Problem 7.2 Express the current waveform

i(t) = −0.2cos(6π ×109t +60) mA

in standard cosine form and then determine the following:

(a) Its amplitude, frequency, and phase angle.

(b) i(t) at t = 0.1 ns.

Solution:

i(t) = −0.2cos(6π ×109t +60) (mA)

= 0.2cos(6π ×109t +60−180) (mA)

= 0.2cos(6π ×109t −120) (mA).

(a) amplitude= 0.2 mA; f = 3×109 Hz = 3 GHz; φ = −120.(b)

i(0.1 ns) = 0.2cos(6π ×109×0.1×10−9−120) (mA)

= 0.2cos(0.6π −120) (mA)

= 0.2cos(108−120) (mA)

= 0.196 mA.

Problem 7.8 A multiplier circuit has two input ports, designatedv1 andv2, and oneoutput port whose voltagevout is equal to the product ofv1 andv2. Assume

v1 = 10cos2π f1t V,

v2 = 10cos2π f2t V,

(a) Obtain an expression forvout in terms of the sum and difference frequencies,fs = f1 + f2 and fd = f1− f2.

(b) Plot its waveform over the time interval [0, 2 s], given thatf1 = 3 Hz andf2 = 2 Hz.

Solution:(a)

vout(t) = v1(t) ·v2(t)

= (10cos2π f1t)(10cos2π f2t)

= 100cos2π f1t cos2π f2t.

Using the identity cosxcosy = 12[cos(x+y)+cos(x−y)],

vout(t) =1002

cos[2π( f1 + f2)t]+cos[2π( f1− f2)t]

= 50(cos2π fst +cos2π fdt).

(b) For f1 = 3 Hz andf2 = 2 Hz,

vout(t) = 50(cos10πt +cos2πt) (V).

The plot ofvout(t) is shown in Fig. P7.8.

4 V

2 V

−2 V

−4 V

010.80.60.40.2

v1

vout

t (s)

v2

Figure P7.8: Plots ofv1(t), v2(t) andvout(t).

Section 7-2: Complex Numbers

Problem 7.10 Express the following complex numbers in polar form:

(a) z1 = 3+ j4

(b) z2 = −6+ j8

(c) z3 = −6− j4

(d) z4 = j2

(e) z5 = (2+ j)2

(f) z6 = (3− j2)3

(g) z7 = (−1+ j)1/2

Solution:(a) z1 = 3+ j4 = +

√32 +42 ∠tan−1(4/3) = 5∠53.13.

(b)

z2 = −6+ j8 = −(6− j8) = −[

+√

62 +82 e− j tan−1(8/6)]

= 10e− j53.13 ·ej180

= 10ej126.87 .

(c)

z3 = −6− j4 = −(6+ j4) = −[

+√

62 +42 ej tan−1(4/6)]

= 7.21ej33.7 ·e− j180

= 7.21e− j146.3 .

(d) z4 = j2 = 2ej90 .

(e) z5 = (2+ j)2 =[

+√

22 +12 ej tan−1(1/2)]2

=[√

5 ej26.56]2

= 5ej53.13 .

(f)

z6 = (3− j2)3 =[

+√

32 +22 e− j tan−1(2/3)]3

=[

3.61e− j33.69]3

= 46.9e− j101.1 .

(g)

z7 = (−1+ j)1/2 = [−(1− j)]1/2

=[

− +√

12 +12 e− j tan−1 1]1/2

= [−1.414e− j45 ]1/2

= [1.414e− j45 ·ej180 ]1/2

= [1.414ej135 ]1/2

= ±1.19ej67.5 .

Problem 7.14 If z = −8+ j6, determine the following quantities:(a) |z|2(b) z2, in polar form(c) 1/z, in polar form(d) z−3, in polar form(e) Re(1/z2)

(f) Im(z∗)(g) Im[(z∗)2]

(h) Re[(z∗)−1/2]

Solution:(a) |z|2 = (−8+ j6)(−8− j6) = 64+36= 100.(b)

z2 = (−8+ j6)(−8+ j6) = (64−36)− j96=+√

282 +962 e− j tan−1(96/28)

= 100e− j73.74 .

(c)

1z

=1

−8+ j6=

−8− j6(−8+ j6)(−8− j6)

=−(8+ j6)

64+36=

e− j18010ej tan−1(6/8)

100

= 0.1e− j143.13 .

(d)

z−3 =1z3 =

(

1z

)3

= (0.1e− j143.13)3 = 10−3e− j429.39

= 10−3e− j429.39 ·ej360 = 10−3e− j69.39 .

(e)

Re

(

1z2

)

= Re(0.1e− j143.13)2 = Re(0.01e− j286.26)

= 0.01cos286.26 = 2.8×10−3.

(f)

Im(z∗) = Im(−8− j6) = Im[−(8+ j6)] = [e− j180 ·10ej36.87 ]

= Im[10e− j143.13 ]

= −10sin143.13 = −6.

(g)

Im[(z∗)2] = Im[(10e− j143.13)2] = Im[100e− j286.26 ] = −100sin286.26 = 96.

(h)

Re[(z∗)−1/2] = Re

[

(

110e− j143.13

)1/2]

= Re[0.316ej71.56 ]

= 0.316cos71.56 = 0.1.

Problem 7.18 Simplify the following expressions and express the result in polarform:

(a) A =5e− j30

2+ j3− j4

(b) B =(−20∠45)(3− j4)

(2− j)+(2+ j)

(c) C =j4

(3+ j2)−2(1− j)+

11+ j4

(d) D =

∣

∣

∣

∣

(2− j) −(3+ j4)−(3+ j4) (2+ j)

∣

∣

∣

∣

(e) E =

∣

∣

∣

∣

5∠30 −2∠45

−2∠45 4∠60

∣

∣

∣

∣

Solution:(a)

A =5e− j30

2+ j3− j4

=(5cos30− j5sin30)− j4(2+ j3)

2+ j3

=4.33− j2.5− j8+12

2+ j3

=(16.33− j10.5)

2+ j3· 2− j32− j3

=(32.66−31.5)− j(21+48.99)

4+9= 0.089− j5.38= 5.38e− j89.05 .

(b)

B =(−20∠45)(3− j4)

2− j+2+ j

=(−20cos45− j20sin45)(3− j4)+(2+ j)(2− j)

2− j

=(−14.14− j14.14)(3− j4)+5

2− j

=−(42.42+56.56)+ j(56.56−42.42)+5

2− j

=−98.98+ j14.14

2− j· (2+ j)(2+ j)

=−202.1− j65.7

5= e− j180(42.50ej18.01) = 42.50e− j161.99 .

(c) C =j4

(3+ j2)−2(1− j)+

11+ j4

=j4

1+ j4+

11+ j4

=1+ j41+ j4

= 1.

(d)

D =

∣

∣

∣

∣

(2− j) −(3+ j4)−(3+ j4) (2+ j)

∣

∣

∣

∣

Problem 7.31 In response to an input signal voltagevs(t) = 24cos2000πt, theinput current in the circuit of Fig. P7.31 was measured asi(t) = 6cos(2000πt −60)mA. Determine the equivalent input impedanceZ of the circuit.

Z

a

b

Passive

circuitvs(t)

i(t)

+_

Figure P7.31:Configuration for Problem 7.31.

Solution:

Vs = 24 V,

I = 6e− j60 (mA).

Z =VI

=24

6e− j60 ×10−3

= 4ej60 kΩ

= (4cos60 + j4sin60) kΩ = (2+ j3.46) kΩ.

Problem 7.36 The input circuit shown in Fig. P7.36 contains two sources, given by

is(t) = 2cos103t A,

vs(t) = 8sin103t V.

This input circuit is to be connected to a load circuit that provides optimumperformance when the impedanceZ of the input circuit is purely real. The circuitincludes a “matching” element whose type and magnitude should be chosen to realizethat condition. What should those attributes be?

a

Z

b

2 Ω

0.5 mF

Matching element

?

Load

circuit4 mH

is(t)

vs(t)

6 Ω

+_

Figure P7.36:Circuit for Problem 7.36.

Solution:

a

Z

b

2 Ω

?

Is Is(6 + j4) Ω ‘

‘‘

−j2 Ω

a

Z

b

?

Is Z‘

‘+_

a

Z

b

?

Vs

Z‘

I s = 2 A

Vs = − j8 V

I ′s =Vs

2=

− j82

= − j4 A

I ′′s = I s+ I ′s = (2− j4) A

Z′ = (6+ j4) ‖ (− j2) ‖ 2 = (1.1− j0.76) Ω

VTh = V′s = I ′′sZ′ = (2− j4)(1.1− j0.76) = −(0.84+ j5.92) V.

The matching elementZx has to cancel the imaginary part ofZ′. Hence

Zx = + j0.76 Ω.

So it has to be an inductorL such that

ωL = 0.76,

or

L =0.76103 = 0.76 mH.

Problem 7.47 Apply nodal analysis in the phasor domain to determineiC(t) in thecircuit of Fig. P7.47.

5 Ω

5 Ω

5 Ω

12 cos (400t − 30o) V

20 mH20 mH

+_

iC

mF1

1.6

Figure P7.47:Circuit for Problem 7.47.

Solution:

5 Ω

5 Ω

5 Ω

+_

IC

V1

V2

12 V−30o

−j4 Ω

j8 Ω j8 Ω

ZL = jωL = j400×20×10−3 = j8 Ω

ZC =− jωC

=− j

400× 11.6 ×10−3

= − j4 Ω

NodeV1:V1

− j4+

V1−V2

5+

V1−V2

j8= 0

NodeV2:V2−V1

5+

V2−V1

j8+

V2−12e− j30

5+

V2

5+ j8= 0

Simultaneous solution leads to:

V1 = 4.986e− j96.352 , V2 = 4.986e− j32.341 ,

IC =V1

− j4=

j4

V1 = 1.25e− j6.532 ,

iC(t) = 1.25cos(400t −6.352) (A).