Current in time domain, voltage source with 3 sin components, unable to understand solution

Discussion in 'Homework Help' started by Lomtrur, Sep 14, 2014.

  1. Lomtrur

    Thread Starter New Member

    Sep 14, 2014
    this is from a previous exam and I do have a solution for it, but I do not understand it.

    "Determine the current of the circuit in the time domain when Z_l is implemented as

    And the current source has been replaced with the following one:
    <br />
U_{in}(t) = 2*(sin(500*2{\pi}t) + sin(1000*2{\pi}t) + sin(1500*2{\pi}t))<br />

    Okay, what I know:
    Current in time domain should be something like this:

    <br />
i(t) = sin({\omega}t + {\psi}) (I have also seen \phi used instead of \psi, am I confusing something here? The book I am using "Understandable Electronic Circuits" by Meizhong Wang uses \psi

    From the previous task (this one here is 2.2, 2.1 uses the same circuit) I have:
    <br />
f_{lower} = 500 Hz<br />
f_{higher} = 1500 Hz<br />
    Z_l was supposed to have the value of 50 Ohms in order to get f_0 = 1 kHz, I am unsure if I am supposed to re-use that here.

    Anyway, onwards:
    Looking at the voltage source, I notice the following:
    <br />
v(t) = V_{max} * sin({\omega}t)<br />

    So I ASSUME, I do not know if this applies, please tell me, but I ASSUME V_{max} = 2V.
    Reasoning: I look at the voltage source. I see 3 sines and something "in front", the 2. If I use the definition above (taken from "Serway Physics for Scientists and Engineers", p.1035) then it has to be 2.
    BUT and this confuses me, if I use, say, t = 7 and plug it in and use a calculator, I get 4.XX, which is greater than 2. So is V_{max} = 2V correct? I assume it has to be because of what follows, but how can I get 4.XX? This is probably very basic stuff but I really don't understand this. I am also unsure if I can get the maximum voltage that way, because I have 3 sines, instead of just 1.

    <br />
sin({\omega}t)<br />
    \omega is defined as 2*{\pi}*f
    So if I look at the 3 sines, I see that the frequencies for those 3 are 500, 1000, and 15000.
    And 500 is the lower cutoff frequency, and 1500 the higher cutoff frequency. What can I use this knowledge for?

    Getting I_{max} I understand, V_{max}/Z = I_{max}, and if I use
    R = Z_l = Z_C
    and then Z = sqrt(R^2 + (Z_C - Z_l)^2)
    I get Z = 50 Ohms so I_{max} = 2/50 (using the V_max from above), and according to my solution that's correct. So I guess I am allowed to use maximum voltage = 2V from above? It might be a trap or a crazy coincidence, but I am confused why I can get >2 from that equation.

    This is as far as it made sense to me, assuming my logic is correct.
    However, according to my solution, the following follows:

    "The signal has 3 sinusoidal components which happen to be at the cut off frequencies and
    the resonant frequency. Thus:"
    <br />
i(t) = (\frac{sqrt(2)}{2} * {I_{max}}  sin(500*2{\pi}t + {\phi}_1) +{I_{max}}  sin(1000*2{\pi}t) + \frac{sqrt(2)}{2} * {I_{max}}  sin(1500*2{\pi}t +{\phi}_2)<br />
    With I_{max} = \frac{2}{50} A

    From Serway Physics p.1043:
    <br />
i = I_{max}sin({\omega}t - {\phi})<br />
    (In RLC circuits)
    This seems different from the solution, the \phi has a negative sign in the book, while in the solution it has a positive one? Is that because order of elements is different? In the book it is R-L-C in series, in the circuit here it is R-C-L. Why does that change the sign of the phi though?

    Next part:
    Current = Voltage / Resistance, in AC it is Current = Voltage / Impedance
    I am going to skip a bit of the solution because I understand it, it is just simply expanding a fraction and simplyifing it, eventually it reaches this:

    <br />
I = {\frac{U_{max}(R - j({\omega}L - \frac{1}{{\omega}C}))}{R^2 +({\omega}L - \frac{1}{{\omega}C})^2}} \Rightarrow tan({\phi}({\omega})) = \frac{\frac{1}{{\omega}C} - {\omega}L}{R}<br />

    How does that follow?
    From the impedance triangle you get that phase angle between current and voltage is tan^(-1)(\frac{X_l -X_C}{R}), but here X_l and X_C are swapped, why?
    This has gotten quite lengthy so thanks for your time.

    I am unable to use the "Upload a File" button, so I am using Imgur for the circuit instead:
  2. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    If I understand things correctly the question is asking you to:

    1. Determine the inductance value which would give series resonance at 1000Hz. Hence the circuit resistance is 50 ohms at that condition.
    2. Given the input U(t) what the output time domain response would be.
    You state:
    For the given circuit values the bandwidth would be 1000Hz - giving upper and lower 3dB points at 1500Hz and 500Hz.
    I find a value of L to give 1000Hz resonance as 7.9577mH.
    Stricly speaking whilst giving a bandwidth of 1000Hz this does not mean the upper and lower 3dB frequencies (for the current response) are actually 1500Hz and 500Hz. In fact they are 1618.041Hz and 618.035Hz. I guess for the sake of simplifying the question we can assume 500Hz and 1500Hz are "close enough".
    Is this a correct interpretation thus far?
    It would be good if you could give a concise statement of the actual question as your post requires some interpretation.

    Regarding phase angles of individual current terms.
    • At resonance current is in phase with source voltage.
    • Below resonant frequency the circuit is capacitive. Hence current will lead the source voltage in phase.
    • Above resonant frequency the circuit is inductive. Hence current will lag the source voltage in phase.
    • At the -3dB frequencies the phase lag or lead should be 45 degrees. Close attention to the math will reveal this condition is not met at either 500Hz or 1500Hz for the circuit under consideration.
    Last edited: Sep 15, 2014
  3. MrAl

    Distinguished Member

    Jun 17, 2014

    I'd like to ask a question of the OP myself too...

    1. Are you saying that the circuit shown now has impedance Z equal to an unknown inductance?
    2. Are you also saying that the current source is now a voltage source (and presuming it's polarity is positive upward for now) and that source is of the form:
    V(t)=A*[(sin(w*t)+sin(2*w*t)+sin(3*w*t)] ?

    If the answer to #2 is 'yes' then the current response must contain all of the frequency components if the circuit elements are all linear:

    It can not be simply of the form:
    I(t)=B*sin(w*t+ph) or

    because we do not loose any frequency components so we can not end up with a single frequency response. The lower and upper frequencies may have more or less amplitude then the center frequency but they will still be present.
    Last edited: Sep 15, 2014
  4. Lomtrur

    Thread Starter New Member

    Sep 14, 2014
    Whoops sorry for being unclear!

    The circuit I posted is used in 3 tasks, 2.1, 2.2. and 2.3, the one I struggle with is 2.2
    In task 2.1 Z_L "is just there", it is not specified as inductance.
    Task 2.1 is:
    "What is the value of Z_l in order to have a resonance at in circuit 3?
    Determine the following quantities:
    • quality factor ,
    • the bandwidth, and
    • the lower and higher cutoff frequencies."
    A few short calculations using known formulas, the problem I have is with task 2.2:
    "Assume that the source of circuit 3 is replaced by the following one:"
    Here goes the u(t) = ... I posted above
    "Determine the current of the circuit in the time domain when Z_l is implemented as

    I assume that's because I said I was unsure if I was supposed to re-use Z_l = 50 Ohms from the previous task (2.1)?
    I am still not 100% if I am "allowed" to use that because 2.1 did not say it is an inductance but 2.2 says so, but if I use it I get Z = 50 Ohms and can use I_max=V_max/Z = 2/50 A, which is what I am supposed to get. I was just wondering if there is any other way to get I_max = 2/50 A, in case I am not supposed to assume Z_l = 50 Ohms, though I can't think of a good reason not to use that, just being paranoid I guess.

    Sorry I must have mistyped it somewhere, voltage source is replaced with another

    Thanks, I am going to take a look at it!
  5. MrAl

    Distinguished Member

    Jun 17, 2014
    Another what?