# Relation between crossover frequency of loop gain and closed loop's corner frequency when feedback is not unity

In this Control Loop Cookbook from TI, in the red box the author said that the crossover frequency (indicated in figure 2) of the open-loop is same as 3dB down frequency or corner frequency of the closed loop.

I don't think this statement is correct with the specific case in figure 2. The statement is only correct if feedback factor Kfb is unity, but from the closed-loop gain, we can infer that Kfb is not 1 (Kfb = 0.5).

What do you think about this? Is the article wrong? In the general case, the crossover frequency of open-loop fc (I think calling it "loop gain" is more accurate) makes the closed loop corner frequency at Kfb*fc.

• It doesn't make much sense to me either. Commented Mar 7, 2018 at 13:08
• So do you agree with my conclusion? With that system above, Kfb =0.5 and the crossover frequency of loop gain fc, the closed-loop gain 3dB corner frequency should be Kfb*fc or 0.5fc not fc as the article. Commented Mar 7, 2018 at 13:16
• I don’t think the figure 2 is that great either because they show the closed loop gain remaining higher at frequencies higher than the open loop gain can sustain. Commented Mar 7, 2018 at 13:34

It is not a mistake, the author is correct.

..at $$\\ f_c\$$ (where $$\\ T=1\$$, with associated $$\ 90^{\circ}\$$. phase lag), the closed loop gain $$\\ G(s)\$$ is 3dB down..

Let $$\\ T=1\angle90^{\circ}=\cos(90^{\circ})+j\sin(90^{\circ})=j\$$

The closed-loop transfer function is:

$$\\ G(s) = \frac{1}{K_{FB}} \cdot \frac{T}{1+T}\$$

Substituting $$\\ T=j\$$ and solving for the gain:

$$\\ |G(s)|= \frac{1}{K_{FB}} \cdot |\frac{j}{1+j}|=\frac{1}{K_{FB}} \cdot |\frac{1}{2}+\frac{1}{2}j|=\frac{1}{K_{FB}} \cdot \sqrt{(\frac{1}{2})^2+(\frac{1}{2})^2}=\frac{1}{K_{FB}} \cdot \frac{1}{\sqrt{2}}\$$

$$\\ 20 \cdot \log_{10}(\frac{1}{\sqrt{2}})=-3 \$$ [dB]

As the author stated; when the open loop gain $$\\ T(s)\$$ equals one, with associated $$\ 90^{\circ}\$$ phase lag, the closed loop gain $$\\ G(s)\$$ is down 3dB, or $$\\ \frac{1}{\sqrt{2}}\$$

• I can't follow your maths but with a closed loop gain of 2 the closed loop gain is down 3dB when the open loop gain is also equal to 2 (not 1).
– user173271
Commented Jan 21, 2020 at 10:32
• G(s) = T/(1+(0.5T))
– user173271
Commented Jan 21, 2020 at 10:40
• No, not true, the closed loop gain is down 3db when the open loop gain T(s), as defined by the author, is equal to 1 with an angle of 90deg, or in other words when T(s) =j..
– user173292
Commented Jan 21, 2020 at 11:18
• Ah! I see where the confusion is. "As defined by the author" is the crux. What the author calls the "open loop" gain is usually referred to as the "loop" gain. By conventional definition the "open loop" gain is the forward loop gain or Vout/Vdiff for an op amp and the "loop" gain is the gain right around the loop. I agree that where the loop gain in the usual sense, (open loop gain in the author's sense), is unity, the closed loop gain is down -3dB whatever the value of the low frequency closed loop gain.
– user173271
Commented Jan 21, 2020 at 16:43
• Yes exactly, as I wrote; when $\ T=1\angle 90^{\circ}$ then $\ G(s)=\frac{1}{\sqrt{2}}G_0$.. $\ T$ is what the author calles the open loop gain, and I agree that I would have called this the loop gain. So the statements the author make are all completely true and accurate, and the math holds true, the confusion comes from two facts; the author used the term "open loop gain" instead of "loop gain" and also the author uses the term "gain" when he means "transfer function" the gain is not the transfer function it self but the magnitude of the transfer function.
– user173292
Commented Jan 21, 2020 at 16:49

I think your confusion comes from the definition of the closed-loop bandwidth.

The closed-loop bandwidth is defined as the frequency where the magnitude of the closed loop gain is -3 db down from the intended value. For root power quantities, this corresponds to a factor of $$\1/\sqrt{2} = 0.7071\$$ times the intended value.

In this example, the intended magnitude is 2 so the bandwidth is defined as the frequency where the magnitude falls to $$\ 2 \cdot 1/\sqrt{2} = 1.414.\$$

The depiction in figure 2 aligns with this definition, but it is not fully accurate since the magnitude is a little higher than 1.414 at the designated bandwidth. Please pardon the author for this inaccuracy.

The definition of the closed-loop bandwidth given by the author is correct. The closed-loop transfer function is

$$G_{cl} = \frac{G_f}{1 + G_f \cdot G_b}$$

with the forward transfer function (controller and plant) $$\G_f = K_{EA} \cdot K_{MOD} \cdot K_{PWR} \cdot K_{LC}\$$ and the backward transfer function (feedback network) $$\G_b = K_{FB}.\$$

We can substitute the open-loop transfer function $$\G_{ol} = G_f \cdot G_b\$$, which the author calls $$\T\$$, into the equation. This gives

$$G_{cl} = \frac{\frac{G_{ol}}{G_b}}{1 + G_{ol}} = \frac{1}{G_b} \cdot \frac{G_{ol}}{1 + G_{ol}} \equiv \frac{1}{K_{FB}} \cdot \frac{T}{1 + T}.$$

The static component of $$\1 / G_b\$$ defines the intended magnitude. Since $$\G_b\$$ does not include any dynamic component in this example and therefore does not change with frequency, only the term $$\frac{G_{ol}}{1 + G_{ol}}$$ defines the bandwidth when $$\frac{|G_{ol}|}{|1 + G_{ol}|} = \frac{1}{\sqrt{2}}.$$

$$\G_{ol}\$$ is complex so the calculation is usually not straightforward. The author therefore chooses a simplified example where the phase is -90° when $$\|G_{ol}| = 1\$$. This means that $$\G_{ol} = -1j\$$ in this example. This results in $$\frac{|G_{ol}|}{|1 + G_{ol}|} = \frac{|-1j|}{|1 - 1j|} = \frac{1}{\sqrt{2}}$$ at the open-loop magnitude crossover frequency.

As you can see, the definition of the bandwidth is fulfilled in this example and it only depends on $$\G_{ol}\$$, not on $$\G_{b}\$$ (at least not separate from $$\G_{ol}\$$ 😉). This changes when the phase of $$\G_{ol}\$$ is not close to -90° at the crossover frequency or $$\G_{b}\$$ includes dynamic components that are significant around the crossover frequency. This will result in $$\|G_b| \cdot |1 + G_{ol}| \neq \sqrt{2}\$$ when $$\|G_{ol}| = 1\$$. The closed-loop bandwidth will therefore not equal the open-loop magnitude crossover frequency under different conditions than $$\|G_b| = 1\$$ and $$\\arg(G_{ol}) = -90°\$$ when $$\|G_{ol}| = 1\$$.

The open-loop magnitude crossover frequency being equivalent to the closed-loop bandwidth is therefore only true for one specific case. It is a rule of thumb for different typical cases. It gives a good approximation when the open-loop crossover phase is close to -90°. When you try to stick to the other rule of thumb, that the phase margin should be 60° or better, this will usually result in a good approximation.

Please point out mistakes if you find any.