telecommunications

TELECOMMUNICATIONS

Dr. Hugh Blanton

ENTC 4307/ENTC 5307

Complex Numbers

Dr. Blanton - ENTC 4307 - Complex Numbers 3

Complex numbers

M = A + jB

Where j2 = -1 or j = √-1

|M| = √(A2 + B2)

and tan = B/A

B

A

M

Real

Imaginary

ARGAND diagram


j notation

Refers to the expression

Z = R + jX

• X is not imaginary• Physically the j term refers to +j = 90o lead and -j = 90o lag


Complex Number DefinitionsComplex Number Definitions

• Rectangular Coordinate System: • Real (x) and Imaginary (y)

components, A = x +jy

• Complex Conjugate (AA*) refers to the same real part but the negative of the imaginary part.• If A = x + jy, then A* = x jy.

xx

+jy+jy

jyjy

+x+x

62624212 .. jjyxA

46324221 .. jjyxB

1

2

1

2

1 2


Complex Number DefinitionsComplex Number Definitions

• Polar Coordinates: Magnitude and Angle• Complex conjugate has the same magnitude but the

negative of the angle.• If A =M90, then A*=M-90


• Rectangular to Polar Conversion

• By trigonometry, the phase angle “” is,

• Polar to Rectangular Conversion • y = imaginary part= M(sin )• x = real part = M(cos )

2222 imaginaryrealyxM

x

yarctan


+jy+jy

jyjy

+x+x

62624212 .. jjyxA

46324221 .. jjyxB

1

2

1

2

1 2


Vector Addition & SubtractionVector Addition & Subtraction

• Vector addition and subtraction of complex numbers are conveniently done in the rectangular coordinate system, by adding or subtracting their corresponding real and imaginary parts.• If A = 2 + j1 and B = 1 – j2:

• Then their sum is:• A + B = (2+1) + j(1 – 2) = 3 – j1

• and the difference is:• A - B = (2 1) + j(1 (– 2)) = 1 + j3


• For vector multiplication use polar form.• The magnitudes (MA,MB) are multiplied

together while the angles () are added.• MuItiplying “A” and “B”:

• AB = (2.24 26.60)(2.24 63.40) = 5 36.8


• Vector division requires the ratio of magnitudes and the differences of the angles:

901463626242

242

463242

626242

...

.

..

..

B

A


+jy+jy

jyjy

+x+x

62624212 .. jjyxA46324221 .. jjyxB

1

2

1

2

1 22 1

A-B

A+B


Complex Impedance SystemComplex Impedance System

• RF components are frequently defined by their terminal impedances or admittances in the complex rectangular coordinate system.• Complex impedance is the vector

sum of resistance and reactance.• Impedance = Resistance ± j Reactance

)( jXRZ

RR

+jX+jX

jXjX

+R+R

inductiveinductive

11 jXRZ

1R

1jX

1jX

capacitivecapacitive

11 jXRZ


• Series connections are handled most conveniently in the impedance system.

2121

221121

jXjXRR

jXRjXRZZZT

)(

)(

1Z 2Z


Complex Admittance SystemComplex Admittance System

• Parallel circuit descriptions may be viewed in the complex admittance system• Complex impedance is the vector sum

of conductance and susceptance.• Admittance = Conductance ± j Susceptance

• where and

SiemensorMhosjBGZ

Y ,)( 1 GG

+jB+jB

jBjB

+G+G

inductiveinductive

11 jBGY

1G

1jB

1jB

capacitivecapacitive

11 jBGY R

G1

jX

jB

1


• Parallel connections are handled most conveniently in the admittance system.

2121

221121

jBjBGG

jBGjBGYYYT

)(

)(1Y 2Y


Z dependence on (RCL )

1 2 5 10. 20. 50. 100.frequency1

510

50100

5001000

Impedance

o parallel

series


Currrent dependence on

1 2 5 10. 20. 50. 100.frequency

100

500

1000

Current (ma)

o

o

Imin

Imaxx√2

parallel

series


• At RF, particularly at high power levels, it is very important to maximize power transfer through careful impedance matching.• Improperly matched component

connection leads to “mismatch loss.”


RF Components & Related Issues

RF Components & Related Issues

• Unique component problems at RF:• Parasitics change behavior• Primary and secondary resonances• Distributed vs. lumped models• Limited range of practical values• Tolerance effects• Measurements and test fixtures• Grounding and coupling effects• PC-board effects


V and I Phase relationships

VS VR2 VL VC 2

and

tan VL VC

VR

Io Vo / Z

VL

IVR

VS

VC

VL-VC


R, XC and Z relationships

Z R2 XL XC 2 (in )

and

tan XL XC

R

XL

IR

Z

XC

XL-XC


Example 1

Consider this circuit with = 105 rad s-1

1 k 0.01 F

XL 1

C

1

105.10 8103

Z R jXC 103 j10 3

Z [(10 3 )2 (10 3)2 ] 2 103

tan X / R 1 or 45o


Example 2

5 20 10 10

Z R jXC 5 20j 1010j

15 10j

Z [(15)2 ( 10)2] 18

tan X / R10/15 or 33.7o


Example 2 Cont’d

5 20 10 10

~~

200 V

I VZ

20015 10 j

200

15 10 j

15 10 j

15 10 j

200 (15 10 j)

225 100

I 8 / 13 (15 10 j)

I 813

(225 100 ) 11.1 A


Example 2 Cont’d

VL

I

VR

VS

VC

VL-VC

XL

I

R

Z

XC

XL-XC

Z = 15 - 10jI = 8/13(15 + 10j)


General procedures

• convert all reactances to ohms• express impedance in j notation• determine Z using absolute value• determine I using complex conjugate• draw phasor diagram• Note: j = -1/j so

R + (1/jC) = R - j/C


Example 3

5 10 20 15

- express the impedance in j notation- determine Z (in s) and - determine I for a voltage of 24V


Example 3 Cont’d

5 10 20 5

Z R jXC 5 10j 2015j

255j

Z [(25)2 (5)2] 25.5

tan X / R5/ 25 or 11.3o


Example 3 Cont’d

5 20 10 10

~~

24V

I V

Z

24

25 5 j

2425 5 j

25 5 j25 5 j

24(25 5 j)625 25

I 0.0369 (25 5 j)

I 0.0369 (625 25) 0.94 A


Example 4

Construct a circuit which contains at least one L and one C components which could be represented by:

Z = 10 - 30j


1Z

1Z1

1Z2

or Z Z1 Z 2

Z1 Z 2

Z (10 30 j ) ( 20 10 j )(10 30 j ) ( 20 10 j )

100 700 j

30 40 j

100 700 j30 40 j

30 40 j30 40 j

10 10 j

Parallel circuits

10 - 30j

20 - 10j

*

*


Z (20 30 j ) ( 20 30 j )( 20 30 j ) ( 20 30 j )

500 1200 j

40 60 j

500 1200 j

40 60 j

40 60 j

40 60 j

15200

(52000 78000 )

10 15 j

Parallel circuits

20 - 30j

20 - 30j

*

*

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