EENG31400 / EENGM1400 DIGITAL FILTERS AND SPECTRAL ANALYSIS

Hello, if you have any need, please feel free to consult us, this is my wechat: wx91due

EENG31400 / EENGM1400

DIGITAL FILTERS AND SPECTRAL ANALYSIS

Q1 Figure 1 shows the magnitude of the 32-point DFT X[k], k=0,....31 of discrete time signal x[n]= e,!。! for the case of window w[n]:

(a)Sketch the DTFT X(Ω) of x[n]

(b)Comment on the shape of the spectrum X[k] provided by the DFT relative to that of X(Ω). Why are there so many frequency components in the DFT spectrum? Explain.


The DTFT spectrum of signal x[n] = e,!。! consists of a single impulse at frequency Ω。. 

Due to the windowing process that takes place when we calculate the DFT,  the spectrum sampled by the DFT is the DTFT spectrum of x[n] x w[n], i.e. the spectrum of the rectangular window convolved  with  the  spectrum  of  x[n]  (impulse  at  frequency Ω。).  

As  a  result  multiple frequency components appear. (8 marks)


(c)Which normalised radial frequency Ω do the following indices k correspond to:

• k = 16 • k = 6

K = 16: Ω 1 =16x2τ/32= τ K = 6: Ω2 = 6x2τ/32 (4 marks)

(d) Which of the two actions described below would bring the shape of the DFT spectrum X[k] closer to that of X(Ω,

a)   Increasing the length of the DFT to 256

b)   Increasing the length of window w[n] to 32 Justify your answer

The answer is (b). Increasing the length of the DFT  (equivalent to padding the windowed signal with zeroes) will not change the shape of the spectrum.  

Increasing the length of the DFT means sampling the DTFT spectrum of the windowed signal more densely. The shape of the DTFT spectrum (described above) remains the same, as it depends on the length and type of window used. 

The longer the window the smaller the main lobe width in the spectrum. (6 marks)


Q2 

(a)Determine whether each of the following statements is true or false (where u is the unit step function). Justify your answers.



i.      The signal x(t) = u(t+T0) - u(t-T0) will not suffer from aliasing if it is sampled with a sampling period T<2T0



ii.       The signal x(t) with Fourier transform X(“) = u(“+“0)- u(“-“0) will not suffer from aliasing if it is sampled with a sampling period T < z/“。



iii.      The  signal x(t)  with  Fourier  transform X(“) = u(“)- u(“-“0) will not suffer from aliasing if it is sampled with a sampling period T < 2z/“。 (9 marks)



(b)Consider an ideal discrete-time band-stop filter with impulse response h[n]  for which the frequency response in the interval -z ≤ Ω ≤ z is: Determine the frequency response of the filter whose impulse response is h[2n] (4 marks)

(c)Consider the system of Figure 2 for changing the sampling rate of signal x[n] by a non-integer factor. Determine the output x! [τ] if x[n] = cos(3πn/4) , L=6 and M=7.



Suppose that a continuous time speech signal sc(t) with the Fourier transform spectrum Sc(ω) shown in Figure 3(a) is processed by the system shown in Figure 3(b) to produce the discrete time signal sr [n]. H(Ω) in Figure 3(b) is an ideal discrete-time low pass filter with a cut-off frequency of Ωc and a gain of L in the passband. The signal sr [n] will be used as input to a speech encoder that operates correctly only on discrete time speech signals sampled at an 8- kHz rate. Choose values of L, M and Ωc that produce the correct input signal sr [n] for the speech encoder.

The sampling frequency is equal to 44.1 KHz. We want the output to have an 8KHz rate hence M/L = 44.1/8 = 441/80 . We choose L=80, M=441 and Ωc=π/441 (4 marks)


Q3 

(a)Following the windowing method of FIR filter design, use a Hamming window to design a 3-tap FIR lowpass filter with a cut-off frequency of 800 Hz and a sampling rate of 8 kHz.


发表评论

电子邮件地址不会被公开。 必填项已用*标注