# Circuit Theory/Convolution Integral/Examples/example48

find voltage across 1 ohm resistor ... which clearly is half Vt or Vr .. and ultimately Vt is the voltage across a 1 ohm resistor

Given that Vs(t) = 3μ(t) mV, and the IL(0-) = 1 ma, find the voltage across the one home resistor.

Outline:

• find transfer function/differential equation
• find homogeneous solution
• find particular solution for step function
• evaluate initial conditions
• find impulse solution from step function solution
• use convolution integral to find general solution
• evaluate integration constant

## Contents

### Transfer Function

From inspection can see the following:

$V_{TR} = V_R = \frac{V_T}{2}$

So:

$H(s) = \frac{V_{tr}}{V_s} = \frac{V_{tr}}{V_s} = \frac{V_T}{V_s}\frac{V_{tr}}{V_T} = \frac{1}{1+s} * \frac{1}{2}$

### Homogeneous Solution

Setting the denominator equal to zero, can see that the time constant is -1. Thus the solution is of the form:

$V_{TR_h} = A*e^{-t} + C_1$

### Particular Solution

After a long period of time hooked to a DC voltage source, an inductor appears as a short. This means that all 3mV is going to drop across the equivalent of a 1 ohm resistor. Half of this is going be VTR. So the particular solution is going to be:

$V_{TR_p} = 1.5 mV$

### Initial Conditions

The initial current is going to be 1 mA. This is going to be dropped by the 1 ohm resistor creating VT, thus VT(0+) = 1 mV and VTR is going to be 0.5 mV. This means that:

$V_{TR}(t) = A*e^{-t} + C_1 + 1.5 mV$
$V_{TR}(0) = 0.5 mV = A + C_1 + 1.5mV$
$A+C_1 = -1 mV$

The inductor obliges the rest of the circuit and drops the other 2mV so the source is happy. This is the source of the other initial condition. First have to turn VTR into an expression of total current which is the inductor current:

$i_{TR} = \frac{V_{TR}}{1} = V_{TR}$

Because the current is split in half has it goes through the resistors, the total current is twice that of i_{TR}:

$i_L = 2*V_{TR} = 2*(A*e^{-t} + C_1 + 1.5m)$

From the inductor terminal relation, know that the dervivative of this times inductance is going to equal the initial 2 mV across the inductor:

$V_L(t) = L*{d i_L \over dt} = L*-2Ae^{-t}$
$V_L(0) = 2m = L*-2A = -2*A$
$A = -1m$

This means that C1 is zero. So the answer is:

$V_{TR} = 1.5 - e^{-t} mV$