CMPE306 Lecture24 Frequency Response 2 101201

7/27/2019 CMPE306 Lecture24 Frequency Response 2 101201

1/17

Find the load impedance that maximizes the powertransfer.

n e max mum power n e oa mpe ance

1-1

UMBC ENEE423 Communication Systems

Course Notes E F C LaBerge, 2008 All rights reserved.Portions Copyright Prentice, Hall, Inc. All rights reserved

Week 1


2/17

!"

Chapter 14.1Frequency Response

Fundamentals of Electric Circuits 3nded.,

2007) 14.1-4

1-2

UMBC ENEE423 Communication Systems


Week 1


3/17

Example (Synthesis)- - .

What is the gain at 2 90 Hz? at 2 150 Hz?

C

Single pole filter, two options: or .

filter, take the output voltage to be across t

RL RC

RL he inductor.

0

1 260 sec but 10 mH is 2.2" 1.2", and the smalle

10 mH

R

L

we make , the smaller we have to make . Not a good answer.L R

0

er, a e e ou pu vo age across e res s or

1

160 sec , 0.1 is 2mm 1.25mmF

1-3UMBC ENEE423 Communication Systems


Week 1

.


4/17

0( j RC

V

H

rps / 31.8( )

1 r s 1 / 31.8

s

j jff

H

2

/ 31.8( )

1 ( / 31.8)

ff

H

90 Hz, ( ) 0.9428

150 Hz 0.9782

f f

H

H

50 k , 0.1 FC

1

f)|

0 50 100 1500

.|H



Week 1

f in Hz


5/17

Series Resonance (1)

and inductive reactances are equal in magnitude, thereby resultingin a purely resistive impedance (at a specific frequency)

1( ) s R j L

j C

V

Z HI

Resonance occurs at when

R j LC

Z

0 0

0

Im( ) 1 0

resonance frequency

L C

Z

1. is purely resistiveRZ2. and are in phase, pf 1s V I

0

1 rad/s

LC

. e magn u e o smaximum

4. The voltage of the inductor and



Week 15 of 20

0 0 2 Hz2

fLC

Chapter 14.5

the capacitor can be more than that

of the source


6/17

Series Resonance (2) 1L

Z

Z

C

Z

L

1

R ( 1/ ) ,

1tanZ

Z L C

L C

1 0LC R Z

Cv

s

I V Z0

0

( ),R R( )

s sL Cj L

j C

V VV V

m V Z

II 0m

L m

VL QV

R V

2 2R ( 1/ )

m mIZ L C

I

0( )

mC m

VQV

R C V



Week 16 of 20

( ),( )

s sL Cj L

j C

V V

Z Z

Chapter 14.5

0 0

R Q


7/17


221

2 2

1( ) m

VP I R R

1 2

2

Half maximum power dissipation at ,

1 1voltage or current power

22

1 2 0

22

( ) ( ) ( )2 2 2mP P R

.

2 2

m m

R

1 2

2 2

1,2 ( ) R ( 1/ ) 2Z L C R



Week 17 of 20Chapter 14.5


8/17

Series resonance (3a)

2 2

1,2( ) R ( 1/ ) 2Z L C R 1 1

vo tage or current power22

1

1

2 2

R

L L LC

1 1voltage or current power

22

2

2

1

2 2

R R

L L LC

0 1 2

0 1 2 0

2 1 bandwidthB



Week 1


9/17


max stored in the circuitQuality factor Q 2dissipated in one period at resonance

E

E

210 02

In a series RLC circuit

2 f L LLI

212 00

(1/ )

1

R RRI f

L Resonant circuit

0

20

R CR

R

characterized by five

related parameters:

For a hig

L Q

h- circuit ( 10)Q Q

1 2 0, , , ,




1 0 2 0,


10/17

Practice 14.7

A series-connected circuit has 4 , L 25 mH, and 50R Q

300 3

(50)(4)(a) 8 10 rad/s

25 10

L QRQ

R L

0 2 6 3

0

3

1 1 10.625 F

(64 10 )(25 10 )

8 10o

CLLC

1 0

50Since Q 10

/ 2 8000 80 7920 rad/s

Q

B

2 0

2

0

/ 2 8000 80 8080 rad/s

100(c) At ,

2

m

B

VP

R

2

1.25 kW

8

2

1

2

At , 0.5 0.625 kW2

At 0.5 0.625 kW

m

m

VP

R

VP



Week 110 of 20

2R

Chapter 14.5


11/17

Parallel Resonance (1)

11

)( LjCjR V

I

HY

)(atwhenoccursResonance 0

LCj

R

Y

frequencyResonance

01)Im(

0

00 LC

Y 0At1 R

H

rad/s

1

0LC

m

mL m

I Rj jQI

V

I

LjCj

RIm

V

HYV

0 0

0 0( )C m mC jI R C jQI I V



Week 111 of 20

Lj CL

Chapter 14.6


12/17

Parallel Resonance (2)

2

1

Ha - requency

1 1 1

0 1 2

20 1 2 0

2

1 1 1

2 2RC RC LC

2 1

1B

RC

0

0

2

01

Q RCB L

1 0

1 0

2

0 2 0

2 2 21

1

Q QQ

B




2 2Q Q


13/17

Practice 14.8

0 1 2

A parallel resonant circuit has 100k , 20mH, and 5nF

Calculate , , , , and

L C

Q B

5

0-3 -9

10 rad/s(20 10 )(5 10 )LC

5 -3

0

100 10 50(10 )(20 10 )

QL

0B 2000rad/s50Q

5

1 0

5

/ 2 10 -1000 99 krad/sB



Week 1

13 of 20Chapter 14.5

2 0


14/17

Practice 14.9fre uencresonanttheCalculate

AT RESONANCE, THE

10 / 0.2j

REACTANCE IS ZERO!

.10 1/ 0.2j

2 2

10 1 2 10(1 2 ) 10 20j j jj j j

20 0 00 1 (2 ) 2020Im ( ) 0 0

j jj

H

0 0

0



Week 1


2 2

0 0 01 4 20 2.179 rad/s4 2


15/17

Decibels: Definition

210

1

log , with a unit is ,P belsP

210

1

more commonly 10log expressed in (dB).P

decibelsP

210

number of bels log P

G Some special cases:

0.1 ; the quantity corresponds to the factor expressed in dB10 x yy

1

2dB 10

1

10log (1/10 of bels)P

GP

2 1

2 1

( / ) 1 0 dB

( / ) 10 10 dB

P P y x

P P y x

1 2 dB 10

1 2 dB 10

10log 1 0 dB

2 10log 2 3 dB

P P G

P P G

2 1

2 1

( / ) 0.1

/ 2 3.010...

10 dB

3 dB

y x

P P

P P y x

x



Week 1

1 2 dB 10

15 of 23

When engineers refer to 3 dB gain or 3 dB loss, they mean a factor of 2!


16/17

Decibels: Absolute units

Engineers will often normalize with respect to a fixed value

EXAMPLE: dBm = power relative to 1 mW

=

0 dBm 1 mW 30 dBW

10 dBm 10 mW 20 dBW

30 dBm 1 W 0 dBW

10 dBm 0.1 mW 40 dBW

3 dBm 2 mW 27 dBW

Many different normalizations are used in different fields!



Week 1



17/17

The Decibel Scale in Circuit Theory

.

20log is used for voltage or current

2. dB is logarithmic measurement of the

ratio of one variable to another of the

same kind

. ,

negative signs and angles are handled

separately.2

11

1input power

VP

1

2

22

2

1output power

2

VP

R

2

2 2 1dB 10 10 10

1 1 2

10log 10log 10logP V R

GP V R



Week 1


2 21 2 dB 10 10

1 1

If 10log 20logV V

R R GV V

CMPE306 Lecture24 Frequency Response 2 101201

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