Find output voltage as a function of time

1. The problem statement, all variables and given/known data

The op amp in the circuit seen in the figure is ideal. Find the steady-state expression for vo(t) = vg = 1cos(106t) V. Express your answer in terms of t, where t is in microseconds. Enter the phase angle in radians.

Hints: (1) Let the node above the capacitor be Node a, and the node at the inverting (i.e., minus) input be Node b. (2) Write KCL equations at Node a and at Node b. Hints: This gives equations in terms of Va and Vo. Also note that the 40 K-Ohms load resistor is not part of the equations

2. Relevant equations

Current = 0 into input terminals of op amp
ZC = 1/(jωC), where j=[itex]\sqrt{-1}[/itex]
KCL
V = IZ

3. The attempt at a solution

Vg is given in the format: v(t) = Vmaxcos(ωt+ø), where ω is the frequency and ø is the phase angle.

Vb = 5V

For KCL, ∑ientering = ∑ileaving

Node a:
i100kΩ + i20kΩ = i5kΩ + i100pF

Using V=IZ,

[itex]\frac{V_o – V_a}{100k}[/itex]+[itex]\frac{V_b – V_a}{20k}[/itex] = [itex]\frac{V_a – V_g}{5k}[/itex] + Va*(jωC)

[itex]\frac{V_o – V_a}{100k}[/itex] + [itex]\frac{5 – V_a}{20k}[/itex] = [itex]\frac{V_a-1}{5k}[/itex] + Va*j*10^6*100*10^(-12)

Node b:
i10pF = i20k

Vb*(jωC) = [itex]\frac{V_b-V_a}{20k}[/itex]
Vb*(jωC) = [itex]\frac{5-V_a}{20k}[/itex]

Solving, I get Va = 5-j and Vo = 95+j24

Converting Vo to polar, Vo = 98∠14.2° V
Converting to radians, Vo = 98∠0.25 V
Putting this in terms of t,

Vo(t) = 98cos(10^6t+0.25) V

Converting to microseconds,

Vo(t) = 98cos(10t+0.25) V which is incorrect. Not exactly sure what I’ve done wrong here.

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