How do I plot a symbolic equation that I have created?

Hello everyone. I am about a month into a uni project and I am stuck on something that I cannot figure out, despite weeks of little bits of test code and Googling. I won't share the whole thing, as it's a project and currently about 350 lines. In general though, I have built something that solves the differential equations for an RLC circuit, showing all the working using symbolic maths.

So I get it down to the point where I have my symbol (Yt) which is equal to: val=1.0 - 0.2*exp(-1.0*t)*sin(5.0*t) - 1.0*exp(-1.0*t)*cos(5.0*t)

I have then successfully used fplot to draw this equation. However, I am now trying to demonstrate time invariance. The only way that I can find to do this is by creating a time vector, and then plotting the equation against time+1, similar to the test code below.

InputStartTime = 1
t=0:.005:5;
hold on;
test3 = -1.0*exp(-1.0*t).*cos(5.0*t)- 0.2*exp(-1.0*t).*sin(5.0*t) + 1

firstPoint = test3(1) % start point of function

plot(t+InputStartTime,test3)
plot([0 (t(1)+InputStartTime)], [0 firstPoint]) % Plot line from origin to start point of delayed function

However, and this is the key, test3 above is hardcoded. I cannot find a way of getting the sym into a format that I can use. If I try something like string, MATLAB then complains that the matrices are the "Incorrect dimensions for matrix multiplication". I then had to change the '*' to '.*' in the above equation. I have no idea how to take my solved equation from Symbolic maths and use it like this. Can anyone help please? This is how the formula looks once I have solved it, with yt being a "1x1 sym".

performing a string() of this gives: "1.0 - 0.2*exp(-1.0*t)*sin(5.0*t) - 1.0*exp(-1.0*t)*cos(5.0*t)" which shows that there is a * between exp() and sin/cos rather than the required .*

[Torsten]

Use "fplot".

[Paul]

Hi Robyn,

We have a function yt(t) = .... that I presume is the output of the circuit in response to a particular input. How can time invariance of the system be demonstrated by plotting yt(t) ?

[Robyn Galliers]

Hi Paul. This formula is the overall response to a stepinput. So if I initiate the stepinput at t=1 rather than t=0, I am showing that the response is the same but time-shifted, i.e. demonstrating time invariance.

The script links to a Simulink model which plots the response on the scope, delayed to t=[StepInputStart]. I am now just trying to create a similar graph in the Livescript side to match, showing that what I have solved analytically is the same as that created from the Simulink model. And despite messing around with plot() and fplot() for a few weeks now I cannot yet find a method to plot my solved equation with a delayed start.

[Paul]

You can certainly use plot or fplot to just shift the plot of Yt to the right by StepInputStart seconds and that will match the Simulink plot generated with the step input delayed by the same time. But that's not really showing that the analytical solution shifts in accordance with the time invariance property. To do that, you'd have to analytically solve for the output of the system in response to the delayed input, and compare that to delaying the output that is generated in response to the undelayed input. I guess the approach to take depends on the specific wording of the question.

[Robyn Galliers]

Hi Paul,

I just wanted to thank you for your input. It was the seed that I needed to take the next step, and after another day of trying things I have now managed to solve it analytically by adding another symbol (tau) into my equation for y(t).

My test code below then allowed me to plot the graph below, which is exactly what I wanted.

test1 = subs(yt, tau, InputStartTime)
test2 = subs(yt, tau, 0)

hold on;
fplot(test1,[InputStartTime 5])
fplot(test2,[0 5])
plot(tme,unitstep)

Answer 1

[John D'Errico]

By the way, there is NOTHING symbolic in the code you wrote.

syms t % this makes t symbolic

% no need to put decimal points in there.
test3 = -exp(-t).*cos(5*t)- 0.2*exp(-t).*sin(5*t) + 1;

Now, test3 is a function of t. I'm not sure what you are thinking about in terms of time invariance. If you change t or the limits of what you plot, then what you see will change shape.

t0 = 1;
fplot(test3,t0 + [0,5])

Perhaps what you are thinking about is to shift the x axis, but plot the same curve, merely arbitrarily translated along the x axis? Not using syms at all, you might create a function handle.

test3 = @(t,t0) -exp(-(t - t0)).*cos(5*(t - t0))- 0.2*exp(-(t - t0)).*sin(5*(t - t0)) + 1;

Now plot this curve, starting at the point t0, over a width of 5 units.

t0 = 1;
fplot(@(t) test3(t,t0),t0 + [0,5])
hold on

t0 = 3;
fplot(@(t) test3(t,t0),t0 + [0,5])

There we see the same basic shape, but merely shifted in time. However, I don't see how this demonstrates anything about time invariance.

[Robyn Galliers]

Hi John,

:) yes I know, the code above is just test code to try get this working. My symbolic code is all in my project, which over 250 lines of code solves the RLC equation's roots, coefficients, damping ratio etc. resulting in yt.

As I said above, I have not been able to find anyway other than the hack above to try show that the function plotted with a step input at t=1 is the same as if the step occurred at t=0.

I am unsure why you say this doesn't indicate time invariance, as I get the same graph shape starting at (t-\tau) as I do at (t)

Regards,

[Robyn Galliers]

John, I also wanted to thank you for your answer. Although it wasn't exactly what I was after, it was food for thought and helped lead me to the eventual solution. Thanks for your time in answering.