ganago student lab11

7/29/2019 Ganago Student Lab11

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Lab 11: Fourier Analysis

LAB EXPERIMENTS USING

NI ELVIS II

AND NI MULTISIM

Alexander Ganago

Jason Lee Sleight

University of Michigan

Ann Arbor

Lab 11

Fourier Analysis

2010 A. Ganago Introduction Page 1 of 19


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Goals for Lab 11

Learn about:

The Fourier seriesa powerful tool for analysis of periodic signals The Fourier spectra of standard signalssine, square, triangular, ramp The application of Fourier analysis to building a power supply The necessity and role of a Low-Pass filter in addition to a rectifier in a power

supply

The effects of the load resistance and the filters capacitance on the amplitude ofripple voltages in the output of a power supply

In the pre-lab:

Calculate the amplitudes of Fourier spectra of standard signals Use Arbitrary Waveform Generator to create standard signalssine, square,

triangular, ramp and negative rampas sums of sinusoids

Simulate a half-wave rectifier circuit, its output waveforms and spectra for 3casesno filter, a weak Low-Pass filter, and a strong Low-Pass filter

In the lab:

Measure the waveforms and Fourier spectra of standard signals Build a half-wave rectifier circuit without any filter and with a Low-Pass filter,

with small and large capacitance

Measure its output waveforms and spectra in 3 casesno filter, a weak Low-Passfilter, and a strong Low-Pass filter

In the post-lab:

Compare the simulated and measured waveforms and Fourier spectra of standardsignals

Compare the amplitudes of ripple voltages in the output of a power supplycalculated from the approximate formula, obtained in simulations, and measured

in the lab

Explain the role of the filters capacitance on the amplitude of ripple voltages,based on your lab data

Relate the time-domain analysis of the power supply output waveforms to thefrequency-domain analysis of the power supply output spectra

Optional goals:

Explore the role of the load resistance and its effect on the amplitude of ripples ina power supply

Measure the output waveforms and spectra of your power supply with a smallerload resistance

Explain the role of the load resistance on the amplitude of ripple voltages, basedon your lab data.



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Introduction

The Fourier series is a powerful technique for analysis of periodic signals, which

complements their description as time-dependent voltages or currents (waveforms) with

an equally complete description of signals as spectra (intensity versus frequency).

In this lab, you will use the Fourier series both ways:

1. For the creation of signals as sums of sine waves (in the pre-lab), and2. For decomposition, or analysis, of a signal into the sum of sine waves (in the

pre-lab and in the lab).

In the beginning, you will work with standard waveformssquare, triangular, and ramp;

then, you will apply the Fourier analysis to a very practical job of building a good power

supply that converts AC voltage into DC voltage. This application helps you bringtogether the two views on the same signal the time-dependent voltage, or waveform,

and the Fourier spectrum, or intensity as function of frequency; it will also encourage youto refresh the knowledge of filters.

Periodic Signals as Sums of Sinusoids

According to the Fourier theorem, a periodic signal can be represented as the

following sum, or series:

( )sv t

0 0

1 1

( ) cos(2 ) sin(2 )s n n

n n

v t a a n f t b n f t

= =

= + + 0

Here, is the constant term, independent of frequency (it can be very important, like in

the case of your power supply model, as explained below);

0a

0f is the frequency (in Hz) of

the periodic signal under study (it is also called the fundamental frequency); is the

number of the cosinusoidal or sinusoidal component of the original signal; is the peak

amplitude of the cosinusoidal component, and is the peak amplitude of the

sinusoidal component.

n

na

thn nb thn

An alternative form of the Fourier series involves phase shifts:

vs(t) = a

0+ Cn cos(2 n f0 t+ n )

n=1

Here, is the phase shift of the cosinusoidal component. Note that, due to the

presence of phase shifts, sinusoidal components are not needed in this sum.

n

thn



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According to the Fourier theorem, all sine (and cosine) waves in the series have either the

fundamental frequency 0f or its multiple frequencies 0n f ; the waves at multiplefrequencies are called the harmonics. A familiar example is the pure musical tone and its

overtones.

In particular cases, some of the Fourier components may vanish (if the signal isrepresented with an even or an odd function of time) or be suppressed by filtering.

For instance, a signal at 125 Hz is expected to have the Fourier components:

at zero frequency (the DC component), at 125 Hz (the fundamental), as well as the 2PndP harmonic at 250 Hz, the 3PrdP harmonic at 375 Hz, etc., the 10PthP harmonic at 1,250 Hz, and so forth.

Depending on the waveform, amplitudes of some of those harmonics may vanish (see

examples below). If this signal is passed through a filter that blocks everything above

1 kHz, then only the DC, the fundamental, and harmonics up to the 8 Pth

P will pass, while allthe higher ones will be suppressed by the filter.

The Fourier Series Formulas for Standard Waveforms

The formulas for calculations of Fourier series of standard signals, which you will beusing in this lab, are well known and listed below.

For simplicity, only a few periods of an endless waveform are shown (all waveforms areendless for truly periodic signals), and only a few components out of the endless series

for each Fourier spectrum are shown (the only exception is the Fourier spectrum of a sine

wave, which truly has one component).



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The Sine Wave

The sine wave at frequency f and peak amplitude A , with zero average, has only one

componentitself, which is exactly matching the fundamental.

Figure 11-1. A sketch of a sinusoidal waveform with zero average.

Figure 11-1. A sketch of the Fourier spectrum of a sinusoidal waveform with

zero average.



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The Square Wave

The square wave at the frequency f , peak amplitude and 50% duty cycle (which is an

odd function with zero average), shown in Figure 11-3, has only odd-numbered spectralcomponents, shown as a sketch in Figure 11-4.

Figure 11-3. A sketch of a square waveform with zero average and 50% duty cycle.

Figure 11-4. A sketch of the Fourier spectrum of a square waveform with zero average

and 50% duty cycle.



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Notice that the spectral components (on the right sketch) are seen as sharp peaks (spikes)

at discrete frequencies.

Every sketch of a Fourier spectrum in this Introduction is very approximate: it shows the

positions of the spectral components, but does not provide accurate information about

their amplitudes.

The Fourier series formula for a square wave is given below.

Square wave of peak amplitude A =

=4A

n sin(2 n f

0 t)

n = odd integer

=

=4A

sin(2 f

0 t) +

+

4A

3 sin(2 3 f0 t) +

+4A

5 sin(2 5 f

0 t) +

+4A

7 sin(2 7 f

0 t) + ...

Thus, in our example of a signal at 125 Hz passed through a Low-Pass filter with cutoff

frequency of 1 kHz, for the square wave with zero average, we will obtain the following

Fourier components:

the fundamental at 125 Hz the 3PrdP harmonic at 375 Hz the 5PthP harmonic at 625 Hz the 7th harmonic at 875 Hz, and nothing else.

In the pre-lab, you will construct the square wave from its Fourier components andobserve how well the sum of components listed above represents the original signal.

Note that, if the cutoff frequency of a Low-Pass filter were below 375 Hz, then the passedsignal would be just a sine wave. This is no joke: in practice, when you need to ensure

steep rising/falling edges of a digital signal (very similar to a square wave) at f GHz,

your circuit should be able to pass at least the bandwidth of 5 f GHz so that the outputsignal looks has the edges steeper than those of a sine wave.



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The Triangular Wave

The spectrum of a triangular wave at the frequency f and peak amplitude A (which is

an even function with zero average) has only odd-numbered components.

Figure 11-5 shows a triangular waveform, and Figure 11-6 shows a sketch of its Fourierspectrum.

Figure 11-5. A sketch of a triangular waveform with zero average.

Figure 11-6. A sketch of the Fourier spectrum of a triangular waveform with zeroaverage.



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The Fourier series formula for a triangular wave is given below.

Triangular wave of peak amplitude A =

=8A

n( )2

cos(2 n f0

t)

n = odd integer =

=8A

2

cos(2 f0

t) +

+8A

2 9

cos(2 3 f0

t) +

+8A

2 25

cos(2 5 f0

t) +

+8A

2 49

cos(2 7 f0

t) + ...

Note the similarities and important distinction between the Fourier spectra of the squarewave and the triangular wave. Both signals have only odd-numbered harmonics but the

amplitudes of harmonics decrease much faster in the triangular wave spectrum: the

harmonic number n is squared in the denominator.



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The Ramp, or Saw-Tooth Wave

The spectrum of a saw-tooth wave at the frequency f and peak amplitude A (which has

zero average) has both odd-numbered and even-numbered components. Figure 11-7

shows this signal as a waveform, and Figure 11-8 shows a sketch of its Fourier spectrum.

Figure 11-7. A sketch of a ramp, or saw-tooth waveform with zero average.

Figure 11-8. A sketch of the Fourier spectrum of a saw-tooth (ramp) with zero average.

The Fourier series formula for a saw-tooth wave is given below.



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Saw-tooth wave of peak amplitude A =

= 1( )n+1 2A n

sin(2 n f0

t)

n = integer

=

=2A

sin(2 f0

t) +

+2A

2 sin(2 2 f

0 t+ 180o ) +

+2A

3 sin(2 3 f

0 t) +

+2A

4 sin(2 4 f

0 t+ 180o ) + ...

The role of phase shifts is very important. In the pre-lab, you will simulate this signal

with and without the phase shifts and observe a significant difference between their

waveforms. Namely, without the phase shifts you will obtain the negative rampwaveform, as shown in Figure 11-9.

Figure 11-9. A sketch of a negative ramp waveform with zero average.

Note that, if your spectrum analyzer measures only the amplitudes but does not measure

the phase shifts, the results of spectral analysis of both waveforms will be identical.



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An Important Comment on the Amplitudes in Fourier Spectra

On the sketches shown in Figures 11-2, 11-4, and 11-6, you could have noticed that the

vertical scale is not in volts but in dBV, which stands for decibel-Volts. This is a

logarithmic measure of amplitudes, widely used because it allows us to accurately show

on the same plot the peaks whose intensities differ by several orders of magnitudes.

According to the definition, the amplitude in dBV equals:

,

10Amplitude in dBV 20 log1

S RMS

RMS

V

V

=

For a sinusoidal signal with peak amplitude of (volts peak), the amplitude

in volts RMS (root-mean square) can be found from the simple relationship:

,S peakV ,S RMSV

,

, 2

S peak

S RMS

V

V = . Note that, in the lab, you often measure peak-to-peak amplitudes, whichare, for sine waves with zero average, twice as large as their peak amplitudes.

Here is an example of calculations. Consider a sine wave whose peak-to-peak amplitude

equals 30 Vpp (volts peak-to-peak). Its peak amplitude equals 15 Vp (volts peak), and its

RMS amplitude equals15

10.612

peak

RMS

VV= . The same amplitude expressed in dBV can

be found as: ( )10 1010.61

20 log 20 log 10.61 20 1.026 20.521

RMS

RMS

VdBV

V

= = =

Note the convenience of using the logarithmic scale: two signals whose peak amplitudesdiffer by a factor of 10 have the amplitudes different by 20 dB. It means that a plot with alogarithmic vertical scale that has 8 divisions 10 dB each (typical in the lab) can

accurately display signals whose amplitudes differ by 60 dB or more, in other words, the

ratios of peak amplitudes at or above a factor of 1,000an impossible feat for a plot witha linear scale.

In practice, the logarithmic scale is very convenient for measurements of signal-to-noiseratios, which are therefore expressed in dB.



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Application of Fourier Analysis:

How to get DC Output Voltage from AC Input Voltage

As you know, our power distribution grid uses AC voltages (Figure 11-10) because thisminimizes losses in transmission lines. At the same time, electronic appliances need DC

voltages (Figure 11-11). Thus we have to obtain DC out of the AC input voltages.

Devices that do this job are called power supplies. In this lab you will study a model of a

power supply.

Figure 11-10. Sinusoidal voltage is the input for a power supply circuit.

Figure 11-11. The output voltage of a perfect power supply is constant, or DC.



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If you apply a sinusoidal input voltage to the circuit with a semiconductor diode such as

shown in Figure 11-12, the output voltage will be always positive, as shown in Figure11-13, because the diode conducts large currents only when the voltage across it is

positive. This circuit is called a half-wave rectifier, because only one half of the input

waveform reaches the output.

Figure 11-12. A half-wave rectifier circuit.

Figure 11-13. The output voltage of a half-wave rectifier.

From comparison of Figures 11-8 and 11-10, it becomes clear that a rectifier alone is notenough to obtain real DC output. The waveform in Figure 11-10 does indeed havepositive average but its amplitude varies from zero to VBMB (for simplicity, here we neglect

the voltage drop across the diode).

Let us consider the spectra of waveforms shown in Figures 11-11 and 11-13.



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Figure 11-14 shows a sketch of the Fourier spectrum of a perfect DC voltage, the desired

output of a power supply. For comparison, Figure 11-15 shows a sketch of the Fourierspectrum of a half-wave rectifiers output voltage.

Figure 11-14. Output of a perfect power supply has only one spectral componentthe

DC.

Figure 11-15. Output of a half-wave rectifier has many spectral componentsthe DC,the fundamental, and all harmonics.

Comparison of Figures 11-14 and 11-15 clearly shows the solution for our problem: we

need to keep the DC voltage but get rid of the fundamental and all harmonics.



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Fortunately, we already know what circuit can do the job: its a Low-Pass filter. Recall

that one of the key parameters of a Low-Pass filter is its cutoff frequency Cf (Figure

11-16).

Figure 11-16. A sketch of the transfer function of a Low-Pass filter.

Basically, we change the block diagram of the power supply to include a Low-Pass filter

in addition to the rectifier (Figure 11-17). The rectifier input is sinusoidal AC voltagewhose polarity changes twice every period; out of this input, the rectifier creates voltage

of constant polarity, but it is the Low-Pass filter that creates the output voltage of (nearly)

constant amplitude.

Figure 11-17. Block diagram of a power supply includes a Low-Pass filter in addition tothe rectifier.



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An arbitrary Low-Pass filter will pass the DC component of the rectified voltage and

suppress its higher harmonics but it may pass the fundamental and lower harmonics, asshown in Figure 11-18, where the 2P

ndP and 3P

rdP harmonics are barely suppressed. This is a

poor filter whose output is not perfect DC voltage.

Figure 11-18. An example of a poor filter that passes the fundamental and several

harmonics of the rectified voltage.

A much more desirable case is shown in Figure 11-19, where the filter cuts off the

fundamental and all harmonics. In this case the Low-Pass filter is excellent and we may

expect the perfect DC output.

Figure 11-19. An example of an excellent filter that passes only the DC component of the

rectified voltage and suppresses the fundamental and several harmonics.



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Many real-world power supplies fall in-between the extreme cases shown in Figures

11-18 and 11-19; their output voltages have ripples, as shown in Figure 11-20.

Figure 11-20. A sketch of the power supplys output voltage with ripples.

Of course, we wish to minimize the ripples, and Fourier analysisalong with our

knowledge of filter circuitscan help us find the way to do so.

First of all, let us consider the Low-Pass filter circuit, similar to what you already studied.

Probably, the simplest filter is a capacitor connected in parallel with the load resistor, asshown in Figure 11-21.

Figure 11-21. A simple Low-Pass filter in the power supply circuit.

The cutoff frequency of this filter equals1

2C

LOAD

fR C

=

.



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In order to achieve perfect filtering (Figure 11-19), we need to reduce this frequency; in

other words, increase the product of load resistance and capacitance. Clearly, largercapacitances lead to better filtering. A disadvantage is easy to foresee: really large

capacitors can get really bulky.

Note that larger load resistances mean lower power, because2

LOAD

LOAD

VPowerR

= .

In other words, it is easier to build a good filter for loads that consume less power.

As an Exploration for extra credit in this Lab, you will be offered an opportunity to study

the effect of the load resistance on the output voltage.

Of course, you can recall what you learned about filters in Lab 9 and foresee that higher-

order filters would ensure better suppression of the fundamental and harmonics and thus

reduce the ripples.

Finally, let us consider a simple formula for the amplitude of ripple voltages sketched inFigure 11-20. This formula is derived for a half-wave rectifier with the first-order filtershown in Figure 11-21.

0

1RIPPLES PEAK PEAK

LOAD LOAD

TV V V

R C f R =

C

Here, T is the period of the input sine wave, 0f is its frequency, and is the peak

output voltage of a half-wave rectifier.

PEAKV

This formula is approximate but conservative. It means that the calculated ripple voltages

will be larger than those obtained in a real circuit; in other words, your power supply willactually work better than you could expect from calculations.



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Pre-Lab:

Part 1. Standard Waveforms

A.Complete the following table for a 100 Hz, 5 V

B

PPKB

sine wave:

Frequency

Component

Frequency

(Hz)

Amplitude

(dbVBRMSB)

Amplitude

(VBRMSB)

Amplitude

(VBPKB)

Fundamental

B. Complete the following table for a 100 Hz, 5 VBPPKB square wave:Frequency

Component

Frequency

(Hz)

Amplitude

(dbVBRMSB)

Amplitude

(VBRMSB)

Amplitude

(VBPKB)

Fundamental

2Pnd

P Harmonic

3Prd

P Harmonic

4Pth

P Harmonic

5Pth

P Harmonic

C.Complete the following table for a 100 Hz, 5 VBPPKB triangle wave:Frequency

Component

Frequency

(Hz)

Amplitude

(dbVBRMSB)

Amplitude

(VBRMSB)

Amplitude

(VBPKB)

Fundamental

2Pnd

P Harmonic

3Prd

P Harmonic

4Pth

P Harmonic

5Pth

P Harmonic

2010 A. Ganago Pre-Lab Page 1 of 7


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D.Complete the following table for a 100 Hz, 5 VBPPKB ramp wave:Frequency

Component

Frequency

(Hz)

Amplitude

(dbVBRMSB)

Amplitude

(VBRMSB)

Amplitude

(VBPKB)

Fundamental

2Pnd

P Harmonic

3Prd

P Harmonic

4Pth

P Harmonic

5Pth

P Harmonic

E. Use the NI ELVISmx Arbitrary Waveform generator to explore how sine waves can beadded to create the following 100 Hz, 5 VBPPKB signals:

Square wave (up to first 5 non-zero frequency components) (Pre-Lab Printout #1) Triangle wave (up to first 5 non-zero frequency components) (Pre-Lab

Printout #2)

Negative ramp (up to first 5 non-zero frequency components) (Pre-LabPrintout #3). In order to create this plot, disregard all phase shifts.

Positive ramp (up to first 5 non-zero frequency components) (Pre-LabPrintout #4). In order to create this plot, include the phase shift as shown in the

Introduction.

Set the duration of the segment to be 30 ms. Create a new component for the segment.

Set the amplitude and frequency to the fundamental component amplitude and

frequency. One at a time, create 4 more components. Each component will

correspond to a frequency component of the waveform (notice how the 2 Pnd

P5Pth

P

components display as +Sine, this illustrates that they are being added together). Set

the frequency, amplitude, and phase shift before adding the next component. You

should gradually see the correct waveform take shape. Once you have all 5 non-zero

frequency components, create a screenshot of your waveform (Pre-Lab Printouts #14

as listed above). Repeat this process for the other waveforms.

F. Use Multisim to simulate the frequency spectrum analysis of a 100 Hz, 5 V BPPKB sinewave from the FGEN. Use the NI ELVISmx Dynamic Signal Analyzer (DSA) to

obtain the frequency analysis. On the DSA, set the frequency span to be 700 and the



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Window to be Hanning. Allow the VI to run for a while to allow the output to stabilize.

Create a printout of the DSA output. (Pre-Lab Printout #5).

Qualitatively discuss the agreement/disagreement between your theoretical expectations

and your simulation results.

G.Use Multisim to simulate the frequency spectrum analysis of a 100 Hz, 5 V BPPKB squarewave from the FGEN. Use the NI ELVISmx Dynamic Signal Analyzer (DSA) to






H.Use Multisim to simulate the frequency spectrum analysis of a 100 Hz, 5 V BPPKB trianglewave from the FGEN. Use the NI ELVISmx Dynamic Signal Analyzer (DSA) to






Continued on the next page



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Part 2. Rectified Waveforms

Begin with the following circuit (without any capacitor):

For this part, you will use the following circuit:

Where RBLOADB = 5 k

Use a 1N4933 diode

Use a 100 Hz, 5 VBPPKB sine wave as the input signalFor each problem, use Multisim to simulate the circuit. Create two printouts for each problem.

First, use the OSCOPE to view the input and output waveforms in the time domain. Second, use

the DSA to view the output waveform in the frequency domain (use a 700 Hz frequency span

and a Hanning window).

Approximate the dBVBRMSB at each of the frequencies to complete the table on the next page.



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I. OSCOPE plot Pre-Lab Printout #8DSA plot Pre-Lab Printout #9

Frequency

(Hz)

Amplitude

(dbVBRMSB)

DC

100

200

300

400

500

600

700




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J.As discussed in the Introduction, a rectifier itself does not produce DC output. Improve thecircuit by adding a capacitor in parallel with the load resistor.

Use C = 4.7 F

OSCOPE plot Pre-Lab Printout #10

DSA plot Pre-Lab Printout #11Frequency

(Hz)

Amplitude

(dbVBRMSB)

DC

100

200

300

400

500

600

700




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K.Keep using the same circuit but change the capacitance 100-fold.

Use C = 470 F

OSCOPE plot Pre-Lab Printout #12

DSA plot Pre-Lab Printout #13

Frequency(Hz)

Amplitude(dbVBRMSB)

DC

100

200

300

400

500

600

700

L. Discuss what changes in the output as the result of using a larger capacitor.

M. Calculate the ripple voltage using the formula given in the Introduction for both 4.7 F and470 F capacitors; compare the results of calculations with the results of Multisim simulations.



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In-Lab Work

Part 1: Standard Waveform Frequency Analysis

Part 1.1 Sine Wave Turn on the NI ELVIS II.

Open the NI ELVISmx Instrument Launcher.

Launch the FGEN and Dynamic Signal Analyzer (DSA) VIs.

Connect the output from the FGEN directly to AI 0 (along with ground to the

terminal).

Power on the PB.

On the FGEN, create a 100 Hz, 5 V BPPKB sine wave. Run the FGEN VI.

On the DSA VI, set the source channel to be AI 0, the voltage range to be +/ 5V, thefrequency span to be 700 Hz, and the window to be Hanning. Leave the other settings

unchanged. Run the DSA VI to obtain the frequency analysis (you should allow the

DSA VI to run for several seconds in order to allow the output to stabilize).

Stop the DSA VI.

Approximate the dBVBRMSB at each of the following frequencies:

Frequency (Hz) Voltage (dBVBRMSB)

DC

100

200

300

400

500

600

700

Create a printout of the DSA output. (In-Lab Printout #1).

2010 A. Ganago In-Lab Page 1 of 13


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Part 1.2 Square Wave On the FGEN, create a 100 Hz, 5 V BPPKB square wave. Run the FGEN VI.

Run the DSA VI to obtain the frequency analysis (you should allow the DSA VI to runfor several seconds in order to allow the output to stabilize).

Stop the DSA VI.



DC

100

200

300

400

500

600

700




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Part 1.3 Triangle Wave On the FGEN, create a 100 Hz, 5 V BPPKB triangle wave. Run the FGEN VI.


Stop the DSA VI.



DC

100

200

300

400

500

600

700


Power off the PB.



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Part 2: Rectifier Frequency Analysis

Part 2.1 No Filter

On the FGEN, create a 100 Hz, 5 V BPPKB sine wave. Run the FGEN VI.

Build the following circuit:

Here the input voltage source is the function generator.

Use R = 5 k

Use a 1N4933 diode.

Wire VBINB and VBOUTB to AI 0 and 1 respectively.

Launch the OSCOPE VI.

Use the OSCOPE to view the input and output waveforms. Adjust the OSCOPEparameters so that you can clearly view both waveforms.

Create a printout which clearly displays the waveforms. (In-Lab Printout #4)

On the DSA VI, set the Source Channel to be AI 1.

Run the DSA VI to obtain the frequency analysis (you should allow the DSA VI to run

for several seconds in order to allow the output to stabilize).

Continued on the next page.



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DC

100

200

300

400

500

600

700


Power off the PB.



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Part 2.2 Weak Filter (C = 4.7 F)

Add a 4.7 F capacitor to the circuit from Part 2.1. This capacitor, in parallel with the

load resistor, will act as a Low-Pass filter for your output voltage.








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DC

100

200

300

400

500

600

700


Power off the PB.



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Part 2.3 Strong Filter (C = 470 F)

Continue to use the circuit from Part 2.2, except replace the 4.7 F capacitor with a

470 F capacitor.

Use the OSCOPE to view the input and output waveforms. Adjust the OSCOPE

parameters so that you can clearly view both waveforms.






DC

100

200

300

400

500

600

700


Power off the PB.

If you wish to do Explorations, for extra credit, continue with the assignment on the

next page.

If you do not wish to do Explorations, this is the end of In-Lab work.

Turn off all instruments and clean your workplace.



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Part 3: Explorations, for Extra Credit:

A Different Load Resistance

Part 3.1 No Filter; A Different Load ResistanceRemove the capacitor from your circuit, as shown below.

Replace your load resistor with R = 100









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DC

100

200

300

400

500

600

700


Power off the PB.



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Explorations (continued)

Part 3.2 Weak Filter (C = 4.7 F); A Different Load Resistance

Add a 4.7 F capacitor to your circuit, as shown below. Note that, due to the differentload resistor, the parameters of your Low-Pass filter will be different from what you

observed in Part 2.2.

Continue to use 100 load resistor.








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DC

100

200

300

400

500

600

700


Power off the PB.



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Explorations (continued)

Part 3.3 Strong Filter (C = 470 F); A Different Load Resistance

Replace the 4.7 F capacitor in your circuit with a 470 F capacitor.

Continue to use 100 load resistor.







DC

100

200

300

400

500

600

700


Power off the PB.

This is the end of lab work.

Turn off all instruments and clean your workplace.



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Post-Lab:

1.Standard WaveformsA.

Discuss the role of phase shifts for the positive/negative ramp waveforms whichyou created in Pre-Lab Part 1.E.

B.Discuss the agreement/disagreement between your simulation of the sine wavefrequency spectrum (Pre-Lab Printout #5) and your experimental sine wave

frequency spectrum (In-Lab Printout #1).

C.Discuss the agreement/disagreement between your simulation of the square wavefrequency spectrum (Pre-Lab Printout #6) and your experimental square wave


D.Discuss the agreement/disagreement between your simulation of the triangularwave frequency spectrum (Pre-Lab Printout #7) and your experimental triangular


2.Rectified WaveformsA.Discuss the agreement/disagreement between your simulation of the no-capacitor

half wave rectifier (Pre-Lab Printouts #89) to your experimental results of the

no-capacitor half wave rectifier (In-Lab Printouts #45).

B.Discuss the agreement/disagreement between your simulation of the 4.7Fcapacitor half wave rectifier (Pre-Lab Printouts #1011) to your experimental

results of the 4.7F capacitor half wave rectifier (In-Lab Printouts #67).

C.Discuss the agreement/disagreement between your simulation of the 470 Fcapacitor half wave rectifier (Pre-Lab Printouts #1213) to your experimental

results of the 470 F capacitor half wave rectifier (In-Lab Printouts #89).

D.Discuss the role of the capacitor and the role it has in determining the output ofthe rectifier.

E. For each of your half wave rectifier experiments (no capacitor, 4.7F capacitor,and 470F capacitor), calculate the average power of the output in two ways: first

using the dBV measurements from the DSA (In-Lab Printouts #4,6,8), second

using the time domain measurements from the OSCOPE (In-Lab Printouts

#5,7,9). Discuss the agreement/disagreement between the two methods for each

experiment.

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F. Discuss the amplitude of ripples in the 4.7 F and 470 F experiments from twostandpoints: time-domain (for how long the capacitor is discharged, etc.) and

frequency-domain (which frequency components are passed by the filter, etc.).

3.Explorations: Varying ResistanceA.Compare your results for the 100 resistor experiments to the respective 5 k

resistor experiment in Part 2 (for each of the three experiments). What are the

similarities/disparities between the results?

B.Explain the effect the resistor has on the output waveform from two standpoints:time-domain (for how long the capacitor is discharged, etc.) and frequency-

domain (which frequency components are passed by the filter, etc.).

ganago student lab11

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