Efficiency of DSBFC Modulator vs. Balanced Modulator: A Comparison

[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.

 

[Fisher92]

Yep, I had Pavg = Avg[(Ec^2 sin^2(wc t)] * [1 + avg{m^2 sin^2(wi t)}] but i also tried the integration with yours Pavg = Avg[(Ec^2 cos^2(wc t)] * [1 + avg{m^2 sin^2(wi t)}] directly, with no success.

 

[rude man]

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?

How could you possibly get a mess in the denominator? P = V^2. Should be no denominator at all except for a constant.

 

[Fisher92]

In the next step where its integrated to get the average is when I get a mess. Unless I've missed something with this step I can't get it to come out neatly?

 

[Fisher92]

Hey RM, sorry for harassing you about this but I have to submit this in 24hrs and I can't for the life of me figure out how you went from the avg formula in post #16 to the power?

Once I have the power, I think the SNR is given by P/(2N_0*f_IF)?. I can put that in dBm pretty easily.
So using the power from your post 24, I get SNR =(Ec^2 * m^2/4 * G_T * C)/(2N_0*f_IF)?

Thanks

 

[rude man]

Hey RM, sorry for harassing you about this but I have to submit this in 24hrs and I can't for the life of me figure out how you went from the avg formula in post #16 to the power?

Once I have the power, I think the SNR is given by P/(2N_0*f_IF)?. I can put that in dBm pretty easily.
So using the power from your post 24, I get SNR =(Ec^2 * m^2/4 * G_T * C)/(2N_0*f_IF)?

Thanks

That is correct. Assuming I myself got the power correct, but I'm pretty sure I did.

I really think you can derive the power number yourself. Use avg[f(t)] = ∫f(t)dt/∫dt over an integer number of cycles.

E.g. avg[sin²(wt)] = ∫₀^Tsin²(ωt)dt/T where T = 2π/ω?

 

[Fisher92]

'E.g. avg[sin2(wt)] = ∫0Tsin2(ωt)dt/T where T = 2π/ω?'

I get pi/w as my result but this can't be right?
http://www.wolframalpha.com/input/?i=integrate+%28sin%28w*t%29%29%5E2+dt+from+0+to+2pi%2Fw

Just never seems to come out - I've been using my nspire, not wolfram but I am getting results with w and pi which shouldn't be in the answer?

 

[Fisher92]

Hey RM, several hours later and still no success with deriving the power expression. Not sure what timezone your in but I'm on the east coast of Australia so it's getting towards bed time here. As I said this assignment is due tomorrow night and the pesky rotation of the Earth means that there is probably only a few hours in the next 12 when we are both awake. -There are still two parts left once I figure out the power derivation:

d. If the total power amplification through the receiver stages is GR find the output power of the receiver.
As you pointed out a few days ago, the power I am trying to find for a. and c. is not actually the power that will be utilised by the superhet.
-It will obviously be limited to the established bandwidth. so -?

e. If the noise power at the output is identical to the noise power at the input of the receiver, find the NF for the receiver
The Noise Figure is essentially the degradation of the SNR throughout the superhet chain. Still not sure where to go from here though?

I may end up not being able to answer d & e, even so I would like to understand it.

I really wan't to figure out the power though as that's let's me answer a and c properly so rehealy hoping you can offer some more pointers when you see this - just can't get it to work out from the Pavg expression for some reason, I'll keep trying though.

Thanks

 

[rude man]

From trig,
sin^2(2πft) = 1/2 - 1/2 cos(4πft)
The average of cos (4πft) over 1 or more integer periods is zero (1 period = 1/2f).
So the average of sin^2(2πft) = 1/2.

You also need to use the general fact that
avg(a + b) = avg(a) + avg(b) and avg(ab) = avg(a) * avg(b). a and b are assumed independent of each other, which is the case since a = sin(wi t) and b = cos(wc t) here.

Expand [1 + m sin(wi t)]^2] out as 1 + 2m sin(wi t) + m^2 sin^2(wi t), take averages as I've shown, then you wind up with the expression for power as Ec^2 * m^2/4. For your SNR ratio, set m = 1.

I'm in time zone GMT-7 (Arizona USA) all year. (No dst).

 

[rude man]

'E.g. avg[sin2(wt)] = ∫0Tsin2(ωt)dt/T where T = 2π/ω?'

I get pi/w as my result but this can't be right?
http://www.wolframalpha.com/input/?i=integrate+%28sin%28w*t%29%29%5E2+dt+from+0+to+2pi%2Fw

Just never seems to come out - I've been using my nspire, not wolfram but I am getting results with w and pi which shouldn't be in the answer?

The integral is correct but you need to divide by the period which is 2pi/w.

 

[rude man]

Since there is no indication that the receiver itself generates meaningful noise, and since
NF = input SNR/output SNR expressed in dB, what would NF be?

I should point out one other thing: power gain = (voltage gain)^2. So you should recheck if G_T and C were voltage or power gain numbers. If voltage gain you need to use G_T^2 and C^2 and also G_R^2 in your computations (except you don't need G_R at all).

The power gain I gave you was the power at the output of the modulator, before G_T and C.

EDIT: looking at an earlier post of yours, I would assume G_T is a voltage gain and C definitely a power gain.

 

[Fisher92]

Thanks RM, very close with the power now. I've got:
\frac{(E_c^2*1/2*(1+m^2/2)*G_T^2)}{R_T}

The only problem I have is how to get rid of that 1 because I end up with:
P=\frac{0.25E_c^2*(m^2+2)*G^2}{R}*C

?

Thanks

 

[Fisher92]

You also need to use the general fact that
avg(a + b) = avg(a) + avg(b) and avg(ab) = avg(a) * avg(b). a and b are assumed independent of each other, which is the case since a = sin(wi t) and b = cos(wc t) here.

Expand [1 + m sin(wi t)]^2] out as 1 + 2m sin(wi t) + m^2 sin^2(wi t), take averages as I've shown, then you wind up with the expression for power as Ec^2 * m^2/4.

I think this is where my problem comes in:
avg(1)=1
avg(2m sin(wi t))=0 as the integral of sin(wi t) from 0 to 2pi/w =0
avg m^2 sin^2(wi t) = m^2/2

so avg = 1+m^2/2

Need to get rid of that 1 somehow.

 

[rude man]

Thanks RM, very close with the power now. I've got:
\frac{(E_c^2*1/2*(1+m^2/2)*G_T^2)}{R_T}

This is correct.
[quote}

The only problem I have is how to get rid of that 1 because I end up with:
P=\frac{0.25E_c^2*(m^2+2)*G^2}{R}*C

?
Thanks[/QUOTE]

Why's that a problem? It's correct!

The carrier power is Ec^2/2R and the info power is Ec^2/4R if m=1. This is the power at the modulator output. Multiply by (G_T^2)*C to get the power available at the receiver input.

This power includes the carrier of course. In your SNR calculations you will want to use only the modulated signal (info) power.

 

[Fisher92]

Ok, thanks so for the SNR I would essentially have the power above with a (4R) as the denominator of the numerator...and 2fIFNo as the denominator?

 

[Fisher92]

I have:
SNR=\frac{\frac{E_c^2*1/4*(m^2+2)*G_T^2}{4R_T} *C))}{2f_{IF} N_0}
?

 

[rude man]

Ok, thanks so for the SNR I would essentially have the power above with a (4R) as the denominator of the numerator...and 2fIFNo as the denominator?

For the SNR your numerator is (Ec^2/4R)(G_T^2)C and the denominator is 2No*f_IF.

 

[rude man]

I have:
SNR=\frac{\frac{E_c^2*1/4*(m^2+2)*G_T^2}{4R_T} *C))}{2f_{IF} N_0}
?

No, you can't count the carrier power as signal power.

 

[Fisher92]

Ok, first mistake is not setting m to 1 - then I need to remove the carrier power. I am going on an eqn I found in the text, Esf=mEc/2 sp Pinfo=(Ec^2/4)/R - Is that how you derived the same?

 

[Fisher92]

SNR=\frac{\frac{E_c^2 G_T^2}{4R_T}*C}{2f_{IF}*N_0}

If the total power amplification through the receiver stages is GR find the output power of the receiver.
As we've established the power at the input is the information or sidebands power
P_{out}=\frac{E_c^2 G_T^2*G_R^2}{4R_T}*C
Assuming the gains are voltage, & ill make a comment on that

e. If the noise power at the output is identical to the noise power at the input of the receiver, find the NF for the receiver

NF = input SNR/output SNR

If the above, power output is correct, this is easy enough - algebra may simplify but my calculator will figure that out

?

Thanks

 

[rude man]

SNR=\frac{\frac{E_c^2 G_T^2}{4R_T}*C}{2f_{IF}*N_0}

If the total power amplification through the receiver stages is GR find the output power of the receiver.
As we've established the power at the input is the information or sidebands power
P_{out}=\frac{E_c^2 G_T^2*G_R^2}{4R_T}*C
Assuming the gains are voltage, & ill make a comment on that

I think that's good.

e. If the noise power at the output is identical to the noise power at the input of the receiver, find the NF for the receiver

NF = input SNR/output SNR

If the above, power output is correct, this is easy enough - algebra may simplify but my calculator will figure that out

?

Thanks

Shouldn't be too hard! And I speak euphemistically! Let me know what you came up with?

 

[Fisher92]

I'm in time zone GMT-7 (Arizona USA) all year. (No dst).

-Gotta love the internet!

For part d, the noise figure, I ended up getting N_0=\frac{1}{G_R^2}

Probably not right (doesn't make much sense I don't think) but its only one part of a question.

Thanks for all your help with these two questions RM!

 

[rude man]

-Gotta love the internet!

For part d, the noise figure, I ended up getting N_0=\frac{1}{G_R^2}

Probably not right (doesn't make much sense I don't think) but its only one part of a question.

Thanks for all your help with these two questions RM!

No, the NF = 0 since the receiver itself does not introduce any noise of its own.

Let me know what if any feedback you get from your instructor for any part or parts of this problem? Especially about the power at the receiver input.

 

[Fisher92]

Thanks, will do - the uni has a two week turn around for assignment marking.

 

[rude man]

Thanks, will do - the uni has a two week turn around for assignment marking.

As the less literate here in Arizona would say - 10-4!

PS where in NZ are you? My in-laws visited NZ some years ago and raved about it.

 

[Fisher92]

Central QLD, Australia (3/4 the way up (North is up) on the east coast - the coast is still central for some reason -because everyone in Australia lives on the coast I guess)