# Digital Signal Processing(3) by pptfiles

VIEWS: 3 PAGES: 22

• pg 1
```									 Digital to Analogue Conversion
• Natural                x(t )  sin(2ft)
signals tend to
be analogue

x(n)  sin(2fnt s )
• Need to
convert to
digital
Sampling
• Need to hold the signal steady for long enough
to enable the A to D converter to generate an
output

Sample      Analogue
and        To digital
Hold       converter
• Quantization converts continuous value to
a discrete (usually integer) value

• Output value is rounded so accuracy lost

• Maximum quantization error of ±0.5 lsb

• Error is combined with signal as noise
Quantization accuracy
• Least significant bit determines accuracy. So for
a 2 Volt peak to peak signal, an 8 bit converter
can accurately represent multiples of 7.81mV
but anything in between will be rounded

If input range is  1 volt then for 8 and 16 bit converters:
full voltage range 2 volts
lsb value          num bits
    8
 7.81mV
2              2
or
full voltage range 2 volts
lsb value          num bits
 16  30.5V
2              2
Quantization error

Error is at most
±1/2 an lsb, or
±3.905 mV for the 8
bit converter or
±15.25µV in the 16
bit case
Quantization error
• Relatively small signal changes are
subject to severe quantization errors
• Quantization error
creates steps

• Steps create
distortion which is
visible in the
frequency domain

• Noise shown on dB
scale as it is
relatively small
compared to the
signal
Sampling theory
• Signal needs to be sampled at twice the
speed of the fastest change to be captured
• Shannon or the Nyquist sampling theorem,
(authors of 1940s papers)
• Theorem states that a continuous signal
can be properly sampled, only if it does
not contain frequency components above
one-half of the sampling rate
Correctly sampled signal
• Signal frequency is 0.09 of the sample rate
(i.e. sample rate is about 11x signal freq)

• e.g. 90Hz signal sampled at 1kHz
Sample rate still ok
• Signal frequency is 0.31 of the sample rate
(i.e. sample rate about 3x signal freq)

• 3.2 samples / cycle but freq still preserved
Improper Sampling
• Signal frequency is 0.95 of the sample rate (i.e.
sample rate only slightly higher than signal freq)

• Only 1.05 samples per cycle. Produces a 0.05Hz
alias signal which is mixed with the original
Sidebands
• Sampling a signal is effectively
multiplication of signals in the time domain

• Multiples of the sample frequency are
produced as well as sum and difference
frequencies (sidebands)
No sampling no sidebands
• Time domain to frequency domain of an
analogue signal
Sampled signal produces
sidebands
Incorrectly sampled signal
• Breaching Nyquist causes aliasing with
overlapping sidebands
Simulated sampling
• Using a sample rate of 1kHz, the frequency
spectrum with noise was calculated from:
f (n)  sin(2 50 nts )  sin(2 120 nts )  sin(2 300 nts )

• Then modified to illustrate aliasing by changing
300Hz signal to 800Hz:
f (n)  sin(2 50 nts )  sin(2 120 nts )  sin(2 800 nts )
Correct sample rate

f (n)  sin(2 50 nts )  sin(2 120 nts )  sin(2 300 nts )
Aliasing at 200Hz

f (n)  sin(2 50 nts )  sin(2 120 nts )  sin(2 800 nts )
Anti-alias filter
• Frequencies higher than those of interest (such
as noise) need to be blocked before sampling.
Use an analogue low pass filter
DSP system
Analogue     Sample   Analogue              Digital      Analogue
Anti-alias    and     to digital         to analogue   Reconstruction
DSP
Filter      Hold    converter           converter        Filter

• Low pass input filter removes F > 0.5F(S)
• Reconstruction filter removes high
frequency F(S) multiples
Sound Blaster block diagram

```
To top