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About DAE

-MFra-
21-Topaz II

About DAE

Hello everyone!
I reworked a worksheet presented by a member of the community I think in 2018 related to the solution of a system of algebraic differential equations with a periodic sawtooth excitation function. I find quite different results depending on how the excitation function is defined. I wonder who's wrong. I enclose the version presented by the member (.mcdx) and my own (dae + 13032023.xmcd).
I await your opinion.

DAE 0.jpgDAE 1.jpgDAE 2.jpgDAE 3.jpgDAE 4.jpgDAE 5.jpgDAE 6.jpgDAE 7.jpgDAE 8.jpg

 

1 ACCEPTED SOLUTION

Accepted Solutions
ttokoro
20-Turquoise
(To:ttokoro)

Make Fourier series of input current wave form.
Get each transient response by inverse Laplace.

Add all response and you can get the answer.

 

image.pngimage.pngimage.png

View solution in original post

5 REPLIES 5
-MFra-
21-Topaz II
(To:-MFra-)

I apologize,

I have attached the wrong file. I am attaching the file that I meant to attach. In my version (dae + 13032023.xmcd), it is necessary to set the duty cycle equal to 9.091% and   d.sawth=1.1ms, so that there is correspondence between the two worksheets.

ttokoro
20-Turquoise
(To:-MFra-)

Why is(t) not equal itot(t)? Kirchhoff's Current Law must true even at transient response.

 

image.png

ttokoro
20-Turquoise
(To:ttokoro)

Assuming current source as follows.

Transfer function of the circuit.

Output voltage of one to five cycles.

Current of each branch circuits. Their total is as same as input current source. 

image.pngimage.pngimage.pngimage.png

ttokoro
20-Turquoise
(To:ttokoro)

Make Fourier series of input current wave form.
Get each transient response by inverse Laplace.

Add all response and you can get the answer.

 

image.pngimage.pngimage.png

ttokoro
20-Turquoise
(To:ttokoro)

3. Convolution of signals.

Sampling rate: 10 ns.

This method can solve for even 1 Hz of input current signal. (But only for few ms as shown in last figure.) 

h(t) is almost impulse for 1 Hz input current and output voltage is iin*R3, therefore, 1/10 of input current waveform.

This convolution result is larger than 0.5V at the peak of current signal. This is because the sampling rate is larger to evaluate the output signal more correctly. (Last one use 100ns for sampling rate and convolute 100000 data.)  

 

image.pngimage.pngimage.pngimage.png

image.png

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