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Let x represent the airspeed of the plane, and let y represent the speed of the wind, both in kilometers per hour When the plane flies against the wind, the wind takes away from its groundspeed Therefore x y = 750 Let s put this equation into SI form Subtracting x from each side gives us y = x + 750 Multiplying through by 1 tells us that y = x 750 When the plane flies with the wind, the wind adds to its groundspeed That means x + y = 990 To morph this into SI form, we can subtract x from each side, getting y = x + 990 Now let s mix the right sides of these two SI equations: x 750 = x + 990 When we add 750 to each side, we obtain x = x + 1,740 We can add x to each side to get 2x = 1,740 Finally we divide through by 2, discovering that x = 870 Now we know that the airspeed of the plane is 870 km/h Let s plug this value into one of the SI equations We can use the first one, getting y = x 750 = 870 750 = 120 This tells us that the wind is blowing at 120 km/h with respect to the earth

In pulse amplitude modulation (PAM), the strength of each individual pulse varies according to the modulating waveform In this respect, PAM is very much like ordinary amplitude modulation An amplitude-versus-time graph of a hypothetical PAM signal is shown in Fig 26-12A

First, we morph In both of these equations, a multiple of y can be separated out and placed alone on the left side of the equals sign, producing quadratic functions of x In the first equation, we can subtract 2y from each side and then transpose the sides to get

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Theorem (logic equation) X+0 = X X*1 = X X+1 = 1 X*0 = 0 X+X = X X*X = X ( X) = X X+( X) = 1 X*( X) = 0 X+Y = Y+X X*Y = Y*X X+(X*Y) = X X*( Y)+Y = X+Y X+Y+Z = (X+Y)+Z = X+(Y+Z) X*Y*Z = (X*Y)*Z = X*(Y*Z) X*(Y+Z) = (X*Y)+(X*Z) (X+Y) = ( X)*( Y) (X*Y) = ( X)+( Y) What it s called OR identity AND identity

2y = 4x 2 + 6x + 8 Dividing through by 2, we get the function y = 2x 2 3x 4 In the second original equation, we can subtract y from each side and then transpose the sides, getting y = 3x 2 + 5x 11 Multiplying through by 1, we get y = 3x 2 5x + 11

486 Data transmission Normally, the pulse amplitude increases as the instantaneous modulating-signal level increases But this can be reversed, so that higher audio levels cause the pulse amplitude to go down Then the signal pulses are at their strongest when there is no modulation Either the positive or negative PAM methods provide good results The transmitter works a little harder if the negative modulation method is used

Next, we mix When we directly mix the right sides of the above two quadratic functions, we get a single equation in one variable:

head pointing up ( ) or by an ampersand (&), and the OR operation is denoted by a detached arrowhead pointing down ( )

In PAM, the pulses all last for the same length of time The effective transmitted power is varied by changing the actual peak amplitude of the pulses Another way to change the transmitter output is to vary the duration, or width, of the pulses This scheme is called pulse duration modulation (PDM) or pulse width modulation (PWM), shown in Fig 26-12B Normally, the pulse duration increases as the instantaneous modulating-signal level increases But, as with PAM, this can be reversed The transmitter must work harder to accomplish negative PDM Regardless of whether positive or negative PDM is employed, the peak pulse amplitude remains constant

2x 2 3x 4 = 3x 2 5x + 11 We can add the quantity (3x 2 + 5x 11) to each side, obtaining x 2 + 2x 15 = 0 which is a quadratic equation in polynomial standard form

Next, we solve We now have an equation that can be easily factored It does not take long to figure out that the above quadratic is equivalent to

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