The step response of a circuit is its response to a step input, u(t). Knowing the step response of a linear circuit is extremely valuable as we can figure out the circuit's stability. In this problem we will analyze the circuit in Figure 1 for the output voltage, vO(t) given an initial condition. Then you will find the output response of two input steps--a current input, iI(t), and a voltage step, vI(t).
Figure 1
The element values are as follows: R1=1kΩ, R2=1kΩ, R3=2kΩ, CA=0.5μF, and CB=1.5μF.
To begin we note that the capacitor bridge composed of four capacitors can be replaced with one capacitor, CTOT. What is the value of CTOT (in microfarads)?
unanswered
First we consider the circuit in Figure 1 when iI(t)=0A, vI(t)=0V, and vO(0)=1V. What are the time constant, τ, of the output voltage and the output voltage, vO(t), at time t=0.3ms?
τ (in milliseconds)
unanswered
vO(t) (in Volts) at time t=0.3ms?
unanswered
For the remainder of this problem we consider the step response where iI(t)=1u(t)mA, vI(t)=1u(t)V, and vO(0)=0V. Remember that u(t) is the unit step function:
5 answers
sorry, 8.01 and 6.0 anything are not subjects I respond to. Been there, done that. You are all on your own.
Moreover, I can not see the circuit !
to add capacitors in parallel
Ceq = C1 + C2 because C = Q/V
V the same, charges add
to add capacitors in series
q is the same on both
in general V = C Q
here V = V1 + V2 = C1 q + C2 q
V/q = (C1+C2)
so Ceq = q/V = 1/(C1 + C2)
Ceq = C1 + C2 because C = Q/V
V the same, charges add
to add capacitors in series
q is the same on both
in general V = C Q
here V = V1 + V2 = C1 q + C2 q
V/q = (C1+C2)
so Ceq = q/V = 1/(C1 + C2)
in general q/C
here V = V1 + V2 = q/C1 + q/C2
V/q = (1/C1+1/C2)
so Ceq = q/V = 1/(1/C1 + 1/C2)
= C1 C2 /(C1+C2)
here V = V1 + V2 = q/C1 + q/C2
V/q = (1/C1+1/C2)
so Ceq = q/V = 1/(1/C1 + 1/C2)
= C1 C2 /(C1+C2)
Here we analyze the time response of the circuit in Figure 3 which contains capacitive and resistive elements and a voltage source, VS. The voltage source is switched between two nodes in the circuit, and we observe the voltage response across C1. All answers in this problem are looking for expressions in terms of VS, C1, C2, R1, and R2.
Initially for t<0 the switch connects the voltage source, VS to node 1 and the circuit is allowed to reach steady-state. What is the steady-state voltage, v(t), across C1 for t<0?
At t=0, the switch is repositioned to connect the voltage source to node 2 and the circuit is allowed to reach steady-state again. What is the voltage, v(t), across C1 after the circuit reaches steady-state?
We can use differential equations to write an expression for the voltage, v(t), across C1 for t>0. The homogeneous and particular solutions can be used to express the voltage, v(t)=H∗e−t/τ+P. Give expressions for H, P and τ.
τ
P
H
@Damon plz help me for this question if u got it correct ??
Initially for t<0 the switch connects the voltage source, VS to node 1 and the circuit is allowed to reach steady-state. What is the steady-state voltage, v(t), across C1 for t<0?
At t=0, the switch is repositioned to connect the voltage source to node 2 and the circuit is allowed to reach steady-state again. What is the voltage, v(t), across C1 after the circuit reaches steady-state?
We can use differential equations to write an expression for the voltage, v(t), across C1 for t>0. The homogeneous and particular solutions can be used to express the voltage, v(t)=H∗e−t/τ+P. Give expressions for H, P and τ.
τ
P
H
@Damon plz help me for this question if u got it correct ??
1 answer