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    <p align="right">12/25/03</p>
    <h2>Operational Amplifiers</h2>
    <p>An operational amplifier with 
    <i>voltage gain</i>
    , 
    <i>A</i>
    , can be represented by the following symbol:</p>
    <p style='text-indent:.5in'>
      <img src="./opamps_files/image001.gif" />
    </p>
    <p>The voltages to be amplified are 
    <m:math>
      <m:semantics>
        <m:mrow>
          <m:msub>
            <m:mi>v</m:mi>
            <m:mo>&#8722;</m:mo>
          </m:msub>
          <m:mrow>
            <m:mo>(</m:mo>
            <m:mi>t</m:mi>
            <m:mo>)</m:mo>
          </m:mrow>
        </m:mrow>
      </m:semantics>
    </m:math>
    and 
    <m:math>
      <m:semantics>
        <m:mrow>
          <m:msub>
            <m:mi>v</m:mi>
            <m:mo>+</m:mo>
          </m:msub>
          <m:mrow>
            <m:mo>(</m:mo>
            <m:mi>t</m:mi>
            <m:mo>)</m:mo>
          </m:mrow>
        </m:mrow>
      </m:semantics>
    </m:math>
    , which are the 
    <i>inverting</i>
    (minus) and 
    <i>non-inverting</i>
    (plus) inputs, respectively. The amplified voltage output is 
    <m:math>
      <m:semantics>
        <m:mrow>
          <m:msub>
            <m:mi>v</m:mi>
            <m:mrow>
              <m:mi>o</m:mi>
              <m:mi>u</m:mi>
              <m:mi>t</m:mi>
            </m:mrow>
          </m:msub>
          <m:mrow>
            <m:mo>(</m:mo>
            <m:mi>t</m:mi>
            <m:mo>)</m:mo>
          </m:mrow>
        </m:mrow>
      </m:semantics>
    </m:math>
    .</p>
    <p>An 
    <i>ideal</i>
    operational amplifier is one for which</p>
    <ul style='margin-top:0in' type="circle">
      <li>the output voltage is 
      <m:math>
        <m:semantics>
          <m:mrow>
            <m:msub>
              <m:mi>v</m:mi>
              <m:mrow>
                <m:mi>o</m:mi>
                <m:mi>u</m:mi>
                <m:mi>t</m:mi>
              </m:mrow>
            </m:msub>
            <m:mrow>
              <m:mo>(</m:mo>
              <m:mi>t</m:mi>
              <m:mo>)</m:mo>
            </m:mrow>
            <m:mtext>&#8201;</m:mtext>
            <m:mtext>&#8201;</m:mtext>
            <m:mo>=</m:mo>
            <m:mtext>&#8201;</m:mtext>
            <m:mi>A</m:mi>
            <m:mrow>
              <m:mo>(</m:mo>
              <m:mrow>
                <m:msub>
                  <m:mi>v</m:mi>
                  <m:mo>+</m:mo>
                </m:msub>
                <m:mrow>
                  <m:mo>(</m:mo>
                  <m:mi>t</m:mi>
                  <m:mo>)</m:mo>
                </m:mrow>
                <m:mtext>&#8201;</m:mtext>
                <m:mo>&#8722;</m:mo>
                <m:mtext>&#8201;</m:mtext>
                <m:msub>
                  <m:mi>v</m:mi>
                  <m:mo>&#8722;</m:mo>
                </m:msub>
                <m:mrow>
                  <m:mo>(</m:mo>
                  <m:mi>t</m:mi>
                  <m:mo>)</m:mo>
                </m:mrow>
              </m:mrow>
              <m:mo>)</m:mo>
            </m:mrow>
          </m:mrow>
        </m:semantics>
      </m:math>
      </li>
      <li>the currents at the "plus" and "minus" input terminals
      are absolutely zero</li>
      <li>the voltage gain 
      <m:math>
        <m:semantics>
          <m:mrow>
            <m:mi>A</m:mi>
            <m:mtext>&#8201;</m:mtext>
            <m:mo>&#8594;</m:mo>
            <m:mtext>&#8201;</m:mtext>
            <m:mi>&#8734;</m:mi>
          </m:mrow>
        </m:semantics>
      </m:math>
      .</li>
    </ul>
    <p>The operational amplifier is an example of a 
    <i>differential</i>
    amplifier in that its output voltage, 
    <m:math>
      <m:semantics>
        <m:mrow>
          <m:msub>
            <m:mi>v</m:mi>
            <m:mrow>
              <m:mi>o</m:mi>
              <m:mi>u</m:mi>
              <m:mi>t</m:mi>
            </m:mrow>
          </m:msub>
          <m:mrow>
            <m:mo>(</m:mo>
            <m:mi>t</m:mi>
            <m:mo>)</m:mo>
          </m:mrow>
        </m:mrow>
      </m:semantics>
    </m:math>
    , is proportional to the 
    <i>difference</i>
    between its two input voltages, 
    <m:math>
      <m:semantics>
        <m:mrow>
          <m:msub>
            <m:mi>v</m:mi>
            <m:mo>+</m:mo>
          </m:msub>
          <m:mrow>
            <m:mo>(</m:mo>
            <m:mi>t</m:mi>
            <m:mo>)</m:mo>
          </m:mrow>
        </m:mrow>
      </m:semantics>
    </m:math>
    and 
    <m:math>
      <m:semantics>
        <m:mrow>
          <m:msub>
            <m:mi>v</m:mi>
            <m:mo>&#8722;</m:mo>
          </m:msub>
          <m:mrow>
            <m:mo>(</m:mo>
            <m:mi>t</m:mi>
            <m:mo>)</m:mo>
          </m:mrow>
        </m:mrow>
      </m:semantics>
    </m:math>
    . Practical operational amplifiers approximate the behavior
    of ideal amplifiers fairly well over the audio frequency
    range.</p>
    <p>As our first example application of operational
    amplifiers, consider the following circuit:</p>
    <p style='text-indent:.5in'>
      <img src="./opamps_files/image002.gif" />
    </p>
    <p>In this circuit, 
    <m:math>
      <m:semantics>
        <m:mrow>
          <m:mi>R</m:mi>
        </m:mrow>
      </m:semantics>
    </m:math>
    and 
    <m:math>
      <m:semantics>
        <m:mrow>
          <m:msub>
            <m:mi>v</m:mi>
            <m:mrow>
              <m:mi>i</m:mi>
              <m:mi>n</m:mi>
            </m:mrow>
          </m:msub>
          <m:mrow>
            <m:mo>(</m:mo>
            <m:mi>t</m:mi>
            <m:mo>)</m:mo>
          </m:mrow>
        </m:mrow>
      </m:semantics>
    </m:math>
    represent the source of the input voltage to the amplifier,
    which might be, for example, a microphone or some other
    sensor, or even another amplifier. Note that this circuit
    feeds the entire output voltage back into the negative input
    terminal of the operational amplifier. This circuit therefore
    employs maximum negative voltage feedback.</p>
    <p>To analyze this circuit, recall that, for ideal
    operational amplifiers,</p>
    <p style='text-indent:.5in'>
      <m:math>
        <m:semantics>
          <m:mrow>
            <m:msub>
              <m:mi>v</m:mi>
              <m:mrow>
                <m:mi>o</m:mi>
                <m:mi>u</m:mi>
                <m:mi>t</m:mi>
              </m:mrow>
            </m:msub>
            <m:mrow>
              <m:mo>(</m:mo>
              <m:mi>t</m:mi>
              <m:mo>)</m:mo>
            </m:mrow>
            <m:mtext>&#8201;</m:mtext>
            <m:mtext>&#8201;</m:mtext>
            <m:mo>=</m:mo>
            <m:mtext>&#8201;</m:mtext>
            <m:mi>A</m:mi>
            <m:mrow>
              <m:mo>(</m:mo>
              <m:mrow>
                <m:msub>
                  <m:mi>v</m:mi>
                  <m:mo>+</m:mo>
                </m:msub>
                <m:mrow>
                  <m:mo>(</m:mo>
                  <m:mi>t</m:mi>
                  <m:mo>)</m:mo>
                </m:mrow>
                <m:mtext>&#8201;</m:mtext>
                <m:mo>&#8722;</m:mo>
                <m:mtext>&#8201;</m:mtext>
                <m:msub>
                  <m:mi>v</m:mi>
                  <m:mo>&#8722;</m:mo>
                </m:msub>
                <m:mrow>
                  <m:mo>(</m:mo>
                  <m:mi>t</m:mi>
                  <m:mo>)</m:mo>
                </m:mrow>
              </m:mrow>
              <m:mo>)</m:mo>
            </m:mrow>
          </m:mrow>
        </m:semantics>
      </m:math>
    </p>
    <p>For our particular circuit, 
    <m:math>
      <m:semantics>
        <m:mrow>
          <m:msub>
            <m:mi>v</m:mi>
            <m:mo>&#8722;</m:mo>
          </m:msub>
        </m:mrow>
      </m:semantics>
    </m:math>
    is the same as 
    <m:math>
      <m:semantics>
        <m:mrow>
          <m:msub>
            <m:mi>v</m:mi>
            <m:mrow>
              <m:mi>o</m:mi>
              <m:mi>u</m:mi>
              <m:mi>t</m:mi>
            </m:mrow>
          </m:msub>
        </m:mrow>
      </m:semantics>
    </m:math>
    :</p>
    <p style='text-indent:.5in'>
      <m:math>
        <m:semantics>
          <m:mrow>
            <m:msub>
              <m:mi>v</m:mi>
              <m:mo>&#8722;</m:mo>
            </m:msub>
            <m:mtext>&#8201;</m:mtext>
            <m:mo>&#8801;</m:mo>
            <m:mtext>&#8201;</m:mtext>
            <m:msub>
              <m:mi>v</m:mi>
              <m:mrow>
                <m:mi>o</m:mi>
                <m:mi>u</m:mi>
                <m:mi>t</m:mi>
              </m:mrow>
            </m:msub>
          </m:mrow>
        </m:semantics>
      </m:math>
    </p>
    <p>Because the ideal operational amplifier draws no current
    through its input terminals, (and hence develops no voltage
    drop across the input resistor, 
    <m:math>
      <m:semantics>
        <m:mrow>
          <m:mi>R</m:mi>
        </m:mrow>
      </m:semantics>
    </m:math>
    ), then</p>
    <p style='text-indent:.5in'>
      <m:math>
        <m:semantics>
          <m:mrow>
            <m:msub>
              <m:mi>v</m:mi>
              <m:mo>+</m:mo>
            </m:msub>
            <m:mtext>&#8201;</m:mtext>
            <m:mo>&#8801;</m:mo>
            <m:mtext>&#8201;</m:mtext>
            <m:msub>
              <m:mi>v</m:mi>
              <m:mrow>
                <m:mi>i</m:mi>
                <m:mi>n</m:mi>
              </m:mrow>
            </m:msub>
          </m:mrow>
        </m:semantics>
      </m:math>
    </p>
    <p>so that</p>
    <p style='text-indent:.5in'>
      <m:math>
        <m:semantics>
          <m:mrow>
            <m:msub>
              <m:mi>v</m:mi>
              <m:mrow>
                <m:mi>o</m:mi>
                <m:mi>u</m:mi>
                <m:mi>t</m:mi>
              </m:mrow>
            </m:msub>
            <m:mrow>
              <m:mo>(</m:mo>
              <m:mi>t</m:mi>
              <m:mo>)</m:mo>
            </m:mrow>
            <m:mtext>&#8201;</m:mtext>
            <m:mtext>&#8201;</m:mtext>
            <m:mo>=</m:mo>
            <m:mtext>&#8201;</m:mtext>
            <m:mi>A</m:mi>
            <m:mrow>
              <m:mo>(</m:mo>
              <m:mrow>
                <m:msub>
                  <m:mi>v</m:mi>
                  <m:mrow>
                    <m:mi>i</m:mi>
                    <m:mi>n</m:mi>
                  </m:mrow>
                </m:msub>
                <m:mrow>
                  <m:mo>(</m:mo>
                  <m:mi>t</m:mi>
                  <m:mo>)</m:mo>
                </m:mrow>
                <m:mtext>&#8201;</m:mtext>
                <m:mo>&#8722;</m:mo>
                <m:mtext>&#8201;</m:mtext>
                <m:msub>
                  <m:mi>v</m:mi>
                  <m:mrow>
                    <m:mi>o</m:mi>
                    <m:mi>u</m:mi>
                    <m:mi>t</m:mi>
                  </m:mrow>
                </m:msub>
                <m:mrow>
                  <m:mo>(</m:mo>
                  <m:mi>t</m:mi>
                  <m:mo>)</m:mo>
                </m:mrow>
              </m:mrow>
              <m:mo>)</m:mo>
            </m:mrow>
          </m:mrow>
        </m:semantics>
      </m:math>
    </p>
    <p>or</p>
    <p style='text-indent:.5in'>
      <m:math>
        <m:semantics>
          <m:mrow>
            <m:mrow>
              <m:mo>(</m:mo>
              <m:mrow>
                <m:mn>1</m:mn>
                <m:mtext>&#8201;</m:mtext>
                <m:mo>+</m:mo>
                <m:mtext>&#8201;</m:mtext>
                <m:mi>A</m:mi>
              </m:mrow>
              <m:mo>)</m:mo>
            </m:mrow>
            <m:mtext>&#8201;</m:mtext>
            <m:msub>
              <m:mi>v</m:mi>
              <m:mrow>
                <m:mi>o</m:mi>
                <m:mi>u</m:mi>
                <m:mi>t</m:mi>
              </m:mrow>
            </m:msub>
            <m:mrow>
              <m:mo>(</m:mo>
              <m:mi>t</m:mi>
              <m:mo>)</m:mo>
            </m:mrow>
            <m:mtext>&#8201;</m:mtext>
            <m:mtext>&#8201;</m:mtext>
            <m:mo>=</m:mo>
            <m:mtext>&#8201;</m:mtext>
            <m:mi>A</m:mi>
            <m:mtext>&#8201;</m:mtext>
            <m:msub>
              <m:mi>v</m:mi>
              <m:mrow>
                <m:mi>i</m:mi>
                <m:mi>n</m:mi>
              </m:mrow>
            </m:msub>
            <m:mrow>
              <m:mo>(</m:mo>
              <m:mi>t</m:mi>
              <m:mo>)</m:mo>
            </m:mrow>
          </m:mrow>
        </m:semantics>
      </m:math>
    </p>
    <p>so that</p>
    <p style='text-indent:.5in'>
      <m:math>
        <m:semantics>
          <m:mrow>
            <m:mtext>&#8201;</m:mtext>
            <m:msub>
              <m:mi>v</m:mi>
              <m:mrow>
                <m:mi>o</m:mi>
                <m:mi>u</m:mi>
                <m:mi>t</m:mi>
              </m:mrow>
            </m:msub>
            <m:mrow>
              <m:mo>(</m:mo>
              <m:mi>t</m:mi>
              <m:mo>)</m:mo>
            </m:mrow>
            <m:mtext>&#8201;</m:mtext>
            <m:mtext>&#8201;</m:mtext>
            <m:mo>=</m:mo>
            <m:mtext>&#8201;</m:mtext>
            <m:mfrac>
              <m:mi>A</m:mi>
              <m:mrow>
                <m:mrow>
                  <m:mo>(</m:mo>
                  <m:mrow>
                    <m:mi>A</m:mi>
                    <m:mtext>&#8201;</m:mtext>
                    <m:mo>+</m:mo>
                    <m:mtext>&#8201;</m:mtext>
                    <m:mn>1</m:mn>
                  </m:mrow>
                  <m:mo>)</m:mo>
                </m:mrow>
              </m:mrow>
            </m:mfrac>
            <m:mtext>&#8201;</m:mtext>
            <m:msub>
              <m:mi>v</m:mi>
              <m:mrow>
                <m:mi>i</m:mi>
                <m:mi>n</m:mi>
              </m:mrow>
            </m:msub>
            <m:mrow>
              <m:mo>(</m:mo>
              <m:mi>t</m:mi>
              <m:mo>)</m:mo>
            </m:mrow>
          </m:mrow>
        </m:semantics>
      </m:math>
    </p>
    <p>As 
    <m:math>
      <m:semantics>
        <m:mrow>
          <m:mi>A</m:mi>
          <m:mtext>&#8201;</m:mtext>
          <m:mo>&#8594;</m:mo>
          <m:mtext>&#8201;</m:mtext>
          <m:mi>&#8734;</m:mi>
        </m:mrow>
      </m:semantics>
    </m:math>
    ,</p>
    <p style='text-indent:.5in'>
      <m:math>
        <m:semantics>
          <m:mrow>
            <m:mtext>&#8201;</m:mtext>
            <m:msub>
              <m:mi>v</m:mi>
              <m:mrow>
                <m:mi>o</m:mi>
                <m:mi>u</m:mi>
                <m:mi>t</m:mi>
              </m:mrow>
            </m:msub>
            <m:mrow>
              <m:mo>(</m:mo>
              <m:mi>t</m:mi>
              <m:mo>)</m:mo>
            </m:mrow>
            <m:mtext>&#8201;</m:mtext>
            <m:mtext>&#8201;</m:mtext>
            <m:mo>=</m:mo>
            <m:mtext>&#8201;</m:mtext>
            <m:mtext>&#8201;</m:mtext>
            <m:msub>
              <m:mi>v</m:mi>
              <m:mrow>
                <m:mi>i</m:mi>
                <m:mi>n</m:mi>
              </m:mrow>
            </m:msub>
            <m:mrow>
              <m:mo>(</m:mo>
              <m:mi>t</m:mi>
              <m:mo>)</m:mo>
            </m:mrow>
          </m:mrow>
        </m:semantics>
      </m:math>
    </p>
    <p>The fact that the output voltage follows the input voltage
    shows that our circuit is an example of a voltage follower.
    The main desirable feature of such circuits is that they draw
    little current at their inputs, but can supply hefty currents
    at their output terminals. If it weren't for this advantage,
    we could accomplish just as much with a piece of wire.</p>
    <p>As a second example, consider the following circuit:</p>
    <p style='text-indent:.5in'>
      <img src="./opamps_files/image003.gif" />
    </p>
    <p>In comparison with the first circuit, not that only a
    fraction of the output voltage is fed back into the negative
    input terminal of the operational amplifier in this circuit.
    Again, recall</p>
    <p style='text-indent:.5in'>
      <m:math>
        <m:semantics>
          <m:mrow>
            <m:msub>
              <m:mi>v</m:mi>
              <m:mrow>
                <m:mi>o</m:mi>
                <m:mi>u</m:mi>
                <m:mi>t</m:mi>
              </m:mrow>
            </m:msub>
            <m:mrow>
              <m:mo>(</m:mo>
              <m:mi>t</m:mi>
              <m:mo>)</m:mo>
            </m:mrow>
            <m:mtext>&#8201;</m:mtext>
            <m:mtext>&#8201;</m:mtext>
            <m:mo>=</m:mo>
            <m:mtext>&#8201;</m:mtext>
            <m:mi>A</m:mi>
            <m:mrow>
              <m:mo>(</m:mo>
              <m:mrow>
                <m:msub>
                  <m:mi>v</m:mi>
                  <m:mo>+</m:mo>
                </m:msub>
                <m:mrow>
                  <m:mo>(</m:mo>
                  <m:mi>t</m:mi>
                  <m:mo>)</m:mo>
                </m:mrow>
                <m:mtext>&#8201;</m:mtext>
                <m:mo>&#8722;</m:mo>
                <m:mtext>&#8201;</m:mtext>
                <m:msub>
                  <m:mi>v</m:mi>
                  <m:mo>&#8722;</m:mo>
                </m:msub>
                <m:mrow>
                  <m:mo>(</m:mo>
                  <m:mi>t</m:mi>
                  <m:mo>)</m:mo>
                </m:mrow>
              </m:mrow>
              <m:mo>)</m:mo>
            </m:mrow>
          </m:mrow>
        </m:semantics>
      </m:math>
    </p>
    <p>Now, as above,</p>
    <p style='text-indent:.5in'>
      <m:math>
        <m:semantics>
          <m:mrow>
            <m:msub>
              <m:mi>v</m:mi>
              <m:mo>+</m:mo>
            </m:msub>
            <m:mtext>&#8201;</m:mtext>
            <m:mo>&#8801;</m:mo>
            <m:mtext>&#8201;</m:mtext>
            <m:msub>
              <m:mi>v</m:mi>
              <m:mrow>
                <m:mi>i</m:mi>
                <m:mi>n</m:mi>
              </m:mrow>
            </m:msub>
          </m:mrow>
        </m:semantics>
      </m:math>
    </p>
    <p>Because the ideal operational amplifier draws no current
    at its input terminals, 
    <m:math>
      <m:semantics>
        <m:mrow>
          <m:msub>
            <m:mi>v</m:mi>
            <m:mrow>
              <m:mi>-</m:mi>
            </m:mrow>
          </m:msub>
          <m:mrow>
            <m:mo>(</m:mo>
            <m:mi>t</m:mi>
            <m:mo>)</m:mo>
          </m:mrow>
        </m:mrow>
      </m:semantics>
    </m:math>
    is simply a voltage-divided copy of 
    <m:math>
      <m:semantics>
        <m:mrow>
          <m:msub>
            <m:mi>v</m:mi>
            <m:mrow>
              <m:mi>o</m:mi>
              <m:mi>u</m:mi>
              <m:mi>t</m:mi>
            </m:mrow>
          </m:msub>
          <m:mrow>
            <m:mo>(</m:mo>
            <m:mi>t</m:mi>
            <m:mo>)</m:mo>
          </m:mrow>
        </m:mrow>
      </m:semantics>
    </m:math>
    :</p>
    <p style='text-indent:.5in'>
      <m:math>
        <m:semantics>
          <m:mrow>
            <m:msub>
              <m:mi>v</m:mi>
              <m:mo>&#8722;</m:mo>
            </m:msub>
            <m:mrow>
              <m:mo>(</m:mo>
              <m:mi>t</m:mi>
              <m:mo>)</m:mo>
            </m:mrow>
            <m:mtext>&#8201;</m:mtext>
            <m:mo>=</m:mo>
            <m:mtext>&#8201;</m:mtext>
            <m:mfrac>
              <m:mrow>
                <m:msub>
                  <m:mi>R</m:mi>
                  <m:mn>1</m:mn>
                </m:msub>
              </m:mrow>
              <m:mrow>
                <m:msub>
                  <m:mi>R</m:mi>
                  <m:mn>1</m:mn>
                </m:msub>
                <m:mtext>&#8201;</m:mtext>
                <m:mo>+</m:mo>
                <m:mtext>&#8201;</m:mtext>
                <m:msub>
                  <m:mi>R</m:mi>
                  <m:mn>2</m:mn>
                </m:msub>
              </m:mrow>
            </m:mfrac>
            <m:mtext>&#8201;</m:mtext>
            <m:msub>
              <m:mi>v</m:mi>
              <m:mrow>
                <m:mi>o</m:mi>
                <m:mi>u</m:mi>
                <m:mi>t</m:mi>
              </m:mrow>
            </m:msub>
            <m:mrow>
              <m:mo>(</m:mo>
              <m:mi>t</m:mi>
              <m:mo>)</m:mo>
            </m:mrow>
          </m:mrow>
        </m:semantics>
      </m:math>
    </p>
    <p>Thus,</p>
    <p style='text-indent:.5in'>
      <m:math>
        <m:semantics>
          <m:mrow>
            <m:msub>
              <m:mi>v</m:mi>
              <m:mrow>
                <m:mi>o</m:mi>
                <m:mi>u</m:mi>
                <m:mi>t</m:mi>
              </m:mrow>
            </m:msub>
            <m:mrow>
              <m:mo>(</m:mo>
              <m:mi>t</m:mi>
              <m:mo>)</m:mo>
            </m:mrow>
            <m:mtext>&#8201;</m:mtext>
            <m:mo>=</m:mo>
            <m:mtext>&#8201;</m:mtext>
            <m:mi>A</m:mi>
            <m:mtext>&#8201;</m:mtext>
            <m:mrow>
              <m:mo>(</m:mo>
              <m:mrow>
                <m:msub>
                  <m:mi>v</m:mi>
                  <m:mrow>
                    <m:mi>i</m:mi>
                    <m:mi>n</m:mi>
                  </m:mrow>
                </m:msub>
                <m:mrow>
                  <m:mo>(</m:mo>
                  <m:mi>t</m:mi>
                  <m:mo>)</m:mo>
                </m:mrow>
                <m:mtext>&#8201;</m:mtext>
                <m:mo>&#8722;</m:mo>
                <m:mtext>&#8201;</m:mtext>
                <m:mfrac>
                  <m:mrow>
                    <m:msub>
                      <m:mi>R</m:mi>
                      <m:mn>1</m:mn>
                    </m:msub>
                  </m:mrow>
                  <m:mrow>
                    <m:msub>
                      <m:mi>R</m:mi>
                      <m:mn>1</m:mn>
                    </m:msub>
                    <m:mtext>&#8201;</m:mtext>
                    <m:mo>+</m:mo>
                    <m:mtext>&#8201;</m:mtext>
                    <m:msub>
                      <m:mi>R</m:mi>
                      <m:mn>2</m:mn>
                    </m:msub>
                  </m:mrow>
                </m:mfrac>
                <m:mtext>&#8201;</m:mtext>
                <m:msub>
                  <m:mi>v</m:mi>
                  <m:mrow>
                    <m:mi>o</m:mi>
                    <m:mi>u</m:mi>
                    <m:mi>t</m:mi>
                  </m:mrow>
                </m:msub>
                <m:mrow>
                  <m:mo>(</m:mo>
                  <m:mi>t</m:mi>
                  <m:mo>)</m:mo>
                </m:mrow>
              </m:mrow>
              <m:mo>)</m:mo>
            </m:mrow>
          </m:mrow>
        </m:semantics>
      </m:math>
    </p>
    <p style='text-indent:.5in'>
      <m:math>
        <m:semantics>
          <m:mrow>
            <m:msub>
              <m:mi>v</m:mi>
              <m:mrow>
                <m:mi>o</m:mi>
                <m:mi>u</m:mi>
                <m:mi>t</m:mi>
              </m:mrow>
            </m:msub>
            <m:mrow>
              <m:mo>(</m:mo>
              <m:mi>t</m:mi>
              <m:mo>)</m:mo>
            </m:mrow>
            <m:mtext>&#8201;</m:mtext>
            <m:mrow>
              <m:mo>(</m:mo>
              <m:mrow>
                <m:mn>1</m:mn>
                <m:mtext>&#8201;</m:mtext>
                <m:mo>+</m:mo>
                <m:mtext>&#8201;</m:mtext>
                <m:mi>A</m:mi>
                <m:mtext>&#8201;</m:mtext>
                <m:mfrac>
                  <m:mrow>
                    <m:msub>
                      <m:mi>R</m:mi>
                      <m:mn>1</m:mn>
                    </m:msub>
                  </m:mrow>
                  <m:mrow>
                    <m:msub>
                      <m:mi>R</m:mi>
                      <m:mn>1</m:mn>
                    </m:msub>
                    <m:mtext>&#8201;</m:mtext>
                    <m:mo>+</m:mo>
                    <m:mtext>&#8201;</m:mtext>
                    <m:msub>
                      <m:mi>R</m:mi>
                      <m:mn>2</m:mn>
                    </m:msub>
                  </m:mrow>
                </m:mfrac>
                <m:mtext>&#8201;</m:mtext>
                <m:msub>
                  <m:mi>v</m:mi>
                  <m:mrow>
                    <m:mi>o</m:mi>
                    <m:mi>u</m:mi>
                    <m:mi>t</m:mi>
                  </m:mrow>
                </m:msub>
                <m:mrow>
                  <m:mo>(</m:mo>
                  <m:mi>t</m:mi>
                  <m:mo>)</m:mo>
                </m:mrow>
              </m:mrow>
              <m:mo>)</m:mo>
            </m:mrow>
            <m:mtext>&#8201;</m:mtext>
            <m:mo>=</m:mo>
            <m:mtext>&#8201;</m:mtext>
            <m:mi>A</m:mi>
            <m:mtext>&#8201;</m:mtext>
            <m:msub>
              <m:mi>v</m:mi>
              <m:mrow>
                <m:mi>i</m:mi>
                <m:mi>n</m:mi>
              </m:mrow>
            </m:msub>
            <m:mrow>
              <m:mo>(</m:mo>
              <m:mi>t</m:mi>
              <m:mo>)</m:mo>
            </m:mrow>
          </m:mrow>
        </m:semantics>
      </m:math>
    </p>
    <p style='text-indent:.5in'>
      <m:math>
        <m:semantics>
          <m:mrow>
            <m:msub>
              <m:mi>v</m:mi>
              <m:mrow>
                <m:mi>o</m:mi>
                <m:mi>u</m:mi>
                <m:mi>t</m:mi>
              </m:mrow>
            </m:msub>
            <m:mrow>
              <m:mo>(</m:mo>
              <m:mi>t</m:mi>
              <m:mo>)</m:mo>
            </m:mrow>
            <m:mtext>&#8201;</m:mtext>
            <m:mo>=</m:mo>
            <m:mtext>&#8201;</m:mtext>
            <m:mfrac>
              <m:mi>A</m:mi>
              <m:mrow>
                <m:mn>1</m:mn>
                <m:mtext>&#8201;</m:mtext>
                <m:mo>+</m:mo>
                <m:mtext>&#8201;</m:mtext>
                <m:mi>A</m:mi>
                <m:mtext>&#8201;</m:mtext>
                <m:mfrac>
                  <m:mrow>
                    <m:msub>
                      <m:mi>R</m:mi>
                      <m:mn>1</m:mn>
                    </m:msub>
                  </m:mrow>
                  <m:mrow>
                    <m:msub>
                      <m:mi>R</m:mi>
                      <m:mn>1</m:mn>
                    </m:msub>
                    <m:mtext>&#8201;</m:mtext>
                    <m:mo>+</m:mo>
                    <m:mtext>&#8201;</m:mtext>
                    <m:msub>
                      <m:mi>R</m:mi>
                      <m:mn>2</m:mn>
                    </m:msub>
                  </m:mrow>
                </m:mfrac>
              </m:mrow>
            </m:mfrac>
            <m:mtext>&#8201;</m:mtext>
            <m:msub>
              <m:mi>v</m:mi>
              <m:mrow>
                <m:mi>i</m:mi>
                <m:mi>n</m:mi>
              </m:mrow>
            </m:msub>
            <m:mrow>
              <m:mo>(</m:mo>
              <m:mi>t</m:mi>
              <m:mo>)</m:mo>
            </m:mrow>
          </m:mrow>
        </m:semantics>
      </m:math>
    </p>
    <p style='text-indent:.5in'>
      <m:math>
        <m:semantics>
          <m:mrow>
            <m:msub>
              <m:mi>v</m:mi>
              <m:mrow>
                <m:mi>o</m:mi>
                <m:mi>u</m:mi>
                <m:mi>t</m:mi>
              </m:mrow>
            </m:msub>
            <m:mrow>
              <m:mo>(</m:mo>
              <m:mi>t</m:mi>
              <m:mo>)</m:mo>
            </m:mrow>
            <m:mtext>&#8201;</m:mtext>
            <m:mtext>&#8201;</m:mtext>
            <m:mo>=</m:mo>
            <m:mtext>&#8201;</m:mtext>
            <m:mfrac>
              <m:mrow>
                <m:mn>1</m:mn>
                <m:mtext>&#8201;</m:mtext>
              </m:mrow>
              <m:mrow>
                <m:mtext>&#8201;</m:mtext>
                <m:mfrac>
                  <m:mrow>
                    <m:msub>
                      <m:mi>R</m:mi>
                      <m:mn>1</m:mn>
                    </m:msub>
                  </m:mrow>
                  <m:mrow>
                    <m:msub>
                      <m:mi>R</m:mi>
                      <m:mn>1</m:mn>
                    </m:msub>
                    <m:mtext>&#8201;</m:mtext>
                    <m:mo>+</m:mo>
                    <m:mtext>&#8201;</m:mtext>
                    <m:msub>
                      <m:mi>R</m:mi>
                      <m:mn>2</m:mn>
                    </m:msub>
                  </m:mrow>
                </m:mfrac>
                <m:mtext>&#8201;</m:mtext>
              </m:mrow>
            </m:mfrac>
            <m:msub>
              <m:mi>v</m:mi>
              <m:mrow>
                <m:mi>i</m:mi>
                <m:mi>n</m:mi>
              </m:mrow>
            </m:msub>
            <m:mrow>
              <m:mo>(</m:mo>
              <m:mi>t</m:mi>
              <m:mo>)</m:mo>
            </m:mrow>
          </m:mrow>
        </m:semantics>
      </m:math>
    </p>
    <p style='text-indent:.5in'>
      <m:math>
        <m:semantics>
          <m:mrow>
            <m:msub>
              <m:mi>v</m:mi>
              <m:mrow>
                <m:mi>o</m:mi>
                <m:mi>u</m:mi>
                <m:mi>t</m:mi>
              </m:mrow>
            </m:msub>
            <m:mrow>
              <m:mo>(</m:mo>
              <m:mi>t</m:mi>
              <m:mo>)</m:mo>
            </m:mrow>
            <m:mtext>&#8201;</m:mtext>
            <m:mtext>&#8201;</m:mtext>
            <m:mo>=</m:mo>
            <m:mtext>&#8201;</m:mtext>
            <m:mrow>
              <m:mo>(</m:mo>
              <m:mrow>
                <m:mn>1</m:mn>
                <m:mtext>&#8201;</m:mtext>
                <m:mo>+</m:mo>
                <m:mtext>&#8201;</m:mtext>
                <m:mfrac>
                  <m:mrow>
                    <m:msub>
                      <m:mi>R</m:mi>
                      <m:mn>2</m:mn>
                    </m:msub>
                  </m:mrow>
                  <m:mrow>
                    <m:msub>
                      <m:mi>R</m:mi>
                      <m:mn>1</m:mn>
                    </m:msub>
                  </m:mrow>
                </m:mfrac>
              </m:mrow>
              <m:mo>)</m:mo>
            </m:mrow>
            <m:mtext>&#8201;</m:mtext>
            <m:msub>
              <m:mi>v</m:mi>
              <m:mrow>
                <m:mi>i</m:mi>
                <m:mi>n</m:mi>
              </m:mrow>
            </m:msub>
            <m:mrow>
              <m:mo>(</m:mo>
              <m:mi>t</m:mi>
              <m:mo>)</m:mo>
            </m:mrow>
          </m:mrow>
        </m:semantics>
      </m:math>
    </p>
    <p>Because the input and output voltages have the same sign
    (move up and down together), this amplifier is called a 
    <i>non-inverting amplifier</i>
    . The voltage gain, 
    <m:math>
      <m:semantics>
        <m:mrow>
          <m:mi>G</m:mi>
        </m:mrow>
      </m:semantics>
    </m:math>
    , of this amplifier is:</p>
    <p style='text-indent:.5in'>
      <m:math>
        <m:semantics>
          <m:mrow>
            <m:mi>G</m:mi>
            <m:mtext>&#8201;</m:mtext>
            <m:mo>=</m:mo>
            <m:mtext>&#8201;</m:mtext>
            <m:mn>1</m:mn>
            <m:mtext>&#8201;</m:mtext>
            <m:mo>+</m:mo>
            <m:mtext>&#8201;</m:mtext>
            <m:mfrac>
              <m:mrow>
                <m:msub>
                  <m:mi>R</m:mi>
                  <m:mn>2</m:mn>
                </m:msub>
              </m:mrow>
              <m:mrow>
                <m:msub>
                  <m:mi>R</m:mi>
                  <m:mn>1</m:mn>
                </m:msub>
              </m:mrow>
            </m:mfrac>
          </m:mrow>
        </m:semantics>
      </m:math>
    </p>
    <p>Note that we can use an operational amplifier to realize
    an amplifier of any gain we wish simply by adjusting the
    ratio of a couple of resistors external to the 
    <i>IC</i>
    operational amplifier chip. The significance of this result
    is as an example of how an operational amplifier chip, which
    includes the overwhelming majority of the circuit components
    (approaching, perhaps, 
    <i>100</i>
    ) in the total circuit, can be mass produced at a very low
    cost and then customized for a specific purpose (realizing an
    amplifier with a particular gain, in this case) by the
    addition of a few external components. When the application
    of operational amplifiers was limited (in so-called analog
    computers) mainly to synthesizing integrators, their cost was
    several hundred dollars, each. Now, because we realize their
    versatility and employ them in many diverse applications,
    demand is large enough to drive the price for some common
    types below one dollar.</p>
    <p>Operational amplifiers are, indeed, wonderful devices. We
    should note, however, that available operational amplifiers
    handle mainly low power, audio frequency, signals. At high
    frequencies and high powers, amplifiers that use several
    individual transistors reign.</p>
    <p>As our third example, consider the following circuit:</p>
    <p style='text-indent:.5in'>
      <img src="./opamps_files/image004.gif" />
    </p>
    <p>As 
    <m:math>
      <m:semantics>
        <m:mrow>
          <m:mi>A</m:mi>
          <m:mtext>&#8201;</m:mtext>
          <m:mo>&#8594;</m:mo>
          <m:mtext>&#8201;</m:mtext>
          <m:mi>&#8734;</m:mi>
        </m:mrow>
      </m:semantics>
    </m:math>
    , analysis shows that</p>
    <p style='text-indent:.5in'>
      <m:math>
        <m:semantics>
        <m:mrow>
          <m:msub>
            <m:mi>v</m:mi>
            <m:mrow>
              <m:mi>o</m:mi>
              <m:mi>u</m:mi>
              <m:mi>t</m:mi>
            </m:mrow>
          </m:msub>
          <m:mtext>&#8201;</m:mtext>
          <m:mo>=</m:mo>
          <m:mtext>&#8201;</m:mtext>
          <m:mo>&#8722;</m:mo>
          <m:mtext>&#8201;</m:mtext>
          <m:mfrac>
            <m:mrow>
              <m:msub>
                <m:mi>R</m:mi>
                <m:mi>F</m:mi>
              </m:msub>
            </m:mrow>
            <m:mrow>
              <m:mi>R</m:mi>
              <m:mtext>&#8201;</m:mtext>
              <m:mo>+</m:mo>
              <m:mtext>&#8201;</m:mtext>
              <m:msub>
                <m:mi>R</m:mi>
                <m:mn>1</m:mn>
              </m:msub>
            </m:mrow>
          </m:mfrac>
          <m:mtext>&#8201;</m:mtext>
          <m:msub>
            <m:mi>v</m:mi>
            <m:mrow>
              <m:mi>i</m:mi>
              <m:mi>n</m:mi>
            </m:mrow>
          </m:msub>
        </m:mrow>
        F</m:semantics>
      </m:math>
    </p>
    <p>Because the output and input voltages for this amplifier
    have opposite signs, this amplifier is called an 
    <i>inverting</i>
    amplifier. The gain, 
    <m:math>
      <m:semantics>
        <m:mrow>
          <m:mi>G</m:mi>
        </m:mrow>
      </m:semantics>
    </m:math>
    , of the circuit is</p>
    <p style='text-indent:.5in'>
      <m:math>
        <m:semantics>
          <m:mrow>
            <m:mi>G</m:mi>
            <m:mtext>&#8201;</m:mtext>
            <m:mo>=</m:mo>
            <m:mtext>&#8201;</m:mtext>
            <m:mo>&#8722;</m:mo>
            <m:mtext>&#8201;</m:mtext>
            <m:mfrac>
              <m:mrow>
                <m:msub>
                  <m:mi>R</m:mi>
                  <m:mi>F</m:mi>
                </m:msub>
              </m:mrow>
              <m:mrow>
                <m:mi>R</m:mi>
                <m:mtext>&#8201;</m:mtext>
                <m:mo>+</m:mo>
                <m:mtext>&#8201;</m:mtext>
                <m:msub>
                  <m:mi>R</m:mi>
                  <m:mn>1</m:mn>
                </m:msub>
              </m:mrow>
            </m:mfrac>
          </m:mrow>
        </m:semantics>
      </m:math>
    </p>
    <p>Once again, we can realize any reasonable gain by choosing
    suitable values for the resistors in the circuit.</p>
    <p>As our fourth example, consider:</p>
    <p style='text-indent:.5in'>
      <img src="./opamps_files/image005.gif" />
    </p>
    <p>This circuit is an inverting amplifier with two input
    voltages. As 
    <m:math>
      <m:semantics>
        <m:mrow>
          <m:mi>A</m:mi>
          <m:mtext>&#8201;</m:mtext>
          <m:mo>&#8594;</m:mo>
          <m:mtext>&#8201;</m:mtext>
          <m:mi>&#8734;</m:mi>
        </m:mrow>
      </m:semantics>
    </m:math>
    , analysis shows</p>
    <p style='text-indent:.5in'>
      <m:math>
        <m:semantics>
          <m:mrow>
            <m:msub>
              <m:mi>v</m:mi>
              <m:mrow>
                <m:mi>o</m:mi>
                <m:mi>u</m:mi>
                <m:mi>t</m:mi>
              </m:mrow>
            </m:msub>
            <m:mtext>&#8201;</m:mtext>
            <m:mo>=</m:mo>
            <m:mtext>&#8201;</m:mtext>
            <m:mo>&#8722;</m:mo>
            <m:mtext>&#8201;</m:mtext>
            <m:mrow>
              <m:mo>(</m:mo>
              <m:mrow>
                <m:mfrac>
                  <m:mrow>
                    <m:msub>
                      <m:mi>R</m:mi>
                      <m:mi>F</m:mi>
                    </m:msub>
                  </m:mrow>
                  <m:mrow>
                    <m:msub>
                      <m:mi>R</m:mi>
                      <m:mn>1</m:mn>
                    </m:msub>
                  </m:mrow>
                </m:mfrac>
                <m:mtext>&#8201;</m:mtext>
                <m:msub>
                  <m:mi>v</m:mi>
                  <m:mn>1</m:mn>
                </m:msub>
                <m:mtext>&#8201;</m:mtext>
                <m:mo>+</m:mo>
                <m:mtext>&#8201;</m:mtext>
                <m:mfrac>
                  <m:mrow>
                    <m:msub>
                      <m:mi>R</m:mi>
                      <m:mi>F</m:mi>
                    </m:msub>
                  </m:mrow>
                  <m:mrow>
                    <m:msub>
                      <m:mi>R</m:mi>
                      <m:mn>2</m:mn>
                    </m:msub>
                  </m:mrow>
                </m:mfrac>
                <m:mtext>&#8201;</m:mtext>
                <m:msub>
                  <m:mi>v</m:mi>
                  <m:mn>2</m:mn>
                </m:msub>
              </m:mrow>
              <m:mo>)</m:mo>
            </m:mrow>
          </m:mrow>
        </m:semantics>
      </m:math>
    </p>
    <p>This configuration, therefore, is an inverting 
    <i>summing</i>
    amplifier: its output is a linear combination of the two
    inputs, 
    <m:math>
      <m:semantics>
        <m:mrow>
          <m:msub>
            <m:mi>v</m:mi>
            <m:mrow>
              <m:mi>1</m:mi>
            </m:mrow>
          </m:msub>
        </m:mrow>
      </m:semantics>
    </m:math>
    and 
    <m:math>
      <m:semantics>
        <m:mrow>
          <m:msub>
            <m:mi>v</m:mi>
            <m:mrow>
              <m:mi>2</m:mi>
            </m:mrow>
          </m:msub>
        </m:mrow>
      </m:semantics>
    </m:math>
    . (Summing amplifiers are not easily made with the
    non-inverting amplifier configuration. The "virtual ground"
    at the "minus" terminal of the inverting configuration turns
    out to be essential for the summing process.) These results
    are easily generalized to include inverting amplifiers with
    three or more inputs. An interesting application of such a
    multiple-input summing amplifier is in mixing audio signals
    from microphones used by, for example, a mixed group of
    vocalists and instrumentalists. The individual voices or
    instruments can be emphasized or de-emphasized by adjusting
    the individual resistors, 
    <m:math>
      <m:semantics>
        <m:mrow>
          <m:msub>
            <m:mi>R</m:mi>
            <m:mrow>
              <m:mi>1</m:mi>
            </m:mrow>
          </m:msub>
        </m:mrow>
      </m:semantics>
    </m:math>
    and 
    <m:math>
      <m:semantics>
        <m:mrow>
          <m:msub>
            <m:mi>R</m:mi>
            <m:mrow>
              <m:mi>2</m:mi>
            </m:mrow>
          </m:msub>
        </m:mrow>
      </m:semantics>
    </m:math>
    , In practical audio mixers, these resistors take the form of
    "pots" (potentiometers), resistors whose value can be varied
    by turning their knobs. Once the individual resistors have
    been adjusted to give the appropriate audio mix, the
    resistor, 
    <m:math>
      <m:semantics>
        <m:mrow>
          <m:msub>
            <m:mi>R</m:mi>
            <m:mrow>
              <m:mi>F</m:mi>
            </m:mrow>
          </m:msub>
        </m:mrow>
      </m:semantics>
    </m:math>
    , serves as a master control to adjust the overall sound
    level.</p>
    <p>As our final example, consider the following operational
    amplifier circuit:</p>
    <p style='text-indent:.5in'>
      <img src="./opamps_files/image006.gif" />
    </p>
    <p>If the capacitor is initially discharged so that 
    <m:math>
      <m:semantics>
        <m:mrow>
          <m:msub>
            <m:mi>v</m:mi>
            <m:mrow>
              <m:mi>o</m:mi>
              <m:mi>u</m:mi>
              <m:mi>t</m:mi>
            </m:mrow>
          </m:msub>
          <m:mrow>
            <m:mo>(</m:mo>
            <m:mn>0</m:mn>
            <m:mo>)</m:mo>
          </m:mrow>
          <m:mtext>&#8201;</m:mtext>
          <m:mo>=</m:mo>
          <m:mtext>&#8201;</m:mtext>
          <m:mn>0</m:mn>
        </m:mrow>
      </m:semantics>
    </m:math>
    , then analysis shows that</p>
    <p style='text-indent:.5in'>
      <m:math>
        <m:semantics>
          <m:mrow>
            <m:msub>
              <m:mi>v</m:mi>
              <m:mrow>
                <m:mi>o</m:mi>
                <m:mi>u</m:mi>
                <m:mi>t</m:mi>
              </m:mrow>
            </m:msub>
            <m:mrow>
              <m:mo>(</m:mo>
              <m:mi>t</m:mi>
              <m:mo>)</m:mo>
            </m:mrow>
            <m:mtext>&#8201;</m:mtext>
            <m:mo>=</m:mo>
            <m:mtext>&#8201;</m:mtext>
            <m:mo>&#8722;</m:mo>
            <m:mtext>&#8201;</m:mtext>
            <m:mfrac>
              <m:mn>1</m:mn>
              <m:mrow>
                <m:mrow>
                  <m:mo>(</m:mo>
                  <m:mrow>
                    <m:mi>R</m:mi>
                    <m:mtext>&#8201;</m:mtext>
                    <m:mo>+</m:mo>
                    <m:mtext>&#8201;</m:mtext>
                    <m:msub>
                      <m:mi>R</m:mi>
                      <m:mn>1</m:mn>
                    </m:msub>
                  </m:mrow>
                  <m:mo>)</m:mo>
                </m:mrow>
                <m:mtext>&#8201;</m:mtext>
                <m:mi>C</m:mi>
              </m:mrow>
            </m:mfrac>
            <m:mtext>&#8201;</m:mtext>
            <m:mrow>
              <m:msubsup>
                <m:mo>&#8747;</m:mo>
                <m:mrow>
                  <m:mtext>&#8201;</m:mtext>
                  <m:mn>0</m:mn>
                </m:mrow>
                <m:mrow>
                  <m:mtext>&#8201;</m:mtext>
                  <m:mi>t</m:mi>
                </m:mrow>
              </m:msubsup>
              <m:mrow>
                <m:msub>
                  <m:mi>v</m:mi>
                  <m:mrow>
                    <m:mi>i</m:mi>
                    <m:mi>n</m:mi>
                  </m:mrow>
                </m:msub>
              </m:mrow>
            </m:mrow>
            <m:mrow>
              <m:mo>(</m:mo>
              <m:mi>&#964;</m:mi>
              <m:mo>)</m:mo>
            </m:mrow>
            <m:mtext>&#8201;</m:mtext>
            <m:mi>d</m:mi>
            <m:mi>&#964;</m:mi>
          </m:mrow>
        </m:semantics>
      </m:math>
    </p>
    <p>For this circuit, the output voltage clearly is
    proportional to the integral of the input voltage. As a
    consequence, this circuit is called an 
    <i>integrator</i>
    . Such circuits formed the basis of analog computers. During
    the 1940's and 1950's, such computers were engineers' main
    tools for solving the coupled sets of differential equations
    that described complicated engineering systems. Ultimately,
    digital computers replaced analog computers because digital
    computers offer much more accuracy and flexibility. During
    World War II, however, all of the newly sophisticated control
    systems for guns, radar and aircraft were designed with
    analog computers.</p>
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