I want add the following value 5e^(-25/2t)u(t) as a Vin to my circuit in LTSpice. However I don't know how to do this - can someone help?
Use a voltage b-source with this:
V = 5*exp(-12.5*time)
U(t) is not needed because it is a step input and goes to 1 as soon as the simulation starts (at t=0 the b-source is considered on)
If you wanted a time offset (like 0.3s for the step function going from 0 to 1) then use an if statement)
You can get very close using just the built-in EXP source
From the Help file:
Syntax: Vxxx n+ n- EXP(V1 V2 Td1 Tau1 Td2 Tau2) Time-dependent exponential voltage source Name Description Units V1 Initial value V V2 Pulsed value V Td1 Rise delay time sec Tau1 Rise-time constant sec Td2 Fall delay time sec Tau2 Fall-time constant sec For times less than Td1, the output voltage is V1. For times between Td1 and Td2 the voltage is given by V1+(V2-V1)*(1-exp(-(time-Td1)/Tau1)) For times after Td2 the voltage is given by V1+(V2-V1)*(1-exp(-(time-Td1)/Tau1))-(V2-V1)*(1-exp(-(time-Td2)/Tau2))
If you make
Tau1 very short, and
Td2 - Td1 maybe 5 or 6x
Tau1, you get very close to the exponential pulse you want:
In any practical circuit, the fact that this model has a non-zero rise time will not cause any significant difference in the results compared to using the ideal exponential pulse. And if it does, it's likely that the results from this model are closer to reality than from the model using zero rise time.