Question 3 of 5: Causal System Function — Step Response, Inverse Filtering and Sinusoidal Steady State

Question 3: Causal System Function — Step Response, Inverse Filtering and Sinusoidal Steady State (25 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A causal LTI system with $H(z) = (1-z^{-1})/[(1-0.6z^{-1})(1+0.6z^{-1})] = (1-z^{-1})/(1-0.36z^{-2})$. In positive powers, $H(z) = z(z-1)/[(z-0.6)(z+0.6)]$: zeros at $z = 0$ and $z = 1$, poles at $z = \pm 0.6$. Causality fixes the region of convergence at $|z| \gt 0.6$, which contains the unit circle, so the system is also stable and a frequency response exists.

Find. (a) the step response $y[n]$; (b) the input that produces the first-difference output $\delta[n]-\delta[n-1]$; (c) the steady-state response to $\cos(\pi n/3)$ applied for all $n$.

Approach. In (a) multiply by the step transform and watch the zero at z = 1 cancel the step pole, then invert by partial fractions; in (b) invert the system algebraically, which is legal here because the inverse is FIR; in (c) evaluate H on the unit circle at the input frequency and use the real-sinusoid eigenfunction result.

1. Form the output transform for a step input. With $X(z) = 1/(1-z^{-1})$ for $|z| \gt 1$, $$Y(z) = H(z)X(z) = \frac{1-z^{-1}}{(1-0.6z^{-1})(1+0.6z^{-1})}\cdot \frac{1}{1-z^{-1}} = \frac{1}{1-0.36z^{-2}},\qquad |z| \gt 0.6 .$$ The differencing zero at $z=1$ cancels the step pole exactly — the signature of a highpass system, whose step response must decay to zero rather than settle on a constant.
2. Expand in partial fractions and invert. Writing $\dfrac{1}{(1-0.6z^{-1})(1+0.6z^{-1})} = \dfrac{A}{1-0.6z^{-1}} + \dfrac{B}{1+0.6z^{-1}}$ and evaluating the residues at $z^{-1} = 1/0.6$ and $z^{-1} = -1/0.6$ gives $A = B = \tfrac12$. Each right-sided term inverts to a decaying exponential, so $$\boxed{\,y[n] = \tfrac12\left[(0.6)^{n} + (-0.6)^{n}\right]u[n]\,}$$ which can equally be written as $y[n] = (0.6)^{n}$ for even $n \ge 0$ and $y[n] = 0$ for odd $n$.
3. Sanity-check the first few samples. Long division of $1/(1-0.36z^{-2})$ gives $1 + 0.36z^{-2} + 0.1296z^{-4}+\dots$, i.e. $y[0]=1$, $y[1]=0$, $y[2]=0.36$, $y[3]=0$, $y[4]=0.1296$, $y[6]=0.046656$; the same values follow from the difference equation $y[n] = x[n]-x[n-1]+0.36\,y[n-2]$ driven by a step. The alternating zeros are the visible consequence of the two poles being equal and opposite.

Part (b) reverses the roles of input and output. Because the required output is itself the first difference, the algebra is short.

4. Invert the system function. The requirement $Y(z) = H(z)X(z)$ with $Y(z) = 1 - z^{-1}$ gives $$X(z) = \frac{Y(z)}{H(z)} = \left(1-z^{-1}\right)\cdot\frac{(1-0.6z^{-1})(1+0.6z^{-1})}{1-z^{-1}} = 1 - 0.36z^{-2}.$$ The common factor $1-z^{-1}$ cancels, leaving a two-term polynomial in $z^{-1}$, hence $$\boxed{\,x[n] = \delta[n] - 0.36\,\delta[n-2]\,}$$ a finite-length, causal input.
5. Note why the answer is unique and legitimate. $X(z)$ is a polynomial in $z^{-1}$, so its region of convergence is the whole plane except $z=0$ and the inverse transform is unique; no ROC ambiguity arises even though $H(z)$ has a zero on the unit circle. Direct substitution confirms it: driving $y[n] = x[n]-x[n-1]+0.36y[n-2]$ with $x = \{1, 0, -0.36\}$ yields $y[0]=1$, $y[1]=-1$ and $y[n]=0$ thereafter.

Part (c) applies a sinusoid that has been present since n = −∞, so only the steady-state (frequency-response) term survives — there is no transient to compute.

6. Evaluate the frequency response at $\omega = \pi/3$. Setting $z = e^{j\pi/3}$, the numerator is $1 - e^{-j\pi/3} = 1\,\angle\,60^\circ$ and the denominator is $1 - 0.36e^{-j2\pi/3} = 1.18 + j0.3118 = 1.2205\,\angle\,14.80^\circ$. Dividing, $$H(e^{j\pi/3}) = \frac{1\angle 60^\circ}{1.2205\angle 14.80^\circ} = 0.8193\,\angle\,45.20^\circ .$$
7. Apply the real-sinusoid steady-state rule. For a real, stable LTI system driven by $x[n]=\cos(\omega_0 n)$ for all $n$, $y[n] = |H(e^{j\omega_0})|\cos\bigl(\omega_0 n + \angle H(e^{j\omega_0})\bigr)$. Hence $$\boxed{\,y[n] = 0.8193\,\cos\!\left(\frac{\pi}{3}n + 0.7889\ \text{rad}\right) = 0.8193\,\cos\!\left(\frac{\pi}{3}n + 45.20^\circ\right)\,}$$ for all $n$. Iterating the difference equation from rest for many samples reproduces exactly this amplitude and phase once the transient has decayed, which is a useful independent check.