حساب z-transformوتمثيل النتائج باستخدام ezpiotوتنظيم الرسوماتفي مصفوفه فرعيه باستخدام subplot

syms n z a b

% 1. Define the time-domain functions
f1 = heaviside(n);                    % u(n)
f2 = a^n * heaviside(n);              % a^n * u(n)
f3 = cos(a*n) * heaviside(n);         % cos(an) * u(n)
f4 = sin(a*n) * heaviside(n);         % sin(an) * u(n)
f5 = b^n * sin(a*n) * heaviside(n);   % b^n * sin(an) * u(n)

% 2. Compute the Z-transforms
F1_Z = ztrans(f1, n, z);
F2_Z = ztrans(f2, n, z);
F3_Z = ztrans(f3, n, z);
F4_Z = ztrans(f4, n, z);
F5_Z = ztrans(f5, n, z);

% 3. Display Z-transforms in readable format
disp('F1_Z ='); pretty(F1_Z)

F1_Z =
  1     1
----- + -
z - 1   2

disp('F2_Z ='); pretty(F2_Z)

F2_Z =
1     a
- - -----
2   a - z

disp('F3_Z ='); pretty(F3_Z)

F3_Z =
ztrans(cos(a (n + 1)), n, z)   1
---------------------------- + -
              z                2

disp('F4_Z ='); pretty(F4_Z)

F4_Z =
ztrans(sin(a (n + 1)), n, z)
----------------------------
              z

disp('F5_Z ='); pretty(F5_Z)

F5_Z =
        /                    z \
b ztrans| sin(a (n + 1)), n, - |
        \                    b /
--------------------------------
                z

% 4. Plot using ezplot (with given a and b values)
a_val = 0.5;
b_val = 0.1;

subplot(3,2,1)
ezplot(subs(F1_Z), [0 10])
title('F1\_Z = ZT{u(n)}')

subplot(3,2,2)
ezplot(subs(F2_Z, a, a_val), [0 10])
title('F2\_Z = ZT{a^n u(n)}')

subplot(3,2,3)
ezplot(subs(F3_Z, a, a_val), [0 10])
title('F3\_Z = ZT{cos(an) u(n)}')

subplot(3,2,4)
ezplot(subs(F4_Z, a, a_val), [0 10])
title('F4\_Z = ZT{sin(an) u(n)}')

subplot(3,2,5)
ezplot(subs(F5_Z, [a b], [a_val b_val]), [0 10])
title('F5\_Z = ZT{b^n sin(an) u(n)}')

[Walter Roberson]

Approximate translation:

Calculate z-transform, plot results using ezpiot, and organize graphs into a subarray using subplot.

[Les Beckham]

It looks like OP has already done just that.

@Shahed: Do you have a question?

[Walter Roberson]

I notice that the ztrans for F3, F4, and F5 all use the indeterminate form ztrans() instead of particular calculations. @Shahed might be asking about that part, perhaps.

[Paul]

I will speculate that all of the results are unexpected because the default for heaviside(0) = 1/2, whereas u(0) = 1.

ztrans computes the unilateral z-transform, so there is no need to explicitly multiply by u(n) for any of these cases.

[Walter Roberson]

syms n z a b
f3 = cos(a*n) * heaviside(n)

f3 = 

sympref('HeavisideAtOrigin', 0);
F3_0 = ztrans(f3, n, z);
sympref('HeavisideAtOrigin', 1/2);
F3_12 = ztrans(f3, n, z);
sympref('HeavisideAtOrigin', 1);
F3_1 = ztrans(f3, n, z);

[F3_0; F3_12; F3_1]

ans = 

so the results are all the same, except with the HeavisideAtOrigin added as a constant.

[Walter Roberson]

You get different results if you assume n > 0... which should cause the heaviside() call to return 1

syms n z a b
assume(n>0)
f3 = cos(a*n) * heaviside(n)

f3 = 

sympref('HeavisideAtOrigin', 0);
F3_0 = ztrans(f3, n, z);
sympref('HeavisideAtOrigin', 1/2);
F3_12 = ztrans(f3, n, z);
sympref('HeavisideAtOrigin', 1);
F3_1 = ztrans(f3, n, z);

[F3_0; F3_12; F3_1]

ans = 

[Paul]

Unclear to me how, or even if, ztrans is accounting for that assumption on n. That is, if n must be positive, then I don't know how the z-transform of f(n) = cos(a*n) is defined insofar as the z-transform sum starts at n = 0.

I would proceed with one of the following options. Maybe ztrans should do better for the latter two.

syms a n z
f(n) = cos(a*n);
ztrans(f(n))

ans = 

sympref('default');
u(n) = heaviside(n) + kroneckerDelta(n)/2;
f(n) = cos(a*n)*u(n);
ztrans(f(n))

ans = 

simplify(ans)

ans = 

sympref('HeavisideAtOrigin',1);
f(n) = cos(a*n)*heaviside(n);
ztrans(f(n))

ans = 

simplify(ans)

ans = 

However, I don't understand this:

ztrans(f(n))   % (1)

ans = 

As we saw above, that ans can be simplified. But it doesn't simplify after subtracting 1?

simplify(ans-1)  % (2)

ans = 

How can (1) simplify but (2) does not?

Answer 1

[Paul]

It does simplify after taking enough steps

sympref('HeavisideAtOrigin',1);
syms a
syms n integer
f(n) = cos(a*n)*heaviside(n);
ztrans(f(n))   % (1)

ans = 

simplify(ans-1,200)

ans = 

[num,den] = numden(ans);
num/den

ans =