11
SPECTRUM F DIGITAL SIGNALS 1 Objectie The aim is to graphically portray the fundamental effects on digital signal within a frequency spectrum and to understand the effects of filters on digital signal. We also aim to understand the behaviour digital under frequency domain and time domain. 2 Task 1 RMS of a signal 50Hz sin wave Figure1: sine wave 50Hz Vrms=1.61V Multimeter reading Theoretical calculation 1.59Vrms V V V peak rms 61 . 1 2 = = The variation between the calculated and measured RMS values is very small, thus we approach almost ideal cases. In most instances there is a large variation between calculated values and measured values because practical components are defective.

Spectrum of a Digital Signal

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Analysis of digital signals at high frequency

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Page 1: Spectrum of a Digital Signal

SPECTRUM OF DIGITAL SIGNALS

1 Objective

The aim is to graphically portray the fundamental effects on digital signal within a frequency

spectrum and to understand the effects of filters on digital signal. We also aim to understand

the behaviour digital under frequency domain and time domain.

2 Task 1

RMS of a signal

50Hz sin wave

Figure1: sine wave 50Hz

Vrms=1.61V

Multimeter reading

Theoretical calculation

1.59Vrms V

VV

peak

rms 61.12

==

The variation between the calculated and measured RMS values is very small, thus we

approach almost ideal cases. In most instances there is a large variation between calculated

values and measured values because practical components are defective.

Vpeak=2.275V

Vrms=1.61V

measured with a

digital oscilloscope

Vpeak to peak=4.56

measured with a

digital oscilloscope.

Page 2: Spectrum of a Digital Signal

SPECTRUM OF DIGITAL SIGNALS

RMS voltage with zero DC offset

50Hz block wave

Figure2: 50Hz square wave

Multimeter reading Vrms=2.72V

Theoretical computation Vpeak=Vrms=2.51, only for square waves.

3 Task 2

-3dB Filter Point

Filter Design

The required bandwidth is 100 kHz; a capacitor of 1.6nf was chosen to meet bandwidth

specifications.

Ω=×××

==−

7.994100000106.12

1

2

19

ππfcR , 1 k Ω was chosen.

Figure4: RC low pass filter circuit

Vout

Vpeak-peak=5.08V

Vpeak=2.51V

Page 3: Spectrum of a Digital Signal

SPECTRUM OF DIGITAL SIGNALS

Katlego Mohlala Electronics 4A01 Page 3

Table 1

freq(Hz) Vin(V) Vout(V) Vout/Vin log(freq)dB

10 1.5 1.53 1.02 1

20 1.59 1.59 1 1.30103

31 1.58 1.58 1 1.49136169

40 1.61 1.61 1 1.60205999

50 1.6 1.6 1 1.69897

60 1.61 1.61 1 1.77815125

70 1.63 1.63 1 1.84509804

80.9 1.66 1.66 1 1.90794852

90 1.63 1.64 1.006135 1.95424251

100 1.65 1.65 1 2

110 1.67 1.67 1 2.04139269

120 1.68 1.69 1.005952 2.07918125

130 1.62 1.62 1 2.11394335

140 1.61 1.62 1.006211 2.14612804

150 1.66 1.66 1 2.17609126

160 1.66 1.67 1.006024 2.20411998

170 1.61 1.62 1.006211 2.23044892

180 1.6 1.6 1 2.25527251

190 1.63 1.64 1.006135 2.2787536

200 1.64 1.65 1.006098 2.30103

251 1.61 1.62 1.006211 2.39967372

300 1.59 1.6 1.006289 2.47712125

350 1.59 1.6 1.006289 2.54406804

400 1.61 1.62 1.006211 2.60205999

450 1.6 1.61 1.00625 2.65321251

500 1.59 1.6 1.006289 2.69897

600 1.59 1.6 1.006289 2.77815125

700 1.6 1.61 1.00625 2.84509804

800 1.59 1.6 1.006289 2.90308999

900 1.6 1.61 1.00625 2.95424251

1000 1.6 1.61 1.00625 3

1500 1.59 1.6 1.006289 3.17609126

2000 1.59 1.6 1.006289 3.30103

2500 1.59 1.6 1.006289 3.39794001

3000 1.59 1.6 1.006289 3.47712125

3500 1.59 1.6 1.006289 3.54406804

4500 1.59 1.6 1.006289 3.65321251

5500 1.59 1.6 1.006289 3.74036269

6500 1.59 1.6 1.006289 3.81291336

7500 1.59 1.59 1 3.87506126

8500 1.59 1.59 1 3.92941893

9500 1.59 1.59 1 3.97772361

10000 1.59 1.59 1 4

11000 1.59 1.58 0.993711 4.04139269

12000 1.59 1.58 0.993711 4.07918125

13000 1.59 1.58 0.993711 4.11394335

14000 1.59 1.57 0.987421 4.14612804

Page 4: Spectrum of a Digital Signal

SPECTRUM OF DIGITAL SIGNALS

15000 1.59 1.57 0.987421 4.17609126

16000 1.58 1.57 0.993671 4.20411998

17000 1.58 1.56 0.987342 4.23044892

18000 1.58 1.56 0.987342 4.25527251

19000 1.58 1.56 0.987342 4.2787536

20000 1.58 1.55 0.981013 4.30103

21000 1.58 1.55 0.981013 4.32221929

22000 1.58 1.54 0.974684 4.34242268

23000 1.58 1.54 0.974684 4.36172784

24000 1.58 1.53 0.968354 4.38021124

25500 1.57 1.52 0.968153 4.40654018

26500 1.58 1.52 0.962025 4.42324587

27500 1.57 1.52 0.968153 4.43933269

28500 1.58 1.51 0.955696 4.45484486

30000 1.57 1.5 0.955414 4.47712125

35000 1.57 1.48 0.942675 4.54406804

45000 1.57 1.42 0.904459 4.65321251

50000 1.56 1.4 0.897436 4.69897

55000 1.56 1.36 0.871795 4.74036269

60000 1.56 1.33 0.852564 4.77815125

65000 1.56 1.29 0.826923 4.81291336

70000 1.55 1.26 0.812903 4.84509804

75000 1.55 1.22 0.787097 4.87506126

80000 1.55 1.19 0.767742 4.90308999

85000 1.55 1.16 0.748387 4.92941893

90000 1.54 1.13 0.733766 4.95424251

95000 1.54 1.1 0.714286 4.97772361

100000 1.51 1.05 0.695364 5

105000 1.54 1.05 0.681818 5.0211893

110000 1.54 1.02 0.662338 5.04139269

115000 1.54 0.992 0.644156 5.06069784

125000 1.54 0.941 0.611039 5.09691001

135000 1.53 0.889 0.581046 5.13033377

145000 1.53 0.856 0.559477 5.161368

165000 1.53 0.778 0.508497 5.21748394

Figure5: input-output waveforms of a sine wave through a filter.

Input

Output

Page 5: Spectrum of a Digital Signal

SPECTRUM OF DIGITAL SIGNALS

Figure6: bode plot

Power comparison

Assume a load resistance of 1Ω.

WR

VP out

out 24.101

2.32

===

The out power is about a sixth of the input power.

Task 3

Spectrum of a Digital Signal

Time domain spectrum at a fundamental frequency of 500 Hz:

Figure7: Time domain representation of square wave with 2.5V DC offset at 500 Hz.

0

1

2

3

4

5

6

0 0.2 0.4 0.6 0.8 1 1.2

Vout/Vin

Log(frequency)(dB)

Bode plot of trnsfer function vs

dB frequency

-3dB Point

Conner frequency

WR

VPin 81.16

1

1.4 22

===

2.5V dc

offset

Page 6: Spectrum of a Digital Signal

SPECTRUM OF DIGITAL SIGNALS

Figure8: frequency spectrum of 500 Hz square wave with harmonics placed odd multiples of

the fundamental frequency.

Figure9: Frequency spectrum outline of 500 Hz square wave.

With reference to figure8, the pulse harmonics are decreasing due to the fact that a Fourier

transforms sums decreasing pulse amplitude with an increase in frequency.

Knee frequency

st rise

6

1 10656.1−

×= , the rise time measured with digital oscilloscope?

kHzT

krise

nee 73.19210656.1

116

=××

=−

ππ

Dc component

at 0Hz

1st Harmonic

at

500 Hz

2nd

Harmonic

at 700 Hz

Page 7: Spectrum of a Digital Signal

SPECTRUM OF DIGITAL SIGNALS

4 Task 4

Filtering a Digital Signal

Filter design

Fundamental frequency was chosen as 500 Hz.

Capacitance, c=4.7µf, we therefore calculate the required resistance.

Ω≅Ω=×

==−

6466.63)107.4)(500(2

1

2

16

ππfcR , however a resistance of 63.66Ω was not

available during the practical therefore we used a 68Ω resistor. We then designed a low pass

RC filter from these parameters.

a)

Figure10: filter input output wave forms of the RC filter. Time domain spectra.

b)

figure11 (a) (b)

Rise

time tr Fall time

tf

Rise

time tr

Fall

time tf

Page 8: Spectrum of a Digital Signal

SPECTRUM OF DIGITAL SIGNALS

Katlego Mohlala Electronics 4A01 Page 8

Figure11: (a) depicts the FFT of the input waveform and (b) if the FFT of low pass filter

output. Frequency domain spectra.

With reference to figure11 (b) we can conclude that a low pass filter bypasses high frequency

components and suppresses low frequency components and the dc value remains unchanged.

Table 2

Input Output

Rise time 415.5μs 708μs

RMS

voltage 1.74V 1.04V

Fall time 503.5μs 776.6μs

Vpeak-

peak 4.48V 3.44V

Hzt

krise

frequencynee 7661

==−

π, input signal.

Hzt

krise

frequencynee 95.4491

==−

π, output signal.

Verification calculations

Filter rise time:

usf

t rise 6.636500

11=

×==

ππ

Measured rise time value=415.5µs. This could have been caused by the variation in the

resistance value, we designed for 63.6Ω but we only had 68Ω available, however the

difference is good enough for practical purposes.

st rise

6

1 10656.1 −×=

Output rise time= ustt riserise 5.415)( 2

2

2

1 =+ which is the same as the measured value.

Page 9: Spectrum of a Digital Signal

SPECTRUM OF DIGITAL SIGNALS

5 Task 5

5.1 Effect of bandwidth on RC filters

We cascade two filters as follows:

Figure12: series cascaded filters

Filter1 has a bandwidth of 500Hz and filter2 has a bandwidth of 250Hz

a) Transfer function

+

+==

11 11

1)(

CsRV

VsH

in

out

b)

Figure12: Input and output of cascaded filters. Same analogy

SPECTRUM OF DIGITAL SIGNALS

Effect of bandwidth on RC filters

Filter1 has a bandwidth of 500Hz and filter2 has a bandwidth of 250Hz .

+ 22

1

CsR

Figure12: Input and output of cascaded filters. Same analogy in figure10 applies.

Page 10: Spectrum of a Digital Signal

SPECTRUM OF DIGITAL SIGNALS

figure13: FFT (a) input (b) output

c)

Table 3

Input Output

Rise time 3.608μs 760μs

RMS

voltage 2.09V 548mV

Fall time 3.608μs 756μs

Vpeak-

peak 4.56V 1.92V

kHzt

krise

frequencynee 8810608.3

116

=××

==−−

ππ

Verification Calculations

Rise time 500Hz filter= 636.6µs

Rise time 250Hz filter= ms27.1250

1=

×π

Conclusion

The practical was succesful we managed to draw a graphical interface into the insight of

filters affect digital signals and how signals are effect by variation in frequency. The -3dB

point was observed and and we saw how the FFT functions relates to spectrum.

Input having

high

frequency

components

High frequency

components fall outside

the bandwidth, other

frequencies are

bypassed

Page 11: Spectrum of a Digital Signal

SPECTRUM OF DIGITAL SIGNALS

Katlego Mohlala Electronics 4A01 Page 11