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Figure 14-1 rise). The opposite is true when MR < MC. Thus total pro ts are maximized where MR = MC. The pro t-maximizing or best level of output for this monopolist can also be seen in Figure 14-1 (obtained by plotting the value of columns 1, 2, 4, 6, and 7 of Table 14.1). In this gure, the best level of output is at the point where MR = MC. At this best output level of 2.5 units, the monopolist makes a pro t of $1.50 per unit (the vertical distance between D and AC at 2.5 units of output) and $3.75 in total (2.5 units times the $1.50 pro t per unit). Note that since P > MR where MR = MC, the rising portion of the MC curve above the AVC does not represent the monopolist supply curve. In the long run, the monopolist can adjust the scale of plant, and pro ts may persist because of blocked or restricted entry.

To test for invertibility, we may show that a system is invertible by designing an inverse system that uniquely recovers the input from the output, or we may show that a system is not invertible by finding two different inputs that produce the same output. Each system defined above will be tested for invertibility using one of these two methods.

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( a ) This system is clearly invertible because, given the output y ( n ) , we may recover the input using x ( n ) = 0 . 5 y ( n ) . ( h ) This system is not invertible, because the value of x ( n ) at 11 = 0 cannot be recovered from y ( n ) . For example, the response of the system to X ( R )and to x l ( n ) = x ( n ) a & n ) will be the same for any a .

(c) Due to the differencing between two successive input values, this system will not be invertible. For example, note that the inputs x ( n ) and x ( n ) c will produce the same output for any value of c.

Summary

(6) This system corresponds to an integrator and is an invertible system. To show that i t is invertible, we may construct the inverse system, which is .u(n) = y ( n ) - y ( n - I) To show that this is the inverse system, note that

(e) Invertibility must hold for complex as well as real-valued signals. Therefore, this system is noninvertible because

0 ' x(n). 1

(a) If both SI and S2 are linear, shift-invariant, stable, and causal, will the cascade also be linear,

( a ) Linearity, shift-invariance, stability, and causality are easily shown to be preserved in a cascade. For example, the response of S I to the input nxl ( n ) h x z ( n ) will be a w l ( n ) b w 2 ( n )due to the linearity of S,. With this as the input to S2, the response will be, again by linearity, a y , ( n ) hy7(n). Therefore, if both S I and S2 are linear, the cascade will be linear.

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