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Ok, quick question guys (frequency domain analysis in circuits). I have a circuit (actual layout is irrelevant) with a calculated voltage transmission (Vout/Vin) of:

T = jRWC/(1+jRWC)

I have derived the frequency W at which the magnitude of the transmittance is 1/sqrt2 and get RWC=1 (this is correct).

I'm now asked to do the following:

"Find the phase shift (phase of the output - phase of the input) at that frequency"

Am I correct in thinking that it's this:

Phase of Vout = arctan (WCR/0) = pi/2

Phase of Vin = arctan (WCR/1) = (1/1) = pi/4

Thus the phase shift is = pi/2 - pi/4

=pi/4?

My tutor has scribbled some random comments over my original work (which was completely wrong) and ends up with the shift = arctabn (1/WCR) which is essentialy the same (gives the same answer, but his method's different). Is what I'm doing correct, or is just a fluke I have the same answer?!

T = jRWC/(1+jRWC)

I have derived the frequency W at which the magnitude of the transmittance is 1/sqrt2 and get RWC=1 (this is correct).

I'm now asked to do the following:

"Find the phase shift (phase of the output - phase of the input) at that frequency"

Am I correct in thinking that it's this:

Phase of Vout = arctan (WCR/0) = pi/2

Phase of Vin = arctan (WCR/1) = (1/1) = pi/4

Thus the phase shift is = pi/2 - pi/4

=pi/4?

My tutor has scribbled some random comments over my original work (which was completely wrong) and ends up with the shift = arctabn (1/WCR) which is essentialy the same (gives the same answer, but his method's different). Is what I'm doing correct, or is just a fluke I have the same answer?!

well i looked at it like: Vout/Vin = T. We can pick Vin to be whatever phase we want at any given moment cos we can set Vin=Ve^(jwt) for example

So the phase shift will just be the argument of the T = pi/4

I dont understand what you've done in your derivation nor how your tutor arrived at arctan(1/WRC)....I just do argument of T, which is you do a but of jiggery pokery comes out to be arctan(1) whatever W is.......which makes me think I'M WRONG about something. But i got the right answer

So the phase shift will just be the argument of the T = pi/4

I dont understand what you've done in your derivation nor how your tutor arrived at arctan(1/WRC)....I just do argument of T, which is you do a but of jiggery pokery comes out to be arctan(1) whatever W is.......which makes me think I'M WRONG about something. But i got the right answer

Willa

are those supposed to be Imaginary part and Real Part of V.....in which case what are you taking to be V....

I worked out the Im and Re parts of T, which are both exactly the same (RWC/1+(RWC)^2) so what is she taking as Vin and Vout then?

I worked out the Im and Re parts of T, which are both exactly the same (RWC/1+(RWC)^2) so what is she taking as Vin and Vout then?

Yeah, for some reason it only gives you R and I instead of Re and Im. She was taking Vout = jwRC and Vin = 1 + jwRC

well then yes her method is valid because the question allows Vin to be taken as anything you wish....if you choose Vin to be 1+jwRC then Vout is correctly Vout. My method just happened to use that Vin = Some unspecified value, and hence Vout=TVin, so the phase shift is just arg T...which comes out to arctan(1). The problem that arises is that is the phase shift really dependant on W?

shiny

T

= jRWC / (1 + jWRC)

= jRWC (1-jRWC) / (1+jRWC)(1-jRWC)

= (RWC)^2 + jRWC / (1 + (RWC)^2)

= jRWC / (1 + jWRC)

= jRWC (1-jRWC) / (1+jRWC)(1-jRWC)

= (RWC)^2 + jRWC / (1 + (RWC)^2)

oops i take everything back, i forgot to square it!

in which case phase shift is: arctan(Im/Re) = arctan(RWC/RWC^2) = arctan(1/RWC) which is what the tutor has scribbled. And in the specific case RWC = 1, so the phase shift is pi/4.

sorry, this is exactly what i do all the time: make stupid mistakes!

since we're on circuits..has anyone got an answer for me on my circuit question on the last page...the purpose of the large resistor...i'm still stumped and i need to understand for my report! thanks!

Sorry, still don't have a clue about that large resistor! I'm sure it's something to do with a practical we did in Electronics lab - Nick/Andy/Phil? any ideas?

Also, another question (well I had two, but I emailed my tutor about the other!) I have a circuit (see attachment) and have an emf is applied which is zero for all times before t=0, then steps up abruptly to E at t=0, and remains at E for all subsequent times. I need to find the current I(t).

Right, branch equations:

E - V1 = L.dI/dt

V1 - 0 (I - I2)R = L.dI2/dt + I2.R

so I = (L/R)*dI2/dt + 2I2

differentiating we get dI/dt = (L/R)*(d^2I2/dt^2) + 2.dI2dt

Then somehow I got the following expression...

E - L.dI2/dt - I2.R = L.dI/dt

...but I'm not sure where it's come from! Can someone explain? Thanks. Until I get that, I can't go onto the next bit!

(Apologies for my terribly diagram and the fact I can't use latex - how do I get latex?!)

Also, another question (well I had two, but I emailed my tutor about the other!) I have a circuit (see attachment) and have an emf is applied which is zero for all times before t=0, then steps up abruptly to E at t=0, and remains at E for all subsequent times. I need to find the current I(t).

Right, branch equations:

E - V1 = L.dI/dt

V1 - 0 (I - I2)R = L.dI2/dt + I2.R

so I = (L/R)*dI2/dt + 2I2

differentiating we get dI/dt = (L/R)*(d^2I2/dt^2) + 2.dI2dt

Then somehow I got the following expression...

E - L.dI2/dt - I2.R = L.dI/dt

...but I'm not sure where it's come from! Can someone explain? Thanks. Until I get that, I can't go onto the next bit!

(Apologies for my terribly diagram and the fact I can't use latex - how do I get latex?!)

well you can either substitute in the dI/dt = (L/R)*(d^2I2/dt^2) + 2.dI2dt and then solve the ODE for I2 and then plug that back into get I1 or rearrange differently to begin with and form an ODE for I1 and solve. To find the boundary conditions, we know that just before the step up, I=0 and immediately after I is still zero and t=0. We also know that because the current flowing afterwards is initially v.small, most of the voltage drop occurs over the inductor so we can estimae E=L.dI/dt at t=0.

Hey guys, a wave question for you (Andy & Nick, it's Q2a from the problem sheet Waves 2).

We have a string of uniform linear density p which is stretched to a tension T and it's ends are fixed at x=0 and x=L. Using separation of variables, you end up with the general solution:

y(x,t) = (Acos.kx + Bsin.kx)(Ccos.kct + Dsin.kct)

But since y(0,t)=0 and y(L,t)=0, we get A=0 and kL=r.pi

thus y(x,t) = Bsin(rx.pi/L)[Ccos(rct.pi/L)+ Dsin(rct.pi/L)]

I know have to show that the following are both solutions, obeying the boundary conditions:

y(x,t) = Esin(rx.pi/L)sin(rct.pi/L)

y(x,t) = Fsin(rx.pi/L)cos(rct.pi/L)

I can get the second one by using the fact that it's initially at rest, so dy/dt=0, thus in my earlier solution, D=0 and F=BC. How do I obtain the first solution however? I clearly have to use different boundary conditions, but I can't see what they should be. Suggestions?

We have a string of uniform linear density p which is stretched to a tension T and it's ends are fixed at x=0 and x=L. Using separation of variables, you end up with the general solution:

y(x,t) = (Acos.kx + Bsin.kx)(Ccos.kct + Dsin.kct)

But since y(0,t)=0 and y(L,t)=0, we get A=0 and kL=r.pi

thus y(x,t) = Bsin(rx.pi/L)[Ccos(rct.pi/L)+ Dsin(rct.pi/L)]

I know have to show that the following are both solutions, obeying the boundary conditions:

y(x,t) = Esin(rx.pi/L)sin(rct.pi/L)

y(x,t) = Fsin(rx.pi/L)cos(rct.pi/L)

I can get the second one by using the fact that it's initially at rest, so dy/dt=0, thus in my earlier solution, D=0 and F=BC. How do I obtain the first solution however? I clearly have to use different boundary conditions, but I can't see what they should be. Suggestions?

Hoofbeat

Hey guys, a wave question for you (Andy & Nick, it's Q2a from the problem sheet Waves 2).

We have a string of uniform linear density p which is stretched to a tension T and it's ends are fixed at x=0 and x=L. Using separation of variables, you end up with the general solution:

y(x,t) = (Acos.kx + Bsin.kx)(Ccos.kct + Dsin.kct)

But since y(0,t)=0 and y(L,t)=0, we get A=0 and kL=r.pi

thus y(x,t) = Bsin(rx.pi/L)[Ccos(rct.pi/L)+ Dsin(rct.pi/L)]

I know have to show that the following are both solutions, obeying the boundary conditions:

y(x,t) = Esin(rx.pi/L)sin(rct.pi/L)

y(x,t) = Fsin(rx.pi/L)cos(rct.pi/L)

I can get the second one by using the fact that it's initially at rest, so dy/dt=0, thus in my earlier solution, D=0 and F=BC. How do I obtain the first solution however? I clearly have to use different boundary conditions, but I can't see what they should be. Suggestions?

We have a string of uniform linear density p which is stretched to a tension T and it's ends are fixed at x=0 and x=L. Using separation of variables, you end up with the general solution:

y(x,t) = (Acos.kx + Bsin.kx)(Ccos.kct + Dsin.kct)

But since y(0,t)=0 and y(L,t)=0, we get A=0 and kL=r.pi

thus y(x,t) = Bsin(rx.pi/L)[Ccos(rct.pi/L)+ Dsin(rct.pi/L)]

I know have to show that the following are both solutions, obeying the boundary conditions:

y(x,t) = Esin(rx.pi/L)sin(rct.pi/L)

y(x,t) = Fsin(rx.pi/L)cos(rct.pi/L)

I can get the second one by using the fact that it's initially at rest, so dy/dt=0, thus in my earlier solution, D=0 and F=BC. How do I obtain the first solution however? I clearly have to use different boundary conditions, but I can't see what they should be. Suggestions?

i havent gone over wave theory yet, but I'm guessing the first solution is describing the case where the string is sort of "hit" so that it's dy/dt is not 0 at time 0, but it's acceleration IS 0...because the y=0 for all points on the string initially

The second case describes a string pulled to a starting position and then let go at time t=0...hence why y follows some sort of cos-sin superimposed curve at time t=0

so have you tried solving the solving one by doing a double differentiation. If you give me an hour I can try and solve it properly

actually why cant you just say C=0 and E=BD.....cos that then works. The boundary conditions are all contained in the general equation you derived in the first part, so you dont need to worry about anything else surely?

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