The Unit-Step Function, and drivern RL/RC circuits

i am readin about Unit step function, and driven rl/rc circuits, and have some problems with

some exercise, help me please.


here is first exercise: At t = 2, find the value of 2u(1-t) - 3u(t-1) - 4u(t+1)

as i know,the Unit step function has this form, Au(t - t0), after t0 the value of function is

A.

if t0 = 0, then function's value is A, always after 0 seconds.

if functions is Au(-t), then its value is A only on negative side of t axis.



so, first i am a little bit stuck ... how to interpret 2u(1 - t)??? in this case t0 = 1? or

-1?




here is second exercise where i need help: The voltage source 60 - 40u(t) is in series with a

10 ohm resistor, 50mH inductor. find inductor current and voltage at t equal to: a) 0- , b)

0+ , c) infinity, d) 3ms

well, -40u(t) says to me that, after 0 seconds this function's value is -40, so... 60 - 40 =

20v

so in the case of a) 0-, the time is before 0... at this perioud function is 0, then 60 - 0 =

60v

and the current is 60v/10 ohm = 6A this is correct answer... but in the answer they write

that voltage is 0v!! why?

here is answer to this exercise: a)6A, 0v b) 6A, 40v c) 2A, 0v d) 20A, 22v

could u help me with these exercises please, and explain... thanks

The definition of a unit step function is:

u(t) = 1 for t>=0
= 0 for t<0

so for:

2u(1-t) - 3u(t-1) - 4u(t+1)

we have

2u(1-t) = 2 for 1-t>=0 or t<=1
= 0 for 1-t<0 or t>1

3u(t-1) = 3 for t>=1
= 0 for t<1

4u(t+1) = 4 for t>=-1
= 0 for t<-1

So the functions value at t =2 becomes:
0 - 3 - 4 = -7

The second problem at 0- is a 60V source with a 10 ohm resistor and 50 mH Inductor. This is the at steady state for t=0- so a current of 60/10 = 6A is flowing and since the inductor is ideal no voltage is drop across it, therefore voltage is 0V.
At 0+ voltage is steps from 60 to 20 volt. But the inductor will not allow instantaneous current change. So 6A needs to flow in the resistor. So the voltage at the inductor will go down by 40V to keep the voltage across the resistor the same = 60V so the voltage across the inductor now will be -40V.
At infinity we again have steady state for a voltage of 20V so now a current of 2A is flowing and there is again no drop across the inductor.
For the 3ms case you have to solve the circuit using Laplace or differential equations. But I think the answer is wrong since the current will always be between 6A and 2A.

Regarding the second exercise: you are asked to calculate the INDUCTOR voltage.
Since the inductor has zero resistance, the voltage across it is zero at 0-. The current is as you said, 6A.

b) the voltage applied is a negative-going step, of amplitude 40V, from 60V to 20V. Since the inductor current cannot change instantaneously, the inductor current at 0+ is 6A. Since the current does not change, it follows that the voltage across the resistor does not change, so the entire voltage step will be found across the inductor. It really should read -40V, since its direction is opposite that of the current.

c) At infinity, the current is, of course, the remaining voltage / resistance, that is 2A. The inductor voltage is again zero, since its resistance is zero.

d) The inductor current decays exponentially from 6A to 2A, in accord with:
i(t)=i(0+)-(i(0+)-i(∞))*(1-e^(-t/τ)), where τ=L/R=5ms

Therefore,
i(3ms)=6-(6-2)*(1-e^(-3/5))=4.195A

Hence, the resistor voltage is 41.95V, so the inductor voltage is:
Vind=u(3ms)-R*i(3ms)=20-41.95V=-21.95V

The sign is just a matter of convention, but it tells you the voltage is opposite the direction of the current flow.

So I guess the book was wrong on the last point. Or, it's a typo, where the current was supposed to read 4.20A and then the voltage would have been 22V.

Note that the equation should provide the correct answers for any time t. My explanations were meant to give you guidance in quickly checking the results and providing some understanding of what was happening.

aryajur
problem is... that the answer for the first exercise is 1, not -7...

aryajur gave you the right answer. Isn't the exercise 2u(1-t)-3u(t-1)+4u(t+1) ?

well... i checked ERRATA for the book, answer should be 1... i dont know why. and by the way... he solved it using base u(t) function.. but dont u think that 2u(1 - t) looks more like u(-t) function??? then answer should be actually -5... but not 1 anyway...strange


and now about second exercise...

VVV

i understood, except one thing, can u explain please following equation in more details:

i(t)=i(0+)-(i(0+)-i(∞))*(1-e^(-t/τ)), where τ=L/R=5ms

i dont get why i must substract from i(0+) sum of i(0+) and -i(∞) multiplied to (1 - e^(-t/τ)??? can u show me the base equation, or logically explain this?

cuz the only thing i understood here is that the current decays expotentially from 6A to2A..

thanks

BlackOps said:

well... i checked ERRATA for the book, answer should be 1... i dont know why. and by the way... he solved it using base u(t) function.. but dont u think that 2u(1 - t) looks more like u(-t) function??? then answer should be actually -5... but not 1 anyway...strange

Click to expand...

u(t) basically defines u(-t) functions also (f(x) defines f(-x) - Basic maths) So u(-t) is simply the mirrored version of u(t) across the y axis. and u(1-t) will be the mirrored version of u(t-1) and so the equations I wrote are still consistent. Still you should not get -5 !
I think it is a misprint in the book, like maxwellequ said the question must have intended to have a '+' before the 4 instead of a '-'

The differential equation from which you start is really Kirchhoff's law for the single mesh of the circuit.

Vind+Vres=Vsource.

Vind=L*di(t)/dt
Vres=R*i(t)

The quation becomes:
L*di(t)/dt+R*i(t)=V

Divide by R and you get:

L/R*di(t)dt+i(t)=I
L/R=τ, the timeconstant; whenever you have a resistor/ capacitor rcircuit, the time constant is τ=RC, with an inductor and a resistor, τ=L/R

Math teaches us that the solution of this equation is the solution of the homogenous equation, plus one particular solution.

The solution of the homogenous equation τ*di(t)/dt+i(t)=0 is:
i2(t)=A*e^(-t/τ)

The solution of the first equation is i2(t)+const

A particular solution is easily found for t=∞
i(∞)=i1=I

So the solution becomes:
i(t)=i2(t)+i1

i(t)=i(≈)+(i(0)-i(≈))*e^(-t/τ)

That would be the solution you find in a textbook.


Personally, I went further, adding and subtracting i(0):

i(t)=i(0)-i(0)+i(≈)+((i(0)-i(≈))*e^(-t/τ)=i(0)-(i(0)-i(≈))*(1-e^(-t/τ))

That allows me to see that I start at i(0), by setting t=0

Another way to solve this problem is by superposition: you assume zero initial conditions and consider that a -40V voltage step is applied to the circuit. The solution is:
i(t)=-ΔV/R*(1-e(-t/τ)).

This is an exponential that starts from zero and ends at -4A. Since you know that the initial conditions were not zero, add them to get the final solution.
i(t)=i(0)-ΔV/R*(1-e^(-t/τ)).

But ΔV=(60V-20V) And ΔV/R=60/R-20/R=i(0)-i(≈)
So:
i(t)=i(0)-((i(0)-i(∞))*(1-e^(-t/τ))

In fact this is the faster way to find the solution.