Determine the Length of One Oscillation

Hi everyone,

My Z-transform function is H(z) = (-3*z^2 + 4*z)/(8*z^3 -14*z^2 + 8*z -2). After plotted its response frequency, how do I determine the length of one oscillation? Any your help or hint is really appreciated!

Answer 1

[Walter Roberson]

That transfer function does not oscillate. It has an infinite gain over time, so if you feed in even a single non-zero value and the rest 0, it will reach an arbitrarily large output with the same sign() as the single input.

The inverse laplacian gives

(1/4)*exp(t) - (5/8)*cos((1/8)*sqrt(7)*t)*exp((3/8)*t) - (11/56)*sqrt(7)*exp((3/8)*t)*sin((1/8)*sqrt(7)*t)

and although the sin() of sqrt(7)/8 * t implies there is a layer of oscillation of period 8*2*Pi/sqrt(7), that is overwhelmed by the exp(t)/r

[Ben Le]

Thank you so much Walter! there was no oscillate at all in my transfer function unless take inverse laplacian to get the period of the oscillation. Thanks!

[Walter Roberson]

My calculation finds no oscillation even in the laplacian. The differentiation is everywhere positive -- it is just how positive it is that oscillates a bit.

Answer 2

[Paul]

I think this problem is supposed to be attacked like this.

Define H(z)

syms 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.

c = children(H(z));
c1(z) = c{1};
c2(z) = c{2};
[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

a = sym(1)/sym(2);
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

syms k1 k2
[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

syms n ingteger
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.

nval = 0:10;
figure
stem(nval,h(nval))
hold on
stem(nval,subs(iztrans(H(z)),n,nval),'x')

figure
stem(nval,h(nval)-subs(iztrans(H(z)),n,nval))