EEE336 期末试题

大学学习

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2014-15 期末试题

Q1

a)

$y(t)=g(t)g(t)\iff Y(f)=G(f)*G(f)$

According to the convolusion theorem, the bandwidth becomes twice,

$\therefore \Omega=2\Omega_m$

b)

$x[n]=x_a(t)|_{t=nT}=x_a(nT)$
$T=\frac{1}{f}=0.02s$
$x[n]=3\sin(0.2\pi n)-4\cos(0.24\pi n)+6\sin(0.8\pi n)$

c)

(1)

$x(t)=\delta(t+1)+4\delta(t)+\delta(t-1)$
$1\iff \delta(t)$

$\delta(t+1)\iff e^{j\omega}$
$4\delta(t)\iff 4$
$\delta(t-1)\iff e^{-j\omega}$

$\begin{aligned} X(\omega)&=e^{j\omega}+4+e^{-j\omega}\\ &=\cos(\omega)+j\sin(\omega)+4+\cos(\omega)-j\sin(\omega)\\ &=4+2\cos(\omega) \end{aligned}$

(2)

$x(t)=\frac{\sin(80\pi(t-3))}{\pi(t-3)}=80\cdot\frac{\sin(80\pi(t-3))}{80\pi(t-3)}=80\cdot\text{sinc}(80\pi(t-3))$

$80\cdot\text{sinc}(80\pi(t-3))\iff \frac{1}{\pi}\cdot\text{rect}(\frac{\omega}{160\pi^2})\cdot e^{-j3\omega}$

Q2

a)

$H(e^{j\omega})=\sin(3\omega)$

$e^{j3\omega}=\cos 3\omega+j\sin 3\omega$
$e^{-j3\omega}=\cos 3\omega-j\sin 3\omega$
$\dfrac{e^{j3\omega}-e^{-j3\omega}}{2j}=\sin 3\omega$
$h[n]=\dfrac{\delta[n+3]-\delta[n-3]}{2j}$

b)

(1)

correct answer:
$X[k]=1,k=0,1,\dots,9$

wrong answer:
$X[k]=\left\{ \begin{matrix} 1 & k=0\\ 0 & k=1,2,\dots,9 \end{matrix} \right.$

(2)

$\begin{aligned} X_2[k]=&1+e^{-j0.2\pi k}+e^{-j0.4\pi k}+e^{-j0.6\pi k}+e^{-j0.8\pi k}\\ &+e^{-j\pi k}+e^{-j1.2\pi k}+e^{-j1.4\pi k}+e^{-j1.6\pi k}+e^{-j1.8\pi k} \end{aligned}$
$X_2[0]=10$

c)

d)

(1)

$y[n]=\frac{1}{4}(x[n]+x[n-1]+x[n-2]+x[n-3])$

$Y[z]=\dfrac{1}{4}(x[z]+z^{-1}x[z]+z^{-2}x[z]+z^{-3}x[z])$
$\delta[n]\iff 1$
$H[z]=\dfrac{1}{4}(1+z^{-1}+z^{-2}+z^{-3})$
$h[n]=\dfrac{1}{4}(\delta[n]+\delta[n-1]+\delta[n-2]+\delta[n-3])$

(2)

$H(z)=\dfrac{Y[z]}{X[z]}=\dfrac{1}{4}(1+z^{-1}+z^{-2}+z^{-3})$

(3)

$H(e^{j\omega})=\dfrac{1}{4}(1+e^{-j\omega}+e^{j2\omega}+e^{j3\omega})$

Q3

a)

no limitation for $n$, so this is IIR
non-causal filter

b)

symmetrical, length = 5, N = 4, $h[n]=h[N-n]$
Type 1 linear phase FIR filter

$H(z)=h[0]+h[1]z^{-1}+h[2]z^{-2}+h[3]z^{-3}+h[4]z^{-4}$
$\begin{aligned} H(e^{j\omega})&=h[0]+h[1]e^{j\omega}+h[2]e^{j2\omega}+h[3]e^{j3\omega}+h[4]e^{j4\omega}\\ &=a+be^{j\omega}+ce^{j2\omega}+be^{j3\omega}+ae^{j4\omega}\\ &=e^{-j2\omega}[2a\cos 2\omega+2b\cos\omega+c] \end{aligned}$

where $e^{-j2\omega}$ is linear phase

c)

Four zeros:

$2+j2$, $2-j2$, $\dfrac{1}{4}-\dfrac{j}{4}$, $\dfrac{1}{4}+\dfrac{j}{4}$
($\dfrac{1}{2+j2}=\dfrac{1}{4}-\dfrac{j}{4}$)
$\begin{aligned} H(z)&=a+bz^{-1}+cz^{-2}+bz^{-3}+az^{-4}\\ &=\dfrac{az^4+bz^3+cz^2+bz+a}{z^4}\\ &=\dfrac{(z-(2+j2))(z-(2-j2))(z-(\frac{1}{4}-\frac{j}{4}))(z-(\frac{1}{4}+\frac{j}{4}))}{z^4}\\ &=\dfrac{a^4-4.5z^3+10.125z^2-4.5z+1}{z^4}\\ &=1-4.5z^{-1}+10.125z^{-2}-4.5z^{-3}+z^{-4} \end{aligned}$

d)

e)

cut-off frequency $\omega_c=2\pi\cdot\dfrac{f_c}{R_s}=0.2\pi$
5-point, $M=2$

Step 1: Find unwindowed impulse response:
$h_d[n]=\dfrac{1}{n\pi}\sin(0.2\pi n)=\dfrac{\omega_c}{\pi}\text{sinc}(n\omega_c)$

Step 2: Find window samples for $-2\leq n\leq 2$
Rectangular: $w[n]=1$ (selected)
Hann: $w[n]=0.5+0.5\cos(2\pi\frac{n}{2M+1})$
Hamming: $w[n]=0.54+0.46\cos(2\pi\frac{n}{2M+1})$

Step 3: Apply window:
$h_w[n]=h_d[n]\cdot w[n]=\dfrac{0.2\pi}{\pi}\text{sinc}(0.2\pi n)$

Step 4: Make causal:
$h[n]=h_w[n-2]$, $n=0,1,2,3,4$

Q4

a)

$H(s)$ is stable if and only if the poles all lie in the left half-plane: $Re(c_k)\leq 0$ for all $k$.

b)

unity cut-off frequency => $\Omega_c=1$
$p_l=\Omega_ce^{j\frac{N+2l-1}{2N}}$
poles location:
$p_1=e^{j\frac{3}{4}\pi}$
$p_2=e^{j\frac{5}{4}\pi}$

$H_a(s)=\dfrac{1}{s^2+\sqrt 2s+1}$

c)

Bilinear transformation:

$$s=\frac{2}{T}\left(\frac{1-z^{-1}}{1+z^{-1}}\right)$$

where $T$ is the sampling period, $T=2$ (given)
$H_a(s)=\dfrac{1}{s^2+\sqrt 2s+1}$
$G(z)=\dfrac{(1-\frac{\sqrt 2}{2})+(2-\sqrt 2)z^{-1}+(1-\frac{\sqrt 2}{2})z^{-2}}{1+(3-\sqrt 2)z^{-2}}$

d)

$p_0=1-\frac{\sqrt 2}{2}$
$p_1=2-\sqrt 2$
$p_2=1-\frac{\sqrt 2}{2}$

$d_1=0$
$d_2=3-\sqrt 2$

Direct Form 1 (noncanonic)

Direct Form 2 (canonic)

e)

$H(z)$ is stable if all the poles lie inside the unit circle

$$|\lambda_k|<1$$

$(3z+7)=3(z+\frac{7}{3})$, $\frac{7}{3}>1$
$\therefore$ this IIR is unstable