Check range for power function

I am looking for a unit test vectors which checks the range of input values for power function so that I can implement equivalent of C type in fix point

A = fi([1:2^5/32:2^5],0,32); % A > 0, so it starting with lowest positive fraction value
B = fi([-2:1e-2:2],1,32);
% C = power(A,B)  % seeking help here for C type power implementation but
% gettting error
% % Error using fixed.internal.power.validateK (line 13)
% % Exponent input to 'power' must be a real scalar and the value must be a non-negative integer.
% % 
% % Error in fixed.internal.power.powImpl>validateInputs (line 39)
% % fixed.internal.power.validateK(k);
% % 
% % Error in fixed.internal.power.powImpl (line 9)
% % validateInputs(a, k);
% % 
% % Error in  .^  (line 32)
% % y = fixed.internal.power.powImpl(a, k);

figure(100);
A = [1:32];
B = [-2:2];
hold on; grid on;
plot(power(A(1),B(1:end)),'r-.'); 
plot(power(A(2),B(1:end)),'g-.'); 
plot(power(A(3),B(1:end)),'b-.'); 
plot(power(A(4),B(1:end)),'m-.'); 
%.
%.
%.
%.
plot(power(A(32),B(1:end)),'k-.','linewidth', 2.0);

Answer 1

[Walter Roberson]

When you use fi, the power must be a positive integer scalar.

Also, log(A) does not appear to be supported so you cannot use the usual exp(log(A) .* B)

You will need to implement your own fractional power, perhaps using a CORDIC method.

[Life is Wonderful]

Thanks @Walter Roberson, fi was used for ensuring Q format but this can be deleted. Perhaps I need clue for implementing own fractional power i.e. [value^(1/2) == sqrt(value)]

81(3/4) = sqrt(81)*power3 which is 27

y = power(sqrt(sqrt(81)),3)

y = 27

C = power(1,-2)

C = 1

For me it is not clear , if you can confirm cordiccexp(log(A) .* B) , you mean ?? I am missing what

A = fi([1:2^5/32:2^5],0,32); % A > 0, so it starting with lowest positive fraction value
B = fi([-2:1e-2:2],1,32);
y = cordiccexp(log(A) .* B);

Check for incorrect argument data type or missing argument in call to function 'log'.

[Life is Wonderful]

y = cordiccexp(log(1) .* -2);
y

y = 1.0000 - 0.0000i

y = power(2,cordiccexp(log(2) .* -2))

y = 0.8820 - 0.7153i

z = power(2 ,-2)

z = 0.2500

y = nthroot(1,-2);power(2,y) % IS THIS YOUR INTERPRETATION OWN FRACTIONAL POWER

ans = 2

[Life is Wonderful]

I could see following case for computing the power with base and exponent condition

A is integer and B is exponent

if A < 0 && B < 0
        C = power(A,B);
elseif A > 0 && B < 0
        C = power(A,B);
elseif A == 0 && B < 0
        C = power(A,B);
elseif A > 0 && B == 0
        C = power(A,B);
elseif A > 0 && B > 0 
        C = power(A,B);
end