ECEN 301Discussion #16 – Frequency Response1 Easier to Maintain Than to Retake Alma 59:9 9 And now...

ECEN 301 Discussion #16 – Frequency Response 1

Easier to Maintain Than to Retake

Alma 59:9

9 And now as Moroni had supposed that there should be men sent to the city of Nephihah, to the assistance of the people to maintain that city, and knowing that it was easier to keep the city from falling into the hands of the Lamanites than to retake it from them, he supposed that they would easily maintain that city.


Lecture 16 – Frequency Response

Back to AC Circuits and Phasors

• Frequency Response

• Filters


Frequency Response

Frequency Response H(jω): a measure of how the voltage/current/impedance of a load responds to the voltage/current of a source

)()(

)(

jVjV

jHS

LV

)()(

)(

jIjI

jHS

LI

)()(

)(

jIjV

jHS

LZ


Frequency Response

VL(jω) is a phase-shifted and amplitude-scaled version of VS(jω)

)()()(

)()()(

jVjHjV

jHjVjV

SVL

VS

L


Frequency Response

VL(jω) is a phase-shifted and amplitude-scaled version of VS(jω)

)(

)()()(

)()()(

SVL

SVL

VHjSV

jL

VjS

HjV

jL

SVL

VS

L

eVHeV

eVeHeV

jVjHjV

jHjVjV

SVL VHV SVL VH

AAA

eAA Aj

as expressed becan

numbercomplex any :

A

NB

Amplitude scaled Phase shifted


Frequency Response

Example 1: compute the frequency response HV(jω)

R1 = 1kΩ, RL = 10kΩ, C = 10uF

R1

Cvs(t)+–

+RL

–


Frequency Response


R1 = 1kΩ, RL = 10kΩ, C = 10uF

R1

Cvs(t)+–

+RL

–

1. Note frequencies of AC sources

Only one AC source so frequency response HV(jω) will be the function of a single frequency


Frequency Response


R1 = 1kΩ, RL = 10kΩ, C = 10uF

R1

Cvs(t)+–

+RL

–

1. Note frequencies of AC sources2. Convert to phasor domain

Z1 = R1

ZLD=RL

ZC=1/jωC

Vs+–~


Frequency Response


R1 = 1kΩ, RL = 10kΩ, C = 10uF1. Note frequencies of AC sources2. Convert to phasor domain3. Solve using network analysis

• Thévenin equivalent

Z1 = R1

ZC=1/jωC

CT ZZZ ||1


Frequency Response


R1 = 1kΩ, RL = 10kΩ, C = 10uF

C

CST ZZ

ZjVjV

1

)()(

1. Note frequencies of AC sources2. Convert to phasor domain3. Solve using network analysis

• Thévenin equivalent

Z1 = R1

+VT

–

ZC=1/jωC

Vs+–~

CT ZZZ ||1


Frequency Response


R1 = 1kΩ, RL = 10kΩ, C = 10uF

C

CST ZZ

ZjVjV

1

)()(


• Thévenin equivalent4. Find an expression for the load voltage

CT ZZZ ||1

ZT

ZLDVT

+–~

+VL

–

LDC

LD

C

CS

LDT

LDTL

ZZZ

Z

ZZ

ZjV

ZZ

ZjVjV

||)(

)()(

11


Frequency Response


R1 = 1kΩ, RL = 10kΩ, C = 10uF5. Find an expression for the frequency

response ZT

ZLDVT

+–~

+VL

–

LDC

LD

C

C

S

LV

ZZZ

Z

ZZ

Z

jV

jVjH

||

)(

)()(

11


Frequency Response


R1 = 1kΩ, RL = 10kΩ, C = 10uF5. Find an expression for the frequency

response ZT

ZLDVT

+–~

+VL

–

110arctan

110

100

110

100

)10/(10)10/(11010

)10/(10

||)(

22

53534

54

11

11

j

jj

j

ZZZZZ

ZZ

ZZZ

Z

ZZ

ZjH

CCLD

CLD

LDC

LD

C

CV


Frequency Response


R1 = 1kΩ, RL = 10kΩ, C = 10uF

2010.0)(

jHV

5. Find an expression for the frequency response• Look at response for low frequencies (ω = 10)

and high frequencies (ω = 10000)

ZT

ZLDVT

+–~

+VL

–

110arctan

110

100)(

22

jHV

0907.0905.0)( jHV

ω = 10ω = 10000

Lowpass filter

Frequency Response

Lowpass filter


Frequency Response

Example 2: compute the frequency response HZ(jω)

R1 = 1kΩ, RL = 4kΩ, L = 2mH

R1is(t)

+RL

–

L


Frequency Response


R1 = 1kΩ, RL = 4kΩ, L = 2mH

R1is(t)

+RL

–

L 1. Note frequencies of AC sources

Only one AC source so frequency response HZ(jω) will be the function of a single frequency


Frequency Response


R1 = 1kΩ, RL = 4kΩ, L = 2mH

R1is(t)

+RL

–

L

1. Note frequencies of AC sources2. Convert to phasor domain

Is(jω)

ZLD = RL

ZL = jωL

Z1=R1

+ VL

–


Frequency Response


R1 = 1kΩ, RL = 4kΩ, L = 2mH

LDLDL

LDLS

LDLL

ZZZ

ZZZjI

ZjIjV

)(

)(||)(

)()(

1


• Current divider & Ohm’s Law

Is(jω)

ZLD = RL

ZL = jωL

Z1=R1

+ VL

–

IL(jω)

)(

)(||)()( 1

LDL

LDLSL ZZ

ZZZjIjI


Frequency Response


R1 = 1kΩ, RL = 4kΩ, L = 2mH

6

11

1

104.01

800

//1

)(

)(||

)(

)()(

j

RLjRR

R

ZZZ

ZZZ

jI

jVjH

L

L

LDLDL

LDL

S

LZ

Is(jω)

ZLD = RL

ZL = jωL

Z1=R1

+ VL

–

IL(jω)

4. Find an expression for the frequency response


Frequency Response


R1 = 1kΩ, RL = 4kΩ, L = 2mH

262

6

6

)104.0(1

)104.01(800

104.01

800)(

j

jjH ZIs(jω)

ZLD = RL

ZL = jωL

Z1=R1

+ VL

–

IL(jω)

4. Find an expression for the frequency response• Look at response for low frequencies (ω = 10)

and high frequencies (ω = 10000)

0.4800)( jH Z 0040.0194)( jH Z

ω = 10 ω = 10000

Lowpass filter


1st and 2nd Order RLC Filters

Graphing in Frequency DomainFilter Orders

Resonant FrequenciesBasic Filters


Frequency Domain

Graphing in the frequency domain: helpful in order to understand filters

-1.5

0.0

1.5

0.00 2.00 4.00

time

x(t

)

0.00

3.50

-600.0 0.0 600.0

w

X(j

w)

-ω ω

π π

cos(ω0t) π[δ(ω – ω0) + δ(ω – ω0)]

Time domain Frequency domainFourier Transform


Frequency Domain

Graphing in the frequency domain: helpful in order to understand filters

0.0

-10.00 0.00 10.00

wX

(jw

)-W W

1

sinc(ω0t)

Time domain Frequency domain

-0.5-10.00 0.00 10.00

t

x(t

)

W/π

π/W

W

WjX

||,0

||,1)(

Fourier Transform


Basic Filters

Electric circuit filter: attenuates (reduces) or eliminates signals at unwanted frequencies

0.00. 00 10. 00

.

Low-pass

0.00. 00 10. 00

.

High-pass

0.00.00 10.00

.

Band-pass

0.00. 00 10. 00

.

Band-stop

ω ω

ω ω


Filter Orders

Higher filter orders provide a higher quality filter

-150

-130

-110

-90

-70

-50

-30

-10

10

30

50

0.1 1 10 100

Frequency (w)

Gai

n (

dB

) 1st Order

2nd Order

3rd Order

4th Order

32nd Order

ECEN 301 Discussion #13 – Phasors 27

Impedance

LjjZL )(RjZR )(

+L–

+C–

Re

Im

-π/2

π/2R

-1/ωC

ωL

ZR

ZC

ZL

Phasor domain

+R–

CjjZC 1

)(

Vs(jω) +–~

+VZ(jω)

–

I(jω)

Z

Impedance of resistors, inductors, and capacitors


Impedance

+C–

Re

Im

-π/2

-1/ωC

ZC

CjjZC 1

)(

Impedance of capacitors

0

1)(

0 as

jZC

+C–

0

1)(

as

jZC

+C–


Impedance

LjjZL )(

+L–Re

Im

π/2

ωL

ZL

Impedance of inductors

0)(

0 as

jZC

+L–

+L–

)(

as

jZC


Resonant Frequency

Resonant Frequency (ωn): the frequency at which capacitive impedance and inductive impedance are equal and opposite (in 2nd order filters)

LCn

1

C+

vi(t)–

+vo(t)

–

L

C+

vi(t)–

+vo(t)

–L

Impedances in series Impedances in parallelCjLj

jZjZ CL

1

)()(


Resonant Frequency


C

LCj

LC

LCjL

LC

LCL

LCj

LLC

j

LjZL

1

1

ZL=jωL

+Vi(jω)

–

+Vo(jω)

–

ZC=1/jωC

C

LCj

jC

LC

CLC

j

CjZC

1/1

/1 Impedances in series


Resonant Frequency


0)( jVo

C

LCjZL

+Vi(jω)

–

+Vo(jω)

–

ZEQ=0C

LCjZC

0

CLEQ ZZZ

Impedances in series


Resonant Frequency


C

LCj

LC

LCjL

LC

LCL

LCj

LLC

j

LjZL

1

1

C

LCj

jC

LC

CLC

j

CjZC

1/1

/1 Impedances in parallel

ZL=jωL+Vi(jω)

–

+Vo(jω)

–

ZC=1/jωC


Resonant Frequency


)()( jVjV io

C

LCjZL

+Vi(jω)

–

+Vo(jω)

–

ZEQ=∞

C

LCjZC

0

/

||

CL

ZZ

ZZ

ZZZ

CL

CL

CLEQ

Impedances in parallel


Low-Pass FiltersLow-pass Filters: only allow signals under the cutoff

frequency (ω0) to pass

)cos()( 1ttvo

0.00. 00 10. 00

.

Low-pass

ω1 ω0 ω2 ω3

)cos(

)cos(

)cos()(

3

2

1

t

t

ttvi

HL(jω)

R

C+

vi(t)–

+vo(t)

–

R

C+

vi(t)–

+vo(t)

–

L

1st Order

2nd Order


Low-Pass Filters

1st Order Low-pass Filters:

0)( tvo

ZR

+Vi(jω)

–

+Vo(jω)

–

ZC=1/jωC

ZR

+Vi(jω)

–

+Vo(jω)

–

ω1: ω → 0 ω3: ω → ∞

)()( tvtv io

0.00. 00 10. 00

.

Low-pass

ω1 ω0 ω3

ZR

+Vi(jω)

–

+Vo(jω)

–


Low-Pass Filters

2nd Order Low-pass Filters:

ZR

+Vi(jω)

–

+Vo(jω)

–

ZC=1/jωC

ZL=jωL

ω1: ω → 0

)()( tvtv io

ZR

+Vi(jω)

–

+Vo(jω)

–

ω2: ω = ωn

0)( tvo

ω3: ω → ∞

0)( tvo

ZR

+Vi(jω)

–

+Vo(jω)

–

0.00. 00 10. 00

.

Low-pass

ω1 ω0 ω2 ω3

ZR

+Vi(jω)

–

+Vo(jω)

–

Resonantfrequency


High-Pass FiltersHigh-pass Filters: only allow signals above the cutoff

frequency (ω0) to pass

)cos()( 3ttvo )cos(

)cos(

)cos()(

3

2

1

t

t

ttvi

HH(jω)

0.00. 00 10. 00

.

High-pass

ω1 ω0ω2 ω3

+vi(t)

–

+vo(t)

–

1st Order

RC

R

+vi(t)

–

+vo(t)

–

2nd Order

C L


High-Pass Filters

1st Order High-pass Filters:

)()( tvtv io

ZC=1/jωC

+Vi(jω)

–

+Vo(jω)

–

ZR=R

ZR

+Vi(jω)

–

+Vo(jω)

–

ω1: ω → 0 ω3: ω → ∞

0)( tvo

0.00. 00 10. 00

.

High-pass

ω1 ω0 ω3

ZR

+Vi(jω)

–

+Vo(jω)

–


High-Pass Filters

2nd Order High-pass Filters:

ZR

+Vi(jω)

–

+Vo(jω)

–

ZC=1/jωC

ZL=jωL

ω1: ω → 0

0)( tvo

ω2: ω = ωn

0)( tvo

ω3: ω → ∞

)()( tvtv io

ZR

+Vi(jω)

–

+Vo(jω)

–

0.00. 00 10. 00

.

High-pass

ω1 ω0ω2 ω3

ZR+

Vi(jω)–

+Vo(jω)

–

ZR+

Vi(jω)–

+Vo(jω)

–

Resonantfrequency


Band-Pass FiltersBand-pass Filters: only allow signals between the

passband (ωa to ωb) to pass

)cos()( 2ttvo )cos(

)cos(

)cos()(

3

2

1

t

t

ttvi

HB(jω)

0.00.00 10.00

.

Band-pass

ω1 ωa ω2 ω3ωb

+vi(t)

–

+vo(t)

–

2nd Order

RC

L


Band-Pass Filters

2nd Order Band-pass Filters:

ZR

+Vi(jω)

–

+Vo(jω)

–

ZC=1/jωC

ZL=jωL

ω1: ω → 0

0)( tvo

ω2: ω = ωn

)()( tvtv io

ω3: ω → ∞

0)( tvo

0.00.00 10.00

.

Band-pass

ω1 ωa ω2 ω3ωb

ZR

+Vi(jω)

–

+Vo(jω)

–

+Vi(jω)

– ZR

+Vo(jω)

–

ZR

+Vi(jω)

–

+Vo(jω)

–

Resonantfrequency


Band-Stop FiltersBand-stop Filters: allow signals except those between

the passband (ωa to ωb) to pass

)cos(

)cos()(

3

1

t

ttvo

)cos(

)cos(

)cos()(

3

2

1

t

t

ttvi

HN(jω)

+vi(t)

–

+vo(t)

–

2nd Order

RC

L

0.00. 00 10. 00

.

Band-stop

ω1 ωa ω2 ω3ωb


Band-Stop Filters

2nd Order Band-stop Filters:

ω1: ω → 0

)()( tvtv io

ω2: ω = ωn

0)( tvo

ω3: ω → ∞

)()( tvtv io

ZR

+Vi(jω)

–

+Vo(jω)

– ZC=1/jωC

ZL=jωL

0.00. 00 10. 00

.

Band-stop

ω1 ωa ω2 ω3ωb

ZR

+Vi(jω)

–

+Vo(jω)

– ZR

+Vi(jω)

–

+Vo(jω)

– ZR

+Vi(jω)

–

+Vo(jω)

–

Resonantfrequency

ECEN 301Discussion #16 – Frequency Response1 Easier to Maintain Than to Retake Alma 59:9 9 And now...

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Transcript of ECEN 301Discussion #16 – Frequency Response1 Easier to Maintain Than to Retake Alma 59:9 9 And now...