Embarrassing Question

Well... just off of the top of my head, if you expected the traces to be exact inverses of each other, you need to fix the scaling. Different from one plot to the next.

No no they should not be the inverse of each other they should be the EXACT same curve

They are the same function

The scaling is off simply because those are not the exact plots i just posted those picture so people could see what the general shape is

The top one is correct like we would expect.

The bottom one shows a decay for some reason and that is incorrect and i don't know why

Given the fact that the laplace inverse is correct

Anyone else?

I mean this is just an RC circuit lol i know there are some smart folks on here

Obviously i should get the same results

Should you, or should you get an RC time constant capacitor discharge curve instead?

RC time constant discharge curve

It's been a while since I had to do the math but the charge and discharge curves are different. Isn't the inverse of a RC time constant charge curve the discharge curve?

yes of course they are different

But this has nothing to do with What I'm asking

My function is the step response of the circuit not the discharge

The only difference is one is laplace and on is in time t

These two functions are suppose to be identical

One is in s the other is in t

They should match up exact

The step response in t is just 1

So Vout_in_t = 1 * (Inverse Laplace of the function)

They should be identical

Yes I knew it was wrong the second I saw it from intuition

Problem was my code per usual....

Thanks to everyone for jogging my memory

doesn't work so well when you forget to give it a 1/s input

Thanks again folks

All one has to remember is that for a first-order system, the response is within 1% of the new setpoint after 5 time constants.

You have to evaluate the functions over the same time limits?

you forgot a constant somewhere that transforms as 1-e^X

First of all-that has to remain between us- I hate Laplace- and the matlab also- since I am an old Power Electrical Engineer and the complex calculation and Microsoft Excel workbook is my usual way to solve a problem. Of course, nobody cares what a hate or like.

Second, since we have a lot of way to connect R and C let's take this one.

Now in summary, let's remember what I learned many years ago:

1) By definition the transformations are [ >>="it corresponds to"]:

d(f(t))/dt>>s*f(s) integral(f(t))>>f(s)/s k>>k/s f(t)>>f(s)

vcap=qcap/C vcap=integral(i(t))/C

2) The voltage drop on the above circuit is then:

vin=R*i(t)+integral(i)/C

By derivation :

dvin/dt=R*di/dt+i/C

Translating:

s*Vin=(s*R+1/C)*i(s) then:

i(s)=s*vin(s)/(s*R+1/C)

vout(s)=i(s)/C/s

vout(s)=vin(s)/(s*R+1/C)/C

Vin=constant[k] then vin(s)=Vin/s and:

vout(s)=Vin/(s^2*R*C+s) or:

vout(s)=Vin/(R*C)/(s^2+s/(R*C))

Let's put a=1/(R*C) then from table:

1/s/(s+a)>>(1-exp(-a*t))/a then finally:

vout(t)=Vin*(1-exp(-t/(R*C)) Q.E.D.