1 – i3)(1 Ω ) + (i1 – i3)(XC)1

XC =

j2π fc

ω = 2π f

j2π fc = jω c

jω = s

1

XC =

scc = 1 F

1XC =

s(1 F)

1XC =

s 1

V1(t) = (i1)(1 Ω ) + (i1 – i2)(1 Ω ) + (i1 – i3)(1 Ω ) + (i1 – i3)

-

+

+

-




Simplifying s

1 1V1(t) = i1 + i1 – i2 + i1 – i3 + i1 - i3

s s

1 1V1(t) = (3 + )i1 – i2 – (1 + ) (i)

s s

Loop i2:

For t ≥ 0

According to KVL

(i2 – i1)(1 Ω ) + i2(XL) = 0

XL = jω L

s = jωXL = s(1 H)

XL = sSubstituting

(i2 – i1)(1 Ω ) + i2(s) = 0Simplifying

i2 – i1 + si2 = 0

(1 + s)i2 – i1 = 0 (ii)

Loop i3:

For t ≥ 0

According to KVLSum of voltage rise = sum of voltage drop (a)

Sum of voltage rise = k 1i1

1

Sum of voltage drop = (i3 – i1)(1 Ω ) + (i3 – i1) + (i3)(1 Ω )

s

Substituting in (a)

1

(i3 – i1)(1 Ω ) + (i3 – i1) + (i3)(1 Ω ) = k 1i1

s

Simplifying 1

(i3 – i1)(1 Ω ) + (i3 – i1) + (i3)(1 Ω ) - k 1i1 = 0

s1 1

i3 – i1 + i3 - i1 + i3 – k 1i1 = 0

s s




1 1

- + k 1 + 1 i1 + 2 + i3 = 0 (iii)

s s

Equations (i), (ii) & (iii) can be written in matrix form

1 1

3 + -1 - 1 + i1 V1

s s

-1 (1 + s) 0 i2

= 0

1 1- 1 + k 1 + 0 2 + i3 0

s s

A X B

Determinant of A =

1 1 1 1

3 + (1 + s) 2 + - (0)(0) - (-1) (-1) 2 + - 1 + k 1 +

s s s s

1 1

(0) + (-) 1 + (-1)0 – (-) 1 + k 1 + (1 + s)

s s

After simplifyingCharacteristic equation:

(5 – k 1)s2 + (6 – 2k 1)s + (2 – k 1) = 0

When k 1 = 0

(5 – 0)s2 + (6 – 2(0))s + (2 – 0) = 0

5s2 + 6s + 2 = 0as2 + bs + c = 0

Here




a 5

b 6

c 2

-b ± √b2 – 4ac

s1, s2 =2a

-6 ± √62 – 4(5)(2)

s1, s2 =

2(5)

-6 ± √36 – 40

s1, s2 =

10

-6 ± √-4s1, s2 =

10

-6 ± √-1√4

s1, s2 =

10

-6 ± j2

s1, s2 =

10

s1, s2 = -0.6 ± j0.2

s1, s2 = (-0.6 + j0.2), (-0.6 - j0.2)

When k 1 = 1

(5 – 1)s2 + (6 – 2(1))s + (2 – 1) = 0

4s2 + 4s + 1 = 0as2 + bs + c = 0

Here

a 4

b 4

c 1

-b ± √b2 – 4ac

s1, s2 =

2a




-4 ± √42 – 4(4)(1)

s1, s2 =

2(4)

-4 ± √16 – 16

s1, s2 =8

-4 ± √0

s1, s2 =

8

-4 ± 0

s1, s2 =

8

s1, s2 = -0.5, -0.5

When k 1 = 2

(5 – 2)s2 + (6 – 2(2))s + (2 – 2) = 0

3s2 + 2s + 0 = 0

as2 + bs + c = 0

Here

a 3

b 2

c 0

-b ± √b2 – 4ac

s1, s2 =2a

-2 ± √22 – 4(3)(0)

s1, s2 =

2(3)

-2 ± √4 – 0

s1, s2 =

6

-2 ± √4

s1, s2 =6

-2 ± 2

s1, s2 =

6

s1, s2 = 0, 0.667




When k 1 = -1

(5 – (-1))s2 + (6 – 2(-1))s + (2 – (-1)) = 0

6s2 + 8s + 3 = 0

as2 + bs + c = 0

Here a 6

b 8

c 3

-b ± √b2 – 4ac

s1, s2 =

2a

-8 ± √82 – 4(6)(3)

s1, s2 =

2(6)

-8 ± √64 – 72

s1, s2 =

12

-8 ± √-8

s1, s2 =

6

-8 ± √-1√8

s1, s2 =

6-8 ± j2.828

s1, s2 =

6

s1, s2 = (-1.334 + j0.472), (-1.334 - j0.472)

Q#6.32: Show that equation 6-121 can be written in the form

i = ke-ζ ω nt cos (ω n√1 - ζ 2 t + φ )

Give the values for k and φ in terms of k 5 and k 6 of Eq. (6-121).

Solution:

Let k 5 = kcosφ (i)

k 6 = -ksinφ (ii)

k = (kcosφ )2 + (-ksinφ )2

k = k 2cos2φ + k 2sin2φ




k = k 2(cos2φ + sin2φ )

k = k 2(1)

k = √k 2 = √k 52 + k 6

2

Dividing Eq. (i) by (ii)

kcosφ k 5

= -cot φ =

-ksinφ k 6

-1 k 5

φ = cot -

k 6

Using the trigonometric identity

cos (x + y) = cos x cos y – sin x sin y

Q#6.33: A switch is closed at t = 0 connecting a battery of voltage V with a series RL

circuit.

(a) Solution:sw

t = 0

R L

V

i

For t ≥ 0

According to KVL

di

V = iR + L

dtDividing both sides by ‘L’

di R V

+ i =

dt L L

This is a linear non-homogeneous equation of the first order and its solution is,

Thus




R

P =

L

V

Q = L

Hence the solution of this equation

i = e-Pt∫ QePtdt + ke-Pt

V

i = e-(R/L)t∫ e(R/L)tdt + ke-(R/L)t

L

Vi = e-(R/L)t ∫ e(R/L)tdt + ke-(R/L)t

L

e(R/L)t

∫ e(R/L)tdt =

d

dt (R/L)t

L e(R/L)t

∫ e(R/L)tdt =

R

Substituting

V L e(R/L)t

i = e-(R/L)t + ke-(R/L)t

L R

V

i = + ke-(R/L)t

R i(0-) = i(0+) = 0

Substituting i = 0 at t = 0

V

0 = + ke-(R/L)(0)

R

e0 = 1




V

k = -

R

Substituting

V -Vi = + e-(R/L)t

R R

V

i = (1 - e-(R/L)t )

R

P = i2R

t

WR = ∫ i

2

R dt0

t V 2

WR = ∫ (1 - e-(R/L)t )2Rdt

0 R

(a - b)2 = a2 + b2 – 2ab

t V2

WR = ∫ (1 + e-2(R/L)t – 2(1)(e-(R/L)t))Rdt

0 R 2

t V2

WR = ∫ (1 + e-2(R/L)t – 2e-(R/L)t)dt

0 R

V2 t t t

WR = ∫ (1)dt + ∫ e-2(R/L)tdt + ∫ (-2e-(R/L)tdt)

R 0 0 0

Simplifying

V2 2L L 3L

WR = t + e

-(R/L)t

-

e

-2(R/L)t

-

R R 2R 2R

(b)

Li2

WL =

2

LV2




WL = (1 - e-(R/L)t )2

2R 2

(c)

At t = 0

V

2

2L L 3LWR = (0) + e-(R/L)(0) - e-2(R/L)(0) -

R R 2R 2R

V2 2L L 3L

WR = (0) + e0 - e0 -

R R 2R 2R

V2 2L L 3L

WR = (1) - (1) -

R R 2R 2R

V2

WR = 0

R

WR = 0 joules

At t = 0

LV2

WL = (1 - e-(R/L)0)2

2R 2

LV2

WL = (1 – e0)2

2R 2

LV2

WL = (1 – 1)2

2R 2

WL = 0 joules

At t = ∞LV2

WL = (1 - e-(R/L)∞)2

2R 2

LV2

WL = (1 – e-∞)2

2R 2




LV2

WL = (1 – 0)2

2R 2

LV2

WL = joules

2R 2

(d)

In steady state total energy supply

W = WR + WL

V2 2L L 3L LV2

W = t + e-(R/L)t - e-2(R/L)t - + (1 – e-(R/L)t)2

R R 2R 2R 2R 2

Q#6.34: In the series RLC circuit shown in the accompanying diagram, the

frequency of the driving force voltage is

(1) ω = ω n

(2) ω = ω n√1 - ζ 2

Solution:

1000 Ω 1 H

100 sin ω t i(t)

1 µ F

For t ≥ 0

According to KVL

di 1

100 sin ω t = L + iR + ∫ idt

dt C

Here

ω = ω n

di 1

100 sin ω nt = L + iR + ∫ idt … (i)

dt C

1

ω n =

+

-




LC

L = 1 H

C = 1 × 10-6 F

1

ω n =(1 H)( 1 × 10-6 F)

After simplifying

ω n = 1000 rad/sec

Substituting in (i) we get

di 1

100 sin 1000t = L + iR + ∫ idt … (i)

dt C

Differentiating both the sides & substituting the values of L & C we get

d2i di i

100 (1000) cos 1000t = (1) + (1000) +dt2 dt 10-6

Simplifying we get

d2i di100000cos 1000t = + (1000) + 1000000i

dt2 dt

The trial solution for the particular integral isi p = A cos 1000t + B sin 1000t

d2i p di p100000cos 1000t = + (1000) + 1000000i p

dt2 dt

(i p)′ = -1000A sin 1000t + B 1000cos 1000t(i p)′ ′ = -1000000A cos 1000t - B 1000000sin 1000t

(i p)′ = Ist derivative

(i p)′ ′ = 2nd derivative

100000cos 1000t = -1000000A cos 1000t - B 1000000sin 1000t + 1000(-1000A sin 1000t

+ B 1000cos 1000t) + 1000000(A cos 1000t + B sin 1000t)Simplifying

100000cos 1000t = -1000000A cos 1000t – 1000000B sin 1000t - 1000000A sin 1000t +

1000000B cos 1000t + 1000000A cos 1000t + 1000000B sin 1000tSimplifying

Equating the coefficients

Cos:100000 = 1000000B

100000

B =1000000

B = 0.1




Sin:

0 = - 1000000B – 1000000A + 1000000B

0 = –1000000A

A = 0

i p = A cos 1000t + B sin 1000t

Substituting the values of A & Bi p = (0) cos 1000t + (0.1) sin 1000t

i p = 0.1 sin 1000t

e jω t – e-jω t

sin ω t =2j

Here ω = 1000

e j1000t – e-j1000t

sin 1000t =

2j

e j1000t – e-j1000t

ip = 0.1 Transient response

2j

In steady state

At resonanceXL = XC

In a series RLC circuit

Z = R + j(XL - XC)Z = R + j(XC - XC)

Z = R

VIm =

Z

100

Im =1000

Im = 0.1 A

(2) ω = ω n√1 - ζ 2

Determine the values of ω n & ζ substitute & simplify

Do yourself.




THE END

37151512 network analysis by van valkenburg solution chap 6

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