1
S.C.S. – Lab 2. Spectrum Analysis of Periodic Signals
Lab 2.
SPECTRUM ANALYSIS OF PERIODIC SIGNALS
2.A.
THEORY
A periodic signal is a wave that repeats itself over time. The periodic signals can be divided into two
subcategories: harmonic signals and non-harmonic signals.
The amplitude spectrum and the phase spectrum show the amplitude and the initial phase respectively as
a function of frequency.
1) The harmonic signals can be written as:
x( t ) = X ⋅ cos ( 2π ft + ϕ )
(2.1)
where X is the amplitude of the signal, f is its frequency and φ is its initial phase.
The single-sided spectrum of a harmonic signal consists of one spectral line at frequency f0.
2) The non-harmonic signals can be written as:
N
x( t ) = A0 +
∑A
n
⋅ cos ( 2π nf0 t + ϕ n )
(2.2)
n =1
where: A0 is the dc component (the average of the signal) and the terms of the sum are the harmonic
components.
The n = 1 component has the same frequency as the periodic signal (f0), which is the fundamental
frequency. The nth frequency component has frequency n times the fundamental, and is referred to as the nth
harmonic. The fundamental frequency gives the period of the signal: T = 1 / f0 .
The relation (2.2) represents the harmonic form of the Fourier series and is
An
used to convert the mathematical description of a time domain waveform into the
A1
frequency domain.
A2
A0
The spectrum of non-harmonic signals is discrete and can consist of several
spectral lines:
A3
γ the dc component, at frequency 0Hz;
γ the fundamental component (or first harmonic) at the fundamental
f0 2f0 3f0
f
frequency f0;
γ a series of harmonics at frequencies nf0 .
φn
As an example, an amplitude spectrum and a phase spectrum of a nonharmonic signal are depicted in Figure 2.1.
φ1
NOTE: If the fundamental component is missing, the fundamental frequency won’t
be the lowest frequency in the spectrum, but the greatest common divisor (GCD)
of the harmonics’ frequencies.
f0
For example, if a signal has 2 harmonics of 100Hz and 150Hz respectively,
the fundamental frequency of the signal is equal to GCD(100, 150) = 50Hz.
φ3
2f0
3f0
φ2
Figure 2.1
f
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S.C.S. – Lab 2. Spectrum Analysis of Periodic Signals
2.B.
PROBLEMS
2.B.1. Sketch the spectrum and compute the periods of the following signals:
a) x( t ) = 3 ⋅ sin ( 2π ⋅ 100t )
(
)
(
)
(
b) x( t ) = 6 + cos 2 ⋅10 3 π t + 0.333 ⋅ cos 6 ⋅ 10 3 π t + π + 0.2 ⋅ cos 10 4 π t
)
2.B.2. Consider the amplitude spectrum of a periodic signal, consisting of a harmonic of 3.5kHz frequency
and 3V amplitude and a harmonic of 5kHz frequency and 5V amplitude. Write the mathematical expression of
the signal and compute its period.
2.C.
LABORATORY WORK
Draw in ORCAD the schematic shown in Figure 2.2.
R1
1k
V1
VOFF = 0
VAMPL = 0
FREQ = 0
PHASE = 0
0
V2
VOFF = 0
VAMPL = 0
FREQ = 0
PHASE = 0
0
0
V3
VOFF = 0
VAMPL = 0
FREQ = 0
PHASE = 0
0
Figure 2.2
For each VSIN source, set all its parameters to 0 then display the PHASE parameter on the schematic, as
shown in the figure.
Create a new simulation profile in Time Domain, and set: Run to time at 200 ms and Maximum step size at
5μs.
Place a voltage marker on the schematic, in order to display the sum of the 3 signals generated by the
sources.
2.C.1. We will first study the amplitude spectrum of a harmonic signal. Modify the parameters of the first
(
source so that it generates the harmonic signal: x( t ) = cos 2π ⋅10 3 t
)
(the initial phase must be set at 90°,
because the VSIN source generates sine waves). Compute the period of x(t).
Run the simulation then adjust the time domain at (0 – 5ms). Notice that the displayed trace is a cosine.
Verify if the period is the same as you have calculated before.
button (FFT). Adjust the frequency axis to (0 – 10kHz).
To display the amplitude spectrum, press the
Notice that the spectrum consists of one spectral line of amplitude 1V, at frequency 1kHz.
2.C.2. Return to the schematic and modify the frequency of the source to 2kHz. Compute the period of the
signal. Run the simulation then adjust the time domain to (0 – 5ms). Verify if the period is the same as you have
calculated.
Go to the frequency domain and adjust the frequency axis to (0 – 10kHz). Notice the differences from the
previous spectrum.
2.C.3. Let’s study now the amplitude spectrum of a non-harmonic signal, consisting of 2 harmonics. Go
back to the schematic and set the parameters of the second source so that the output signal is:
3
S.C.S. – Lab 2. Spectrum Analysis of Periodic Signals
(
)
(
x( t ) = cos 2π ⋅10 3 t + 0.333 ⋅ cos 2π ⋅ 3 ⋅ 10 3 t
)
Compute the fundamental frequency and the period of the signal.
Run the simulation and display the outputs of the 2 sources and their sum. Adjust the time axis to (0 –
5ms). Notice that the sum is a non-harmonic periodic signal.
Measure with the cursor the amplitude of the sum. Notice that in this case, when the 2 signals are in phase
(their initial phases have the same value), the amplitude of their sum is equal to the sum of the amplitudes of the
2 signals.
Go to the spectrum and adjust the frequency axis to (0 – 10kHz). Place the 3 spectrums on different plots.
Notice that the spectrum of the sum contains now 2 spectral lines:
γ the fundamental component, at 1kHz (which is the fundamental frequency);
γ the 3rd harmonic, at 3kHz.
2.C.4. Go back to the schematic and modify the phase of the second source so that the two signals are out of
phase (the phase shift must be 180°).
Run the simulation and display only the sum. Notice that it looks different from the previous case. Also
notice that in this case, when the signals are out of phase, the amplitude of their sum is less than the sum of
their amplitudes.
Go in the frequency domain and notice that the spectrum looks exactly as the spectrum previously
displayed, even if the 2 signals are different. However, where can you see a difference between the two
signals?
2.C.5. We will study now the amplitude spectrum of a non-harmonic signal, consisting of 3 harmonics.
Go back to the schematic and set the third source so that the output of the circuit is:
(
)
(
)
(
x( t ) = cos 2π ⋅10 3 t + 0.333 ⋅ cos 2π ⋅ 3 ⋅10 3 t + π + 0.2 ⋅ cos 2π ⋅ 5 ⋅10 3 t
)
Compute the fundamental frequency and the period of the signal.
Run the simulation, go to the frequency domain and plot only the sum.
2.C.6. Let’s now study the case when the dc components are non-zero.
First take a look at the signal in the time domain. What is the mean value of the signal?
Go back to the schematic and set VOFF as follows: 1V for the 1st source, 2V for the 2nd source and 3V for
the 3rd source.
Run the simulation and plot the outputs of the sources and their sum. Notice that the mean values of the
signals have changed.
Go to the frequency domain and leave only the spectrum of the sum. What is different from the previous
spectrum?
2.C.7. Until now, the signal had the fundamental harmonic (at fundamental frequency 1kHz) and the period of
the signal was: T = 1 / f0 = 1 ms . Let’s see further what happens if the fundamental harmonic is missing.
Modify the amplitude of the first source to 0V. Compute the fundamental frequency and the period of the
signal.
Run the simulation. Verify if the period is the same as you have just computed.
Now go back to the schematic and modify the frequency of the second source to 3.5kHz (instead of 3kHz).
Compute the fundamental frequency and the period of the signal.
Run the simulation and verify if the period is the same as you have calculated.
*****