# Determine the Length of One Oscillation

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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!

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### Accepted Answer

Walter Roberson
on 4 Feb 2017

Edited: Walter Roberson
on 5 Feb 2017

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

##### 2 Comments

Walter Roberson
on 5 Feb 2017

### More Answers (1)

Paul
on 21 Feb 2023

Edited: Paul
on 21 Feb 2023

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))

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 = num/16

den = den/16

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

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

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'))

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)

Compare to the iztrans of the original H(z), which has a more complicated expression

iztrans(H(z))

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))

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