ABE425 Engineering Measurement ABE425 Engineering Measurement SystemsSystems

Measurement Systems with Electrical Signals

Dr. Tony E. Grift

Dept. of Agricultural & Biological EngineeringUniversity of Illinois

Agenda

1. AC and DC signals2. Transducers3. OpAmps4. Active Filters5. Loading error examples

AC and DC signals

Alternating Current (AC)Direct Current (DC)Most signals have both!

DC component (offset) measurement: Put DMM on DCAC component measurement: Put DMM on AC

Scope gives Amplitude and peak-peak valueTo get the scope trace at 0: Put input on gnd.

How is the scope amplitude related to the AC value on the DMM?

Let’s figure out what the AC RMS value is of this signal

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

x 10-3

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

1

11,000

2* *

T ms

f HzT

radpi f

s

RMS value represents Power

2 212

1.44100DC

UP Watt

R

2

2 2

U U

2

2

U

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

x 10-3

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

x 10-3

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1DC: 12 Volt, R = 100 Ω

AC RMS value must give same power as DC of the same value

Mean value

Root Mean Square

2 212

1.4410

RMSAC

UP Watt

R

Question

If an AC Voltmeter shows 12V RMSWhat is the amplitude of this signal?

Simple AC Voltmeters measure the amplitude and divide by sqrt(2)

This only works for Sinusoidal signals!!

True RMS voltmeters measure real Power, Resistance and take the square root of the ratio

This works for ANY signal since it follows the definition of the power equalization of DC and AC signals

The RMS value is NOT the mean of the AC signal. It is the Root of the Mean of the Squared value!

* 2 12* 2 16.97RMSU U V

Digital TRUE RMS meters digitize the signal and compute the RMS value from the definition

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

x 10-3

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Digitize at least one cycle of the signalSquare itCompute mean value

Integrate signalDivide by cycle time

Take square root

A Transducer converts a physical measurand into an electric signal

AntennaCathode Ray Tube. LCD monitorFluorescent light, light bulb. Light Emitting Diode (LEDs)Magnetic stripe cards

Photocells/Light

Combined model of (a) input source, (b) amplifier and (c) output load

You want to prevent loading errorsChoose Ri high

Choose Ro low

Non-inverting OpAmp

Ofi

i

i

O

VRR

RV

VV

VVAV

i

fii

i

fiO

i

fi

i

O

i

fi

i

O

ifi

iO

Ofi

iiO

R

RVV

R

RRV

A

V

RR

R

A

V

V

RR

RA

AV

AVRR

RAV

VRR

RVAV

1lim

1

1

1

1

i

fiO R

RVV 1

Non-inverting OpAmp : Virtual ‘ground’ principle

i

fiO R

RVV 1

i

O

VV

VVVVAV

Since no current is flowing intothe OpAmp:

f

iO

i

i

R

VV

R

V

What is the input resistance for the source ?

iV

OpAmps have a limited band width (741 is about 1 MHz)

Inverting OpAmp

0

V

RR

RV

RR

RVV

VVAV

if

iO

if

fi

O

ii

fO

fi

i

fi

fi

fi

i

fi

fi

O

fi

fi

fi

iO

fi

iO

fi

fiO

VR

RV

A

RR

R

A

RR

RV

RR

RA

RR

RAV

V

RR

RAV

RR

RAV

RR

RV

RR

RVAV

lim

11

1

ii

fO V

R

RV

Inverting OpAmp: Virtual ground principle

0

V

VVVVAVO

ii

fO V

R

RV

i

f

i

O

f

O

i

i

R

R

V

V

R

V

R

Vi

Since no current is flowing intothe OpAmp:

What is the input resistance for the source ?

iV

Capacitor equation

Charge across Capacitor (Coulomb)Capacitance value (Fahrad (German for bicycle))Voltage across Capacitor (Volt)Current through Capacitor (Ampere)

dt

dQti

tVCtQ

CC

cC

*

ti

tV

C

tQ

C

c

C

Integrating Inverting OpAmp : Virtual ground

dttVRC

tV iO

1

0

V

VVVVAVO

dttVRC

tVdt

tdVRCtV

R

tV

dt

tdVC

dt

tdQti

tVtVR

tVti

iOO

i

iOCC

OCi

C

1

,

Differentiating Inverting OpAmp: Virtual ground

0

V

VVVVAVO

dt

tdVRC

dt

tdVRCtV

R

tV

dt

tdVC

dt

tdQti

R

tVti

iCO

OCCC

OC

dt

tdVRCtV i

O

A buffer gives an near infinite input resistance and a near zero output resistance

O

O IN

V A V V V V

V V

An instrumentation amplifier has two high impedance (resistance) inputs

OpAmps have a very high input impedance (resistance)This configuration has superb Common Mode Rejection Ratio (CMRR) up to 70 dB

A simple way to attenuate a signal is by using a voltage divider

2

1 2O i

RV V

R R

Decibel notation

Addition is much simpler than multiplication

Notation in Bel (after Alexander Graham Bell)

For Power

For Voltages (Power ~ Voltage2)

In deciBel (0.1 Bel)

Belin log10 P

log*2 log 10210 UU

(dB) deciBelin log*20Belin log*2 1010 UU

Common filters arrangements are low-pass, high-pass, band-pass and band-stop (notch)

Butterworth filters are smooth, but have a high roll-on roll-off factor.

Chebyshev filters have sharp roll offs but lots of ripple

Bessel filters are tame (no ripples) but a gradual roll off

Active filters combine amplification and filtering in one circuit!

What is the input impedance the source ‘sees’?

Active Low-Pass filter analysis (1st order)

1 1

22

2 22

2

1*

1\ \

1 1

Z R

RRj C

Z Rj C j CRR

j C

20

1

O

i

V RG

V R 2

1

O

i

V ZG

V Z

2 20

1 1

1 1

1 1

Z RG G

Z R j j

Without C In general with impedances

0

1

1dB dB

dB

G Gj

Notation differences

Wheeler / Ganji System

frequency in Hz

Corner frequency

Grift System

frequency in rad/s

Corner frequency rad/s

Time constant (s)

2 f

cyclef Hz

s

RC

1

1G j

j

11 C

1

1 2G j

j f RC

1

2Cf RC

Active High-Pass filter with OpAmp and Capacitor / Resistor pair (1st order)

What is the input impedance the source ‘feels’?

Active High-Pass filter analysis

1 1

2 2

1Z R

j C

Z R

20

1

O

i

V RG

V R 2

1

O

i

V ZG

V Z

2 2 2 20 0

1 1 11

1 1 1

Z R j R C jG G G

Z j RC jRj C

Without C In general with impedances

0 21

1

1dB dB dBdB

G G jj

High-pass and low pass section separated by OpAmp

Bandwidth and Distortion

ABE425 Engineering Measurement ABE425 Engineering Measurement SystemsSystems

Measurement Systems with Electrical Signals

The End

Dept. of Agricultural & Biological EngineeringUniversity of Illinois