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Argument Of Transfer Function - What does arg(T(w)) mean in real life

  • Thread starter thomas49th
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Hi, I have a question something along the lines of

Here is a low pass filter, when the resistor = R and the capacitor = C. There is a sinusoidal voltage input source (V1) and a voltage across the capacitor (V2). The transfer function is T(w) = V2/V1.
Determine the |T(w)| and arg(T(w)) where T(w) is the transfer function. Calculate the magnitude of T(w) when V2 lags V1 by 45°

OKAY! So w = 2*pi*f.

T(w) = [tex]\frac{1-jRwC} {-(RWC)^2 - 1}[/tex]

|T(w)| = [tex]\frac{\sqrt{1+(RwC)^2}}{(RWC)^2 - 1}[/tex]

so am I right in thinking arg(T(w)) is tan(x) = -RwC???

BUT how on earth do I determine the magnitude when V2 lags V1 by 45°. I thought the capcitor only lags current. I'm confused

Thanks
 

Answers and Replies

  • #2
vela
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Hi, I have a question something along the lines of

Here is a low pass filter, when the resistor = R and the capacitor = C. There is a sinusoidal voltage input source (V1) and a voltage across the capacitor (V2). The transfer function is T(w) = V2/V1.
Determine the |T(w)| and arg(T(w)) where T(w) is the transfer function. Calculate the magnitude of T(w) when V2 lags V1 by 45°

OKAY! So w = 2*pi*f.

T(w) = [tex]\frac{1-jRwC} {-(RWC)^2 - 1}[/tex]

|T(w)| = [tex]\frac{\sqrt{1+(RwC)^2}}{(RWC)^2 - 1}[/tex]

so am I right in thinking arg(T(w)) is tan(x) = -RwC?
Yes, that's right.
BUT how on earth do I determine the magnitude when V2 lags V1 by 45°. I thought the capacitor only lags current. I'm confused.
V1(jω), V2(jω), and T(jω) are all complex quantities, so you can write each in polar form, i.e., re. How are the phases of these three complex quantities related?
 
  • #3
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[tex]
r = \sqrt{1+(RwC)^2}
[/tex]
ϕ = arctan(-RwC)

V2/V1 = T is that what you were suggesting? so can you say that something like T(jw) = V1(jw)/V2(j(w-pi/2))?

P.S How did you get to use super and sub script without LaTex?

EDIT: Is any of this useful
V1 = cos(w) + jsin(w)
V1 = e^(jϕ)
V2 = V1/(100jwC + 1)

or am I just spouting obvious facts?

Is there like some magical property such that the phases add up to 90°?

I thought about adding phasers together but If you wanted me to do that you wouldn't of suggested converting into polar notation as that is useful for multiplication and division?

Why have you written j's in front of the w's? Are you talking about the complex part only?

Thanks
Thomas
 
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  • #4
vela
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To get subscripts and superscript, go into advanced mode when posting, and you'll see options above the input area.

I'm not sure how far along you're into this network analysis stuff, but the transfer function is usually written as a function of s, that is, H(s) = Vo(s)/Vi(s). You talk about the s domain and the time domain, blah blah blah. When you're looking at a pure sine wave of frequency ω, that corresponds to the case when s=jω, so you often see quantities written as function of jω. It's just a convention. Maybe your class does it differently. (I'm going to use s instead of jω because it's easier to type. Just replace s with jω or simply ω everywhere you see it.)

What linear systems do is: 1) change the amplitude of a signal, and 2) change the phase of a signal. The factor by which the amplitude changes is called the gain. It corresponds to the magnitude of the transfer function. The shift in phase of the signal, as you can probably guess, is equal to the phase of the transfer function. Hopefully you can see this from the relationship Vo(s) = H(s) Vi(s), but I'll leave it to you to convince yourself that the statements are true.

So what the problem is asking you to find is the frequency at which the phase of the transfer function is -45 degrees and then to find the corresponding gain.
 

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