1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Balanced Modulator AM Radio

  1. Jul 19, 2013 #1
    1. The problem statement, all variables and given/known data

    The figure shows a balanced modulator. Inputs to the top modulator are the information
    signal and the carrier signal and the inputs to the bottom modulator are the inverse of the
    information signal and the carrier signal. Then the output from the bottom modulator is
    inverted again before being added to the output from the top modulator.
    a. Derive an expression for the output signal s(t).
    b. What are the frequency components present in this signal?
    c. Efficiency of an AM modulator is defined as ratio of the power in the side bands
    (actual information) to the total transmitted power. Calculate the maximum efficiency
    of the DSBFC modulator we studied in lectures and the modulator in this problem.
    Discuss the difference in efficiencies of the two systems.

    2. Relevant equations


    3. The attempt at a solution
    The output of the top modulator is:
    [tex] (E_c+E_i*sin(2\pi*f_i*t))*sin(2\pi*f_c*t) [/tex]
    The output of the bottom modulator is:
    [tex] (E_c-E_i*sin(2\pi*f_i*t))*sin(2\pi*f_c*t) [/tex]
    This is then inverted, so:
    The output of the top modulator is:
    [tex] -(E_c+E_i*sin(2\pi*f_i*t))*sin(2\pi*f_c*t) [/tex]
    and summed:
    [tex] s(t)=2E_i*sin(2\pi*f_i*t)*sin(2\pi*f_c*t) [/tex]
    Using Trig Identities this goes to:
    [tex] s(t)=E_i*cos(2\pi(f_i-f_c)t)-cos(2\pi(f_i+f_c)t) [/tex]

    Reaaly not sure about this? seems kind of right since the carrier band has been removed?
    Again not sure but from what I have in a I would say the upper and lower sidebands are the only components - fi-fc and fi+fc?

    Any help welcome.

  2. jcsd
  3. Jul 19, 2013 #2

    rude man

    User Avatar
    Homework Helper
    Gold Member

    Where do you get Ec as your first term?
    The output of the top modulator, being balanced, contains only sum and difference frequencies of fi and fc. Your expression includes a component at fc also.


    EDIT: Each modulator output can also include a dc term.

    Your final expression is however close to right. very close.
    Last edited: Jul 19, 2013
  4. Jul 19, 2013 #3
    Thanks rude man, the final expression wouldn't be missing a 2 per-chance?

    Nothing that I've read (a lot) shows the mathematical output of a balanced modulator, so I've been working from DSB-FC modulators -'standard' AM.

    -Can you point me in the direction of some literature which explains the mathematical output of the balanced modulator? I don't think I can just ignore the first term Ec can I?

  5. Jul 19, 2013 #4

    rude man

    User Avatar
    Homework Helper
    Gold Member

    Hi 92 - I had an equally hard time findng adequate literature on balanced modulators (b.m.). My textbook assumes a DSB-FC input rather than the given sinusoid, so all the derivations are meaningless unless I were to put a lot of effort into reverse-engineering the deriations which I'm not inclined to do.

    I do know that the balanced modulator does not produce a frequency component at fi as you have it. I suggest you interpret the b.m. as just a plain multiplier, which will give you the sidebands at fc - fi and fc + fi. Purely as a practical matter, most b.m.'s also produce a dc output.

    So I suggest you multiply Ecsin(wct) by Eisin(wit) and append a dc voltage V0 to each b.m.'s output. Then use the trig relation you cited to get the sidebands. You will find that the dc terms disappear by virtue of the final summing operation. The output is a DSB-SC modulation, the second "S" standing for "suppressed".

    Actually, most b.m.'s use switching diodes or other switching methods so the "carrier" is really a square wave. By multiplying this waveform by your input sinusoid you can readily see that the actual b.m. output also contains sidebands at sum and difference frequencies of the input sinusoidal frequency and (odd) harmonics of the square wave. In practical applications these higher sidebands are filtered out of existence. This approach gives you the same output as though the carrier were also just a sinusoid. I suppose it's possible that your instructor intended for you to show these higher harmonics in S(t) though.

    Some commercially available modulators act like pure multipliers for small carrier voltage amplitudes but revert to square wave modulation when that voltage amplitude is >> the input signal.
  6. Jul 19, 2013 #5

    rude man

    User Avatar
    Homework Helper
    Gold Member

    Yep, but Ec has to get in there somewhere too ...
  7. Jul 19, 2013 #6
    Thanks, i'll see what happens when I treat the bm as a pure multiplier. Why would I want Ec in the output? Isn't Ec suppressed?
  8. Jul 19, 2013 #7

    rude man

    User Avatar
    Homework Helper
    Gold Member

    Not if the modulator acts like a pure multiplier.

    But if you assume large-amplitude (square-wave) modulation, then yes, Ec disapperas in favor of whatever scale factor the elctronics provides.

    I get the impression from the problem that you are to use simple multiplication, though.

    I suppose it all depends on what kind of circuit you were provided with for a modulator. Got any kind of description or diagram?n It would be the same circuit as for DSB-FC modulation.
  9. Jul 19, 2013 #8
    The problem with pure multiplication is that the two inverters effectively cancel each other out - mathematically they become completely pointless.
    Got an email from my lecturer re this question (paraphrased)
    - 'Modulator is not just a multiplier - the basic modulation DSBFC - In the balanced modulator the modulator blocks are the basic DSBFC modulators. In fact most AM modulators use DSBFC as the basic building block and add other functions to optimise the output signal.'
    I think me and you both misinterpreted the question, the whole block diagram is the balanced modulator, with the two 'modulators' being DSB-FC - stupidly worded question.

    This is an assignment question so hes not going to tell my ya/na- but from that I suspect that I am at least on the right track. The problem is that after reading a lot about DSB-SC I think that I should have E/2 out the front - maybe not though, I do have two (double?) sidebands and a suppressed carrier in the output signal?
    -Again, really not sure because wiki has VmVc/2 -Thoughts?

    I wouldn't mind some pointers on c also if your willing. I know the answer is max efficiency of DSB-FC is 33% and DSB-SC is 100% as there is no carrier. Evidently I can't just say this though - I need to go from the standard expression of DSB-FC&SC and convert those to powers which I am not quite sure how to do?
    dsb-fc =
    [ tex ] E_c*sin(2\pi*f_c*t)+\frac{m*E_c}{2}*cos[2\pi(f_c-f_i)t)-\frac{m*E_c}{2}*cos[2\pi(f_c+f_i)t) [ /tex ]
    The first thing I am thinking is to set m as 1 but I still can't think how to turn this into power?

    [edit] don't know why pf doesn't like my laTex?
  10. Jul 19, 2013 #9

    rude man

    User Avatar
    Homework Helper
    Gold Member

    If we did have pure multiplication:
    Top modulator output = Ei Ec sin(wct) sin(wit) + DC
    Bottom modulator output= -Ei Ec sin(wct sin(wit) + DC
    Bottom modulator output after inversion = Ei Ec sin(wct) sin(wit) - DC
    Summer output = 2 sin(wct) sin(wit) = DSB-SC output.

    However, this is beside the point. I also plead guilty to misinterpreting the question. Each modulator is presumably a simple, basic square-wave modulator, with the whole diagram representing a balanced modulator.

    So OK, the output of each modulator will be kEi[sin(wit){1/2 + 2/pi cos(wct) - 2/3pi cos(3wct) + ...}]. k depends on the particular circuit gain.

    This output contains wi and frequency components (nwc +/- wi), n odd.

    Just use that output and its inverse to get the b.m. output. The output will contain a component at wi and all the difference frequencies. All these terms are usually filtered out so only the two sidebands wc +/- wi appear at the output.

    DSB-SC is only 50% efficient. If you want 100% efficiency you go to SSB-SC.

    [edit] don't know why pf doesn't like my laTex?[/QUOTE]

    Never use it myself! Old dog & new tricks ...
  11. Jul 19, 2013 #10
    Thanks, can you explain why DSB-SC is only 50%.. Given that ssb is 100% it follows but I don't know how to show that? Wiki begs to differ about the efficiency of DSB-SC: http://en.wikipedia.org/wiki/Double-sideband_suppressed-carrier_transmission

    -This probably comes down to the definition of efficiency used but since the question states ' Efficiency of an AM modulator is defined as ratio of the power in the side bands (actual information) to the total transmitted power' and only the sidebands are transmitted how could the efficiency under this definition be less than 100%?

    Back to part a.
    Is k just a different notation for the modulation index - we use m?
    Not considering the series representation is there something wrong with my original answer? Obviously, as it doesn't have a wi term but what have I messed up?
    Considering the modulators as DSB-FC and the simple mathematical representation of that.
  12. Jul 20, 2013 #11

    rude man

    User Avatar
    Homework Helper
    Gold Member

    I need to look at this a bit more myself. I need to figure out what kind of modulator can take sin(wi t) as an input and come up with a 'standard' AM sigal of the form A[1 + m sin(wi t)]cos(wc t). Right now it seems to me any modulator I know of would need to have the input signal be 1 + Ei sin(wi t) rather than just Ei sin(wi t).
  13. Jul 20, 2013 #12
    Thanks r m, grateful for your help. -I should probably start a new thread for this but since you know your radio COMMS maybe you could help me out on what should be a much easier question.

    A superhet receiver is receiving a signal from an AM modulator where the carrier
    signal has an amplitude Ec and carrier frequency is fc. Frequency of the information signal is fi.

    The modulator has the following parameters.
    * output (load) resistance - RT,
    * modulation index - m,
    * total amplification through the modulator stages - GT

    During the propagation through the air signal power is attenuated by a factor C (C < 1).

    and I want to know the power received by the superhet.
    I would have thought that I needed to know what type of modulator is on the transmission end but that info is not given.
    -I'm thinking that I should just take the 'standard' equation given for AM (DSB-FC) and multiply it by the gain, square it and divide by R. Something about it doesn't seem right though?
    [tex] P=((E_c*sin(2\pi*f_c*t)+\frac{mE_c}{2}cos[2\pi(f_c-f_i)]-\frac{mE_c}{2}cos[2\pi(f_c+f_i)])*G_T)^2/R*C [/tex]
  14. Jul 20, 2013 #13

    rude man

    User Avatar
    Homework Helper
    Gold Member

    Going back to your first problem for just a sec:

    the problem I had is that a basic modulator is already a DSB-SC-output device. This the case with a diode bridge or semiconducor modulator. To get a DSB-FC output you need to input E0 + Ei sin(wi t), not just Ei sin(wi t).

    Anyway, assuming each of the two modulators puts out a DSB-FC signal, I am making a minor correction to your original solution which I think is what was intended, even though to me the circuit makes little sense as an implementation of a DSB-SC modulator:

  15. Jul 20, 2013 #14

    rude man

    User Avatar
    Homework Helper
    Gold Member

    Confusion again! If the receiver impedance is very high, which would be typical, the power received by the receiver would be nearly zero!

    But if we want the available power at the receiver, then I suppose you could take the output signal of the modulator Vo = Ec[1 + m sin(wi t)]cos(wc t), multiply by the gain stages' gain G_T, square, divide by R_T, then multiply that power by C to get to the receiver.

    Which looks like what you did.

    They might want averaged power over time rather than instantaneous power.
  16. Jul 20, 2013 #15
    Thanks again r m.

    back to the original question,
    Taking [tex] s(t)=Ei∗cos(2π(fi−fc)t)−cos(2π(fi+fc)t) [/tex] as the answer to a,
    b. would be the upper and lower sidebads, fi-+fc?
    c. I need to take the answer from a. and turn it into power which I guess is a simple case of V^2/R but that will give me instantaneous power like the second question. How can I get the average powers for these? I figure that the max power occurs when the cos terms go to 1?
  17. Jul 20, 2013 #16

    rude man

    User Avatar
    Homework Helper
    Gold Member

    Usually, fi << fc so let's call them fc +/- fi. Yes, these are the upper (+) and lower (-) sidebands.
    Average power in your standard AM signal, defined by Ec[1 + m sin(wi t)]cos(wc t) can be derived by squaring the signal and computing the time-average thereof. Don't work with the sideband expression, work with the above instead.

    Hint: multiply out [1 + m sin(wi t)]^2 and then take averages of each expanded term. The next step would look like

    Pavg = Avg[(Ec^2 cos^2(wc t)] * [1 + avg{m^2 sin^2(wi t)}]

    You should wind up with two terms, one being the power in the carrier and the other in the modulating signal.

    I'll let you struggle with this & give you a hand if needed. :smile:
  18. Jul 20, 2013 #17
    Thanks, ill have a go now and get back to you. This is going to turn into an integrating exercise isn't it? Also, since I want to know the efficiency of DSB-FC vs DSB-SC I will have to do this for both the standard AM which you said to start with and also the answer for part a aren't I?
  19. Jul 20, 2013 #18

    rude man

    User Avatar
    Homework Helper
    Gold Member

    Yes. But an easy one.
    Yes, but the DSB-SC case is easier I think (though I haven't done it yet myself). When you get the two power terms for the DSB-FC case you can compare the i term with the c term to get the relative power between the two components i and c. I still think efficiency of the DSB-SC case should be 50% and that of the SSB-SC case 100%. Half the power goes to waste for DSB-SC. For the DSB-FC case all the carrier power and half the i power are wasted. Oh well ...
  20. Jul 20, 2013 #19
    had an epiphany - why do I need the average power for this question, the instantaneous and average efficiency should by rights be the same especially mathematically?
    Started with the general signal for AM:
    [tex] E=E_c*sin(\omega_c*t)+\frac{mE_c}{2}*cos(\omega_c-\omega_i)-\frac{mE_c}{2}*cos(\omega_c+\omega_i) [/tex]
    [tex] P=\frac{E_c^2*sin^2(\omega_c*t)}{R}+\frac{(\frac{mE_c}{2})^2*cos^2(\omega _c-\omega_i)}{R}-\frac{(\frac{mE_c}{2})^2*cos^2(\omega_c+\omega_i)}{R} [/tex]
    so the efficiency is
    [tex] \mu = \frac{\frac{(\frac{mE_c}{2})^2*cos^2(\omega _c-\omega_i)}{R}-\frac{(\frac{mE_c}{2})^2*cos^2(\omega_c+\omega_i)}{R}}{\frac{E_c^2*sin^2(\omega_c*t)}{R}+\frac{(\frac{mE_c}{2})^2*cos^2(\omega _c-\omega_i)}{R}-\frac{(\frac{mE_c}{2})^2*cos^2(\omega_c+\omega_i)}{R}} [/tex]
    Setting m,Ec,R and the sin and cos terms to 1 (net +1) will give the absolute max efficiency?
    [tex] \mu = \frac{\frac{(\frac{1}{2})^2*1}{1}+\frac{(\frac{1}{2})^2*1}{1}}{\frac{1*1}{1}+\frac{(\frac{1}{2})^2*1}{1}+\frac{(\frac{1}{2})^2*1}{1}} [/tex]
    This gives the fabeled 33.33% efficiency for DSB-FC.
    Regarding DSB-SC, the same process will obviously give 100% since it is always x/x but I have read some conflicting stuff about its efficiency. Some say 100% and some say 50% inc you -so I was thinking of just saying that it is unnecessary to transmit the info twice - in the two side-bands so it could be 50% and then say something about ssb-sc??

    Regarding the second question i posted, part b asks
    b. If the intermediate frequency of the receiver is f_(IF), what is the maximum RF bandwidth
    the device can have if the receiver is not to suffer from image frequency problem.
    This must just be something about superhets I guess but I can't find it??
  21. Jul 21, 2013 #20

    rude man

    User Avatar
    Homework Helper
    Gold Member

    Yes it is. Let's take a concrete example.

    In the std AM broadcast band the IF is tuned to a constant 455 KHz, meaning that the only input signals that make it thru the IF stage must have a sum or difference frequency with the local oscillator frequency fLO to produce a 455KHz IF. The passband of the IF stage is only about 10 KHz above + below 455 KHz. So the sidebands are limited to +/- 10 KHz.

    So e.g. for a station at fc = 540 KHz the LO is at 540 + 455 = 995 KHz. Now suppose a second station broadcasts at 1450 KHz. 1450 - 995 = 455 KHz also so that 2nd station will also get through the IF stage if the front end doesn't reject that frequency when the radio dial is set to 540 KHz. Any incoming signal below 540 + 2*455 will not get through the IF stage since then the IF signal is below the IF fixed tuned stage.

    I will leave it to you to produce the general answer.
  22. Jul 21, 2013 #21
    So I have an intermediate frequency f_IF (generally 455kHz in the real world?). My bandwidth is (f_IF-f_c) to (f_IF+fc) which would give me a BW of (f_IF+fc)-(f_IF-f_c) = 2fc?

    -From your post BW=fc+2f_IF but I can't seem to get that. BW=upper cutoff-lower cutoff? which I can't get algebraically?
  23. Jul 24, 2013 #22
    Hopefully this thread isn't dead yet, been busy at work.

    I'm still stuck on part b of that question, can't seem to get anything reasonable for the bandwidth?

    Since the post with the question is a fair way back:

    . An superheterodyne receiver is receiving a signal from an AM modulator where the carrier
    signal has an amplitude Ec and carrier frequency if fc. Frequency of the information signal is fi
    The modulator has following parameters.
    * output (load) resistance - RT,
    * modulation index - m,
    * total amplification through the modulator stages - GT
    During the propagation through the air signal power is attenuated by a factor C (C < 1).

    a. Calculate the input power received at the receiver. - done
    b. If the intermediate frequency of the receiver is fIF, what is the maximum RF bandwidth
    the device can have if the receiver is not to suffer from image frequency problem.
    c. If the RF bandwidth from part “b” is used and noise density (noise power per Hz) is
    N0, calculate the SNR at the input of the receiver.

  24. Jul 24, 2013 #23

    rude man

    User Avatar
    Homework Helper
    Gold Member

    At this point, for part b, all I can add without violating the rules of this formum is to hint that the maximum permissible bandwidth has nothing to do with fc. I suggest you re-read my previous post carefully & think about where the image signal becomes a problem.

    I might get back to you on part c.
  25. Jul 25, 2013 #24

    rude man

    User Avatar
    Homework Helper
    Gold Member

    Part c: first, establish the power available at the receiver we discussed previously. This was something like Ec2 * m2/4 * GT * C. Obviously, max signal power is when m = 1.

    Then you still haven't got the answer to part b. But let's call it B Hz.

    Then, if the noise density is N0 W/Hz, what is the total noise power in terms of B?

    Finally, the SNR is by definition signal power/noise power, usually expressed in dB as 10 log10(signal power/noise power).
  26. Jul 25, 2013 #25
    Thanks r m, still confused on b, i know that pf doesn't allow just giving answers but maybe if I tell you what i'm thinking you can point out where I have made mistakes:

    -Image frequency is any unwanted frequency that makes it through the mixer?
    -The mixer output is: (wiki)
    The output of the mixer may include the original RF signal at fRF, the local oscillator signal at fLO, and the two new heterodyne frequencies fRF + fLO and fRF − fLO
    - Ideally, the IF bandpass filter removes all but the desired IF signal at fIF.. Hence the bandwidth should be f_IF? I think not.
    -But....The image frequency is 2 fIF higher (or lower) than f_RF......So I'm thinking that the bandwidth needs to be f_RF-2fIF to f_RF+2IF so BW = 4fIF? where fIF is the intermediate frequency not the image frequency

    This at least meets the condition of having nothing to do with the carrier frequency, directly at least but i am still not convinced.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted