[SOLVED] Simple question in L.P

[omerysmi]

In this LPF the input is step signal, I need to find an experssion of Vo(0+) and of Vo(∞).


My answer is:
Vo(0+) = (E*R2)/(R1+R2)
Vo(∞) = E

Am I right?

 

[andre_luis]

Sure, seems correct.

 

[omerysmi]

Sure, seems correct.

So in fact the transient is not blocked in this circuit? (Unlike standart L.P.H circuit of capatiance in series to resistor that the transient is blocked)

As we can see in the drawing there is a transient of E*R2/R1+R2

 

[D.A.(Tony)Stewart]

The transient (rapid step) is only suppressed by the resistor ratio.

In particular with an RC filter where the Cap has ESR or Rs like R2, this result applies.

 

[omerysmi]

The transient (rapid step) is only suppressed by the resistor ratio.

In particular with an RC filter where the Cap has ESR or Rs like R2, this result applies.

The drawing of this circuit is right?

 

[Venkadesh_M]

Yeah the step response is correct. Why don you try a expression.

 

[D.A.(Tony)Stewart]

Correct.

Now Imagine a steady DC source with Cap with ESR, with PWM load.

The pulsed ripple is the same R ratio except using source and load.

They dont stress the significance of this in education enough.

The ESR or series R or differential series resistance in a semiconductor or any nonlinear part or winding resistance in a choke are all examples that cause non ideal step transient in pulsed loads or sources.

Another way to apply this is to specify load regulation of a source which may have a large Cap on output.

the net effect of a step load is same as a step source, the error is the Rload/(Rload+Rsource) and this is called Load regulation Error.

Where Rsource is usually the total ESR of the semiconductors that drive the output for dc and ESR of the capacitors for pulsed load ripple estimates.

So source to load ratio is important for a voltage source as the voltage source is never ideal=0 Ohms.

 

[omerysmi]

Yeah the step response is correct. Why don you try a expression.

The total resistance in this circuit is R1 + R2 right?

 

[Venkadesh_M]

Yes, but why do you need it. The graph in #5 is completely definable after 0, try to write a expression so that you will get a more deeper understanding.

 

[omerysmi]

Yes, but why do you need it. The graph in #5 is completely definable after 0, try to write a expression so that you will get a more deeper understanding.

The expression is
Vo(0+) = (E*R2)/(R1+R2)
Vo(∞) = E

 

[andre_luis]

The total resistance in this circuit is R1 + R2 right?

You must be aware that the term "real" also has a specific meaning on math.

Although in fact R1+R2 represents the real component of the overall impedance ( composed by a sum of a Real plus an Imaginary part ), for a transient analysis it is useless a separation or them. Anyway, as you pointed before, at the initial condition ( t=0+ ) is true.

 

[Venkadesh_M]

The expression is
Vo(0+) = (E*R2)/(R1+R2)
Vo(∞) = E

Hi,

I don want you to write this equation, that you did in #1, but V(t) a single expression then you can substitute all 0+ values and get the output voltage.

V(t) = ? for ( t = 0+ )

 

[omerysmi]

You must be aware that the term "real" also has a specific meaning on math.

Although in fact R1+R2 represents the real component of the overall impedance ( composed by a sum of a Real plus an Imaginary part ), for a transient analysis it is useless a separation or them. Anyway, as you pointed before, at the initial condition ( t=0+ ) is true.

Look, I have to find the time constant (τ) so i need to find the equivalent resistance in the circuit (Req*C)
so Req is in fact R1+R2

 

[Venkadesh_M]

what is next? i mean final expression. i would like to see the whole expression.