z-transform of cosh(at) sin(bt)

Take sampling period T. Write cosh at · sin bt = ½ (e^(at) + e^(−at)) sin bt.

Use the standard result Z[e^(−αt) sin bt] = z e^(−αT) sin bT / (z² − 2z e^(−αT) cos bT + e^(−2αT)).

Put α = −a and α = +a:

Z[cosh at sin bt] = ½ { z e^(aT) sin bT / (z² − 2z e^(aT) cos bT + e^(2aT)) + z e^(−aT) sin bT / (z² − 2z e^(−aT) cos bT + e^(−2aT)) }

Combined form. Let p = e^(aT), so 1/p = e^(−aT), c = cos bT, s = sin bT. Adding over a common denominator (using p · 1/p = 1):

• Numerator: p(z² − 2zc/p + 1/p²) + (1/p)(z² − 2zcp + p²) = (p + 1/p)(z² + 1) − 4cz = 2 cosh aT (z² + 1) − 4cz
• Denominator: z⁴ − 4c cosh aT · z³ + (2 cosh 2aT + 4c²) z² − 4c cosh aT · z + 1

Answer:

Z[cosh at sin bt] = z sin bT [ cosh aT (z² + 1) − 2z cos bT ] / [ z⁴ − 4 cos bT cosh aT z³ + (2 cosh 2aT + 4 cos² bT) z² − 4 cos bT cosh aT z + 1 ]

valid for |z| > e^(aT).