hi ALL,
ive been stuck for a very long time on these 2 questions for a long time. there just revision questions.
As i have very little knowledge about amps. I would appreciate some help as i feel hopeless. I do NOT expect you guys to answer for me of course since this is homework help section but would like if you could start me of with the right formula and possibly a little information on amps. but mainly help me understand how i can get the correct answer so i can do it myself.

Thanks a million!

1.Three amplifiers are cascaded together. The first stage has a voltage gain of 16 dB and the third stage has a voltage gain of 13 dB. An input signal of 15.5 mV (ptp) results in an output from the third stage of the cascaded amplifier of
2.2 V (ptp), the voltage gain in dB of the second stage is;

A. 12
B. 14
C. 16
D. 18
E. None of the above


2.A circuit consists of two amplifiers with gain 16 dB and 14 dB and a filter with attenuations of -10 dB. If the output voltage is 80 mV, the input voltage is;

A. 4 mV
B. 6 mV
C. 8 mV
D. 10 mV
E. None of the above

 

Q1) What does dB mean?

Q2) If you are told that an amplifier has a gain of 10dB, what does that tell you?

Q3) Given an input voltage and an output voltage (and assuming a linear amplifier), what is the gain of the amplifier, in dB, in terms of Vout and Vin?

Q4) How do gains combine when the output of one amplifier is applied as the input signal to another?

Answer these as best you can, and we can get you started on the right page and proceed from there.

 

Q1) What does dB mean?

Q2) If you are told that an amplifier has a gain of 10dB, what does that tell you?

Q3) Given an input voltage and an output voltage (and assuming a linear amplifier), what is the gain of the amplifier, in dB, in terms of Vout and Vin?

Q4) How do gains combine when the output of one amplifier is applied as the input signal to another?

Answer these as best you can, and we can get you started on the right page and proceed from there.

A1) dB means decibels and its used to measure sound levels. dB is a logarithmic unit used to describe a ratio. The ratio may be power, sound pressure, voltage or intensity or several other things.

A2) the gain of an amplifier is the amount your multiplying the voltage by.

A3) so if the input voltage was 1v and the output was 10v then we will say that this voltage had a gain of 10db.

A4) Dont know this one im afraid.

 

A2) the gain of an amplifier is the amount your multiplying the voltage by.

A3) so if the input voltage was 1v and the output was 10v then we will say that this voltage had a gain of 10db.

A4) Dont know this one im afraid.

You didn't complete the answer to A2.
A 10 time voltage gain is not a gain of 10dB. Look up the definition for dB and see how it relates to a voltage gain.

 

You are fairly close on some, but seem to be avoiding others.

Let's examine each:

1) The bel is defined as the base-10 logarithm of the ratio of two powers. Thus:

\(G\ = \ \log_{10} \left( \frac {P_{out}}{P_{in}} \right) \ \text{bels (or B)}\)

The decibel is one tenth of a bel, thus,

\(G\ = \ 10 \log_{10} \left( \frac {P_{out}}{P_{in}} \right) \ \text{decibels (or dB)}\)

Since power is proportional to the square of the voltage, for a voltage gain we have:

\(G\ = \ 10 \log_{10} \left( \frac {V^2_{out}}{V^2_{in}} \right) \ \text{dB}\)

\(G\ = \ 20 \log_{10} \left( \frac {V_{out}}{V_{in}} \right) \ \text{dB}\)

2) If the gain of an amplifier is 10dB that means that

\(G\ = \ 10dB \ = \ 20 \log_{10} \left( \frac {V_{out}}{V_{in}} \right) \ \text{dB}\)

\(\frac{10}{20} \ = \ 0.5 \ = \ \log_{10} \left( \frac {V_{out}}{V_{in}} \right) \)

\(10^{0.5} \ = \ \frac {V_{out}}{V_{in}} \)

\(\frac {V_{out}}{V_{in}} \ = \ 3.16\)

3) No. If Vin = 1V and Vout = 10V, the voltage gain is 10, which is also 20dB. Remember, dB is always (supposed to be) a ratio of power.

4) Consider three amplifiers, one with a gain of 2, one with a gain of 3, and one with a gain of 4. A 1V signal is applied to the input of the first amplifier.

Q5)What is the output of the first amplifier?

Now apply that signal to the input of the second amplifier.

Q6) What is the output of the second amplifier?

Finally, apply that to the input of the third amplifier.

Q7) What is the output of the third amplifier?

Q8) What is the overall gain from the input of the first amplifier to the output of the final amplifier?

Q9) What is the relationship between the three individual gains given and the answer to Q8?

Q10) What are the gains of each amplifier expressed in dB?

Q11) What is the overall gain expressed in dB?

Q12) What is the relationship between the answers to Q10 and the answer to Q11?

Each of these questions, Q5 through Q12, is looking for a specific numerical answer. Show your work, so that if you make a mistake I can try to see where and why you made the mistake.

 

You are fairly close on some, but seem to be avoiding others.

Let's examine each:

1) The bel is defined as the base-10 logarithm of the ratio of two powers. Thus:

\(G\ = \ \log_{10} \left( \frac {P_{out}}{P_{in}} \right) \ \text{bels (or B)}\)

The decibel is one tenth of a bel, thus,

\(G\ = \ 10 \log_{10} \left( \frac {P_{out}}{P_{in}} \right) \ \text{decibels (or dB)}\)

Since power is proportional to the square of the voltage, for a voltage gain we have:

\(G\ = \ 10 \log_{10} \left( \frac {V^2_{out}}{V^2_{in}} \right) \ \text{dB}\)

\(G\ = \ 20 \log_{10} \left( \frac {V_{out}}{V_{in}} \right) \ \text{dB}\)

2) If the gain of an amplifier is 10dB that means that

\(G\ = \ 10dB \ = \ 20 \log_{10} \left( \frac {V_{out}}{V_{in}} \right) \ \text{dB}\)

\(\frac{10}{20} \ = \ 0.5 \ = \ \log_{10} \left( \frac {V_{out}}{V_{in}} \right) \)

\(10^{0.5} \ = \ \frac {V_{out}}{V_{in}} \)

\(\frac {V_{out}}{V_{in}} \ = \ 3.16\)

3) No. If Vin = 1V and Vout = 10V, the voltage gain is 10, which is also 20dB. Remember, dB is always (supposed to be) a ratio of power.

4) Consider three amplifiers, one with a gain of 2, one with a gain of 3, and one with a gain of 4. A 1V signal is applied to the input of the first amplifier.

Q5)What is the output of the first amplifier?

Now apply that signal to the input of the second amplifier.

Q6) What is the output of the second amplifier?

Finally, apply that to the input of the third amplifier.

Q7) What is the output of the third amplifier?

Q8) What is the overall gain from the input of the first amplifier to the output of the final amplifier?

Q9) What is the relationship between the three individual gains given and the answer to Q8?

Q10) What are the gains of each amplifier expressed in dB?

Q11) What is the overall gain expressed in dB?

Q12) What is the relationship between the answers to Q10 and the answer to Q11?

Each of these questions, Q5 through Q12, is looking for a specific numerical answer. Show your work, so that if you make a mistake I can try to see where and why you made the mistake.

A5) The output of the first amplifier is 20v 1v x a gain of 2 = 20v

A6) The second one is 60v 20 20v x a gain of 3 = 60v

A7) The third is 240v 60v x a gain of 4 = 240v


A8) Total Gain of 288,000? i multiplied each individual gain together.

A9)Not sure about this one.

A10)
1st) \(20 \log_{10} \left( \frac {20{V{out}}}{1V{in}} \right)\) = 26dB ( i feel this is wrong)
2nd) \(20 \log_{10} \left( \frac {60V{out}}{20V{in}} \right)\) = 9.5 dB
3rd) \(20 \log_{10} \left( \frac {240V{out}}{60V{in}} \right)\) = 12dB

A11) 3rd) \(20 \log_{10} \left( \frac {240V{out}}{1V{in}} \right)\) = 47.6dB? not sure about this one.

A12) Not sure about this one either

 

A5) The output of the first amplifier is 20v 1v x a gain of 2 = 2v

A6) The second one is 60v 20 20v x a gain of 3 = 60v

A7) The third is 240v 60v x a gain of 4 = 240v


A8) Total Gain of 288,000? i multiplied each individual gain together.

A9)Not sure about this one.

A10)
1st) \(20 \log_{10} \left( \frac {20{V{out}}}{1V{in}} \right)\) = 26dB ( i feel this is wrong)
2nd) \(20 \log_{10} \left( \frac {60V{out}}{20V{in}} \right)\) = 9.5 dB
3rd) \(20 \log_{10} \left( \frac {240V{out}}{60V{in}} \right)\) = 12dB

A11) 3rd) \(20 \log_{10} \left( \frac {240V{out}}{1V{in}} \right)\) = 47.6dB? not sure about this one.

A12) Not sure about this one either

PLEASE IGNORE THIS. MADE A STUPID MISTAKE.

CORRECTION:

The output of the first amplifier is 2v 1v x a gain of 2 = 2v

A6) The second one is 6v 20 2v x a gain of 3 = 6v

A7) The third is 24v 6v x a gain of 4 = 24v


A8) Total Gain of 288 i multiplied each individual gain together.

A9)Not sure about this one.

A10)
1st) \(20 \log_{10} \left( \frac {2{V{out}}}{1V{in}} \right)\) = 6dB ( i feel this is wrong)
2nd) \(20 \log_{10} \left( \frac {6V{out}}{2V{in}} \right)\) = 9.5 dB
3rd) \(20 \log_{10} \left( \frac {24V{out}}{6V{in}} \right)\) = 12dB

A11) 3rd) \(20 \log_{10} \left( \frac {24V{out}}{1V{in}} \right)\) = 47.6dB? not sure about this one.

 

PLEASE IGNORE THIS. MADE A STUPID MISTAKE.

CORRECTION:

The output of the first amplifier is 2v 1v x a gain of 2 = 2v correct
A6) The second one is 6v 20 2v x a gain of 3 = 6v correct

A7) The third is 24v 6v x a gain of 4 = 24v correct

A8) Total Gain of 288 i multiplied each individual gain together.
Let's see; 1V in comes out 24V, a gain of 288? I don't think so; try again.

A9)Not sure about this one.
It's because you didn't get the previous answer correct.

A10)
1st) \(20 \log_{10} \left( \frac {2{V{out}}}{1V{in}} \right)\) = 6dB correct
2nd) \(20 \log_{10} \left( \frac {6V{out}}{2V{in}} \right)\) = 9.5 dB correct
3rd) \(20 \log_{10} \left( \frac {24V{out}}{6V{in}} \right)\) = 12dB correct

A11) 3rd) \(20 \log_{10} \left( \frac {24V{out}}{1V{in}} \right)\) = 47.6dB? Try again.

You got a good start.

 

The output of the first amplifier is 2v 1v x a gain of 2 = 2v

A6) The second one is 6v 20 2v x a gain of 3 = 6v

A7) The third is 24v 6v x a gain of 4 = 24v

Correct

A8) Total Gain of 288 i multiplied each individual gain together.

Here is where showing your work would have paid off. You say that you multiplied the individual gains together, which is the correct way, but then you give an answer that is NOT the result of multiplying the individual gains together. Was it just a stupid math blunder? Was it misinterpreting what the individual gains were? Was it something else? If the work was shown, even just like you did it in the first three answers above, we could tell. As it is, we've got to guess. My guess is that you simply grabbed the output voltages and multiplied them together, so you went 2*6*24 and got 288.

If so, this is where I get to stand up on my soap box (again) and preach about traking your units. Had you tracked your units properly, you would have gotten

\(
G_{tot}\ = \ (2V)(6V)(24V)\ = \ 288V^3
\)

You would immediately spotted that the units made no sense, telling you that the answer is wrong. No need to go any further until you find and fix that problem.

Then you probably would have quickly realized that you meant to do

\(
G_{tot}\ = \ (2)(3)(4)\ = \ 24
\)

The next thing you want to get in the habit of doing is asking whether the answer makes sense. In this case, you have the initial input voltage of 1V and the final output voltage of 24V. So does it make sense that the overall gain could be 288? No! In this case you have easy ways to compute the overall gain -- so do it both ways and compare the results. If they don't agree adequately, then something is wrong. No need to go any further until you find and fix that problem.

A9)Not sure about this one.

You actually answered this correctly as the comment made in A8

A10)
1st) \(20 \log_{10} \left( \frac {2{V{out}}}{1V{in}} \right)\) = 6dB ( i feel this is wrong)
2nd) \(20 \log_{10} \left( \frac {6V{out}}{2V{in}} \right)\) = 9.5 dB
3rd) \(20 \log_{10} \left( \frac {24V{out}}{6V{in}} \right)\) = 12dB

Correct, though you could have used the stated gain instead of the specific Vin and Vout values from this example.

One very useful thing to note is that a power gain of 2 equates to 3.01dB (so 3dB is fine for most purposes). Doubling the voltage quadruples the power (or doubles it twice), giving 6dB.

A11) 3rd) \(20 \log_{10} \left( \frac {24V{out}}{1V{in}} \right)\) = 47.6dB? not sure about this one.

The lefthand side is correct, but you made a mistake somewhere when carrying out the operations to get the right hand side.

Again, asking if the answer makes sense comes in handy. In this case, you can easily estimate bounds on the answer and check the result you get against those bounds.

You know that log_10(10) = 1 and log_10(100) = 2. Therefore, log_10(24) has to lie somewhere in between 1 and 2. That means that your result has to lie somewhere between 20dB and 40dB. If it doesn't, you know your answer is wrong. No need to go any further until you find and fix that problem.

 

Thank you for clarifying my mistakes.

A8) Total Gain = 2 x 3 x 4 = 24

Total gain is 24. Very silly of my to mistake the output voltage as the gain.

A9) So the relationship between the 3 gains is 288. got this by multiplying the gains together.

A11) \(20 \log_{10} \left( \frac {24V{out}}{1V{in}} \right)\) = 27.6dB

 

Thank you for clarifying my mistakes.

A8) Total Gain = 2 x 3 x 4 = 24

Total gain is 24. Very silly of my to mistake the output voltage as the gain.

Not so silly, just the kind of mistake that we all make and that you will make many more times. But my point is that if you develop the habit of religiously checking your units, you will catch most of these mistakes right away. I've long since gotten to the point where it is so engrained that I have a very hard time following any work that does not properly track the units because it simply yells out "wrong" at every step.

A9) So the relationship between the 3 gains is 288. got this by multiplying the gains together.

Now we're back to this value of 288. What gives?

A11) \(20 \log_{10} \left( \frac {24V{out}}{1V{in}} \right)\) = 27.6dB

Correct.

Now, what is the relationship between the overall gain, in dB, and the gains of the individual amplifiers, in dB?

 

Q8) What is the overall gain from the input of the first amplifier to the output of the final amplifier?

Q9) What is the relationship between the three individual gains given and the answer to Q8?

A8) Total Gain = 2 x 3 x 4 = 24

Total gain is 24. Very silly of my to mistake the output voltage as the gain.

A9) So the relationship between the 3 gains is 288. got this by multiplying the gains together.

A11) \(20 \log_{10} \left( \frac {24V{out}}{1V{in}} \right)\) = 27.6dB

That is not the answer to Q9. If the total gain is 24 (as in Q8), why would it be different for Q9?
As you have previously stated, you multiply the individual gains together to get the total gain.

What wasn't asked was how the 3 gains (in dB) relate to the total gain (in dB).
How does 6dB, 9.5dB, and 12dB relate to 27.5dB?

So, let's look at the original questions:

1.Three amplifiers are cascaded together. The first stage has a voltage gain of 16 dB and the third stage has a voltage gain of 13 dB. An input signal of 15.5 mV (ptp) results in an output from the third stage of the cascaded amplifier of 2.2 V (ptp), the voltage gain in dB of the second stage is;

A. 12
B. 14
C. 16
D. 18
E. None of the above

2.A circuit consists of two amplifiers with gain 16 dB and 14 dB and a filter with attenuations of -10 dB. If the output voltage is 80 mV, the input voltage is;

A. 4 mV
B. 6 mV
C. 8 mV
D. 10 mV
E. None of the above

1. What we are told:
Input: 15.5mV (P-P)
Output: 2.2mV (P-P)
Stage 1 gain: 16dB
Stage 3 gain: 13dB
Need to find:
Total gain (from input signal to output signal)
Stage 2 gain

2, What we are told:
Stage 1 gain: 16dB
Stage 2 gain: 14dB
Filter (gain): -10dB
Output signal: 80mV
Need to find:
Total gain
Input signal

 

What wasn't asked was how the 3 gains (in dB) relate to the total gain (in dB).
How does 6dB, 9.5dB, and 12dB relate to 27.5dB?

Ah, but it WAS asked. See Q12. It just wasn't answered.

 

A9) Sorry my bad i copied and pasted my previous answer. i should have just said that i multiplied each individual gain to get the total gain of the 3 cascaded amplifiers.

A12) It seems that the total gain (dB) is the addition of the 3 individual gain (dB)

so for working out of my original questions.

1.Three amplifiers are cascaded together. The first stage has a voltage gain of 16 dB and the third stage has a voltage gain of 13 dB. An input signal of 15.5 mV (ptp) results in an output from the third stage of the cascaded amplifier of 2.2 V (ptp), the voltage gain in dB of the second stage is;

A. 12
B. 14
C. 16
D. 18
E. None of the above

input Voltage: 15.5mV (0.0155V)
Stage 1: 16dB
Stage 2: __dB
Stage 3: 13dB
Output voltage: 2.2V

What i will do is convert the dB's to voltage gain so then i can multiply the input volt by the gain to get the input volt of the next amp.

Conversion to voltage gain
formula: 10^(dB/20)
amp 1
10^(16/20) = 6.3(Av)
amp 3
10^(13/20) = 4.5(Av)

Now to find input voltage of each input via multiplying the input by the Av.

Amp 1
Vout= Vin*AV
0.0155V x a gain of 6.3 = 0.0976V

Amp 3
Vin=Vout/Av
2.2v/a gain of 4.5 = 0.4889V

Now i can find out the voltage gain for the second stage now that we know its
input from stage 1's output of 0.0976V And its output from stage 3's input of 0.04889V

voltage gain= Vout / Vin

0.4889v/ 0.0976v = 5(Av)

so now i know that stage 2 has a voltage gain of 5, now i will see what that is in dB.

20log(0.4889v/0.0967v) = 14dB

answer is B!

Now for question 2. i understand what Atennuation means, its if Vout is smaller than Vin. A=Vout/Vin.

how do i implement this in finding the answer for question 2

 

I have decided to attempt question 2.

I understand that Attenuation and gain is the same thing apart from the fact that attenuation is the negative and gain is the positive. whether a signal has lost power or voltage or was amplified.

2.A circuit consists of two amplifiers with gain 16 dB and 14 dB and a filter with attenuations of -10 dB. If the output voltage is 80 mV, the input voltage is;

A. 4 mV
B. 6 mV
C. 8 mV
D. 10 mV
E. None of the above

2, What we are told:
Stage 1 gain: 16dB
Stage 2 gain: 14dB
Filter (gain): -10dB
Output signal: 80mV

i need to find:
Total gain
Input signal


Input signal(Vin)= Vout/Av

Vout=80mV
Total gain(AV)= 16dB + 14dB = 30dB
however - 10dB is being filtered out which means 30dB - 10dB= 20dB

now converting dB to Voltage gain = 10^(20/20)= a total gain of 10

now i can do formula to find Vin

Vin = 20dB/80mV = 8mV

The answer seems to be C!

Very proud of myself for doing this. Thought id never understand these things. Thank you guys for expert help!

Am i correct in my calculations? and is there any shortcut formulas. i feel i took the long route for some.

 

now i can do formula to find Vin

Vin = 20dB/80mV = 8mV

Where did you get this? How can you divide mV into dB to get mV? That doesn't make any sense. Maybe you were thinking of something different, got interrupted, and wrote this down. you will want to rearrange:
Vout = Vin * gain

 

Where did you get this? How can you divide mV into dB to get mV? That doesn't make any sense. Maybe you were thinking of something different, got interrupted, and wrote this down. you will want to rearrange:
Vout = Vin * gain

oops simple mistake i was distracted. was meant to divide it by the total gain i got from the working about above the mistake

10/80mV= 8v

 

oops simple mistake i was distracted. was meant to divide it by the total gain i got from the working about above the mistake

10/80mV= 8v

And another mistake.

I'll let you figure this one out.

 

A12) It seems that the total gain (dB) is the addition of the 3 individual gain (dB)

Now let's see why this is the case.

Gain_total = Gain_A*Gain_B*GainC

Convert this to dB:

Gain_total_dB = 20dB*log(Gain_A*Gain_B*GainC)

We know that log(A*B) = log(A) + log(B), so

Gain_total_dB = 20dB*[log(Gain_A)+log(Gain_B)+log(GainC)]

Gain_total_dB = 20dB*log(Gain_A) + 20dB*log(Gain_B) + 20dB*log(GainC)

Gain_total_dB = Gain_A_dB + Gain_B_dB + GainC_dB

so for working out of my original questions.

What i will do is convert the dB's to voltage gain so then i can multiply the input volt by the gain to get the input volt of the next amp.

This approach is valid and you got the correct answer. So congratulations. But I think the point of the problem was more to work with the gains in dB directly. So let's see how that would work.

Convert the overall gain to dB:

Gain = 20*log10(Vout/Vin)dB
Gain = 20*log10(2.2V/15.5mV)dB
Gain = 43.04dB ~= 43dB

Now use the fact that the total gain, in dB, is the sum of the individual gains:

Gain_Total = 16dB + Gain_B + 13dB = 43dB
Gain_B = 43dB - 16dB - 13dB
Gain_B = 14dB

See how much simpler that was? Other than taking the one logarithm, the math could all be done in your head.

Conversion to voltage gain
formula: 10^(dB/20)
amp 1
10^(16/20) = 6.3(Av)
amp 3
10^(13/20) = 4.5(Av)

A word on notation: The way you have written (Av) above makes it look like it is either a factor multiplying the gain or it is a note that the units are "Av". What you shoudl have done is something like the following:

formula: Av = 10^(dB/20)
amp 1
Av_1 = 10^(16dB/20) = 6.3
amp 3
Av_3 = 10^(13dB/20) = 4.5

 

I have decided to attempt question 2.

Good! Dig in and try even if you aren't sure you've got it down.

I understand that Attenuation and gain is the same thing apart from the fact that attenuation is the negative and gain is the positive. whether a signal has lost power or voltage or was amplified.

The idea is right, but the wording could be better. Attenuation is simply when the magnitude of the gain is <1. So a gain of 0.5 is an attentuation, while a gain of -50 is not.

When you convert to dB, THEN an attenutation is negative. Oh, and a subtle point that has been brought out is that gain in dB is ALWAYS based on the magntidue of the gain, which comes directly from the fact that the voltage is squared to convert it to power.

Vin = 20dB/80mV = 8mV

Something should have been screaming at you right here!

Av = Vout/Vin

Vin = Vout/Av

Vin = 80mV/10 = 8mV

It will take time and effort to get there, especially if your instructors don't care about enforcing units (which most, sadly, do not), but it's pretty clear that you did the calculation at one point correctly, probably on a sheet of paper somewhere, and then just wrote down the wrong expression here. That happens. But it also happens that you will make such a mistake before doing the calculation and proceed to get 0.25mV unless you are working the units, and not just tacking them on at the end based on what you expect them to be.

and is there any shortcut formulas. i feel i took the long route for some.

You definitely took the long route on the first one, but I addressed that in the prior post.

Congratulations!

It's a refreshing change to have someone that is willing to put in the effort to take the hints we give and do the work to discover the answers themselves.