transient_steady state analysis

7/25/2019 Transient_Steady State Analysis

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5Ptud4'.llHow to do first-order transient problems

Here is a step by step procedure on how to approach first-order transient problems that haveone switching event (like all the problems on hw4). First-order means that there is only one inductor ORone capacitor in the circuit. The procedure can be broken up into two parts. The first part applies to thecircuit before the switching event (t


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circuit. Zero-out any independent sources as you would do for a Rrhevenin calculation (replace voltage

sourceswithshortcircuitsandcurrentsourceswithopencircuits)..t,a , r , .. .,J,:

6. Find the initial condition of any other desired voltages/currents:

L: replace the inductorwith a current source equalto i(t=O)and find x(t=0)

C: replace the capacitor with a voltage source equal to v.(t=O) and find x(t=0)

As I said earlier, make sure you're consistent with how you defined i1 and V. in step 2 when you replace

the inductor/capacitor with a current/voltage source.

7. Find the final condition of any other desired voltages/currents by looking at the circuit in steadystate. You're doing the same thing you did in step 2, but this time you're looking at the circuit

after the switching event (120):

L: replace the inductor with a short circuit and find x(t=m1

C: replace the capacitor with an open circuit and find x(t=oo)

Note that if there are no independent sources present in the circuit for 120, the inductor or capacitor will

release all of its energy and all final conditions will be 0.

8. Use the following equation to express your desired voltage/current as a function of time for t20:

x(t) =x(oo) + [x(0) - x(a)]e-t/'

Other thines to note

The voltage drop across an element is always defined as being in the direction of the current

flowing through it. So, say you have an inductor that is drawn vertically in a circuit and you define theinductor current, ip to be pointing down. The (+) terminal of vl will be on top and the (-) terminal onbottom.

The voltage across an inductor is vr(t) = L+dil(t)/dt and the current through a capacitor is i.(t) =C*dv.(t)/dt. These equations are helpful if, for example, you have i(t) and want to find v(t) withoutcalculating v,.(t=O) and v(t=m). Simply differentiate is(t) with respect to time and multiply it by theinductance.


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transient_steady state analysis

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