The progress I made is that I used a XOR gate with the complement of R, to be fed as input to each of the first 4 inputs of the decrementer. It doesn't seem to work for all input combinations. I would appreciate some suggestions on modification, as this doesn't lead to correct outputs. enter image description here I tried to find out the logical expressions of the incrementer and decrementer individually to link them together using the control input R. Please have a look at the attached figure.

Kindly refer to the attached figure. The expressions that the output needs to assume in each case (R=0 and R=1) have been derived already by me. I can't find a way to link them together.

  • \$\begingroup\$ I tried to upload an image, but it says that the image size is too big. And yeah, I'm new in this portal. \$\endgroup\$ – Curiosity Sep 9 '17 at 20:47
  • \$\begingroup\$ I use Irfanview to shrink image OR file sizes with compression \$\endgroup\$ – Tony Stewart EE75 Sep 9 '17 at 20:52
  • \$\begingroup\$ I uploaded the image now. Used some online compression tools. Have a look, if you have time. \$\endgroup\$ – Curiosity Sep 9 '17 at 20:54
  • \$\begingroup\$ We can guess but we need a bit more of a definition of the block that you have called a 5 bit decrementer. \$\endgroup\$ – RoyC Sep 10 '17 at 8:54
  • \$\begingroup\$ If the input is X, the output of the decrementer is X-1. That's about it. The internal design can't be modified in this question. \$\endgroup\$ – Curiosity Sep 11 '17 at 8:22

A down counter just inverts all the ouputs.

  • \$\begingroup\$ It doesn't do that always. Suppose you have 11101 as input. The output of the downcounter will be 11100. So, all inputs aren't inverted. \$\endgroup\$ – Curiosity Sep 9 '17 at 20:59
  • \$\begingroup\$ I said invert the outputs. With a Reset or parallel load of 00000 the output will be 11111 and count down. \$\endgroup\$ – Tony Stewart EE75 Sep 9 '17 at 21:09
  • \$\begingroup\$ How'll that work? \$\endgroup\$ – Curiosity Sep 9 '17 at 21:10
  • \$\begingroup\$ Could you elaborate? Suppose the input is 11001. For R=0, we want output = 11010 and for R=1, we want 11000. That's the basic thing. I don't see the result coming from any simple inversion. \$\endgroup\$ – Curiosity Sep 9 '17 at 21:10
  • \$\begingroup\$ You don't have provision for that. You can't use any additional components to reset it. All you have is the decrementer and any number of XOR gates. \$\endgroup\$ – Curiosity Sep 9 '17 at 21:43

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