Control Systems MAde Easy

8/4/2019 Control Systems MAde Easy

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FEEDBACK AND CONTROL SYSTEM

MADE EASY

Prepared by:

ENGR. WARREN K. FLERASDESIGN TEAM LEADER

EMERSON INDUSTRIAL AUTOMATION


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OPEN VS CLOSED LOOP


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OPEN LOOP :

Whats the problem does this approach have?

CLOSE LOOP :

Notice , as the term goes on, similarities between filling glass and real feedback systems:

? At the start, you pour fast, and slow down as you reach goal.

? The more accurate you want to be, the longer time it takes.

Know ahead of time your average flow rate when you pour, andthe detailed geometry of the glass. Then, using a stopwatch, youCLOSE YOUR EYESand pour according to your calculations.

What we always do. We pour water, constantly monitoring progress, until wereach our goal. Then we stop. No matter what combination of flow rate glasssize, pitcher geometry, etc. we get the job done.


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OPEN VS CLOSED LOOP

? the main factor that complicated feedback system design is : DELAY

? In a closed loop:

1) measures the variable to be controlled

2) compare this measured value with the desired value or reference and determine the difference

3) use this difference to adjust the controlled variable so as to reduce the difference

? Open loop is easier to build because system stability is not a major problem. At closed loop, stability is a majorproblem.

? At closed loop, stability is a major problem which may tend to overcorrect errors that can cause oscillations ofconstant changing amplitude.

? Open loop if the input is predictable and no disturbances else use closed loop.

? Closed loop used to adjust input to acquire the desired value


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POSITIVE VS NEGATIVE FEEDBACK


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POSITIVE VS NEGATIVE FEEDBACK

Negative feedback reduces the input to gain the proper or desired output

Positive feedback is also called Regenerative Feedback

Positive feedback has only 2 state: High or Low

While negative feedback Vo is linear with respect to input

Positive feedback Vo is in phase with Vin (Comparator)

Negative feedback Vo is out of phase with Vin (diff Amplifier)


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TIME VS FREQUENCY DOMAIN

Time Domain VS Frequency Domain

Oscilloscope is a sample tool to measure

signals characteristics with respect to

time

Spectrum analyzer shows how much

signals lies within each given frequency

band over a range of frequencies

Amplitude vs time display Phase and Gain vs frequency display

i(t)

=C dv/dt Z(s)= 1/sC but s= 2pf

therefore:

Z(s)= 1/ 2 p fC


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Amplitude VS Time


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Gain Phase VS Frequency


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LAPLACE TRANSFORM

Any given signal/ function can be converted between time and frequency domains w/ a pair of mathematical operators

Called a transform.

TRANSFORMS:

? Fourier Series > for repetitive signals, oscillating system

? Fourier transform > non repetitive signals

? Laplace transform > electronic circuits and control system

?

Z-transform > discrete signals, digital signal processing

LAPLACE

? Use to solve ordinary diff eqn to easily solvable algebraic equation

? Transform time domain to frequency domain where both same input & output are functions of complex angularfrequency or radiant per unit time

? Differentiation and integration becomes multiplication and division respectively by s


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Current in a capacitor is proportional to its Capacitance multiplied by the voltage charging rate

it=Cdv/dt C capacitancei current in C as a function of timev v

t

voltage across capacitor as a function of time

Taking the LAPLACE:

{ df(t)/ dt } = sF(s)-f(0)

Is=C(sVs-Vo)

Where: Is = { it}

Vs = { vt} Final value of V

V0 = vt Initial value of V

t=0

Vs=Is/sC + Vo /s eqn 1

The definition of complex impedance Z is the ratio of complex voltage V divided by the complex current I while

Holding the initial state Vo at zero

Zs= Vs

Is Vo=0

{ It } = { Cdv/dt } = sF(s)-f(0)


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Zs= VsIs Vo=0

substituting eqn 1

0Zs = Is/sC + Vo/s

Is Vo=0

Zs= 1sC but s=? = 2pf

Therefore:

Zs = 1

2p fC


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COMPLEX NUMBERS

Combination of real and imaginary parts is called COMPLEX numbers

j2=-1 j=v -1

Z = x + jy

complex z real part imaginary part

impedance resistance reactance

+j inductive

- j capacitive

- +Resistive Resistive


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Ex : a 3 ohm in series with a 4ohm inductive reactance.

3 + j4 4

3

+j

+

hyp= v 42 + 32 = 5

To get the Voltage phase since at series V is different tothe resistance and reactance

Tan = -Xc/ R or +Xl/ R

= arctan 4/ 3

= 530

3 + j4 = 5 ? 53


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Ex : a 3 ohm in series with a 4ohm capacitivereactance.

3 j4

4

3

-j

+

hyp= v 42 + 32 = 5

To get the Voltage phase since at series V is different tothe resistance and reactance

Tan = -Xc/ R or +Xl/ R

= arctan -3/ 4

= 370

4 + j3 = 5 ? - 37


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POLES and ZEROES

ZEROES POLES

Value of s that makes the transfer function goes to zero Value of s that makes the transfer function goes to infinity

Makes amplitude response roll off Makes amplitude response rise

Shifts phase +900 shifts phase -90o

45o at fc '-45o at fc


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3D representation of with s = 1


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This picture of

|H(s)| shows aninverted cone

whose tip is

located at s=1,with the value of

|H(s)|=0

at that point.


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OP-AMPS BASICSThe operational amplifier is an

example of a differential amplifier

in that its output voltage is

proportional to the difference

between the voltages at its twoinput terminals.

The operational amplifier is an example of a differential amplifierin that its output voltage is

proportional to the differencebetween the voltages at its two input terminals.

To calculate the output voltage of an ideal op amp, we multiply the gain, A , of the ideal op amp

by the differencebetween the voltages at the two input terminals of the op amp. For an ideal op

amp, therefore, we can write: vout(t) ??=?A(v+(t) ?- ?v- (t) )

The gain of op amps is large, typically 10,000or more. Thus, if the voltage difference at the input

terminals is 0.0001 Vand the gain A = 10000, then the output voltage of the op amp is ( 0.0001V)(

10000) = 1V.

An ideal op amp draws absolutely no current at either of its input terminals. The 741 op amp

draws a trickle of current at its inputs, but the currents are so small that, for simplicity, we canneglect them in our calculations without noticeable error.


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POLE ZEROES

Low pass filter High pass filter

Integrator Differentiator

Vo= -1/RC ?Vidt Vo= -RC dVi/dt

-90 phase lag +90 phase lead


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POLE ZEROES

Flat response at dc until reaching fcthen -20dB/decade gain slope

Flat response at dc until reaching fc

then +20dB/decade gain slope

Flat response at dc then after fc/10 (-1 decade)starts sloping down until fc having a -60dB at45o then at 10fc (+1decade) flat response at -90o exhibits

Phase lead of 90o max at -1 decade flatresponse 1 decade to fc sloping up to 45o fcto +1 decade sloping up to 90o and 10fconwards flat response


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BODE PLOTIt is a representation of the relative gain and phase shift response of a circuit (wherein anoutput signal is referenced to the input signal ) over a range of frequencies

GAINIt is the relative increase or attenuation of amplitude and is calculated by dBv=20 log (V/1V)

Ex. If V is equal to 1V is 2V or doubled is 0.5V or half is

PHASEThe angular displacement of the signal measured in degrees ? = tan-1 (reactance/resistance)

FREQUENCYIt is plotted in logarithmic scale to help condense the plot to more reasonable size.

Corner FrequencyThe frequency at which the magnitude of the two impedances equal one anotherFor R & C

fc = 1/ (2? RC)

For L & Cfc= 1/(2 ? vLC)

0dB 6dB -6dB


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BASIC RULES FOR STABILITY:

PHASE MARGINgreater than 45o but less than 315o when loop gain is 0dB

GAIN MARGIN

is -20dB or lower when loop phase reaches 360o


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COMPENSATION

FEEDFORWARD

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Control Systems MAde Easy