Experiment 1

Introduction to the Spectrum Analyzer

Prepared by:

Arman Isayan

Eric Goss

Date:9/9/2010

Introduction:

The purpose of this experiment is to introduce the student to the use of the spectrum

analyzer. The spectrum analyzer is a piece of test equipment that is used to measure the

frequency spectrum of a signal. A spectrum analyzer displays signals in the frequency domain in

the same way that an oscilloscope displays a signal in the time domain. Due to the frequency

basis of this course, a spectrum analyzer will be used frequently.

Equipment List:

Resistors: 1kΩ

Capacitors: 10nF

Equipment: Spectrum Analyzer

Function Generator

Oscilloscope

Software: PSPICE

Theory:

In this experiment, a number of signals will be displayed on the spectrum analyzer. The

student will verify that the spectrum that they observe is close to that which is predicted by

Fourier analysis of the signal.

In this comparison it should be noted that in general the voltages measured will not

correspond in magnitude to the Fourier series coefficients. However, their frequencies will be

accurate and their relative amplitudes will be accurate. For example, say that a signal:

is input to the spectrum analyzer. We know that the Fourier transform of this signal is:

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The spectrum analyzer will display:

While the magnitudes may not be 5 nor 2.5, but the ratio of the two magnitudes will be 2.

Likewise the components at ±1000Hz will have equal amplitudes as will components at ±

2000Hz.

Preliminary Calculations:

Attached

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Procedure:

Part 1: Set-up Procedure

Work through self-guided demo for the spectrum analyzer.

Results:

The spectrum analyzer is truly an amazing piece of testing equipment. It allows the user

to visualize signals in frequency domain, collect all important data, and save all the work onto a

floppy disk.

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4

5

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Part 2: Sine Wave

a) Adjust the function generator to produce a 10 kHz, 1V p-p.

Results:

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b) Set the center frequency of the spectrum analyzer to 0Hz, and adjust the frequency span

to 5 kHz/div.

Results:

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c) Observe and describe the spectrum analyzer’s display as you vary the frequency of the

function generator from 5 kHz to 20 kHz.

Results:

It is clear that as the frequency of the sinusoidal signal increases, the conjugate impulses

get pushed out and away from the DC Gain position (zero frequency)

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Part 3: Square Wave

a) Adjust the function generator to produce a 10 kHz, 1V p-p square wave. Observe this

wave form on the spectrum analyzer.

Results:

b) What frequency are these components at? Why?

Results:

The frequencies at which impulses occur are integer multiples of fundamental frequency

nf 0. However, the reason that even multiples of f 0 can not be detected is their zero magnitudes.

And the reason is periodic functions as such produce harmonics with decaying amplitudes.

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c) Find the ratio of each of the observed components to the first component. Compare this

to the theoretical ratio of the Fourier series components.

Results:

Measured Values Theoretical Value

n=-1,-3A−1

A−3

= 464161.8

=2.87 ≅A−1

A−3

=Sinc(−πτ

2T)

Sinc(−3 πτ

2T)=3

n=-1,-5A−1

A−5

= 46497.04

=4.78 ≅A−1

A−5

=Sinc(−πτ

2T)

Sinc(−5 πτ

2T)=5

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d) Change the vertical scale to 10 dB log. How many dB down is the 2nd non-zero

component from the first? Compare these results to those of part c. Now compare the 3rd

non-zero component to 1st.

Results:

Measured Values (dB) Measured

Values

(mV)

A−1−A−3=6.26−(−2.936)=9.196dB→A−1

A−3

=109.196dB

20dB =2.88≅A−1

A−3

=2.87≅ 3

A−1−A−5=6.26−(−7.331)=13.591dB→A−1

A−5

=1013.591 dB

20dB =4.78≅A−1

A−5

=4.78≅ 5

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Part 4: Triangular Wave

a) Adjust the function generator to produce a 10 kHz, 1V p-p triangular wave. Observe this

wave form on the spectrum analyzer.

Results:

b) What frequency are these components at? Why?

Results:

The frequencies at which impulses occur are integer multiples of fundamental frequency

nf 0. Moreover, all impulses located at multiples of f 0 have non-zero amplitude. And the reason

is periodic functions as such produce harmonics with decaying amplitudes.

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c) Find the ratio of each of the observed components to the first component. Compare this

to the theoretical ratio of the Fourier series components.

Results:

Measured Values Theoretical Value

n=-1,-2A−1

A−2

=232.9120.8

=1.93 ≅A−1

A−2

=Sinc2(−πTf

4)

Sinc2(−2πTf

4)=2

n=-1,-3A−1

A−3

=232.981.04

=2.87 ≅A−1

A−3

=Sinc2(−πTf

4)

Sinc2(−3 πTf

4)=3

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d) Change the vertical scale to 10 dB log. How many dB down is the 2nd non-zero

component from the first? Compare these results to those of part c. Now compare the 3rd

non-zero component to 1st.

Results:

Measured Values (dB) Measured

Values

(mV)

A−1−A−2=.332−(−5.465)=5.797dB→A−1

A−2

=105.797dB

20dB =1.95≅A−1

A−2

=1.93≅ 2

A−1−A−3=.332−(−8.939)=9.271dB→A−1

A−3

=109.271dB

20dB =2.91≅A−1

A−3

=2.87≅ 3

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Part 5: Low Pass Filter

a) Apply a 1 V peak sine wave to the input of the low pass filter below. Vary the input from

1 kHz to 100 kHz, making sure that the amplitude remains constant. Using the

oscilloscope, measure the output voltage (peak value) over this frequency range.

Results:

f=1kHz f=10kHz

f=15.9kHz f=20kHz

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f=50kHz f=100kHz

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b) Plot the measured values of |H ( jf )|=|V o|/|V i| versus frequency. What is the cutoff

frequency?

Results:

Given R=1k Ω

C=10 nF

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|H ( jf )|= 1

√1+(2πfRC)2=1

2→f c=15.9kHz

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Part 6: High Pass Filter

a) Apply a 1 V peak sine wave to the input of the high pass filter below. Vary the input

from 1 kHz to 100 kHz, making sure that the amplitude remains constant. Using the

oscilloscope, measure the output voltage (peak value) over this frequency range.

Results:

f=1kHz f=10kHz

f=15.9kHz f=20kHz

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f=50kHz

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b) Plot the measured values of |H ( jf )|=|V o|/|V i| versus frequency. What is the cutoff

frequency?

Results:

Given R=1k Ω

C=10 nF

|H ( jf )|= (2 πfRC)2

√1+(2πfRC)2=1

2→f c=15.9kHz

Conclusion:

Spectrum analyzer is a powerfull tool in communications since almost every discussion

takes place around frequency domain. A number of different signals were fed into the spectrum

analyzer and crusial data was displayed. The only problem that exists is the high pass filter

tranfer function in log scale, which does not match with prelab predictions.

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