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<h2 class="hd hd-2 unit-title">10.1. Activity: Black's discovery.</h2>
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<h3 class="hd hd-3 problem-header">Black's discovery.</h3><p>
Harold Black was an electrical engineer working for Bell Labs on amplification systems. At that time powerful amplifiers were easy to build, using vacuum tubes. They were very unstable, however – the gain they provided could vary dramatically with temperature or other factors, and their gain curve was not at all flat over any reasonable range of frequencies. </p><p>
In 1927 Black discovered an extremely simple and elegant solution to these problems: “negative feedback." In this activity you will verify the efficacy of his method. If his supervisors had carried out the calculations you will do, they would have found it much easier to accept his idea. In fact it took them a long time to come around, because it seemed to throw away huge potential amplification gains. </p>
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Activity: Question 1
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<p>
Start with a system with system function [mathjaxinline]H(s)[/mathjaxinline] and complex gain [mathjaxinline]H(i\omega )[/mathjaxinline]. Suppose that the gain [mathjaxinline]|H(i\omega )|[/mathjaxinline] varies from a given value &#8211; say 10,000 &#8211; by a factor of 2, depending upon the input frequency, temperature, or other unpredictable conditions. Suppose, for simplicity, that [mathjaxinline]H[/mathjaxinline] is actually just a real constant &#8211; so there's no phase lag in this system, and [mathjaxinline]H[/mathjaxinline] itself is the &#8220;open loop" gain. </p>
<p>
Now add feedback, in the simplest form, indicated by the block diagram </p>
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<img src="/assets/courseware/v1/fbd4f29a17bf3e1ef5fa99e290de828c/asset-v1:OCW+18.031+2019_Spring+type@asset+block/images_c6_blacksformula2.svg" width="300px" style="margin: 0px 10px 10px 10px"/>
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<p>
So the feedback is &#8220;linear," with constant [mathjaxinline]K[/mathjaxinline]. Suppose that [mathjaxinline]K[/mathjaxinline] is quite small, say [mathjaxinline]0.01[/mathjaxinline]. </p>
<p>
What is the closed loop gain when [mathjaxinline]H=10,000[/mathjaxinline]? </p>
<p>
(Enter to the nearest integer.) </p>
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Question 2:
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<p>
Still with [mathjaxinline]K=0.01[/mathjaxinline], suppose the gain [mathjaxinline]H[/mathjaxinline] varies unpredictably above or below its nominal value, so that it is somewhere between [mathjaxinline]H=5000[/mathjaxinline] and [mathjaxinline]H=20000[/mathjaxinline], i.e. a ratio of 4 to 1 between its highest and lowest values. Let [mathjaxinline]H_{\mathrm{cl},10000}[/mathjaxinline] be the value of [mathjaxinline]H_{\mathrm{cl}}[/mathjaxinline] when [mathjaxinline]H=10000[/mathjaxinline]. Likewise for [mathjaxinline]H_{\mathrm{cl},5000}[/mathjaxinline] and [mathjaxinline]H_{\mathrm{cl},20000}[/mathjaxinline]. To get a sense of the range of values of [mathjaxinline]H_{\mathrm{cl}}[/mathjaxinline] compute the following ratios. </p>
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(Enter the results to three decimal place accuracy.) </p>
<p>
<p style="display:inline">Ratio [mathjaxinline]\displaystyle \frac{H_{\mathrm{cl},20000}}{H_{\mathrm{cl},10000}}[/mathjaxinline] when [mathjaxinline]H=20000[/mathjaxinline] : </p>
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<p style="display:inline">Ratio [mathjaxinline]\displaystyle \frac{H_{\mathrm{cl},5000}}{H_{\mathrm{cl},10000}}[/mathjaxinline] when [mathjaxinline]H=5000[/mathjaxinline] : </p>
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Question 3:
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<p>
Assuming [mathjaxinline]H=10000[/mathjaxinline], compute the ratio of [mathjaxinline]H_{\mathrm{cl}}[/mathjaxinline] when [mathjaxinline]K=0.005[/mathjaxinline] to [mathjaxinline]H_{\mathrm{cl}}[/mathjaxinline] when [mathjaxinline]K=0.01[/mathjaxinline]. Compute the same ratio of [mathjaxinline]H_{\mathrm{cl}}[/mathjaxinline] for [mathjaxinline]K=0.02[/mathjaxinline] to [mathjaxinline]H_{cl}[/mathjaxinline] when [mathjaxinline]K=0.01[/mathjaxinline]. </p>
<p>
(Enter the results to two decimal points accuracy.) </p>
<p>
<p style="display:inline">Ratio [mathjaxinline]\displaystyle \frac{H_{\mathrm{cl},K=0.005}}{H_{\mathrm{cl},K=0.01}}[/mathjaxinline]: </p>
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<p style="display:inline">Ratio [mathjaxinline]\displaystyle \frac{H_{\mathrm{cl},K=0.02}}{H_{\mathrm{cl},K=0.01}}[/mathjaxinline]: </p>
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Question 4:
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This is not much of an improvement: If the ratio of biggest to smallest [mathjaxinline]K[/mathjaxinline] is 4, then the same is approximately true for the closed loop gain. But the real point of Black's insight is this: The behavior of the small gain is much easier to control. So, the variation of [mathjaxinline]K[/mathjaxinline] might be on the order of [mathjaxinline]1\%[/mathjaxinline] above or below its nominal value. Let's look at the variation of the closed loop gain as [mathjaxinline]H[/mathjaxinline] and [mathjaxinline]K[/mathjaxinline] vary. </p>
<p>
Find the largest and smallest closed loop gains if [mathjaxinline]5000 \leq H \leq 20000[/mathjaxinline] and [mathjaxinline]0.0099 \leq K \leq 0.0101[/mathjaxinline]. </p>
<p>
(Enter the results to two decimal points accuracy.) </p>
<p>
<p style="display:inline">Max gain: </p>
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<h3 class="hd hd-3 problem-header">General analysis.</h3><p>
Black's theorem gives for the closed loop gain </p><table id="a0000001055" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]H_{\mathrm{cl}}=\frac{H}{1+KH}\, .[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
Assume that [mathjaxinline]KH[/mathjaxinline] is large relative to 1. (We took it to be 100 in the example.) Then the value of [mathjaxinline]H_{\mathrm{cl}}[/mathjaxinline] is quite near to [mathjaxinline]1/K[/mathjaxinline]! </p>
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<h2 class="hd hd-2 unit-title">10.2. Activity: Acoustic feedback.</h2>
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This activity will model acoustic feedback. </p>
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Question 1: Linear amplifier
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<p>
Suppose we have a "linear" amplifier with gain [mathjaxinline]g[/mathjaxinline] &#8212; it simply multiplies the input signal by the factor [mathjaxinline]g[/mathjaxinline]. This is an LTI system; in fact it is modeled by a differential equation, one of degree zero! If [mathjaxinline]x(t)[/mathjaxinline] denotes the system response and [mathjaxinline]y[/mathjaxinline] the input signal, then the equation is </p>
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<td class="equation" style="width:80%; border:none">[mathjax]x = gy[/mathjax]</td>
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This fits our [mathjaxinline]P(D)x=Q(D)y[/mathjaxinline] paradigm, with [mathjaxinline]P(s)=?[/mathjaxinline] and [mathjaxinline]Q(s)=?[/mathjaxinline], and transfer function [mathjaxinline]H(s) = ?[/mathjaxinline] </p>
<p>
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Question 2: Positive feedback.
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Now let's suppose that the output from the speaker &#8212; [mathjaxinline]x(t)[/mathjaxinline] &#8212; is fed back positively to the pickup, but with two changes: </p>
<ol class="enumerate">
<li value="1">
<p>
a time delay of [mathjaxinline]a[/mathjaxinline] seconds ([mathjaxinline]a&gt;0[/mathjaxinline]); and </p>
</li>
<li value="2">
<p>
a decrease in its amplitude, i.e. multiply amplitude by a factor of [mathjaxinline]k[/mathjaxinline] (so [mathjaxinline]0&lt;k&lt;1[/mathjaxinline]). </p>
</li>
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Draw the block diagram representing this feedback system on a piece of paper such that it has a single feedback loop. </p>
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Determine the transfer function of the resulting closed-loop system. </p>
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<p style="display:inline">[mathjaxinline]H_{\mathrm{cl}}(s) =[/mathjaxinline]</p>
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Question 3:
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Take [mathjaxinline]g=4[/mathjaxinline] and [mathjaxinline]k=.125[/mathjaxinline], and make a picture of the pole diagram of the transfer function. Is this system stable or unstable? </p>
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Question 4:
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Suppose that [mathjaxinline]g[/mathjaxinline] is fixed but we can vary the feedback gain [mathjaxinline]k[/mathjaxinline] (for example by moving the speaker). </p>
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What values of [mathjaxinline]k[/mathjaxinline] will render the closed-loop system stable? This is important! An unstable system will have responses to the null input signal that blow up exponentially: this is generally an undesirable acoustic effect, known to musicians as "feedback." </p>
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Question 5
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Suppose that the system is stable but just barely. Some frequencies will be "wolf tones," and get amplified much more than others. Where do you expect to find them? </p>
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(Choose all that apply.) </p>
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<text>wolf tones happen when [mathjaxinline]a\cdot \omega[/mathjaxinline] is close to an integer ([mathjaxinline]\omega[/mathjaxinline] in radians/second)</text>
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<text>wolf tones happen when [mathjaxinline]a\cdot \omega[/mathjaxinline] is close to a multiple of [mathjaxinline]2\pi[/mathjaxinline] ([mathjaxinline]\omega[/mathjaxinline] in radians/second)</text>
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<text>wolf tones happen when [mathjaxinline]a\cdot f[/mathjaxinline] is close to an integer ([mathjaxinline]f[/mathjaxinline] = frequency in hertz)</text>
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<text>wolf tones happen when [mathjaxinline]a\cdot f[/mathjaxinline] is close to a multiple of [mathjaxinline]2\pi[/mathjaxinline] ([mathjaxinline]f[/mathjaxinline] = frequency in hertz)</text>
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<text>wolf tones happen when the delay [mathjaxinline]d = av \approx nv/f = n[/mathjaxinline] ([mathjaxinline]f[/mathjaxinline] = frequency in hertz, [mathjaxinline]v[/mathjaxinline] = velocity, [mathjaxinline]n[/mathjaxinline] is an integer)</text>
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<text>wolf tones happen when the delay [mathjaxinline]d = av \approx nv/f = nL[/mathjaxinline] ([mathjaxinline]f[/mathjaxinline] = frequency in hertz, [mathjaxinline]L[/mathjaxinline] = wavelength, [mathjaxinline]v[/mathjaxinline] = velocity, [mathjaxinline]n[/mathjaxinline] is an integer)</text>
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<h3 class="hd hd-3 problem-header">Conclusions</h3><p>
This activity took us into the realm of LTI systems that are *not* modeled by differential equations: the delay has a nice description in the frequency domain, but it does not correspond to a differential operator. </p><p>
We saw that a wolf tone is one where [mathjaxinline]av[/mathjaxinline] is a multiple of wavelength. A typical musical frequency is 440 Hertz; this is the A to which orchestras tune. The speed of sound is 340 m/s, so a sound wave at pitch A is 77 cm long. The delay in our model comes almost entirely from the time it takes sound to travel from the speaker back to the pickup. Since the distance from speaker to pickup is [mathjaxinline]av[/mathjaxinline], the wolf tone will be an A if that distance is a multiple of 77 cm. </p>
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