1

I'm trying to calculate the 1dB compression point in a second order interception. I did find many examples using third order, however, for a second order, I couldn't find much.

In Matlab I could find the interception point using the following code:

% Parameters
a1=5; a2=4;
G1=20*log10(a1);G2=10*log10(a2^2/2);

% Equations
Pi=-10:0.1:10;
Po=G1+Pi;
P2=G2+2*Pi;

% Interception Point (3)
IPi=(G1-G2);
IPo=G1+IPi;

How can I get the 1dB compression point (referenced to output) using this code? And the representation of the gain.

Enric Blanco
  • 5,741
  • 6
  • 22
  • 40
R user
  • 13
  • 4
  • Welcome to EE.SE. This question is in risk of being closed because you show little research effort and for being unclear what are you asking. A quick google search yields this: http://rfic.eecs.berkeley.edu/~niknejad/ee142_fa05lects/pdf/lect9.pdf – Enric Blanco Feb 22 '17 at 10:37
  • Thanks. I also found that powerpoint. Still doesn't answer my question – R user Feb 22 '17 at 10:48
  • May I ask why not? – Enric Blanco Feb 22 '17 at 10:54
  • From the slides I get IIP2=a1/a2. this is the interception point referenced to the input. I guess in my case 5/4. right? How about the 1dB compression point (referenced to the output)? – R user Feb 22 '17 at 11:05
  • 1
    It seems you're looking for a "straight" answer as in some Matlab code. You will not get such an answer here. Do you understand what a 1dB compression point **means** or an IP3 point ? Do you understand what that included Matlab code does ? Why do you need output referred 1dB cp ? If you knew input referred 1dB cp, how would you determine output 1dB cp ? (It's not a difficult calculation). – Bimpelrekkie Feb 22 '17 at 11:15
  • I'm looking for straight answer. If you want to help, indicate books and other useful resources. thanks – R user Feb 22 '17 at 11:25

2 Answers2

1

Here's the straight answer you're looking for:

You can't calculate \$P_{-1dB}\$ from \$IP_2\$.

Why? Because \$IP_2\$ doesn't contain any information about the cubic term of the non-linearity (\$a_3\$), as it only depends on the linear (\$a_1\$) and quadratic terms (\$a_2\$). Recall that:

$$ s_o(t)=a_1s_i(t)+a_2s_i^2(t)+a_3s_i^3(t) $$

For a single-tone input \$s_i(t)=A_0cos(\omega_0t)\$ where \$\omega_0=2{\pi}f_0\$, this is what you get at the output:

$$ \begin{align} s_o(t)&=s_{o,\ DC}(t)+s_{o,\ f_0}(t)+s_{o,\ 2f_0}(t)+s_{o,\ 3f_0}(t)\\ \text{where:}\\ s_{o,\ DC}(t)&=\frac{1}{2}a_2A_0^2\\ s_{o,\ f_0}(t)&=\left(1+\frac{3}{4}\frac{a_3}{a_1}A_0\right)a_1A_0cos(\omega_0t)\\ s_{o,\ 2f_0}(t)&= \frac{1}{2}a_2A_0^2cos(2\omega_0t)\\ s_{o,\ 3f_0}(t)&=\frac{1}{4}a_3A_0^3cos(3\omega_0t) \end{align} $$

The gain compression is a consequence of the cubic term (more generally speaking: of the odd-order terms). That's why you can calculate \$P_{-1dB}\$ from \$IP_3\$: because it contains information about \$a_3\$ (and about \$a_1\$, as well), but not from \$IP_2\$:

$$ \begin{align} IP_{i,2}&=\frac{a_1}{a_2}\\ IP_{i,3}&=\sqrt{\frac{4}{3}\left\lvert\frac{a_1}{a_3}\right\rvert}\\ \end{align} $$

Also note that your Matlab code doesn't even consider the cubic term:

% Parameters
a1=5; a2=4;
Enric Blanco
  • 5,741
  • 6
  • 22
  • 40
  • Thank you for the answer, however, it's not a "straight" answer. It's a well defined explanation which I thank you. I wasn't looking for a straight answer (I REPEAT). This explanation is more than useful. – R user Feb 23 '17 at 12:23
  • You're welcome. If it's been useful for you, I'm more than happy. – Enric Blanco Feb 23 '17 at 12:25
0

There are famous diagrams for IP2 and IP3; this diagram represents IP2.

schematic

simulate this circuit – Schematic created using CircuitLab

analogsystemsrf
  • 33,703
  • 2
  • 18
  • 46