CPRE281_HW04_solutions

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Cpr E 281 HW04 ELECTRICAL AND COMPUTER ENGINEERING IOWA STATE UNIVERSITY HW4: Sample Solutions P1 (10 points): Given the following Verilog code: Fill in the following K-map: f ab 00 01 11 10 c 0 1 P2 (10 points): Write a Verilog module for a 4-to-1 multiplexer. module P1(f, a, b, c); output f; input a, b, c; reg f; always@(*) if (c==0) f=(~a)&(~b); else f=a&b; endmodule 1 0 0 0 0 0 1 0
Cpr E 281 HW04 ELECTRICAL AND COMPUTER ENGINEERING IOWA STATE UNIVERSITY HW4: Sample Solutions There are many solutions to this problem. module P2(w0, w1, w2, w3, s1, s0, f); input w0, w1, w2, w3, s1, s0; output f; reg f; always@(w0 or w1 or w2 or w3 or s1 or s0) if (s1==0) if (s0==0) f = w0; else f = w1; else if (s0==0) f = w2; else f = w3; endmodule
P3 (20 points): a. Prove Theorem17a of Boolean Algebra ( xy + yz +´ x z = xy +´ x z ) using two K-maps. For instance, show that the K-map for the left-hand side is equivalent to the K- map for the right-hand side. 𝑥? + ?? + 𝑥̅? 00 01 11 10 0 1 𝑥? + 𝑥̅? 00 01 11 10 0 1 b. Prove Theorem17b ( ´ x + z x + y ¿ ( y + z )(´ x + z )=( x + y ) ¿ ) with truth tables. That is, show that the truth table for the left-hand side is the same as the truth table for the right-hand side. x y z x+y y+z ´ x ´ x + z ´ x + z ( x + y ) ¿ ) ( x + y ¿ ( y + z )(´ x + z ) 0 0 0 0 0 1 1 0 0 0 0 1 0 1 1 1 0 0 0 1 0 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 0 0 1 0 0 0 0 0 1 0 1 1 1 0 1 1 1 1 1 0 1 1 0 0 0 0 1 1 1 1 1 0 1 1 1 xy z xy z 0 0 1 0 1 1 1 0 0 0 1 0 1 1 1 0
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P4 (15 points): Produce the simplified SOP expressions for the following K- maps: F 1 F 2 F 3 C AB 00 01 11 10 WX YZ 00 01 11 10 WX YZ 00 01 11 10 0 00 00 1 0 0 1 1 01 01 0 0 0 0 11 11 0 1 1 0 10 10 1 1 1 1 0 1 1 0 1 1 1 1 0 1 0 0 0 1 1 0 0 1 1 0 0 1 1 0
P5 (15 points): Produce the simplified POS expressions for the following K- maps: F 1 0 1 00 01 11 10 0 1 0 0 1 0 0 1 F 2 00 01 11 10 00 01 11 10 1 0 0 1 0 0 0 0 0 0 0 0 1 0 0 1 F 3 00 01 11 10 00 01 11 10 AB C WX YZ WX YZ
1 0 0 1 1 0 0 1 1 1 1 1 1 0 0 1
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P6 (10 points) Use Karnaugh Maps to convert the following expressions to simplified SOP expressions : a. ? 1 (𝐴, 𝐵, 𝐶, ?) = 𝐴𝐵𝐶 ̅ + 𝐴𝐵? + 𝐴 𝐶 ? ̅ + 𝐴𝐵 ̅ 𝐶 ̅ ? b. ? 2 (𝐴, 𝐵, 𝐶, ?) = ∑𝑚(1,5,6,7) Q 1 AB 00 01 11 10 CD 00 01 11 10 Q 2 AB 00 01 11 10 CD 00 0 0 0 0 01 11 10 1 0 0 1 1 1 0 0 0 0 0 0 0 0 1 0 0 0 1 1 0 0 1 0 0 0 1 1
P7 (10 points) Use Karnaugh Maps to convert the following expressions to simplified POS expressions : a. ? 1 (𝐴, 𝐵, 𝐶, ?) = (𝐴 + 𝐵 ̅ + 𝐶)(𝐴 ̅ + 𝐵 ̅ + ?)(𝐴 + 𝐵 + ?)(𝐴 + 𝐵 + ? ̅ ) b. ? 2 (𝐴, 𝐵, 𝐶, ?) = ∏ 𝑀(2,4,7,11) Q 1 AB 00 01 11 10 CD 00 01 11 10 Q 2 AB 00 01 11 10 CD 00 0 1 0 0 01 0 0 0 0 11 0 1 0 1 10 1 0 0 0 0 0 1 0 1 0 1 0 1 0 1 1 0 0 1 1
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P8 (10 points): For each expression below, derive the simplest SOP expression using don’t care terms for simplification wherever possible : a. ? 1 (𝑎, 𝑏, ?) = ∑𝑚(0,3,5,7) + D(1,4) b. ? 2 (𝑎, 𝑏, ?, ?) = ∏ 𝑀(1,5,9,12,15) + D(3,4,7,11) H 1 ab 00 01 11 10 c 0 1 H 2 ab 00 01 11 10 cd 00 1 01 0 11 d 10 1 d 0 1 0 1 0 d 0 d 1 1 1 1 0 0 d d 1 1 1
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