Frequency Response and Output of an RLC Circuit

[beyondlight]

Homework Statement

http://tinypic.com/view.php?pic=doxum1&s=6

Consider a casual LTI system implemented as the RLC circuit above. x(t) is the input. And y(t) is the output across the capacitor.

a) Find the differential equation relating x(t) and y(t)
b) Determine the frequency response of this system by considering the output of the system to inputs of the form x(t)=e^jwt.
c) Determine the output y(t) if x(t) =sin(t)

Homework Equations

i=C\frac{dy(t)}{dt}

The Attempt at a Solution

V_{in} = RCy'(t) + L\frac{di(t)}{dt} + y'(t)

V_{in} = RCy'(t) + LCy''(t) + y(t)

e^{jωt} = RCωjH(jω)e^{jω} - LCω^{2}H(jω)e^{jωt} + H(jω)e^{jωt}

1 = (RCωj - LCω^{2} + 1) H(jω)

H(jω) = \frac{1}{RCωj - LCω^{2} + 1}And for c)

Is then y(t) = H(jω)sin(t) ?

 

[HallsofIvy]

Homework Statement

http://tinypic.com/view.php?pic=doxum1&s=6

Consider a casual LTI system implemented as the RLC circuit above. x(t) is the input. And y(t) is the output across the capacitor.

a) Find the differential equation relating x(t) and y(t)
b) Determine the frequency response of this system by considering the output of the system to inputs of the form x(t)=e^jwt.
c) Determine the output y(t) if x(t) =sin(t)

Homework Equations

i=C\frac{dy(t)}{dt}

The Attempt at a Solution

V_{in} = RCy'(t) + L\frac{di(t)}{dt} + y'(t)

y is the output, not the input. What happened to x? Also, your last y'(t) should be y(t) but you correct that in the next line.

V_{in} = RCy'(t) + LCy''(t) + y(t)

e^{jωt} = RCωjH(jω)e^{jω} - LCω^{2}H(jω)e^{jωt} + H(jω)e^{jωt}

1 = (RCωj - LCω^{2} + 1) H(jω)

H(jω) = \frac{1}{RCωj - LCω^{2} + 1}


And for c)

Is then y(t) = H(jω)sin(t) ?

 

[beyondlight]

x(t) = RCy'(t) + L\frac{di(t)}{dt} + y(t)

Like this?

I think the other lines are OK. This is just mistake on this particular line i believe.

Sorry for confusing V_in with x(t). Both are the same.

 

[NascentOxygen]

Replace di/dt with the second derivative of y, like you showed earlier.

You now have a second order DE, and that's to be expected whenever you mix L and C in any circuit — a second-order system.

Solve for y(t).

 

[beyondlight]

I have done it already:

x(t) = RCy'(t) + LCy''(t) + y(t)

e^{jωt} = RCωjH(jω)e^{jω} - LCω^{2}H(jω)e^{jωt} + H(jω)e^{jωt}

1 = (RCωj - LCω^{2} + 1) H(jω)

H(jω) = \frac{1}{RCωj - LCω^{2} + 1}
So if x(t)=sin(t) is then

y(t) = H(jω)\frac{(e^{jt}-e^{-jt})}{2j} = \frac{1}{RCωj - LCω^{2} + 1}\frac{(e^{jt}-e^{-jt})}{2j}

Is it correct done?

 

[rude man]

I have done it already:

x(t) = RCy'(t) + LCy''(t) + y(t)

e^{jωt} = RCωjH(jω)e^{jω} - LCω^{2}H(jω)e^{jωt} + H(jω)e^{jωt}

1 = (RCωj - LCω^{2} + 1) H(jω)

H(jω) = \frac{1}{RCωj - LCω^{2} + 1}



So if x(t)=sin(t) is then

y(t) = H(jω)\frac{(e^{jt}-e^{-jt})}{2j} = \frac{1}{RCωj - LCω^{2} + 1}\frac{(e^{jt}-e^{-jt})}{2j}

Is it correct done?

Yes. Eliminate the (e^jt -e^-jt)/2j terms to get your H(jw).

And y(t) and x(t) don't belong in that equation any more. The equation should finally read
Y(jw) = H(jw)X(jw). You have transformed from the time domain to the frequency domain.

Also: you don't need to form the Euler equivalent of sin(wt) as you have done:

The process is to assume sine input x(t) → X(jw) = e^jwt, then y(t) derives from
Y(jw) = H(jw)X(jw) in the form y(t) = |H(jw)| with phase angle tan^-1{Im(H)/Re(H)} where Re and I am stand for real part & imaginary part resp.

 

[beyondlight]

Yes. Eliminate the (e^jt -e^-jt)/2j terms to get your H(jw).

And y(t) and x(t) don't belong in that equation any more. The equation should finally read
Y(jw) = H(jw)X(jw). You have transformed from the time domain to the frequency domain.

Also: you don't need to form the Euler equivalent of sin(wt) as you have done:

The process is to assume sine input x(t) → X(jw) = e^jwt, then y(t) derives from
Y(jw) = H(jw)X(jw) in the form y(t) = |H(jw)| with phase angle tan^-1{Im(H)/Re(H)} where Re and I am stand for real part & imaginary part resp.

Why eliminate (e^jt -e^-jt)/2j to get H(jω) when i already got it? I want to express y(t) in terms of H(jω) in the time domain, which i suppose i have don correctly, and of course i don't have to form the Euler equivalent.

But isn't x(t) → X(jω) = π/j(δ(ω-ω0) -δ(ω+ω0)) ?My next question is how to solve the inverse Fourier transform integral for H(jω)? So that i get h(t), the impulse response.

x(t)=\frac{1}{2π}\int_{-∞}^{∞}H(jω)e^{jωt}dω = \frac{1}{2π}\int_{-∞}^{∞}\frac{1}{RCωj - LCω^{2} + 1}e^{jωt}dωIt seems like a tricky integral, do you have any suggestions how to solve it?

EDIT: It's a causal LTI system. Maybe that will help us.

 

[rude man]

You're trying to use the Fourier transform with your X(jw). I'm not very familiar with the Fourier transform, but I do know it's unnecessarily complicated here. That's the because the problem asks for frequency response which implies a steady-state condition with sinusoidal input & output. The Fourier transform handles non-sinusoidal inputs, and that is not needed here.

In this case the method is as I outlined previously. The derivation is based on nothing more than complex algebra - no Fourier transform needed.

y(t) = magnitude |H(jw)| and phase angle tan^-1Im(H)/Re(H).

Problem is I don't know just what approaches your course is taking in solving such problems. It's extremely atypical and very awkward to approach an electrical problem like this one with the Fourier transform. And as I said, I don't use the Fourier transform and neither do most electrical engineers. (We do us the Fourier integral when faced with input pulses of finite duration). We are e^jwt folks when faced with this type of problem, and Laplace transform folks when faced with non-sinusoidal inputs and/or non-steady-state problems!

 

[beyondlight]

I meant of course

h(t)=\frac{1}{2π}\int_{-∞}^{∞}H(jω)e^{jωt}dω = \frac{1}{2π}\int_{-∞}^{∞}\frac{1}{RCωj - LCω^{2} + 1}e^{jωt}dω not x(t) on the left hand side.

Okay. But as i recall the Fourier transform can both handel periodic and aperiodic signals. In the case of aperiodic signals the transform is usually a impulse train.

I wanted to solve the task like this because it was outlined so in my student book. That is by using the inverse Fourier transform of the frequency response H(jw).

 

[rude man]

Okay. But as i recall the Fourier transform can both handel periodic and aperiodic signals. In the case of aperiodic signals the transform is usually a impulse train.

I wanted to solve the task like this because it was outlined so in my student book. That is by using the inverse Fourier transform of the frequency response H(jw).

I understand, and you have the correct integral expression for h(t), but good luck doing that integral! BTW you're supposed to find
y(t), so if you go the Fourier route and find h(t), you're then back in the time domain and have to convolve h(t) with x(t) to get
y(t). If you stick in the frequency domain then you need to solve
y(t) = F^-1{Y(jω)} = ∫X(jω)H(jω)e^jωt. This is of course even worse than the integral for h(t).

A problem in my graduate textbook using the Fourier integral has your circuit exactly except the inductor is shorted, so
H(jω) = 1/(1 + jωT) where T = RC. Even that much simpler transfer function required a table of Fourier integrals to solve the inverse transform integral, an excerpt of which my textbook kindly supplied for this particular case (cited from G.A. Campbell and R.M. Foster, Fourier Integrals for Practical Applications, D. Van Nostrand 1958).

I believe no way are you going to do either inverse-transform integral without a table to look it up. Unfortunately there does not seem to be a freebie table on the Web, least not any I could find.

 

[rude man]

Boy, have I been dumb. I said the integral ∫X(jw)H(jw)dw would be harder still! WRONG! The fact that your input is a sine wave with the transform X(jw) = what you stated, then you just make use of the sampling feature of the δ function:
∫f(x)δ(x-x₀)dx = f(x₀).

So now y(t) = (π/j)∫H(jw){δ(w-w₀) - δ(w+w₀}e^jwtdw where w₀ is your input frequency in radians/s.

You should be able to take it from there. Don't mess with trying to compute h(t) at all.

(Forgive me for writing w instead of ω, it's such a pain ...)

 

[beyondlight]

Boy, have I been dumb. I said the integral ∫X(jw)H(jw)dw would be harder still! WRONG! The fact that your input is a sine wave with the transform X(jw) = what you stated, then you just make use of the sampling feature of the δ function:
∫f(x)δ(x-x₀)dx = f(x₀).

So now y(t) = (π/j)∫H(jw){δ(w-w₀) - δ(w+w₀}e^jwtdw where w₀ is your input frequency in radians/s.

You should be able to take it from there. Don't mess with trying to compute h(t) at all.

(Forgive me for writing w instead of ω, it's such a pain ...)

This should be correct.

y(t)= \frac{π}{j}\int H(jω)(δ(ω-ω_{0})-δ(ω+ω_{0}))e^{jωt}dw=\frac{π}{j}H(jω)(e^{jω_{0}t}-e^{-jω_{0}t})I suppose H(jw) is considered as a constant in this integral even if it depends on w?

 

[rude man]

This should be correct.

y(t)= \frac{π}{j}\int H(jω)(δ(ω-ω_{0})-δ(ω+ω_{0}))e^{jωt}dw=\frac{π}{j}H(jω)(e^{jω_{0}t}-e^{-jω_{0}t})


I suppose H(jw) is considered as a constant in this integral even if it depends on w?

Yoiu're close. But you must know from elementary integration that when you integrate with respect to w there can be no w left after that!

H(jw) must be treated just the way you treated exp(jwt). Then you're home.

 

[beyondlight]

Yoiu're close. But you must know from elementary integration that when you integrate with respect to w there can be no w left after that!

H(jw) must be treated just the way you treated exp(jwt). Then you're home.

So its supposed to be:

y(t)= \frac{π}{j}\int H(jω)(δ(ω-ω_{0})-δ(ω+ω_{0}))e^{jωt}dw=\frac{π}{j}(H(jω_{0})e^{jω_{0}t}-H(-jω_{0})e^{-jω_{0}t})

I guess i don't have to do the antiderivative since the rule of an integration of a delta-pulse tells us so:

∫f(x)δ(x-x0)dx = f(x0)

 

[rude man]

So its supposed to be:

y(t)= \frac{π}{j}\int H(jω)(δ(ω-ω_{0})-δ(ω+ω_{0}))e^{jωt}dw=\frac{π}{j}(H(jω_{0})e^{jω_{0}t}-H(-jω_{0})e^{-jω_{0}t})

I guess i don't have to do the antiderivative since the rule of an integration of a delta-pulse tells us so:

∫f(x)δ(x-x0)dx = f(x0)

You are 100% spot-on. Congrats!

 

[rude man]

You are 100% spot-on. Congrats!

New worries.

I ran this method through using a simpler function than yours, viz. H(jw) = 1/(1+jwT), T constant, and I got the wrong answer. The answer had the right phase but the magnitude needed to be the square root of what I computed: my magnitude was |H(jw)|² = 1/(1+w²T²) whereas it's supposed to be just |H(jw)| = 1/√(1+w²T²). I have been trying to figure out what's wrong but haven't been able to so far.

Also, a minor point, but the inversion integral should have a 1/2π in front of it:
y(t) = (1/2π)∫X(w)H(w)e^jwtdw.

I would really appreciate it if you could report what your teacher gave as the method he wanted to arrive at the answer.

 

[beyondlight]

For the calculation of y(t) i think i was supposed to do it the way i already did. My techer didnt give any suggestion on how to solve it. Neither did the book instruct me to solve it in any particular way.

 

[rude man]

For the calculation of y(t) i think i was supposed to do it the way i already did. My techer didnt give any suggestion on how to solve it. Neither did the book instruct me to solve it in any particular way.

But I thought you said your teacher wanted you to use the Fourier transform?

Anyway, what was your final answer? We could compare notes, I worked it out too - my way.

 

[rude man]

In case you're still interested in this thread: I rechecked my math and everything's OK.

So my post #16 is the end of the story. Sorry for having muddied the water unnecessarily ...