analog-to-digital conversion and digital-to-analog conversion (with dsp) es-3

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Analog-to-digital Conversionand

Digital-to-analog Conversion(with DSP)

ES-3

A very simple (…too simple!)

Digital-to-analog converter8R

4R

2R

R

8R

V1

V2

V4

V8

VOUT

out

outout

outoutoutoutout

outoutoutoutout

vvvvv

vvvvvv

vvvvvvvvv

R

v

R

vv

R

vv

R

vv

R

vv

givesKCL

16

248

15842

)(8)(4)(2

8248

:

1248

8421

8421

8421

scitec.uwichill.edu.bb

Analog Input

A simple

Analog-to-digital converter

+-

scitec.uwichill.edu.bb

Analog Input

A simple

Analog-to-digital converter

+-

3.5v

0v

0 0 0 0 0 0 0 0

high

0 0 0 0 0 0 0 1

1v

0 0 0 0 0 0 1 0

2v

0 0 0 0 0 0 1 1

3v

0 0 0 0 0 1 0 0

4v

low

0 0 0 0 0 0 1 1

3v

0 0 0 0 0 1 0 0

Improve accuracy: scale the DAC

Nyquist Sampling RateThe sampling of data must be at least twice the rate as the maximum

frequency of the analog signal: fs>2fmax

V(t)

Time, t

Use a low pass filter with fc = fmax to remove harmonics sine

Nyquist Sampling RateThe sampling of data must be at least twice the rate as the maximum

frequency of the analog signal : fs>2fmax

V(t)

Time, t

Aliasing

Digital Signal Processing

5 5.5 6 6.5 7 7.5 8 8.5 9

x 10-3

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

time (sec)

sign

al

5 5.5 6 6.5 7 7.5 8 8.5 9

x 10-3

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

5 5.5 6 6.5 7 7.5 8 8.5 9

x 10-3

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

5 5.5 6 6.5 7 7.5 8 8.5 9

x 10-3

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

Running average of 10 pts

Digital FilteringRemoving Noise and Interference from a Digital Signal

Find the average of N points

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2

x 104

10-1

100

101

102

103

104

105

frequency (Hz)

sign

al p

ower

(ar

b.)

sine + noise

filtered noise

FFT of a noisy sine wave and its digitally filtered signal

DSP Example

5 5.5 6 6.5 7 7.5 8 8.5 9

x 10-3

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

Simple Notch Filterto remove 60 Hz interference

t (s)

Vin(n)

fS = 1440 samples per second

1/60 s

60 Hz sine

Vout(n) = Vin(n) + Vin(n - fs / 2fnotch) = Vin(n)+Vin(n-12)

Vin(n)Vin(n-12)

This adds the sample that is 1800 out of phase at 60 Hz

Matlab demo

es3_60Hz.m

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