Find V(s) and V(t) for Vin=45 at an angle of 0o

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The discussion revolves around calculating V(s) and V(t) for a given input voltage Vin=45 at an angle of 0 degrees. The user successfully derives the current and voltage equations, ultimately simplifying Vc to 9/(s+19) and identifying poles at s=-1 and s=-19. Confusion arises regarding the interpretation of Vin in the time domain, with clarification that 45 at 0 degrees represents a phasor in polar form. The notation indicates that V_in(t) can be expressed as 45 cos(ωt), although variations in conventions are acknowledged. The conversation highlights the importance of understanding phasor notation in circuit analysis.
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Homework Statement


everything's in the picture.



Homework Equations



V=I*R

Ic=Iin*R/(R+C)

The Attempt at a Solution


Ok, I think I got the first part alright. I found the current in terms of Vin to be: Vin/(25+(22.5*225/s)/(22.5+225/s)) which simplifies to Vin*(s+10)/(225s+475). Now that I have current I can use the current divider rule to find Ic and then multiply Ic by C to get Vc. By doing that I get the equation [Vin*(s+10)/(225s+475)]*[22.5s/(225/s+22.5)]*225/s. After much simplification I get Vc=9*Vin/(s+19). I divide that by Vin and get 9/(s+19). Setting the bottom equal to zero I get a pole at s=-19. Setting the entire thing equal to zero I get a zero at s=+or- infinity.

Once I get to the 2nd part I'm a bit confused... if Vin=45 at an angle of 0o, is my Vc(t) just simply equal to 45*9/(0+19) since no s is given?
 

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O i accidentally forgot to multiply my 25 by s at the beginning. I actually end up w Vc/Vin=9/(s^2+10s+9) so that makes my poles s=-1,-9 and my zeroes still + or - infinity.
 
yoamocuy said:
Once I get to the 2nd part I'm a bit confused... if Vin=45 at an angle of 0o, is my Vc(t) just simply equal to 45*9/(0+19) since no s is given?
I'm unfamiliar with the notation used. What is "45 at 0 degrees" supposed to mean?
 
well 45 at 0 degrees is still given in the s domain. 45 will be the magnitude and 0 will be the degree that's in the cos or sin if you have a jw anywhere, but since I don't have any imaginary numbers in my s then I shouldn't have a cos or sin anywhere. My problem is, I can solve for my poles and find s=-1, and -19. Does that mean that I should have a voltage for Vin that is = to 45*(e^-t + e^-19t), or is it simply 45?
 
yoamocuy said:
well 45 at 0 degrees is still given in the s domain. 45 will be the magnitude and 0 will be the degree that's in the cos or sin if you have a jw anywhere, but since I don't have any imaginary numbers in my s then I shouldn't have a cos or sin anywhere.
Sorry, that still doesn't make any sense to me.

I'm guessing it means V_{in}=45 e^{i0}=45. In other words, V_{in}(t)=45\delta(t).
 
vela said:
Sorry, that still doesn't make any sense to me.

I'm guessing it means V_{in}=45 e^{i0}=45. In other words, V_{in}(t)=45\delta(t).

Also known as the Steinmetz notation:

<br /> 45 \angle 0^{\circ}\;\texttt{V}<br />

which is basically a shorthand notation for complex numbers in polar form, whereas you in your post have used the exponential form. Both mean the same.
 
Oh, okay. From what I read about the notation, it's specifying V_{in} in the phasor domain, so in the time domain, you have V_{in}(t)=45\sqrt{2} \sin(\omega t+0^\circ).

Edit: I read a bit more, and it seems some people use different conventions. In light of the rest of what the question asks, I think it means

V_{in}(t)=45 \cos(\omega t+0^\circ)=Re(45 e^{i\omega t+0^\circ}).
 
Last edited:
<br /> 45 \cos(\omega t+0^\circ)=45 \angle 0^{\circ}<br />

That's the thing about Steinmetz notation, it doesn't show the angular frequency. It only shows you the magnitude of the corresponding phasor and its phase shift w.r.t. to the reference.

As far as I know, it's much more common amongst engineers than physicists (like the i and j conventions for complex numbers).
 
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