This page defines the term "1dB Compression" and illustrates the definition with a working example.

Consider the primary transfer characteristic of an amplifier...

Assume :

Linear Gain is 20 dB
1 dB compression point, is where Pg = 19 dB
119 W output at 1.5 W input = 19dB = 10Log(119/1.5)

Then :

1dB compression point is 119 Watts

This page defines the term "Decibel (dB)" and illustrates the measure with an example.

A dB is a ratio that we use when we are talking about power

P_i=Power Input
P_o=Power Output
X dB = 10Log₁₀(P_o /P_i )

Example :
A transistor has an output power of 50 watts with 10 watts input power. Its gain in dB is :

10Log₁₀(50 /10) = 10Log₁₀5 = 6.9897 ~ = 7.0 dB

and an attenuator that has an output of 25 watts with an input of 100 watts is :

10Log₁₀(25 /100) = 10Log₁₀0.25 = -6.021 ~ = -6.0 dB

This page defines the term "dBm" and provides some helpful advice.

dBm is an absolute measure of power.

the m stands for milli watts
to convert dBm to watts we divide by ten and then take the inverse log in base 10
so 0dBm = 1 mW

- 0/10 = 0

-10⁰ = 1

10 dBm = 10 mW
20 dBm = 100 mW
30 dBm = 1,000 mW = 1 W
using dBm is helpful when looking at power through circuits, as we can simply add
aa the gain and subtract the attenuation in dB and end up with a final power number
aa in dBm
you can also use dBW

- where 0dBW = 1 Watt

This page defines the term "Q" and provides some helpful advice.

When one thinks of Q one thinks of that upper class British technocrat explaining how the latest weaponry gadget works, to a disinterested James Bond. Who, himself, is more interested in contemplating how to uncover his next female spy. If you want more of that, I am afraid you will have to go elsewhere. We are not here to discuss what might impede fact or fictional relationships, but rather to discuss the real and imaginary relationships of impedances. Q: initially stood for ‘Quality Factor’. This was due to the fact that it was first used to describe the energy storage properties of a circuit in relation to its energy dissipation properties.

Now it is just Q, a dimensionless number. In fact one must be careful not to refer to it as ‘quality factor’ especially around management types. One could just imagine Dilbert proudly telling ‘Pointy-Hair’ that he has just designed an amplifier circuit that has a very low quality factor and promptly getting fired! The next day, as if on a crusade, ‘Pointy-Hair’ starts putting up motivational posters all around the engineering building: ‘Increase the quality factor of your circuit by 100%’. It would become part of management review and a strategic goal… Stranger things have happened.

But where Q really comes into its own is in terms of bandwidth. If bandwidth is f₂-f₁ then :

Where f₀ is the center frequency.

It is now obvious why we may want to design a circuit with a low Q. As we can see from the first equation (or elementary filter design rules) it will mean greater loss. However, for most circuits we want low Q.

(N.B. If we are talking about an element such as an inductor or capacitor we always want a high Q, as a low Q element would be indicative of parasitic resistance.)

Another useful way of looking at Q is on a Smith Chart.

The blue lines indicate a constant Q. If you want to have a circuit with a Q below a certain value, then you must ensure that the transformation does not go outside the Q circle.

This page defines the term "third order intercept" and provides some helpful advice.

What are third order products?

Consider the diagram below :

It is in the frequency domain

3rd Order products are the intermodulation distortion products between the one of fundamental signals and the harmonic of the other signal.

To plot the Third Order Response follow steps 1 - 4 illustrated below :

This page discusses what is meant by a "good match" and provides some helpful advice.

A good match means that the output of one element is matched to the input of another to achieve maximum Power Transfer.

One has to be very careful with analogies in engineering.

However, this is one place that one might help.

Consider the flow of water through a pipe...

Water pressure is analogous to voltage
Water volume is analogous to current
Pipe diameter is analogous to resistance
If you take the output of one pipe and put it against the input of one with a smaller
aadiameter you will maintain your pressure but lose volume. If you put it against a
aapipe with a larger diameter you will not lose volume but you will lose pressure

For maximum power transfer the input and output impedances must be matched.

Now for some Maths (urrrgh)...

Consider the circuit below :

Now Current into the load is I_L

Current Power in to load is P_L

For maximum power

R_L=R_S

You don’t believe us?

You are going to have to look at some calculus!

So if P_L = I_L² R_L

And for maximum power

And P_L = I_L² R_L = V_S² R_L (R_S+ R_L)^-2

Then

So : (R_S+ R_L)^-2 = 2R_L(R_S+ R_L)^-3

1 = 2R_L(R_S+ R_L)^-1

2R_L = R_S+ R_L

R_S= R_L

Here are some useful formulae :

Return Loss = 10Log10(Preflected/Pforward)
Reflection Coefficient ρ (or G) = VR/VF
Return Loss = 20Log10 ρ
VSWR = (1+ρ)/(1-ρ)
ρ = (VSWR-1)/(VSWR+1)