Need help in designing this circuit--- Ramp Model

[mohdfayez]

suppose u have a DC voltage 'x'............anywhere between 0.3V to 7V......design a circuit in such a way that when this voltage x is given to this circuit the out put of the circuit will be a ramp voltage.

lets keep a switch in between the voltage source and this circuit

so basically when the switched is closed only then will the ramping start
the time of the ramping period should be variable...........most probably using a potentiometer...........and the range of the time should be from 0.1 to 10 seconds

so if u give an input voltage of lets say 4V to this circuit , as soon as the switch is closed that 4 volts should start ramping from 0V towards 4V and should attain a steady voltage after the ramping period is over


Can anyone help in this... with explaination coz I am not good at this.. help will be appreciated.. Thank you

Added after 40 minutes:

Required circuit is as given in attachment with output waveform

 

[StephenVG]

What is the application?

 

[Kral]

mohdfayez,
An analog integrator should do the job. Use an op-amp in inverting configuration. Let C be a capacitor from output to Inverting input. Let R be the input resistor from the inverting input connected to a voltage source thru the switch. When the switch is closed, the ramp rate will be -Vin/RC volts per second (R in Ohms, C in Farads). When the switch is opened, the output voltage will "freeze" at the level that exists when the switch is opened.
.
There are some practical problems with this scheme. Since no op-amp is ideal, when the switch is open the output will drift at a rate that depends on both the op-amp bias current and offset voltage. Also, you need a method of resetting the output voltage. It can be reset to zero simply by shorting the feedbacj capacitor.
Regards,
Kral