Chapter 8 Frequency Response Methods

Chapter 8

Frequency Response Methods

莊智清控制工程 Chapter 8 2012 Fall 1

Chapter Contents

1 I t d ti1. Introduction2. Frequency response plots3. Frequency response measurementq y p4. Performance specifications in the frequency domain5. Log magnitude and phase diagrams6 Design examples6. Design examples7. Frequency response methods using control design software8. Sequential design example: disk drive read system9. Summary


Preview

Frequency response: describes how a linear system responds i id l ito sinusoidal inputs.

Use a steady-state sinusoidal input signal and consider the steady-state response of the system as the frequency of the y p y q ysinusoid is varied.

Forms of plotting the frequency response P f f t d it f l t Performance of a system and its frequency response plots Frequency response and time response specifications


8.1 Introduction

System response: described in terms of the complex frequencyi bl d h l i f h l d h lvariable s and the location of the poles and zeros on the s-plane.

Frequency response method

The frequency response of a system is defined as the steady-state response of the system to a sinusoidal input signal. The sinusoid is a unique input signal and the resulting output The sinusoid is a unique input signal, and the resulting output signal for a linear system, as well as signals throughout the system, is sinusoid in the steady state. It differs from the input waveform only in amplitude and phase It differs from the input waveform only in amplitude and phase angles.


Frequency Response

( ) ( ) ( )Y s T s R s

2 2( ) AR s

s( ) sinr t A t Consider with or

s

( ) ( )( )( ) ( )

n

i

m s m sT sq s s p

Assume , then in partial fraction form

12 2

1

( ) n

n

K K sY ss p s p s

1( )ii

p

or

11( ) cos sinnp t p t

ny t K e K e t t

( ) cos siny t t t

If the system is stable, i.e., all pi’s have positive real parts, then as t ,

( )

( ) sin( ( ))

y

A T j t T j

Th t d t t t t i l d d l th it d


The steady-state output signal depends only on the magnitude and phase of T(j) at a specific frequency .

Frequency Response Method

Advantages Availability of sinusoid test signals The design of a system in the frequency domain provides

the designer with control of the bandwidth of the system asthe designer with control of the bandwidth of the system, as well as some measure of the response of the system to undesired noise and disturbances.

Th t f f ti d ibi th i id l t d t t The transfer function describing the sinusoidal steady-state behavior of a system can be obtained by replacing s with jin the system transfer function.

Disadvantages Indirect link between frequency and time responses I iti l t t i t id d Initial state response is not considered.


8.2 Frequency Response Plots

The transfer function of a system G(s) can be described in the f d i bfrequency domain by

( ) ( ) ( ) ( )s jG j G s R jX ( ) Re[ ( )]( ) I [ ( )]

R G jX G j

real part

The transfer function can also be represented by a magnitude

( ) ( ) ( ) ( )s jj j ( ) Im[ ( )]X G jimaginary part

and a phase.( )( ) ( ) ( ) ( )jG j G j e G

magnitude

phase Relationship

phase

1 ( )( ) tan X 2 2 2( ) [ ( )] [ ( )]G R Xd

Two plots will be discussed

1 ( )( ) tan( )R

2 2( ) [ ( )] [ ( )]G R X and


Polar plot: imaginary part versus real part Bode plots: magnitude and phase versus frequency

Polar Plot

Polar plot: the coordinates of the polar plot are the real and i i f G(j )imaginary parts of G(j).

Limitation of polar plot Limitation of polar plot The addition of poles or zeros to an existing system require

the recalculation of the frequency response.


Polar Plot of an RC Filter

The transfer function of a i l RC fil isimple RC filter is

1( )1

G sRCs 1RCs

The sinusoidal steady-state transfer function is1 1 1

The polar plot is obtained from

1

1 1( )1 ( ) 1

G jj RC j

where 11

RC

.

12 2

( ) ( ) ( )1

1 ( ) 1 ( )

G j R jX

j

Also, 2 2

1 11 ( ) 1 ( ) ( ) ( ) ( )G j G

1


12 21

1( )[1 ( ) ]

G 11( ) tan ( )and

Another Polar Plot Example

Consider the transfer function

( )( 1)

KG ss s

( )( 1)

KG jj j

The magnitude and phase are

or

2 4 2 1 2( )( )

KG

1 1( ) tan ( )

The magnitude and phase areand

2

2 4 2 2 4 2( ) ( ) ( ) K KG j R jX

In terms of real and imaginary parts,

( )( )KG

j j p

Let p =1/, then

1( ) KG 1( ) ( ) 90 tan ( )j j p p

The magnitude and phase at 1 are

and


11 1

( )( )

Gj j p

1 1 1 1( ) ( ) 90 tan ( )j j p p and

Another Polar Plot Example

( )( 1)

KG jj j

2

2 4 2 2 4 2( ) K KG j


Bode Plots or Logarithmic Plots

The frequency response is represented as two curves The logarithm of magnitude versus log The phase versus log

Advantages of Bode plot Advantages of Bode plot Bode plots of system in series simply add Bode’s phase-gain relationship is given in terms of logarithms

f h d iof phase and gain A much wider range of system behavior, from low to high

frequency behavior, can be displayed on a single plot Bode plot can be determined experimentally Stability margins (gain and phase margins) can be extracted

from the Bode plotp Dynamic compensation design can be based entirely on

Bode plot


Bode Plot

The transfer function in the frequency domain is

The natural logarithm of G(j) is

( )( ) ( ) jG j G e

The natural logarithm of G(j) isln ( ) ln ( ) ( )G j G j

The logarithm of the magnitude is normally expressed in terms of the logarithm to the base 10, thus

logarithm gain 20log ( )G

where the units are in decibels (dB).

10logarithm gain 20log ( )G

Gain 0.01 0.1 0.2 0.5 1.0 2.0 5.0 10.0 100.0


Gain in dB -40.0 -20.0 -13.98 -6.02 0.0 6.02 13.98 20.0 40.0

Bode Plot Example

For the transfer function1 1( )

1 1G j

j RC j

The logarithm gain is1 2

22

120log 20log 10log(1 ( ) )1 ( )

G

( )1 1

jj RC j

g g

For small frequencies, i.e., <<1/, the logarithm gain is2g g g( ( ) )

1 ( )

20log 10log(1) 0 dBG

For large frequencies, i.e., >>1/ , the logarithm gain is20log 20log dBG

When =1/,

The phase angle is20log 10log 2 3.01dBG

p g

The frequency =1/ is called the break frequency or corner

1( ) tan


q y q yfrequency.

Bode Plot Example

1( )1

G jj

1j

220log 10log(1 ( ) )G

1( ) tan


Bode Plot

Logarithm scale is preferred.g p

For example, when >>1/,

The asymptotic curve is a linear function of log 20log 20log 20logG

y p g

An interval of two frequencies with ratio equal to 10 is called a decade.

The difference between the logarithmic gains, for >>1/, over a decade is 20 dB. Suppose that 0.12 = 1 >>1/, then

20l ( ) 20l ( ) 20l ( 20l )G G1 2 1 2

1

20log ( ) 20log ( ) 20log ( 20log )

20log

G G

That is, the slope of the asymptotic line for the first-order transfer

2

20 dB


function is –20 dB/decade.

Bode Plot

The frequency interval 2 = 21 is called an octave of frequencies. q y 2 1 qThe difference between the logarithmic gains, for >>1/, for an octave, is

The slope of asymptotic line for the first-order transfer function is

11 2

2

20log ( ) 20log ( ) 20log 6.02 dBG G

-6 dB/octave or –20 dB/decade.

The primary advantage of the logarithmic plot is the conversion of multiplicative factors such as (j+1) into additive factor 20(j+1) into additive factor 20 log(j+1).


Bode Plot

Consider the transfer function

The logarithm magnit de of G(j ) is

12

1 1

(1 )( )( ) (1 ) [1 (2 ) ( ) ]

Qb i i

N M Rm m k k k k

K jG jj j j j

The logarithm magnitude of G(j) is

1

20log ( ) 20log 20 log 1 20log ( )Q

Nb i

i

G K j j

1

2

1 1

2 20 log 1 20 log 1 ( ) ( )

i

M Rk

mm k k k

jj j

The Bode diagram can be obtained by adding the plot due to each individual factor.

The separate phase is obtained as The separate phase is obtained as1 1 1

2 21 1 1

2( ) tan (90 ) tan tan ( )Q M R

k ki m

i m k k

N


which is simply the summation of the phase angles due to each individual factor of the transfer function.

k

Bode Plot

Four kinds of factors1. Constant gain2. Poles (or zeros) at the origin3 Poles (or zeros) on the real axis

bK( )j

( 1)j 3. Poles (or zeros) on the real axis4. Complex conjugate poles (or zeros)

We can determine the logarithmic magnitude plot and phase

( 1)j 2[1 (2 ) ( ) ]n nj j

g gangle for these four factors and then utilize them to obtain a Bode diagram for any general form of a transfer function.

Typically the curve for each factor are obtained and then added Typically, the curve for each factor are obtained and then added together graphically to obtain the curves for the complete transfer function.

Furthermore, the procedure can be simplified by using the asymptotic approximation to these curves and obtaining the actual curves only at specific important frequencies.


actual curves only at specific important frequencies.

Bode Plot of a Constant Gain

The logarithmic gain is

and the phase angle is zero.20log constant in dBbK

The gain is simply a horizontal line on the Bode diagram If the gain is negative, the logarithmic gain remains the same

and the phase becomes 1800and the phase becomes -1800.


Bode Plot of Pole/Zero at the Origin

Logarithmic gain phase slope

A simple pole -20 log -900 -20 dB/decade

Multiple pole 20 N log 900 N 20 N dB/decadeMultiple pole -20 N log -900 N -20 N dB/decade

A simple zero 20 log +900 +20 dB/decade


Bode Plot of a Real-Axis Pole

The logarithmic magnitude of the pole factor is 1(1 )j

Th t ti

2 2120log 10log(1 )1 j

10

0

dB

The asymptotic curves For << 1/, 20 log 1 = 0 dB. For >> 1/ -20 log dB

0

-10

-20

The intersection of the two curves occurs

For >> 1/, 20 log dB. The high-frequency slope is -20 dB/decade.-30

01.0

1

1.0

100

10

0 The intersection of the two curves occurs when = 1/, which is the break frequency.

The actual logarithmic gain when = 1/45

is -3 dB for this factor. The phase angle is

1( ) tan10

100

1.0

01.0

1

90


1( ) tan

Bode Plot of a Real-Axis Zero

Bode diagram of a zero factor (1 )j

Obtained in the same manner as that of the pole.

The slope is positive at + 20 dB/decade (beyond the break frequency)

The phase angle is 1( ) tan


Linear Approximation of Real-Axis Pole/Zero


Bode Plot of Complex Conjugate Pole/Zero

The quadratic factor for a pair of complex conjugate poles can be itt i li d fwritten in normalized form as

The logarithm magnitude and the phase angle are respectively 21 (2 ) ( )n nj j 2 1[1 2 ]j u u nuwhere

g g p g p y

2 2 2 220log ( ) 10log((1 ) 4 )G u u

12

2( ) tan ( )1

uu

When u<<1, the magnitude is 0 dB and the phase angle approaches 00. When u>>1, the logarithmic magnitude approaches – 40 log u, which

results in a curve with a slope of 40 dB/decaderesults in a curve with a slope of -40 dB/decade. The phase angle, when u>>1, approaches -1800.

The magnitude asymptotes meet at the 0 dB line when u = / = 1 The magnitude asymptotes meet at the 0-dB line when u = /n = 1.







The maximum value of the frequency response Mp occurs at the resonant frequency r.

When the damping ratio approaches zero, then r approaches n,, the natural frequency.

The resonant frequency is

and the maximum value of the magnitude |G()| is

21 2 , 0.707r n

for a pair of complex poles. 2 1( ) (2 1 ) , 0.707pM G

p p p

When > 0.707, the resonant peak is Mp = 1. The phase at r is 21 2


p r

1 1 2( ) tanr


With respect to 2 2 2 2 2 1 2( ) [( ) (2 ) ]n n nG

we have

2 2 2 2 2 1 2 2 2 2 2( ) 1 [( ) (2 ) ] { 4 ( ) 8 }dG

To have a peak magnitude, 2 2 2 22n n

( ) [( ) (2 ) ] { 4 ( ) 8 }

2 n n n n nd

Thus, the resonant frequency r is 21 2r n




Frequency Response

The frequency response curves can be evaluated on the s-plane by determining the vector lengths and angles at various frequencies along the s=j -axis.

Consider the second-order factor

2

2 2 21( )

( ) 2 1 2nG s

s s s s The transfer function evaluated for real frequency s=j is

( ) 2 1 2n n n ns s s s

2 2

( ) n nG j

where and are the complex conjugate*1s

* *

1 1 1 1

( )( )( ) ( )( )

n nn s jG j

s s s s j s j s

21s jwhere and are the complex conjugate poles.

1s 1 1ns j


Frequency Response

The vectors j-s1 and j-s1*

are vectors from the poles to the frequency j. The magnitude and phase may bemagnitude and phase may be evaluated for various specific frequencies.

The magnitude is

2

( ) nG

and the phase

*

1 1

( )Gj s j s

*1 1( ) ( ) ( )j s j s


Frequency Response


Asymptotic Curves for Basic Terms

( )G j KGain Zero 1( ) 1 /G j j



11( ) 1 /G j j Pole at the originPole ( ) 1/G j j



12( ) 1 2 / and 0 1 1G j j u u u

Two complex poles

( ) 1 2 , / and 0.1 1nG j j u u u


Example: Twin-T Network

Consider the twin-T network where .2

02

( ) ( ) 1( )( ) ( ) 4 1

V s sG sV

RC 2( ) ( ) 4 1inV s s s

planes

poles: 2 3

At = 0 the gain is 1 and the phase is 00

planeszeros: 1j

At = 0, the gain is 1 and the phase is 00.

At =1/, the gain is 0 and the phase angle of the vector from the zero at s=j1 passesof the vector from the zero at s=j1 passes through a transition of 1800.

When approaches the gain becomes 1


When approaches , the gain becomes 1 and the phase is 00 again.

Minimum Phase and Non-minimum Phase

Consider two transfer functions

1( ) s zT ss p

2 ( ) s zT s

s p

and

The magnitudes are the same, while the phases are different.2 2

1 2( ) ( )zT j T j 1 22 2

( ) ( )T j T jp


Minimum Phase and Non-minimum Phase

A transfer function is called a minimum phase transfer function if all its zeros lie in the left-hand s-plane. It is called a nonminimum phasetransfer function if it has zeros in the right-hand s-plane.

The range of phase shift of a minimum phase transfer function is the least possible orfunction is the least possible or minimum corresponding to a given amplitude curve, whereas the range of nonminimum phase curve is greater than the

i i ibl f th iminimum possible for the given amplitude curve.

A transfer function is stable if all its poles lie in the left hand s plane


A transfer function is stable if all its poles lie in the left-hand s-plane. It is unstable if it has poles in the right-hand s-plane.

Example: All Pass Network

An all-pass network is a system whose magnitude is 1 over allwhose magnitude is 1 over all frequencies.

The poles and zeros of an all-pass filter is thus symmetric topass filter is thus symmetric to the j-axis.

Q: What is the transfer function?


An Example of Drawing the Bode Diagram

The Bode diagram of a transfer function G(s), which contains several zeros and poles is obtained by adding the plot due toseveral zeros and poles, is obtained by adding the plot due to each individual pole and zero.

Consider the transfer function5(1 0 1 )j

The factors, in order of their occurrence as frequency increases,

25(1 0.1 )( )

(1 0.5 )(1 0.6( 50) ( 50) )jG j

j j j j The factors, in order of their occurrence as frequency increases,

are1. Constant gain2 Pole at the origin2. Pole at the origin3. Pole at = 24. Zero at = 10.5 A pair of complex5. A pair of complex

poles at = 50.

The total asymptotic magnitude can be plotted by adding the


The total asymptotic magnitude can be plotted by adding the asymptotes due to each factor.

Bode Diagram Example

The exact magnitude curve is obtained by adding the correction terms.

The total phase characteristic () is obtained by adding the phase due to each factor.


Bode Diagram Example

One may obtain approximate curves for the magnitude and phase shift of a transfer function in order to determine the important frequency ranges.

Then, within the relatively small important frequency ranges, the exact magnitude and phase shift can be evaluated by using the exact equation.

On the other hand, computer-aided tools can often be used., p


8.3 Frequency Response Measurements

A sine wave can be used to measure the open-loop frequency response of a control systemresponse of a control system.

In practice, a plot of amplitude versus frequency and phase versus frequency will be obtained.

From these two plots, the open-loop transfer function GH(j)can be deduced.

Similarly the closed-loop frequency response of a control y p q y psystem, T(j), may be obtained and the actual transfer function deduced.

Key for the determination of transfer function based on yfrequency response data Form of the transfer function: poles, zeros, complex or real. Gain Gain Pole and zero locations Damping ratio and resonant frequency.


Frequency Response Measurements

An instrument that can be used for frequency response measurement is the dynamic signal analyzer say HP 25670Ameasurement is the dynamic signal analyzer, say HP 25670A.

Please read the application note control system development using dynamic signal analyzers (for a review of control

i i d th li ti f th i t t)engineering course and the application of the instrument).


Deduction of Transfer Function

Observations: The magnitude declines at The magnitude declines at

about –20dB/decade as increased between 100 and 1000, and because the phase is -450 and the magnitude is –3dB at 300 rad/s, one factor is a pole at p =300p1 =300.

A pair of quadratic zeros with =0.16 exist at n=2450 because that the phasebecause that the phase changes abruptly by nearly +1800, passing through 00 at n=2450.

The proposed transfer function T(s) is

2( ) (2 ) 1s s


2

1 2

( ) (2 ) 1( )( 1)( 1)

n ns sT ss p s p

Deduction of Transfer Function

The difference in magnitude from the corner frequency ( =2450) The difference in magnitude from the corner frequency (n=2450) of the asymptotes to the minimum response is 10 dB, which indicates that = 0.16.indicates that 0.16.

Because the slope of the magnitude returns to 0 dB/decade as exceeds 5000, there is a second pole as well as two zeros.exceeds 5000, there is a second pole as well as two zeros.

This second pole is at p2 =20000 because the magnitude is –3 dB from the asymptote and the phase is -450 at this point.from the asymptote and the phase is 45 at this point.

The transfer function is thus 2( 2450) (0.32 2450) 1( ) s sT

( ) ( )( )( 300 1)( 20000 1)

T ss s


8.4 Performance Specificationin the Frequency Domainin the Frequency Domain

How does the frequency response of a system relate to

For the second-order

the expected transient response of the system?

system, the closed-loop transfer function is

2

( ) sT s

2 2( )

2s

n n

T ss s

At the resonant frequency r, a Resonant peakr

maximum value of the frequency response Mp isattained.

Bandwidth

The bandwidth is the frequency B at which the frequency response has declined 3 dB


from its low frequency value.

Time Response and Frequency Response

As the bandwidth B increases, the rise time of the step response of the system will increaseof the system will increase.

The overshoot to a step input can be related to Mp through the damping ratio .

As the resonant peak M increases in magnitude the overshoot As the resonant peak Mp increases in magnitude, the overshoot to a step input increases.

The magnitude M indicates The magnitude Mp indicates the relative stability of the system.

The bandwidth of a The bandwidth B of a system can be approximately related to the natural frequency of the systemfrequency of the system.

1.1961 1.8508B

n


n

Frequency Response Specifications

Frequency domain specifications

Relative small resonant magnitude: Mp < 1.5 for example

Relative large bandwidth so that the system time constant = 1/B is sufficiently small

The steady-state error for a specific test signal can be related to the gain and number of integrations (poles at the origin) of the open loop transfer function.

The stead state error for a ramp inp t is specified in terms of K The steady-state error for a ramp input is specified in terms of Kv, the velocity constant. The steady-state error is

A

where A is the magnitude of the ramp input

lim ( )

tv

Ae tK


where A is the magnitude of the ramp input.


The velocity constant is 2

This transfer function can be rewritten as

2

0 0lim ( ) lim

( 2 ) 2n n

v s sn

K sG s ss s

2( )( (2 ) 1) ( 1)

n v

n

KG ss s s s

In general, if the open loop transfer function of a feedback system is written as

1(1 )( )Mi i

N QK jG j

then the system is type N and the gain K is the gain constant for the steady state error

1

( )( ) (1 )N Q

k k

G jj j

the steady-state error. For a type-zero system that has an open-loop transfer function

( ) KG j

莊智清控制工程 Chapter 8 2012 Fall 51K = Kp is the low frequency gain on the Bode diagram.

1 2

( )(1 )(1 )

G jj j


The gain constant for a type-one system appears as the gain of the low frequency section of the magnitude characteristic. Indeed, Kv is equal to the frequency at which this portion of the magnitude h t i ti i t t th 0 dB licharacteristic intersects the 0-dB line.

For the open-loop transfer function5(1 )j

22

1

5(1 )( )(1 )(1 0.6( ) ( ) )n n

jG jj j j j

Its velocity constant is Kv = 5. The low frequency asymptote

15( ) ( ) ( ) , 1vKG jj j

intersects the 0-dB line at = 5. j j


8.5 Log Magnitude and Phase Diagrams

Plot the logarithmic magnitude in dB versus the phase angle for a range of frequenciesrange of frequencies.

The diagram can often be obtained from the Bode diagram. The shape of the locus of the frequency response on a log-

magnitude phase diagram is particularly important as the phasemagnitude-phase diagram is particularly important as the phase approaches –1800 and the magnitude approaches 0 dB.


Log Magnitude and Phase Diagrams

5(0.1 1)( ) jGH j

2 2

( )( )(0.5 )(1 0.6( 50) ( 50) )

jGH jj j j j

15( )

(0.5 1)( 6 1)GH j

j j j


8.6 Design Examples: Engraving Machine Control SystemEngraving Machine Control System

Select K to yieldSelect K to yield acceptable step response


Engraving Machine Control System

Approaches Obtain the open loop Bode diagrams Obtain the open-loop Bode diagrams Obtain the closed-loop Bode diagram Use the closed-loop Bode diagram to predict the time

response Open-loop Bode diagram when K=2 is obtained.

2( )( 1)( 2)

G jj j j


Engraving Machine Control System

The closed-loop transfer function2 2

3 2

2( )3 2 2

T ss s s

or

The Bode diagram indicates that 20

3 2

2( )(2 3 ) (2 )

T jj

The Bode diagram indicates that 20 log|T| = 5 dB at r=0.8, that is, Mp is 1.78.

If a second order system approximation If a second-order system approximation is used, we have = 0.29 and, accordingly, n=0.88.

The second order approximation is The second-order approximation is

2

2 2 20.774( )

2 0.51 0.774n

n n

T ss s s s

and the resulting overshoot is 37% with settling time 15.7 seconds.

The actual overshoot for a step input is

n n


The actual overshoot for a step input is 34% and the actual settling time is 17 seconds.

8.8 Sequential Design Example:Disk Drive Read SystemDisk Drive Read System

2

5 0.05 1( ) ( 1)( 1) ( 1) [1 (2 / ) ( / ) ]

G s K s


21 2

( ) ( )( 1) ( 1) [1 (2 / ) ( / ) ]n ns s s s s

Disk Drive Read System

Bode plot when K = 400.


Disk Drive Read System

Open-loop Bode gain Closed-loop Bode gain


Bode Diagrams for Typical Transfer Functions

1 1K

s



1 2( 1)( 1)K

s s



( 1)( 1)( 1)K

s s s 1 2 3( 1)( 1)( 1)s s s



Ks



1( 1)K

s s



1 2( 1)( 1)K

s s s



1 2

( 1)( 1)( 1)

aK ss s s



2Ks



21( 1)

Ks s



12

1

( 1) , ( 1)

aa

K ss s



3Ks



3

( 1)aK ss



3

( 1)( 1)a bK s ss



( 1)( 1)K s s

1 2 3 4

( 1)( 1)( 1)( 1)( 1)( 1)

a bK s ss s s s s



( 1)K s

2

1 2

( 1)( 1)( 1)

aK ss s s


Summary

Frequency response: describes how a linear system responds i id l i i h dto sinusoidal inputs in the steady state.

Polar plot Logarithmic plot (Bode plot) Logarithmic plot (Bode plot) Log magnitude versus phase diagram

Bode plot: Closed-loop performance (maximum magnitude, resonant

frequency) Steady state performance (error constant) Steady state performance (error constant)


Bode Gain Phase Relationship

Let G(s) be a stable and minimum phase transfer function, then th h f G( ) t f b t dthe phase of G(s) at frequency can be computed as

01 ln | ( ) | | |( ) lncoth

2d GG j d

dwhere

The function

2d 0ln( / )

i hti th l f th t f f ti

0

0

| |lncoth ln2

serves as a weighting on the slope of the transfer function.

3

3.5

4

1

1.5

2

2.5


10-1 100 1010

0.5

Documents

Chapter 8 Frequency Response Methods