For first-order LTI systems, the gain-bandwidth constant is often discussed. I've seen the claim that in general, gain and bandwidth don't directly trade off with each other as much as delay. For an example of a discussion where this claim is made, see this Caltech lecture, starting at 6:39 (timestamp in link). Although this is from a circuits class, the relevant discussion has no circuits-specific details and is entirely framed from an LTI systems perspective. For the exact moment where the specific claim that cascading multiple systems with individually smaller gains and individually larger bandwidths to realize the same overall gain with larger overall bandwidth also results in larger delay, go to 12:00.

This claim about the delay confuses me because to my understanding, the delay in a linear system is directly related to the pole frequencies, as these contribute transient terms of the form $Ce^{pt}$. The bandwidth is also a measure of where the lowest frequency poles are, so I don't understand why an overall system with larger bandwidth than another system would have a longer delay. Maybe "delay" is being defined differently than I think it is.

To illustrate my confusion (recreating the example in the lecture video above), suppose I want to realize a transfer function with a gain of 100, and one way is with the transfer function:

$$ H_1(s)=\frac{A_1}{1+\frac{s}{\omega_1}} $$

Suppose that $A_1=100$, $\omega_1=10^8 \text{ rad/s}$, so that for this system the gain-bandwidth product $A_1\omega_1=\omega_u=10^{10} \text{ rad/s}$.

Now suppose that I could also modify the system's gain and bandwidth within the constraint of the gain-bandwidth product, so I could create a new single block:

$$ H_2(s)=\frac{A_2}{1+\frac{s}{\omega_2}} $$

where $A_2=A_1/10=10$ and $\omega_2=10\omega_1=10^9 \text{ rad/s}$. The gain-bandwidth product is the same as for $H_1$. Now I realize the same gain as before by cascading two of the new system, forming $H_C(s)=H_2(s)H_2(s)$.

I've attached plots below of the bode-plots of each system as well as the step responses of each.

bode and step plots

To me this seems to show that the "delay" for the cascaded system is actually lower than the delay of the original system, which is what I would expect since the bandwidth is substantially higher.

Am I misunderstanding what "delay" means in this context?


1 Answer 1


After looking at it a bit more, I think that "delay" here means "phase shift". Since the cascaded system contains more poles, it naturally has more total phase shift than the original system. Plot illustrating this:


So it seems that the tradeoff in making $H_C$ is that for the same gain-bandwidth product, there's more gain across a wider range of frequencies, but the total phase shift is larger for $H_C$ and there's a crossover point at $\omega_2$ where the phase shift of $H_C$ is greater than that of $H_1$. So it seems that for sinusoidal inputs where phase shift translates into a time delay, although $H_C$ has more gain at these higher frequencies it also has more delay. Additionally, I assume that in a feedback systems context this additional phase shift could be problematic. To illustrate the point of "you can get more gain if you wait longer", below are plotted the responses from $H_1$ and $H_C$ to an input of $x(t)=\cos(\omega_{in}t)$ with $\omega_{in}=6\cdot10^9 \text{ rad/s}$. The peak of $H_C$'s output is higher but occurs after the peak of $H_1$'s output.

more gain with more delay


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