Can a linear amplifier have finite bandwidth?

Question

I am reading Sedra and Smith and just want to be sure that the following characterization of an ideal linear amplifier is correct. A linear amplifier was defined as an amplifier for which the output signal is proportional to the input signal given any input signal. Supposing that we define bandwidth as the range of frequencies over which gain is constant; it then seems to me that one can say that an amplifier is linear if and only if it has infinite bandwidth (otherwise you could construct signals which superpose frequency components with different gains and get distortion through the amplifier). Is this correct?

Here is the definition from Sedra and Smith which I quote:

The initial phrase you used was ideal linear amplifier and it will have as much BW as you want it to have. — Andy aka, Sep 17 '21 at 12:33
@Andyaka yes, I am using "linear amplifier" and "ideal linear amplifier" interchangeably. I take it my statement is correct then? Thank you! — EE18, Sep 17 '21 at 12:35
An ideal linear amplifier is not a linear amplifier. LA is a subset of ILA. — Andy aka, Sep 17 '21 at 12:37
@Andyaka I suppose you're saying that a linear amplifier is an amplifier which has a finite bandwidth (and so is linear over that range), while an ILA has infinite bandwidth? — EE18, Sep 17 '21 at 12:38
Low pass filters are linear devices, so clearly not all linear devices have infinite bandwidth. I think what you're missing is that the proportionality constant can itself be a function of frequency (as in a low pass filter). — user1850479, Sep 17 '21 at 14:21
@user1850479 I see. I suppose "linear" is a matter of (arbitrary) definition here and that is why our conclusions diverge. — EE18, Sep 17 '21 at 14:27
Linear actually has a very specific mathematical definition, nothing arbitrary about it. You're just using the colloquial version rather than the math and so coming to the wrong conclusion. S+S is an advanced text, usually you would have taken the prerequisite math before diving into it. If not, worth reviewing. — user1850479, Sep 17 '21 at 14:32
@user1850479 I'm well aware of the standard mathematical definition; I am using the one I quoted from Sedra and Smith. — EE18, Sep 17 '21 at 14:33
"X is a linear amplifier" does not imply that all linear amplifiers are "X". The actual definition of the word "linear" is pretty much as Neil_UK says in his answer, although phase changes are permissible, too. — Matt Timmermans, Sep 17 '21 at 22:11

Neil_UK · Answer 1 · 2021-09-18T06:26:25.973

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TL;DR As engineers, Sedra/Smith make one more assumption than the OP does. Engineers also make one more assumption than do mathematicians.

Mathematicians say a system H is linear iff (if and only if) it obeys the superposition principle, that H(x+y) = H(x) + H(y).

This is weaker than engineers tend to use, as it allows a time-varying system gain, which can generate output harmonics under certain conditions. So time-independent gain implies superposition, but superposition does not imply time-independent gain.

Engineers tend to be lazy and say linear when they mean LTI - Linear Time-Invariant.

An LTI system can still have a frequency response. Engineers tend to say that an amplifier is linear when it has a gain Vout/Vin = k(f), where k can vary with frequency. When excited by any single sinuosoid, the output will look like a perfect scaled copy of the input.

If k varies with frequency, then when excited by multiple sinusoids, the output will not look like the input, but there will be no intermodulation distortion, the only output frequencies will be those present in the input.

Now for the important difference in assumptions. If the input range of frequencies is limited to the flat frequency response part of the spectrum (a condition that is often assumed but not stated), and the amplifier also has linear phase - that is constant signal delay, then the output will look like a scaled version of the input, and finally we can write Vout/Vin = constant.

Plenty of amplifiers and filters exist where phase is not linear even though the gain may be, and the output of those does not look like the input. The output is usually referred to as suffering from phase distortion.

An ideal amplifier will have infinite bandwidth and zero noise, and not suffer clipping up to any arbitrary signal level.

The response of inductors or capacitors is linear, they obey superposition. Networks involving those do not obey Vout/Vin = k, there is a differential or integral operator involved.

Engineers do also use time-varying systems, although they call them mixers. A diode ring pumped with a strong local oscillator (LO) drive changes its gain with the LO period. When a low level signal is passed through, it does obey superposition, and engineers do talk about 'linear' mixers. A good mixer obeys superposition up to higher signal levels than a bad mixer. In a good mixer, the signal passing through does not suffer any intermodulation distortion between its frequency components. There is intermodulation between the LO and the signal components, that's how you get the sum and difference frequencies!

edited Sep 18 '21 at 06:26

answered Sep 17 '21 at 12:49

Neil_UK

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Thanks for the answer. Are you saying that it is not true that an amplifier is linear (defined as $v_{in}=Av_{out}$ for any $v_{in}$) if and only if it has infinite bandwidth? – EE18 Sep 17 '21 at 12:54
Yes. A linear amplifier only has to meet the criterion that vin/vout=k instantaneously, or vin(f)/vout(f) = k(f) where f is the frequency under consideration. It's independent of the frequency response of the amplifier. Beware of insisting that all yellow flowers are daffodils. – Neil_UK Sep 17 '21 at 13:00
(=I am not sure I agree. NB the requirement that the given equation hold for all possible input voltages. I think the if and only if must hold then? – EE18 Sep 17 '21 at 13:23
Of course, I agree with you if the equation need only hold for a subset of signals or frequencies. – EE18 Sep 17 '21 at 13:24
@Neil_UK When referring to the gain, are you talking about open-loop gain? Closed loop-gain? Please be specific. – Carl Sep 17 '21 at 13:28
@Carl you said 'amplifier', ie a 'black box', so the gain I refer to is the amplifier gain, vout/vin of that box's terminals. If you are now making a disctintion between closed loop and open loop gain, it sounds like you're opening that box and seeing how the amplifier is implemented in transistors or opamps with feedback. You can take a high gain amplifier with good or bad linearity, and if you apply feedback to get a low gain, you also improve the linearity of the overall amplifier. – Neil_UK Sep 17 '21 at 14:09
@1729_SR This link at wikipedia looks to be the most thorough and mathematical, and supports my general thrust that a linear system has to obey superposition, if and only if, but need not have a flat frequency response. I am not appealing to the authority of wikipedia per se, but I am sure the mathematical formalism pedants will have corrected any errors in this sort of page. – Neil_UK Sep 17 '21 at 14:18
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@1729_SR Bizarrely, and I've learnt something here, this (last paragraph) says that a linear system can create harmonics and sub-harmonics, if a specific system that obeys superposition (a linear system) is excited by a specific signal. So linearity appears to be an even weaker property than I had imagined. – Neil_UK Sep 17 '21 at 14:23
Neil, thank you very much for all of your commentary. I'm confused though as to how you can say "there is no distortion" in the example you give. If I have an input signal which is a superposition of equal amplitudes of the 1 kHz and the 10 kHz components, and if the amplifier has different gain for the two components, then the output signal is distorted (when we do the inverse transform the result is no longer collinear with the input signal). – EE18 Sep 17 '21 at 14:37
@Neil_UK It has been 20 years since I took that class, but I don't think y(t)=t*cos(t) is a linear function, so possibly that wiki page was recently vandalized. – user1850479 Sep 17 '21 at 15:37
@user1850479 I'm not taking sides on this one, just thinking out loud. y(t)=cos(t) is not linear in t, but y(t)=cos(t).x(t) does obey superposition in x, so is linear is x. It's time varying though, so I'd call that a mixer, which is where the harmonics are coming from. As an engineer, I've used 'linear' mixers, which would obey superposition. – Neil_UK Sep 17 '21 at 15:57
@1729_SR This is more about how the word 'linear' is defined, than how what we think of as linear systems behave. If linear<=>obeys superposition, then what I've said is consistent. A system that obeys superposition does not have to obey vout/vin=constant, consider a linear inductor or capacitor for instance. – Neil_UK Sep 17 '21 at 15:59
@Neil_UK In the wiki article, x(t) is a function with independent variable t, so I am not sure what you mean by "superposition in x"? Just typing in some numbers and plotting in matlab, y(t) in that example does not obey superposition since (for example) y(t+2t) != y(t)+y(2t). – user1850479 Sep 17 '21 at 16:40
@user1850479 x is the signal, t is time. Mucking about with the time scaling will of course give pretty shapes. It's the x that you have to vary, and y is proportional to x at any given time t. – Neil_UK Sep 17 '21 at 16:56
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@Neil_UK I believe mathematicians require two things for linearity: (1) Additivity, which you mention; and, (2) Homogeneity (of degree 1), which maybe you didn't mention. Namely, that: f(A*x) = A*f(x), for all A. You sited the link in comments about this, but I'm not sure (I may have missed it) if you edited your answer to include it. Superposition is another way of saying both. – jonk Sep 17 '21 at 18:27
@jonk I don't have a subtle enough appreciation of mathematics to understand the difference between meeting additivity and meeting homogeneity. A=2 implies f(x+y)=f(x)+f(y) for x=y, and 3 for x=2y etc. The reverse is less obvious to me at the moment and perhaps that's the difference. I'll stay with superposition, I think I understand that. – Neil_UK Sep 17 '21 at 18:57
@jonk: I was wondering if someone would bring that detail up. However, functions that are additive but not linear are very bizarre and pathological things that cannot possibly describe the behavior of any real system, being everywhere discontinuous. It's not even possible to write down an explicit definition of such a function, and even their existence is somewhat questionable, as it depends on which axiomatization of mathematics one uses. See this Wikipedia page for more details. – Ilmari Karonen Sep 18 '21 at 01:18
@IlmariKaronen This is my goto-function for such things! – jonk Sep 18 '21 at 01:59
@jonk: That Weierstrass function is a lot less weird and pathological than non-linear additive functions, though. Heck, it's even continuous and you can draw an (approximate) graph of it! (To graph any non-linear additive function, you'd have to color in the entire plane, since the graph of any such function must be dense in $\mathbb R^2$.) – Ilmari Karonen Sep 18 '21 at 02:04

The Photon · Accepted Answer · 2021-09-17T16:08:28.687

The text you quoted is not a definition of a linear amplifier. It says

Equation (1.4) is a linear relationship; hence the amplifier it describes is a linear amplifier.

Equation (1.4) is not a definition of a linear system or a linear amplifier, it is only one example of a linear relationship.

And an amplifier that follows Equation (1.4) is not the only kind of linear amplifier, it is only one example of a linear amplifier.

An amplifier that follows Equation (1.4) has infinite bandwidth, but other kinds of linear amplifiers do not.

The text continues,

it should be easy to see that if the relationship between \$v_o\$ and \$v_i\$ contains higher powers of \$v_i\$ then the waveform of \$v_o\$ will no longer be identical to that of \$v_i\$. The amplifier is then said to exhibit nonlinear distortion.

Here they do seem to treat the previous equation as definitional.

But you can obtain the same result (higher powers of \$v_i\$ are not allowed in the equation for \$v_o\$) if you use the proper definition of linearity, which has been given in Neil UK's answer.

On the other hand, derivative and integral terms (\$\frac{dv_i}{dt}\$ and \$\int v_i(t) dt\$) are allowed in linear relationships, and integral terms are the ones that result in limited bandwidth.

score 2 · Answer 3 · answered Sep 17 '21 at 12:49

Real op amps (non-ideal) will have finite bandwidths. Perhaps this definition by Allan Hambley can enhance your understanding: -

"Ideal op amps have infinite gain magnitude and unlimited bandwidth. Real op amps have finite open-loop gain magnitude, typically between \$10^4 \$ and \$10^6 \$. Furthermore, the bandwidth of real op amps is limited. The gain of a real op amp is a function of frequency, becoming smaller in magnitude at higher frequencies."

score 2 · Answer 4 · answered Sep 17 '21 at 14:37

Yes, you're right.

We're playing fast and loose with the idea of linearity, here, though. From the perspective of linearity being the "the straightness of the line of the graph of instantaneous input voltage versus instantaneous output voltage", you are exactly right. And, don't forget phase, another contributor to non-linearity in this sense.

If we are talking about linearity in the sense of the equations describing the relationship between input and output being linear, then that's a different story. Here we are referring to the idea that the transfer function obeys the conditions for algebraic linearity, such as commutativity and distributivity, like:

$$ f(x) \cdot g(x) = g(x) \cdot f(x)$$

$$ 2f(x) + 2g(x) = 2(f(x) + g(x)) $$

There is naturally a connection between these two interpretations, but I believe you are referring strictly to the former definition, in which case an ideal amplifier must indeed have infinite bandwidth, for the reasons you identified (and others, such as phase response and output slew rate).

Can a linear amplifier have finite bandwidth?

4 Answers4