Efficiency of DSBFC Modulator vs. Balanced Modulator: A Comparison

[Fisher92]

Homework Statement

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.

Homework Equations

NA

The Attempt at a Solution

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

Reaaly not sure about this? seems kind of right since the carrier band has been removed?
b.
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?
c.
Stuck.

Any help welcome.

Thanks

 

[rude man]

The output of the top modulator is:
$$(E_c+E_i*sin(2\pi*f_i*t))*sin(2\pi*f_c*t)$$

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

[/QUOTE]

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

Your final expression is however close to right. very close.

 

[Fisher92]

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?

Thanks

 

[rude man]

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?

Thanks

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 f_i as you have it. I suggest you interpret the b.m. as just a plain multiplier, which will give you the sidebands at f_c - f_i and f_c + f_i. Purely as a practical matter, most b.m.'s also produce a dc output.

So I suggest you multiply E_csin(w_ct) by E_isin(w_it) 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.

 

[rude man]

Thanks rude man, the final expression wouldn't be missing a 2 per-chance?

Yep, but E_c has to get in there somewhere too ...

 

[Fisher92]

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?

 

[rude man]

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?

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.

 

[Fisher92]

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 he's 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?

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

 

[rude man]

The problem with pure multiplication is that the two inverters effectively cancel each other out - mathematically they become completely pointless.

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 kE_i[sin(w_it){1/2 + 2/pi cos(w_ct) - 2/3pi cos(3w_ct) + ...}]. k depends on the particular circuit gain.

This output contains w_i and frequency components (nw_c +/- w_i), n odd.

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

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

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

Never use it myself! Old dog & new tricks ...

 

[Fisher92]

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.

 

[rude man]

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).

 

[Fisher92]

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?
$$P=((E_c*sin(2\pi*f_c*t)+\frac{mE_c}{2}cos[2\pi(f_c-f_i)]-\frac{mE_c}{2}cos[2\pi(f_c+f_i)])*G_T)^2/R*C$$

 

[rude man]

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:

Homework Statement

3. The Attempt at a Solution
a.
The output of the top modulator is:
$$(E_c+E_i*sin(2\pi*f_i*t))*sin(2\pi*f_c*t)$$
The output of the bottom modulator is:
$$(E_c-E_i*sin(2\pi*f_i*t))*sin(2\pi*f_c*t)$$
This is then inverted, so:
The input to the bottom input to the summing junction is:$$-(E_c - E_i*sin(2\pi*f_i*t))*sin(2\pi*f_c*t)$$
and summed:
$$s(t)=2E_i*sin(2\pi*f_i*t)*sin(2\pi*f_c*t)$$
Using Trig Identities this goes to:
$$s(t)=E_i*cos(2\pi(f_i-f_c)t)-cos(2\pi(f_i+f_c)t)$$

 

[rude man]

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?
$$P=((E_c*sin(2\pi*f_c*t)+\frac{mE_c}{2}cos[2\pi(f_c-f_i)]-\frac{mE_c}{2}cos[2\pi(f_c+f_i)])*G_T)^2/R*C$$

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.

 

[Fisher92]

Thanks again r m.

back to the original question,
Taking $$s(t)=Ei∗cos(2π(fi−fc)t)−cos(2π(fi+fc)t)$$ 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?

 

[rude man]

Thanks again r m.

back to the original question,
Taking $$s(t)=Ei{cos(2π(fi−fc)t)−cos(2π(fi+fc)t)$$ as the answer to a,}
b. would be the upper and lower sidebads, fi-+fc?

Usually, fi << fc so let's call them fc +/- fi. Yes, these are the upper (+) and lower (-) sidebands.

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?

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.

 

[Fisher92]

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?

 

[rude man]

Thanks, ill have a go now and get back to you. This is going to turn into an integrating exercise isn't it?

Yes. But an easy one.

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?

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 ...

 

[Fisher92]

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:
$$E=E_c*sin(\omega_c*t)+\frac{mE_c}{2}*cos(\omega_c-\omega_i)-\frac{mE_c}{2}*cos(\omega_c+\omega_i)$$
P=E^2/R
$$P=\frac{E_c^2*sin^2(\omega_c*t)}{R}+\frac{(\frac{mE_c}{2})^2*cos^2(\omega _c-\omega_i)}{R}-\frac{(\frac{mE_c}{2})^2*cos^2(\omega_c+\omega_i)}{R}$$
so the efficiency is
$$\mu = \frac{\frac{(\frac{mE_c}{2})^2*cos^2(\omega _c-\omega_i)}{R}-\frac{(\frac{mE_c}{2})^2*cos^2(\omega_c+\omega_i)}{R}}{\frac{E_c^2*sin^2(\omega_c*t)}{R}+\frac{(\frac{mE_c}{2})^2*cos^2(\omega _c-\omega_i)}{R}-\frac{(\frac{mE_c}{2})^2*cos^2(\omega_c+\omega_i)}{R}}$$
Setting m,Ec,R and the sin and cos terms to 1 (net +1) will give the absolute max efficiency?
$$\mu = \frac{\frac{(\frac{1}{2})^2*1}{1}+\frac{(\frac{1}{2})^2*1}{1}}{\frac{1*1}{1}+\frac{(\frac{1}{2})^2*1}{1}+\frac{(\frac{1}{2})^2*1}{1}}$$
This gives the fabeled 33.33% efficiency for DSB-FC.
Thoughts?
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??

 

[rude man]

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??

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.

 

[Fisher92]

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?

 

[Fisher92]

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.
-stuck
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.
pointers?

Thanks

 

[rude man]

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?

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.

 

[rude man]

Part c: first, establish the power available at the receiver we discussed previously. This was something like Ec² * m²/4 * G_T * 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).

 

[Fisher92]

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.

 

[rude man]

OK look. Say the dial is set to 540 KHz. We assume the receiver is not capable of receiving stations below 540 KHz. In other words, the receiver bandwidth extends from 540 KHz to some upper frequency TBD. So the IF is at 540 + 455 = 995 KHz. But a signal at 1450 would also get through if the receiver bandwidth were large enough, right? Because 1450 mixes with 955 to give 455 again, which the IF amplifier amplifies, unfortunately.

So if the receiver bandwidth extends to 1450 KHz we are in trouble. Obviously, if the bandwidth extended to 1460 KHz we'd also be in trouble. BUT - if the upper cutoff frequency were say 1440 KHz then what would happen? A signal at 1440 would mix with the 995 KHz LO to give what? And does that frequency get thru the IF amplifier? What about ANY frequency below 1450 KHz?


What if the bandwidth is limited to 1430 Khz on the upper end?

 

[rude man]

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?

Right.

-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

Yes, but forget the original RF signal and the IF oscillator signal. A proper IF amplifier does not pass those.

- Ideally, the IF bandpass filter removes all but the desired IF signal at fIF.. Hence the bandwidth should be f_IF? I think not.

No.

-But...The image frequency is 2 fIF higher (or lower) than f_RF...

That's correct.

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.

I showed you how 1450 - 540 would be trouble. That spread is not 4 IF. What spread is it?

 

[Fisher92]

Thanks a lot for your time rm, really appreciate it/

So from your post I've concluded that the answer to b is BW=2f_IF? |(1450-540)/455=2

I also think that I have an understanding of this now, if you wouldn't mind checking for me

f_c - carrier frequency, 540kHz from your example, this is what would be displayed on a radio?
f_IF - intermediate frequency -455kHz - what the IF amp will amplify, f_C mixes with the LO to produce this, any other signal that the LO will mix with to produce this is an image frequency? In your example, the LO would be set to 995kHz? So if a station was transmitting at 1450kHz a mixing product of the LO would give the IF(455kHz) and would be amplified and allowed through?

To show this algebraically:
fc+2fIF-fc=2IF

*One thing I don't understand is why we assume the receiver can't receive under fc? Taking your example, a station at 85kHZ would also mix with the LO to produce the IF of 455kHz?

Also since this assignment is due soon ill pack in a few questions -
part a.
I had power available at the reciever is:
$$P=((Ec∗sin(2π∗fc∗t)+mEc2cos[2π(fc−fi)]−mEc2cos[2π(fc+fi)])∗GT)2/R∗C$$
-Would the gain be a voltage gain or a power gain?
-assuming the gain is power and not voltage, so the above is wrong and the sin and cos terms go to 1(or -1) to maximise the available instantaneous power I get:
$$P=\frac{C*G*mE_c^2*(m+1)^2}{R}$$ then for part c I would get
? need to look at this a bit more.

 

[rude man]

Thanks a lot for your time rm, really appreciate it/

's OK, I'm retired!

So from your post I've concluded that the answer to b is BW=2f_IF? |(1450-540)/455=2

Looking good!

I also think that I have an understanding of this now, if you wouldn't mind checking for me

f_c - carrier frequency, 540kHz from your example, this is what would be displayed on a radio?

Yes.

f_IF - intermediate frequency -455kHz - what the IF amp will amplify, f_C mixes with the LO to produce this, any other signal that the LO will mix with to produce this is an image frequency? In your example, the LO would be set to 995kHz? So if a station was transmitting at 1450kHz a mixing product of the LO would give the IF(455kHz) and would be amplified and allowed through?

To show this algebraically:
fc+2fIF-fc=2IF

Absolutely right.

*One thing I don't understand is why we assume the receiver can't receive under fc? Taking your example, a station at 85kHZ would also mix with the LO to produce the IF of 455kHz?

How's that? 995 - 85 = 910, 995 + 85 = 1080. Neither is anywhere near 455 KHz.

Also since this assignment is due soon ill pack in a few questions -
part a.
I had power available at the reciever is:
$$P=((Ec∗sin(2π∗fc∗t)+mEc2cos[2π(fc−fi)]−mEc2cos[2π(fc+fi)])∗GT)2/R∗C$$
-Would the gain be a voltage gain or a power gain?
-assuming the gain is power and not voltage, so the above is wrong and the sin and cos terms go to 1(or -1) to maximise the available instantaneous power I get:
$$P=\frac{C*G*mE_c^2*(m+1)^2}{R}$$ then for part c I would get
? need to look at this a bit more.

Need to get back at you tomorrow for the rest. I never did check your power calculations.

 

[Fisher92]

Hm, that 85kHz is nonsense but there is still something I don't understand, the carrier frequency needs to be adjustable so that the receiver can tune into different stations but the intermediate frequency is always the same 455kHz or fIF.
-So the bandwidth is essentially f_c to 910kHz? The bandwidth is probably just 910kHz I guess but what if I wanted to tune into a station running in the MHz band - AM probably never goes this high but I'm trying to understand the concepts.

Thanks

 

[rude man]

If you wanted to tune a station at say 3 MHz your LO could be set to 3.455 MHz. The image would be at (x - 3.455 MHz) = 0.455 MHz or x = 3.910 MHz. So the permissible bandwidth is again 3.910 MHz - 3.000 MHz = 910 KHz = 2IF.

Or, your IF could be below the carrier:
fc = 3 MHz, fIF = 2.545 MHz, image is at x where 2.545 - x = 0.455 or x = 2.090 MHz. Again, 3 - 2.090 = 0.910 MHz = 2 fIF.

The carrier frequency is immaterial. The permissible receiver bandwidth is always 2IF.

PS the AM band extends from around 200 KHz (i.e. well below the std broadcast band, aka long wave, all the way to 30 MHz for short wave at 10m. I believe above 30 MHz it's usually FM but I'm sure Mr. Berkeman might want to comment on this.

 

[Fisher92]

Thanks r m that makes sense - regarding part c,

'If the RF bandwidth from part “b” is used and noise density (noise power per Hz) is
N_0, calculate the SNR at the input of the receiver.'

No has units of W/Hz and I assume total signal attenuation inc noise outside of the bandwidth determined in part b then I have Noise Power =No/2_IF?...Doesn't seem right but I think I want to show is that the noise power is limited to the bandwidth from part b? How can I show this mathematically?
.. Also can you have a look at what i'v e done for the power when you get a chance please. Seems pretty important for the rest of the question as well

SNR=P/(Noise power within in the bandwidth)?

Thanks

 

[rude man]

Thanks r m that makes sense - regarding part c,

'If the RF bandwidth from part “b” is used and noise density (noise power per Hz) is
N_0, calculate the SNR at the input of the receiver.'

No has units of W/Hz and I assume total signal attenuation inc noise outside of the bandwidth determined in part b then I have Noise Power =No/2_IF?...Doesn't seem right but I think I want to show is that the noise power is limited to the bandwidth from part b? How can I show this mathematically?

Your noise power spectral density is No which is in Watts/Hz (1 ohm assumed). So why would you divide by 2f_IF? That would not get you W, would it?

.. Also can you have a look at what i'v e done for the power when you get a chance please. Seems pretty important for the rest of the question as well

SNR=P/(Noise power within in the bandwidth)?

Thanks

Thing is, I suggested avoiding calculating power with the sideband expansion but that's how you did it (your 'epiphany'! ). I calculated power based on modulator output = Ec[1 + m sin wi t]cos(wc t). See my posts 16 and 24.

Doing it my way I get the sideband power I stated in post 24 (carrier power does not count). I'm afraid I'll have to leave it to you to show equivalence between my power number and yours.

Then, SNR = sideband power/noise power.

 

[Fisher92]

I wanted the sidebands so that I could do the efficiency of DSB-SC from memory but that's done now so I can just use the modulator output instead.
When I pick up from post 16 I get a big mess with wc and wi on the denominator and a string of sin and cos on the numerator? Can't seem to get what you have for my power?

 

[rude man]

I wanted the sidebands so that I could do the efficiency of DSB-SC from memory but that's done now so I can just use the modulator output instead.
When I pick up from post 16 I get a big mess with wc and wi on the denominator and a string of sin and cos on the numerator? Can't seem to get what you have for my power?

OK let's go back to post 16. Did you get something that looks like my expression for Pavg? If so it should not be difficult to compute the two averagings I indicated. If not you need to get to that point.