Question 6 of 7: Impulse Response, Frequency Response and Sinusoidal Steady State

Question 6: Impulse Response, Frequency Response and Sinusoidal Steady State (20 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 first-order recursive system $y[n] = \tfrac{1}{2}y[n-1] + x[n] + \tfrac{1}{2}x[n-1]$, known to be causal and stable; in part (c) the input is the everlasting sinusoid $x[n] = \cos\!\left(\tfrac{\pi}{2}n + \tfrac{\pi}{4}\right)$.

Find. The closed-form impulse response, the frequency response function, and the steady-state output sinusoid — amplitude and phase — at $\omega_0 = \pi/2$.

Approach. Transform the difference equation to get $H(z)$, invert it after a long division (the transfer function is not strictly proper), evaluate on the unit circle, and finish with the eigenfunction property of LTI systems, which turns part (c) into a single complex evaluation.

1. (a) Transfer function, then impulse response. Transforming (7),
   $$H(z) = \frac{1 + \tfrac{1}{2}z^{-1}}{1 - \tfrac{1}{2}z^{-1}}, \qquad \text{ROC: } |z| \gt \tfrac{1}{2},$$

   with a pole at $z = \tfrac{1}{2}$ and a zero at $z = -\tfrac{1}{2}$; the pole is inside the unit circle, consistent with the stated stability. Numerator and denominator have the same degree, so a division is needed before partial fractions. Writing $1 + \tfrac{1}{2}z^{-1} = -\left(1 - \tfrac{1}{2}z^{-1}\right) + 2$,

   $$H(z) = -1 + \frac{2}{1 - \tfrac{1}{2}z^{-1}} \;\Rightarrow\; \boxed{h[n] = -\delta[n] + 2\left(\tfrac{1}{2}\right)^{n} u[n].}$$

   Equivalently $h[0] = 1$ and $h[n] = 2(1/2)^{n}$ for $n \ge 1$. The check against the recursion is immediate: $h[0] = 1$, $h[1] = \tfrac{1}{2}(1) + \tfrac{1}{2} = 1$, $h[2] = \tfrac{1}{2}(1) = 0.5$. Omitting the $-\delta[n]$ term is the standard error here — it would make $h[0] = 2$, which the recursion flatly contradicts.

2. (b) Frequency response. The ROC contains the unit circle, so $H(e^{j\omega}) = H(z)\big|_{z = e^{j\omega}}$:
   $$\boxed{H(e^{j\omega}) = \frac{1 + \tfrac{1}{2}e^{-j\omega}}{1 - \tfrac{1}{2}e^{-j\omega}}.}$$

   Multiplying each factor by its conjugate gives a compact magnitude formula:

   $$\left|H(e^{j\omega})\right|^{2} = \frac{1 + \cos\omega + \tfrac{1}{4}}{1 - \cos\omega + \tfrac{1}{4}} = \frac{1.25 + \cos\omega}{1.25 - \cos\omega}.$$

   This is a shelving low-pass characteristic: $\left|H(e^{j0})\right| = 3$ at d.c., falling monotonically to $\left|H(e^{j\pi})\right| = 1/3$ at the Nyquist frequency — a total swing of $9$, or $19.1$ dB. The phase is

   $$\arg H(e^{j\omega}) = -\arctan\!\left(\frac{\tfrac{1}{2}\sin\omega}{1 + \tfrac{1}{2}\cos\omega}\right) - \arctan\!\left(\frac{\tfrac{1}{2}\sin\omega}{1 - \tfrac{1}{2}\cos\omega}\right),$$

   which is zero at both $\omega = 0$ and $\omega = \pi$ (as it must be for a real system whose response is real at those frequencies) and dips to roughly $-53^{\circ}$ in between.

   Magnitude and phase of H(e^jw). The gain shelves from 3 at d.c. down to 1/3 at Nyquist; the dashed line marks the input frequency of part (c), where the gain happens to be exactly unity.
3. (c) Response to the sinusoid, using the eigenfunction property. For a stable LTI system a sinusoid in gives the same sinusoid out, scaled by $\left|H\right|$ and shifted by $\arg H$, both evaluated at the input frequency. Here $\omega_0 = \pi/2$, so $e^{-j\omega_0} = -j$ and
   $$H\!\left(e^{j\pi/2}\right) = \frac{1 - \tfrac{1}{2}j}{1 + \tfrac{1}{2}j}.$$

   Numerator and denominator are complex conjugates of one another, so their magnitudes are equal and the gain is exactly one — consistent with the magnitude formula, since $\cos(\pi/2) = 0$ makes numerator and denominator both $1.25$. The phase is twice the angle of the denominator, with a sign change:

   $$\left|H\right| = 1, \qquad \arg H = -2\arctan\!\left(\tfrac{1}{2}\right) = -0.92730 \text{ rad} = -53.13^{\circ}.$$

   Applying the gain and phase to the input,

   $$\boxed{y[n] = \cos\!\left(\frac{\pi}{2}n + \frac{\pi}{4} - 0.92730\right) = \cos\!\left(\frac{\pi}{2}n - 0.14190\right) = \cos\!\left(\frac{\pi}{2}n - 8.13^{\circ}\right).}$$

   The system passes this particular tone at full amplitude and merely retards it by $8.13^{\circ}$, i.e. by $0.0452$ of a sample period. Because the input is stated without a $u[n]$, it has been present for all time and the answer above is the complete response — there is no transient to add.