Difference between revisions of "ECE438 Week12 Quiz" - Rhea

Revision as of 06:25, 11 November 2010

Quiz Questions Pool for Week 12

Q1. Consider a causal FIR filter of length M = 2 with impulse response

$ h[n]=\delta[n-1]+\delta[n-2]\,\! $

a) Provide a closed-form expression for the 9-pt DFT of $ h[n] $, denoted $ H_9[k] $, as a function of $ k $. Simplify as much as possible.

b) Consider the sequence $ x[n] $ of length 9 below,

$ x[n]=\text{cos}\left(\frac{2\pi}{3}n\right)(u[n]-u[n-9])\,\! $

$ y_9[n] $ is formed by computing $ X_9[k] $ as an 9-pt DFT of $ x[n] $, $ H_9[k] $ as an 9-pt DFT of $ h[n] $, and then $ y_9[n] $ as the 9-pt inverse DFT of $ Y_9[k] = X_9[k]H_9[k] $.

Express the result $ y_9[n] $ as a weighted sum of finite-length sinewaves similar to how $ x[n] $ is written above.

• Solution.

Q2. Consider the discrete-time signal

$ x[n]=6\delta[n]+5 \delta[n-1]+4 \delta[n-2]+3 \delta[n-3]+2 \delta[n-4]+\delta[n-5]. $

a) Obtain the 6-point DFT X[k] of x[n].

b) Obtain the signal y[n] whose DFT is $ W_6^{-2k} X[k] $.

c) Compute six-point circular convolution between x[n] and the signal

$ h[n]=\delta[n]+\delta[n-1]+\delta[n-2]. $

• Same as HW8, Q2 available here.

Q3. Consider the signal

$ x[n] = \begin{cases} cos(\pi n / 8), & n < 0 \\ cos(\pi n / 3), & \mbox{else} \end{cases} $

and assume a rectangular window

$ w[n] = \begin{cases} 1, & |n| < 25 \\ 0, & \mbox{else} \end{cases} $

The STDFT is defined as

$ \begin{align} X(\omega,n) &= \sum_{k} x[k]w[n-k]e^{-j\omega k} \end{align} $

Compute the STDTFT for the following cases:
i. n < -25
ii. n > 25
iii. n = 0

• Solution.

Q4. Consider the STDTFT defined as

$ X(\omega ,n)=\sum_k x[k]w[n-k]e^{-j\omega k} $

where x[n] is the speech signal and w[n] is the window sequence. Prove the following properties:

a. Linearity – if $ v[n]=ax[n]+by[n] $ ,then $ V(\omega ,n)=aX(\omega ,n)+bY(\omega, n) $.

b. Modulation – if $ v[n]=x[n]e^{j\omega_0n} $ ,then $ V(\omega ,n)=X(\omega -\omega_0,n) $.

• Solution.

Q5. Suppose we have two 4-pt sequences x[n] and h[n] described as follows:

$ \begin{align} x[n] &= cos(\frac{\pi n}{2})\text{ ,n=0,1,2,3} \\ h[n] &= 2^n\text{ ,n=0,1,2,3} \end{align} $

a. Compute 4-pt DFT X[k];

b. Compute 4-pt DFT H[k];

c. Compute 4-pt circular convolution directly of $ y[n]=x[n]\circledast_4 h[n] $;(You may use plot to explain your answer)

d. Multiply DFT result of x[n] and h[n]. Then using IDFT to compute y[n] in question c.

• Solution.

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