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Figure 13-4 Passive aircraft receiver schematic quick-board which contains ready-made circuit traces, these are available at Radio Shack stores You may want to lay out the circuit so that the tuning control is at one edge of the board, which will make things a little easier when installing the circuit in an enclosure Building the circuit is pretty straightforward and should only take a few hours to build Refer to Table 13-1 for identifying the resistors and Table 13-2 for identifying the small capacitors in the circuit Remember it is very important to install the diode correctly with respect to its polarity in order for the circuit to work properly The passive air-band receiver also contains a few electrolytic capacitors, and you will remember that these devices have polarity and it must be observed for the circuit to work The aircraft receiver uses a dual op-amp to amplify the signals and you will have to ensure that it is installed correctly An integrated circuit is highly recommended for the op-amp In the event of a circuit failure at some later date, it is much easier to simply unplug an IC rather than trying to unsolder one from a PC board to replace it with a new one When installing the IC into its socket, you will have to make sure that you orient it correctly An IC will typically have a notch or small cut-out at one end of the IC case Pin one (1) will be to the left of the cut-out of notch when viewed from the top of the IC In this project pin 1 of the IC connects to potentiometer R3, while pin 8 of the IC is connected to the power source Note that pin 4 of the IC is connected to the circuit ground You can elect to use a small speaker in this circuit, but in order to make a miniature receiver you may want to simply use a small 1 8 mini jack and just plug into a pair of small earphones The circuit is powered from a standard 9 volt battery; remember the black or minus ( ) battery clip is connected to ground, while the plus (+) or red battery clip lead is connected to capacitor C5 at the plus (+) end You will want to have the tuning capacitor C2 near the edge of the circuit board, so that you can allow the tuning knob to protrude through the case so you can easily tune the receiver Once the receiver has been completed, you should take a few minutes rest After the break, you should inspect the foil side of the circuit board for any cold solder joints Cold solder joints often look dark, dull or blobby If you see any solder joints that look this way, you should resolder the connections You should also take a moment to examine the circuit board for any short-circuits which may be the result of a stray component lead shorting between circuit traces You can make yourself an antenna from a piece of stiff 20 ga wire as a whip antenna to capacitor C1, or install a miniature jack in order to plug in a.

1 This function is stated in binomial factor form To figure out whether its graph opens upward or downward, we must morph it into polynomial standard form by multiplying out the factors: y = (x 3)(4x 1) = 4x 2 + ( x) + ( 12x) + 3 = 4x 2 13x + 3 The coefficient of x 2 is positive That means the parabola opens upward 2 The real zeros of the quadratic function stated in Prob 1 can be found from the factors in the original version We must solve these two first-degree equations: x 3=0 and 4x 1 = 0 In the top equation, we can add 3 to each side, getting x = 3 In the bottom equation, we can add 1 to each side and then divide through by 4, obtaining x = 1/4 The real zeros of the quadratic function are therefore x = 3 or x = 1/4

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It is straightforward to transform this equation to any other frame-ofreference For frame a, = = where Ha = He Re and a Ra = e Re a 01 3 03 1 1 He Re Ra xe + a e a a H x +

3 Because the parabola opens upward, we know that its vertex is an absolute minimum To find the x-coordinate of this point, we average the two zeros: xmin = (3 + 1/4)/2 = (12/4 + 1/4)/2 = (13/4)/2 = 13/8 We can calculate the y-coordinate of this point by plugging in 13/8 for x in the polynomial form of the function: ymin = 4xmin2 13xmin + 3 = 4 (13/8)2 13 13/8 + 3 = 4 169/64 169/8 + 3 = 676/64 1,352/64 + 192/64 = (676 1,352 + 192) / 64 = 484/64 = 121/16 The coordinates of the absolute minimum are (13/8, 121/16) 4 We know the two points where the curve crosses the x axis (representing the zeros of the function) The left-hand x-intercept point is (1/4, 0) The right-hand x-intercept point is (3, 0) We also know that the absolute minimum point is (13/8, 121/16) The graph, based on these three known points, is shown in Fig C-2 On both axes, each increment represents 1 unit 5 Remember the polynomial standard form for a quadratic function of a variable x, where we have coefficients a and b, and a stand-alone constant c If we let the dependent variable be called y, then y = ax 2 + bx + c The function we re interested in is y = 7x 2 + 5x + 2 In the polynomial, we have a > 0, so we know that the graph of the function is a parabola that opens upward When we examine the discriminant d, we find that d = b 2 4ac = 52 4 7 2 = 25 56 = 31 The fact that d is negative tells us that the quadratic equation 7x 2 + 5x + 2 = 0

24 y

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