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10.7 Ratios of lines and areas of regular polygons (a) In two regular polygons having the same number of sides, find the ratio of the lengths of the apothems if the perimeters are in the ratio 5:3. (b) In two regular polygons having the same number of sides, find the length of a side of the smaller if the lengths of the apothems are 20 and 50 and a side of the larger has length 32.5. (c) In two regular polygons having the same number of sides, find the ratio of the areas if the lengths of the sides are in the ratio 1:5. (d) In two regular polygons having the same number of sides, find the area of the smaller if the sides have lengths 4 and 12 and the area of the larger is 10,260.

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h,(n) = 6(n)

(a) By Principle 2, r : r (b) By Principle 2, s : s (c) By Principle 3, (d) By Principle 3, A Ar A Ar p: p 5 : 3. 20 : 50 and s 1 . 25 Q 4 2 R and A 12 1140. 13.

The instantaneous rate of change is evaluated by the derivative. It follows that, for to f (x), so that y f (x) x

) If ~ ( e j " is real and even, then X(ej'") sin(nw) is real and odd. Therefore, when integrated from -n t o n , the integral is zero. Thus, x ( n ) may be written as

A s 2 Q R . Then sr 10,260

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and it follows that x ( n ) is real. Finally, because X ( e j W ) cos(nw) is real and even, x ( n ) is real and even, that is,

C/d. Hence,

Prove the convolution theorem. There are several ways to prove the convolution theorem. One way is by a straightforward manipulation of the DTFT sum. Specifically, if y ( n ) = h ( n ) * .u(n),

Note that the expression in brackets is the DTFT of x(n - 1 ) . Using the delay property of the DTFT, this is equal to x(ejW)e-J'", and the right side of this equation becomes

(19.1)

Approximate values for p are 3.1416 or 3.14 or 7 . Unless you are told otherwise, we shall use 3.14 for p in solving problems. A circle may be regarded as a regular polygon having an infinite number of sides. If a square is inscribed in a circle, and the number of sides is continually doubled (to form an octagon, a 16-gon, and so on), the perimeters of the resulting polygons will get closer and closer to the circumference of the circle (Fig. 10-7).

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which proves the theorem. Another way to prove the convolution theorem is to consider the following cascade of two LSI systems, one with a unit sample response of h ( n ) and the other with a unit sample response of x ( n ) :

Fig. 10-7

If the input to this cascade is a complex exponential, ej"", the output of the first system is H ( e j w ) e j n wBecause this . complex exponential is the input to the second system, the output is H(ej")X(ej")eJn". Therefore, H ( e j " ) X ( e j w ) is the frequency response of the cascade, and because the unit sample response of the cascade is the convolution h ( n ) * x ( n ) , we have the DTFT pair h(n) * x(n) & T D H ( e j w ) x(dw) which establishes the convolution theorem.

1 2 (2pr) (r)

Derive the up-sampling property of the DTFT, which states that if x (ej") is the DTFT of x ( n ) , the DTFT of

19.1 The weekly pro t P, in dollars, of a corporation is determined by the number x of radios produced per week, according to the formula P = 75x 0.03x 2 15 000 (a) Find the rate at which the pro t is changing when the production level x is 1000 radios per week, (b) Find the change in weekly pro t when the production level x is increased to 1001 radios per week.

All circles are similar figures, since they have the same shape. Because they are similar figures, (1) corresponding segments of circles are in proportion and (2) the areas of two circles are to each other as the squares of their radii or circumferences.

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