1-Visitor

Question

RLC Circuit Analysis

Forum|Forum|12 years ago
June 10, 2013
2 replies
39116 views

About this worksheet:

Calculates the impedances across frequencies in a RLC passive crossover network
Applicable in electrical engineering, electronics and high-tech industries, speaker design
Performs using Excel, frequency, resistance, capacitance, inductance, angular frequency, reactance, impedance, etc.

This worksheet using PTC Mathcad includes an example of an engineer designing a hi-fi speaker that must meet a certain frequency response. The horizontal trend must be over the whole spectrum of the audible range with a tolerance of +/- 1dB. Because at least two transducers are needed, you must look at the case in which there is both a low and high frequency driver. You can calculate the impedances of each to determine the overall frequency response. This worksheet uses data from excel and plots the solutions.

Download and explore this worksheet yourself! You can download a free lifetime copy of PTC Mathcad Express and get 30 days of full functionality.

1_RLC_Circuit_Analysis-mcdx.zip

Electrical_Engineering

R

rdliquid

1-Visitor

Hi

I am very new to Matchcad so these questions are very basic:

1. ORIGIN := 1 - What is it for?

2. Why there is an arrow above the absolute values of ZLF, ZHF and ZT?

3. Is "j" a reserve parameter in Matchcad for SQRT(-1)?

Thanks

Dat N

L

LucMeekes

23-Emerald IV

Welcome

ORIGIN defines the starting index of vector and matrix elements. By default it is set to 0.

The arrow is to vectorize an expression. Look that up in the help.. without vectorization, using the |x| operator would give the length of the vector, producing a single number. With vectorization the result of that same operator is a vector with the absolute values of each of the elements.

The value i is the imaginary unit, such that i squared results in -1. You can use j for the same.

Success!

Luc

W

Werner_E

25-Diamond I

According Grids_for_Prime

I think what you use is a routine I posted quite some time ago. But Luc also posted his own version for a grid making routine somewhere in the depth of this forum and as far as I remember his version would also work OK in Prime Express (which does not support programming).

I don't see why you think its a mess 😉

You may just copy the collapsed region into your worksheet and use the commands.

The routines were a quick hack and workaround because inexcusably Prime did not provide a mean to add something so simple a grid to a plot.

The routines create the data for a couple of lines which then when plotted form the grid. The output is a n x 2 matrix, where in the first column are the x-coordinates of the start and end points of the lines and the second column consists of the y-coordinates.

The [NaN NaN] inserted between every two lines is necessary as otherwise the end of one line would be connected with the start of the next one, creating a zig-zag pattern. BTW, NaN means "Not-a-Number".

"Grid" and "logGrid" create only a vertical or horizontal grid, depending on the first function argument being "x" or "y". You may assign each to a separate variable and plot both. So you could have the vertical and the horizontal grid lines in different colors and styles.

Usually this is not necessary and so both are combined using the "stack" command.

The augment command in the programs simply exchange the two columns of the created matrix if axis="y".

K

KarlBogha

14-Alexandrite

Marin,

I know you have an RLC file on the PTC.

I have a circuit problem I have gone over for a week. Could not get the answer.

I forced the process to make the answer work but that is make fit and I know its wrong.

I started on it with the differential equation mesh (loop equation) method.

I did not get matching answers. Then moved to the simpler initial condition method this was where I got closer. Case the answer maybe wrong, not likely.

Basically its the time constants here in the circuit there are two.

If you go by the usual RC and RL time constants usually the simple case is either one or the other is used, here its both. Circuit reduction helps but this circuit's situation is the current in each branch.

If you got can do it easily or are interested to solve it.

I like to see the solution.

Regards,

Karl.

1_RLC-Circuits-Problem_8_27.pdf