I think this problem is supposed to be attacked like this.
Define H(z)
H(z) = (-3*z^2 + 4*z)/(8*z^3 - 14*z^2 + 8*z - 2);
Expand in partial fractions
H(z) = partfrac(H(z))
H(z) =
The first term a delayed unit step and doesn't oscilate. The second term has complex poles and is therefore sinusoidal. Rewrite the second term with leading coefficient of 1 in the denominator.
[num,den] = numden(c2)
num(z) = 
den(z) = 
num = num/16
num(z) =
den = den/16
den(z) =
The denominator can be written in terms of a and w0
w0 = acos(sym(3)/sym(4));
z^2 + 2*a*z*cos(w0) + a^2
ans =
At this point, I suspect the point of the exercise was to find w0 and then determine the length of one period. However, a discrete-time sinusoid with frequency w0 is not periodic, so determining the length of one period is not really feasible.
If we want to recover the full tim domain signal, proceed as follows.
Multiply the numerator by z (this will be important later)
znum = z*num
znum(z) =
Define two constants k1 and k2 and solve for them by comparing coefficients of the z-transform of a^n*(k1*cos(w0*n) + k2*sin(w0*n))*u(n) with the coefficients of znum
[k1,k2] = solve(coeffs(z*(k1*(z - a*cos(w0)) + k2*a*sin(w0)),z,'all') == coeffs(znum,z,'all'))
k1 =
k2 =
Define the discrete-time unit step function
u(n) = kroneckerDelta(n)/2 + heaviside(n);
The inverse z-transform of z*c2(z) is then
zc2(n) = a^n*(k1*cos(w0*n) + k2*sin(w0*n))*u(n);
Therefore the inverse z-transform of H(z) is (recalling that zc2 included a multiply-by-z, so we have to take that back out with a delay)
h(n) = iztrans(c1) + zc2(n-1)
h(n) =
Compare to the iztrans of the original H(z), which has a more complicated expression
iztrans(H(z))
ans =
Plot the two and verify they are equivalent.
stem(nval,subs(iztrans(H(z)),n,nval),'x')
stem(nval,h(nval)-subs(iztrans(H(z)),n,nval))
