Hello, in the picture bellow we have the fundamental plot and the third order intercept plot.
I have few questions regarding this plot
1.the fundamental starts from positive Pout ,why is that?
2.why does the pout point at pin=0 of the funcdamental called gain?
3.why does thethird order intermodulation at pin=0 cut the Pout at negative point?
Thanks.

 

This is on a dB (or dBm) vertical axis, so all power is measured relative to 1 W (dB(W)) or 1 mW (dBm). I'll explain this from the idea that this is most likely dBm, as that's one of the most common power measurements. So, first off, 0 dBm = 1 mW of power. 30 dBm = 1 W of power. So 0 dBm isn't "no power."

Your questions 1 and 2 are related: on the x-axis we have input power (calculated relative to amplitude in volts, actually, rather than straight from mW), and y-axis is the output power with the same units. If you have an amplifier or some other device, then you're going to have gain - "amplification." Gain is typically a >0 dB number (so, in linear units, that's >1 gain - less-than-unity gain is "attenuation" and is negative in the dB scale). So if you have 0 dBm input power (1 mW) and 10 dB of gain, as an example, your output power will be 10 dBm of power (10 mW). Similarly, if your circuit has 3 dB of gain, then your output power for the same input power will be 3 dBm out (or approximately 2 mW). This plot is showing some arbitrary response of an amplifier for a given input power (assuming no gain compression, it looks like), so gain "G" being the height of the output power (in dBm) when Pin = 0 dBm is just to more easily illustrate the gain. That's why the fundamental "starts" at a positive level: it's some amount of power out than is put in. And the amount of power out higher than the amount of power in is the gain. Whoever drew the plot could have arbitrarily put the "starting" input power at a negative value (say, -25 dBm or -30 dBm - the power levels generally considered to be "small signal" [-30 dBm is 1 uW, and -25 dBm is 3.2 uW), and the output power would (most likely) be negative as well. Using the 10 dB gain as an example, if your input power is a negative value like -30 dBm, then your output power will be -20 dBm (or about 10 uW).

The third question relies on the linearity of the device. Typically, amplifiers are not perfectly linear. What that means is, when you provide a pure sinusoid in, you will not get a pure sinusoid out. Instead, you'll get gain for the fundamental tone (the frequency of the signal you put into it), but you'll also get very tiny contributions of the second, third, fourth, etc harmonics. Ideally, these contributions are very tiny - significantly lower than the fundamental. So, for instance, with the same example I gave earlier with the 10 dB amplifier, the second and third harmonics may have contributions of -15 and -20 dB (gain), so the power spectrum would look like this:
Pin = 0 dBm [1 mW]
Pout (fundamental) = 10 dBm [10 mW]
Pout (second harmonic) = -5 dBm [0.316 mW]
Pout (third harmonic) = -10 dBm [0.1 mW]
That's why it's negative: because it has less power in the third harmonic output than was put into the device at the fundamental. Typically, the lower your harmonic contributions are, the more linear your amplifier will be and more pure the output tone will be. TOI (or IM3) is a measure, then, of how linear the circuit is.

With the third harmonic component, however, the slope at which it increases and the power it exhibits at Pin = 0 dBm depends on the device, architecture, and some other factors. Sometimes, you even want to increase the nonlinearity of a device in order to take advantage of certain features (for instance, nonlinear amplifiers can have increased efficiency).

In all of these, keep in mind how dB and logarithms work.

Hopefully this answers your question!

 

Hello ZCochran98, I have a problem understanding the OIP3 formula logic shown in the slide of the link. i know that delta is the power difference between the IM3 and the fundamental. (output)delta decreases by -2dB as input increases by 1dB. i cant see from this grapg why OIP3=P_fundamental+0.5*P_delta ? Thanks.

 

If \(P_1\) is defined to be the fundamental power level point at which we have defined \(\Delta P\), and we know that as input power increases by 1 dBm, so too does output power increase by 1 dBm, we can find the IP3 point by calculating where \(\Delta P = 0\). And because we can see that the difference between the fundamental and third harmonic changes as input (and, correspondingly, output) power changes, we can see that \(\Delta P\) is technically a function of \(P_{out}\) (or \(P_{in}\)). In other words, we have the function (based on the statement that \(\Delta P\) changes by -2 dB per increase in \(P_{in}\) and thus also \(P_{out}\) at the fundamental):
\[\Delta P\left(P_{out}\right) = \Delta P\left(P_1\right) - 2\left(P_{out}-P_1\right)\]
This is a basic line equation. The IP3 point, therefore, is the point where the difference between the first and third harmonic is 0, right? So let's solve for that point (defining \(OIP_3\) to be the output power level at which the intercept occurs):
\[0 = \Delta P\left(P_{out}\right) = \Delta P\left(P_1\right) - 2\left(P_{out}-P_1\right)\]
\[\Delta P\left(P_1\right) = 2\left(P_{out}-P_1\right)\]
\[\frac{1}{2}\Delta P\left(P_1\right) = \left(P_{out}-P_1\right)\]
\[\frac{1}{2}\Delta P\left(P_1\right) + P_1 = P_{out}\]
Given that we define \(OIP_3 \equiv P_{out}\) for this scenario, and simplifying \(\Delta P\left(P_1\right)\) with just \(\Delta P\) for annotation simplicity (and to match the notation used before) we replace it with:
\[\frac{1}{2}\Delta P + P_1 = OIP_3\]
Which gets us to that equation in the figure.
Now, we could generalize this if we wanted. If we knew that \(\Delta P\) decreased by some amount \(\delta\) per 1 dBm increase in \(P_{out}\) (so, previously, this was \(\delta = 2\)), we could say \(OIP_3 = P_1 + \frac{1}{\delta}\Delta P\) instead.