My partner walked off with the sheet so I can't provide you with the table but I have the 6 supposed equations. I need to build a 3-bit up-counter with JK flip-flops. Following are the equations:
J(Q1) = ~Q1 * Q0
K(Q1) = Q1 * Q0
J(Q2) = Q2 * Q0
K(Q2) = Q1 * Q0
J(Q0) = ~Q0
K(Q0) = Q0
I'm at a loss of how to even begin constructing this circuit. Any guidance would be appreciated. I'm using Logisim to construct the circuit. In theory these equations are all I should need to construct this circuit... which is why I'm asking this question here.
in Logisim it's easy,
Goto Window->Combinational Analysis
Then choose the Expressions tab.
Enter your equations one per line (Or use the other tools (Truth Tables/Maps)
Make sure you have your inputs and outputs set up before you start entering equations.
Then click on Build circuit and follow the instructions.
Related
I am currently doing a course called "Modeling of dynamic systems" and have been given the task of modeling a warm water tank in modelica with a distributed temperature description.
Most of the tasks have gone well, and my group is left with the task of introducing the heat flux due to buoyancy effects into the model. Here is where we get stuck.
the equation given is this:
Given PDE
But how do we discretize this into something we can use in modelica?
The discretized version we ended up with was this:
(Qd_pp_b[k+1] - Qd_pp_b[k]) / h_dz = -K_b *(T[k+1] - 2 * T[k] + T[k-1]) / h_dz^2
where Qd_pp_b is the left-hand side variable, ie the heat flux, k is the current slice of the tank and T is the temperature in the slices.
Are we on the right path? or completely wrong?
This doesn't seem to be a differential equation (as is) so this does not make sense without surrounding problem. For the second derivative you should always create auxiliary variables and for each partial derivative a separate equation. I added dummy values for parameters and dummy equations for T[k]. This can be simulated, is this about what you expected?
model test
constant Integer n = 10;
Real[n] Qd_pp_b;
Real[n] dT;
Real[n] T;
parameter Real K_b = 1;
equation
for k in 1:n loop
der(Qd_pp_b[k]) = -K_b *der(dT[k]);
der(T[k]) = dT[k];
T[k] = sin(time+k);
end for;
end test;
I am dealing with a problem which is a variant of a subset-sum problem, and I am hoping that the additional constraint could make it easier to solve than the classical subset-sum problem. I have searched for a problem with this constraint but I have been unable to find a good example with an appropriate algorithm either on StackOverflow or through googling elsewhere.
The problem:
Assume you have two lists of positive numbers A1,A2,A3... and B1,B2,B3... with the same number of elements N. There are two sums Sa and Sb. The problem is to find the simultaneous set Q where |sum (A{Q}) - Sa| <= epsilon and |sum (B{Q}) - Sb| <= epsilon. So, if Q is {1, 5, 7} then A1 + A5 + A7 - Sa <= epsilon and B1 + B5 + B7 - Sb <= epsilon. Epsilon is an arbitrarily small positive constant.
Now, I could solve this as two completely separate subset sum problems, but removing the simultaneity constraint results in the possibility of erroneous solutions (where Qa != Qb). I also suspect that the additional constraint should make this problem easier than the two NP complete problems. I would like to solve an instance with 18+ elements in both lists of numbers, and most subset-sum algorithms have a long run time with this number of elements. I have investigated the pseudo-polynomial run time dynamic programming algorithm, but this has the problems that a) the speed relies on a short bit-depth of the list of numbers (which does not necessarily apply to my instance) and b) it does not take into account the simultaneity constraint.
Any advice on how to use the simultaneity constraint to reduce the run time? Is there a dynamic programming approach I could use to take into account this constraint?
If I understand your description of the problem correctly (I'm confused about why you have the distance symbols around "sum (A{Q}) - Sa" and "sum (B{Q}) - Sb", it doesn't seem to fit the rest of the explanation), then it is in NP.
You can see this by making a reduction from Subset sum (SUB) to Simultaneous subset sum (SIMSUB).
If you have a SUB problem consisting of a set X = {x1,x2,...,xn} and a target called t and you have an algorithm that solves SIMSUB when given two sets A = {a1,a2,...,an} and B = {b1,b2,...,bn}, two intergers Sa and Sb and a value for epsilon then we can solve SUB like this:
Let A = X and let B be a set of length n consisting of only 0's. Set Sa = t, Sb = 0 and epsilon = 0. You can now run the SIMSUB algorithm on this problem and get the solution to your SUB problem.
This shows that SUBSIM is as least as hard as SUB and therefore in NP.
Consider the following module:
module power(input [11-1:0] xi,xq,output [22-1:0] y);
assign y = xi*xi + xq*xq;
endmodule
I know that my single assignment is actually decomposed of 3 steps: 2 squares and one addition. My question is how would the synthesizer decides on the bitwidth of the intermediate steps xi*xi and xq*xq?
I noticed that when running logic equivelance circuit (lec) for the above code, it causes trouble and could only be solved by decomposing the single assignment into three assignments as follows:
module power(input [11-1:0] xi,xq,output [22-1:0] yy);
wire [21-1:0] pi,pq;
assign pi = xi*xi;
assign pq = xq*xq;
assign yy = pi+pq;
endmodule
Here's how your simulator decides on bitwdith for intermediate results.
Verilog Simulation
This expression - assign y = xi*xi + xq*xq; - is an example of a context determined expression. A Verilog simulator takes the widest of all the nets or variables in the expression and uses that. So, in your code, the widest is y at 22 bits wide, so Verilog will use 22 bits throughout.
VHDL Simulation
The behaviour of a VHDL simulator depends on the package used. If you use the numeric_std package, as is recommended, then you would need to obey the following rules:
The width of the sum should be the same as the wider of the two operands.
The width of the product should be the sum of the widths of the operands.
Therefore, your code would compile if translated directly into VHDL:
library IEEE;
use IEEE.std_logic_1164.all;
use IEEE.numeric_std.all;
entity power is
port (xi, xq : in signed(11-1 downto 0);
y : out signed(22-1 downto 0));
end entity power;
architecture A of power is
begin
y <= xi*xi + xq*xq;
end architecture A;
Shouldn't everything be signed?
Given the names of your module (power) and inputs (xi and xq) and having spent 25 years designing radio systems, shouldn't they be signed? Shouldn't your Verilog be:
module power(input signed [11-1:0] xi,xq,output signed [22-1:0] y);
assign y = xi*xi + xq*xq;
endmodule
That is why I chose the signed type from numeric_std, not the unsigned type.
Synthesis
Well, I've waffled on about simulators, but you asked about synthesis. And, to be frank, I don't know what a synthesiser would do. But, given the job of a synthesiser is to design a logic circuit that behaves exactly like the simulation, you would think that any self-respecting synthesiser would use the same bit-widths as the simulator. So, I'm pretty sure that's your answer.
I am reading the book: introduction to the theory of computation and got stuck on this example.
Convert a DFA to an equivalent expression by converting it first to a GNFA(generalized nondeterministic finite automaton) and then convert GNFA to a regular expression.
here is the example:
enter image description here
I should use this recursively to arrive at the the fourth state:
enter image description here
Unfortunately, I cannot understand what is going on from b to c? I only understand that we are trying to get rid of state 2, but how we arrive at c from b?
Thank you very much!
This can be quite tricky at first but I suggest you check definition 1.64 and see the function CONVERT(G) for more clearance. But as a brief explanation using the function for each possible neighbour state:
First from a to b, add a start state and a new accept state;
Afterwards you need to calculate each new path after qrip is removed, in this case state 1;
So, from start to q2, you get only label a from epsilon and a;
Same goes from start to q3, resulting only in b;
Now from q2 to q2 going trough qrip, you have label a to qrip and label a to get back, so you get (aa U b);
Same goes to q3 to q3 through qrip, so resulting in bb, notice that there is no loop in q3 so no union;
Now from q2 to q3 through qrip, you only need to concatenate a and b resulting in ab label;
Lastly the other way around, from q3 to q2 going through qrip, concatenate b and a resulting in ba but this time making the union with the previous path between q3 and q2;
Now choose a new qrip and proceed to do the same algorithm again.
Hope the explanation was clear enough, but as said before refer to the algorithm in the book for a better and more detailed explanation.
The two popular methods for converting a given DFA to its regular expression are-
Arden’s Method
State Elimination Method
Arden’s Theorem states that:
Let P and Q be two regular expressions over ∑.
To use Arden’s Theorem, the following conditions must be satisfied-
The transition diagram must not have any ∈ transitions.
There must be only a single initial state.
Step-01:
Form an equation for each state considering the transitions which come towards that state.
Add ‘∈’ in the equation of the initial state.
Step-02:
Bring the final state in the form R = Q + RP to get the required regular expression.
If P does not contain a null string ∈, then-
R = Q + RP has a unique solution i.e. R = QP*
I am new to Modelica, and I am wondering if it is possible to write a kind of dynamic programming equation. Assume time is discretized by an integer i, and in my specific application x is boolean and f is a boolean function of x.
x(t_i) = f(x(t_{i+d}))
Where d can be a positive or negative integer. Of course, I would initialize x accordingly, either true or false.
Any help or references would be greatly appreciated!
It is possible. In Modelica the discretization in time is usually carried on by the compiler, you have to take care of the equations (continous dynamics). Otherwise, if you want to generate events at discrete time points, you can do it using when statements.
I suggest you to take a look at Introduction to Object-Oriented Modeling and Simulation with OpenModelica (PDF format, 6.6 MB) - a more recent tutorial (2012) by Peter Fritzson. There is a section that on Discrete Events and Hybrid Systems, that should clarify how to implement your equations in Modelica.
Below you can find an example from that tutorial about the model of a bouncing ball, as you can see discretization in time is not considered when you write your dynamic equations. So the continous model of the ball v=der(s), a=der(v) and than the discrete part inside the when clause that handles the contact with the ground:
model BouncingBall "the bouncing ball model"
parameter Real g=9.81; //gravitational acc.
parameter Real c=0.90; //elasticity constant
Real height(start=10),velocity(start=0);
equation
der(height) = velocity;
der(velocity)=-g;
when height<0 then
reinit(velocity, -c*velocity);
end when;
end BouncingBall;
Hope this helps,
Marco
If I understand your question, you want to use the last n evaluations of x to determine the next value of x. If so, this code shows how to do this:
model BooleanHistory
parameter Integer n=10 "How many points to keep";
parameter Modelica.SIunits.Time dt=1e-3;
protected
Boolean x[n];
function f
input Integer n;
input Boolean past[n-1];
output Boolean next;
algorithm
next :=not past[1]; // Example
end f;
initial equation
x = {false for i in 1:n};
equation
when sample(0,dt) then
x[2:n] = pre(x[1:(n-1)]);
x[1] = f(n, x[2:n]);
end when;
end BooleanHistory;