Fourier spectrum
Tina provides another type of spectral analysis called Fourier transform. Fourier transform can also be thought of as an extension of the Fourier series. This means the Fourier transform converts a function of time to a function of frequency, and vice versa. The physical interpretation of Fourier transform is the conversion of a time-domain signal to the AC steady-state frequency spectrum.
F![]()
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However the inverse Fourier transform converts the AC steady-state signal into the time-domain.
F![]()
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The Fourier transform in Tina is a discrete Fourier transform (DFT), where the Fourier integral has been replaced by a nearly equivalent summation formula applied to evenly spaced samples of the signal. To speed up the process the transform is accomplished by a special technique, called fast Fourier transform (FFT). The FFT gets its data from a set of samples with the appropriate number of data points.
Sometimes its useful to define the real spectrums (amplitude-densities):
,
where
and ![]()
Note, that Tina calculates
and displays in the
region only. For the
region applies
,
therefor the amplitude is the same and the phase has opposite sign.
![]()
can
also be calculated directly from the time function.
and ![]()
,
thus
is
an even,
is an odd function.
Example: Lets determine the complex spectrum of a pulse (amplitude 1, width T=1n; fourex3.sch) and compare it with the theoretical result. First carry out a transient analysis then select the pulse wave and choose Fourier spectrum from the Process menu.

After you select Fourier spectrum a dialog box appears, where you have to set the sampling start and end time, the minimum and maximum frequency and the number of samples. You can also choose the diagrams you want to display.

If you press OK Tina performs discrete Fourier transform. The complex amplitude spectrum can be seen on the following figure:

This function is described by the well-known formula:
,
therefore the amplitude spectrum is: ![]()
To check our previous result now we will recalculate the amplitude spectrum using the above formula in Tinas Interpreter (fourex2.ipr). The following function calculates the amplitude spectrum.
Function PulseAmplSpe(f);
Begin
w := 2 * pi * f;
T := 1n;
PulseAmplSpe := 2 / w * Abs(Sin(w * T / 2));
End;
Draw(PulseAmplSpe(Frequency), AmplSpe);
To try out the above example invoke the Interpreter from the Tools
menu, type in the above text or read in fourex2.ipr by selecting the
Open command from the File menu. If you press the Run
button after a short time the following function appears:

You can easily check, that the above spectrum provided by the closed formula and the spectrum provided by Tinas Fourier Spectrum command are exactly the same. Its rather easy to compare these curves using the clipboard. First select the curve which was generated by the Interpreter, then copy this curve onto the clipboard then in the diagram window go back to the previous page and paste the curve back. The curves will be identical, so we do not show here both.
Example: Transform of a sine wave (fourex4.sch)
The Fourier transform of a sine wave results in a Dirac-delta function, that is an infinitely high spike of zero width at the frequency of a sine wave. We can not reproduce that with DFT. We can actually examine sine waves of finite length. The longer the sine wave is, the taller and narrower the spectrum becomes.
Let us first examine a 3ms long sine wave with 1kHz frequency.

If you run Fourier spectrum on this curve you will get the following result.

Although the peak value of this spectrum is correct the shape of the spectrum is quite rough. To get a better spectrum change the simulation time to 21ms and carry out the analysis again. At FFT the frequency resolution is inversely proportional to the duration of the signal.

Note that the actual simulation time in this example was 21ms to get a better spectrum. (Not shown on the figure.)

Let us check this result using the theory of Fourier transform:
F
,
where ![]()


Note, that the main amplitude of the spectrum at
and at
is
.
Let us draw this function with Tinas Interpreter
(fourex3.ipr). In this example we will use a simplified expression,
because this part is significant only regarding the amplitude of the
spectrum. The following simplified function calculates the amplitude spectrum.
Function SinusAmplSpe(f);
Begin
w := 2 * pi * f;
w0 := 2 * pi * 1k;
T := 3m;
SinusAmplSpe := Abs(Sin((w - w0) * T / 2) / (w - w0)
Sin((w + w0) * T / 2) / (w + w0));
End;
Draw(SinusAmplSpe(Frequency), AmplSpe)
To try out the above example invoke the Interpreter from the Tools
menu, type in the above text or read in fourex3.ipr by selecting the
Open command from the File menu. If you press the Run
button after a short time the following function appears

You can easily check, that the above spectrum provided by the closed formula and the spectrum provided by Tinas Fourier Spectrum command are exactly the same. Its rather easy to compare these curves using the clipboard. First select the curve which was generated by the Interpreter, then copy this curve onto the clipboard then in the diagram window go back to the previous page and paste the curve back. The curves will be identical, so we do not show here both.
Example: Amplitude modulation (am.sch).
Let us see the spectrum derived by amplitude modulation. In our
example the amplitude modulated signal is described by the following
formula:
,
where
,
,
m=0.5.

Lets calculate the Fourier spectrum of the modulated waveform:
.
In our case:
.
From this formula its clear, that the spectrum of the
modulated waveform is the sum of the spectrum of three sinusodial
signals. First check our result using Fourier series. Select Fourier
Series from the Process menu and set the Format field to
.
Then press Calculate.

The cosinusodial coefficients (A) are practiacally zero and for the sinusodial coefficients (B) we get the theoretical result.
On the next figure the Fourier spectrum of the same signal can be seen.

You can see, that the amplitude of the smaller spikes is the quarter of the amplitude of the peak at the base frequency, which is 15u=T/2, where T=30u is the length of the modulated wave.
And finally on the last figure you can see the spectrum in case of 210us simulation time.