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I have a filter with Transfer function $H(z)=(1-0.5z^{-1})(1+0.5z^{-1})$ designed for a sampling rate of 800 samples/s.

How to find peak output if a sine of 200Hz and amplitude 4 is applied as input?

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up vote 1 down vote accepted

Your input signal is

$$x(n)=A\sin n\theta_0$$

with $A=4$ and $\theta_0=2\pi f/f_s=2\pi\cdot 200/800=\pi/2$

The frequency response of your system is $H(z)$ evaluated on the unit circle, i.e. for $z=e^{j\theta}$:

$$H(e^{j\theta})=|H(e^{j\theta})|e^{j\phi(\theta)}$$ where $\phi(\theta)$ is the phase response of $H(e^{j\theta})$.

The output signal $y(n)$ is then given by

$$y(n)=|H(e^{j\theta_0})|A\sin (n\theta_0+\phi(\theta_0)\})$$

So the peak amplitude of the output signal is $|H(e^{j\theta_0})|A$.

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