Question 1 of 6: Impulse Response and Direct Form II Realisation

Question 1: Impulse Response and Direct Form II Realisation (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 causal LTI difference equation with coefficients

Find. (a) the closed-form impulse response $h[n]$, i.e. the response to $x[n]=\delta[n]$; and (b) the Direct Form II simulation diagram, which realises the same transfer function with the minimum possible number of delay elements.

Approach. Take the z-transform of the difference equation to obtain the rational transfer function, recognise that its denominator is a perfect square (a repeated pole), expand in partial fractions in the variable $z^{-1}$, and invert term by term using the standard pairs $a^{n}u[n]$ and $(n+1)a^{n}u[n]$.

1. Transform the difference equation. With $\mathcal{Z}\{y[n-k]\}=z^{-k}Y(z)$ under initial rest, $$Y(z)\left(1-\tfrac{1}{2}z^{-1}+\tfrac{1}{16}z^{-2}\right)=X(z)\left(1+\tfrac{1}{2}z^{-1}\right).$$
2. Form the transfer function. Dividing, $$H(z)=\frac{Y(z)}{X(z)}=\frac{1+\tfrac{1}{2}z^{-1}}{1-\tfrac{1}{2}z^{-1}+\tfrac{1}{16}z^{-2}}.$$
3. Factor the denominator. The quadratic in $z^{-1}$ is a perfect square, $1-\tfrac{1}{2}z^{-1}+\tfrac{1}{16}z^{-2}=\left(1-\tfrac{1}{4}z^{-1}\right)^{2}$, so the system has a repeated pole at $z=\tfrac{1}{4}$ and no other poles. Since $|z_p|=0.25\lt 1$ the causal system is BIBO stable.
4. Expand in partial fractions. Writing $u=z^{-1}$ and seeking $\dfrac{1+\tfrac12 u}{(1-\tfrac14 u)^{2}}=\dfrac{A}{1-\tfrac14 u}+\dfrac{B}{(1-\tfrac14 u)^{2}}$ gives $1+\tfrac12 u=A\left(1-\tfrac14 u\right)+B$. Matching the coefficient of $u$ gives $-\tfrac{A}{4}=\tfrac12\Rightarrow A=-2$, and matching the constant gives $A+B=1\Rightarrow B=3$.
5. Invert each term. The causal pairs are $\dfrac{1}{1-a z^{-1}}\leftrightarrow a^{n}u[n]$ and $\dfrac{1}{(1-a z^{-1})^{2}}\leftrightarrow (n+1)a^{n}u[n]$ with $a=\tfrac14$. Therefore $h[n]=-2\left(\tfrac14\right)^{n}u[n]+3(n+1)\left(\tfrac14\right)^{n}u[n]$, and collecting the bracket, $$\boxed{\,h[n]=(3n+1)\left(\tfrac{1}{4}\right)^{n}u[n]\,}$$
6. Check against the recursion. Driving the difference equation with $x[n]=\delta[n]$ gives $h[0]=b_0=1$; $h[1]=\tfrac12 h[0]+b_1=1$; $h[2]=\tfrac12 h[1]-\tfrac{1}{16}h[0]=0.4375$; $h[3]=\tfrac12 h[2]-\tfrac{1}{16}h[1]=0.15625$. The closed form returns $1,\,1,\,\tfrac{7}{16},\,\tfrac{5}{32}$ — identical, so the inversion is correct.

Part (b) asks for the Direct Form II realisation. Direct Form I would use two delay lines (one for the input history, one for the output history); Direct Form II first passes the input through the all-pole section to form an intermediate sequence $w[n]=x[n]+\tfrac12 w[n-1]-\tfrac{1}{16}w[n-2]$, then forms the output from the same delayed samples as $y[n]=w[n]+\tfrac12 w[n-1]$. Because both sections read the same delay chain, only two delay elements are needed — the canonical (minimum-memory) form.