Inverse z of z/(z²+7z+10)

Short answer

Factor denominator and use partial fractions: X(z)= (1/3)[z/(z+2)-z/(z+5)]. Inverse gives x(k)=1/3[(-2)^k-(-5)^k] for k≥0.

Find the inverse z-transform of x(z) = z/(z² + 7z + 10).

Factorise. z² + 7z + 10 = (z + 2)(z + 5).

X(z)/z = 1/[(z + 2)(z + 5)] = A/(z + 2) + B/(z + 5)

A = 1/(−2 + 5) = 1/3
B = 1/(−5 + 2) = −1/3

X(z) = (1/3) [ z/(z + 2) − z/(z + 5) ]

Using Z⁻¹[z/(z − a)] = aᵏ with a = −2 and a = −5:

Answer: x(k) = ⅓ [ (−2)ᵏ − (−5)ᵏ ], k ≥ 0.

Check: x(0) = 0, x(1) = 1, x(2) = −7, x(3) = 39. These also satisfy x(k) = −7x(k−1) − 10x(k−2), which comes from the denominator.

✓ Answer checked by Claude Opus 5.5 (high effort)

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