the question is here
http://i40.tinypic.com/2m2flax.gif

when the diode is letting currect flow threw it
we have
V=4e^(-0.5t)+2t+4

when the diode is not letting currect flow threw it
we have
V=2t

what next
??

 

Hello again,


Since part of the text is missing (that pic cuts off the last part of the text)
i have to guess that they are asking us two questions here:
1. To find the voltage response across the capacitor to a ramp current 'is'
2. To compare the derivative of the result to the current through the capacitor
and decide if the current through the capacitor is really the derivative of the
voltage across it.

In other words, the text would read:

For the circuit shown in Fig P4.24, calculate the ramp response across the capacitor,
that is, the response v due to an input is(t)=r(t) with v(0)=0. Is the cap current
response for the circuit the derivative of the (voltage) ramp response?

If so, then the way to go about it is to first calculate the voltage response as
you already have done (but check your accuracy) and then take the derivative
of that response, call that dv/dt. Then, calculate the current response through
the cap and compare this to the previously calculated dv/dt to see if they are
indeed the same, or different.

The true voltage response is:
v=IS*(4*e^(-0.5*t)+2*t-4)

for 'is' being a ramp r(t)=IS*t.

so you should check your accuracy first, then proceed with the rest.
BTW you should be able to guess the result before you even do any
additional work, but i believe you should verify that too.

Let me know what you come up with ok...

In the future, try to get better pics so parts of the text are not missing.

 

the full text of the question is:
"for the circuit shown in fig p4.24 calculate the ramp response v(.) ,that is the response "v" due to an input is=r(t) with v(0)=0.is the step response for the circuit the derivative of ramp response."
in the end of the equation i got -4 that the only mistake i see.
i cant see were is my mistakes in the getting the ramp response:
Is=Vr/2 +1*Vr
for t>0
t=Vr/2 +1*Vr
(currect equals time because its a ramp graph)
then after making superposition of zero state response with the zero input response ,i get
Vr=C*e^(-0.5t)+2t-4
in I(t)=0 so C=4
and we get
Vr=4*e^(-0.5t)+2t-4

so we need to say is the "is the step response for the circuit the derivative of ramp response"

and
i dont know what equation to take

there is a diode which gives us a different equation for each of state of it.

and i am not sure if my way of thinking is correct

 

Hello again,


Ok, so they are asking if the step response equals the derivative of the
ramp response.

So what you need to do is to find the ramp response (you already did this)
and then find the step response, then take the derivative of the ramp response
and compare that to the step response. If they are the same then the answer
is "yes", but if not, then "no".
I think you should be able to find the step response if you already know how
to find the ramp response, right?

About the diode...

There are two ways of looking at what an "IDEAL" diode really is:
1. An ideal diode using the ideal diode equation that relates current to voltage exponentially.
2. An ideal diode that takes:
Id=0 when Vd<=0 and
Id=I when Vd>0
which simply means the diode is a short when current is flowing in the positive
direction and open otherwise.

You would need to know some background information in order to judge the
context of what type of ideal diode they are talking about, but i would bet it
is #2 above unless this is a more advanced course or something. Also, they
didnt give us any diode parameters like N and IS so we couldnt use the
exponential model anyway.

In that case, as long as there is current flowing from the current source to the
capacitor then that diode is a short circuit...zero ohms. Thus, the equation
you got when the diode is conducting is the correct equation to use for the
comparison.

Proceed like that and see what you can come up with for the step and derivative
of the ramp response.

 

yes i got
Vr=-2e^(-t/2) +2
for Cs diod

and Vr=2 for Open circuit diode

so the idea is to take the equations of each diode situation
and compare each pare

 

Hello again,


Well, the idea is to compare the derivative of Vc with the ramp input and
compare that to the step response of Vc. The diode ends up conducting
all the time for both.

We both got:
Vc(with_ramp_input)=IS*(4*e^(-t/2)+2*t-4)

and the derivative of that is:
dVc/dt=IS*2*(1-e^(-t/2))

and the step response is:
Vc(with_step_input)=IS*2*(1-e^(-t/2))

and since
dVc/dt=Vc(with_step_input)

the answer is "Yes".