20150416114355dsp_midterm_2013
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DSP Midterm 2013-11-12 1. Solve the following difference equation for n ≥ 0 with n n x ) 2 1 ( ] [ = : 1 ] 1 [ , 0 ] 0 [ ], [ ] 2 [ 4 1 ] 1 [ ] [ = = = − + − − y y n x n y n y n y . (15 %) 2. Let y[n] be the sequence obtained by a linear convolution of two causal finite-length sequences h[n] and x[n]. Given y[n] = {-1, -1, 11, -3, 30, 28, 48} and h[n] = {-1, 2, 3, 4}, determine x[n]. The first sample of each sequence starts from n = 0. (10 %) 3. An analog filter with the frequency response shown below is to be replaced by a digital filter. Find the minimum sampling rate of the digital filter. Based on the sample rate you calculate, estimate the values of attenuation (Attn), Ω p , and Ω s . (15 %) Response of digital LPF Ω p Ω s 96 dB Attn 360 kHz 20 kHz Response of analog LPF 4. A discrete causal LTI system has the following frequency response ) 2 cos 3 cos 2 1 ( ) ( 2 Ω + Ω + = Ω Ω − j e H . Sketch a block diagram (containing delays, adders, and constant multiplications) 5. whose Fourier Transform X(ω) is identically Prove that y(t) has the form C 0 + C 1 cos(ω 0 t + φ 0 ), where C 0 , C 1 , and φ 0 are s/s and 48 ks/s is actually difficult to timated by to represent this system. (15 %) Let x(t) be a band-limited signal, zero outside the interval [-1.5ω 0 , 1.5ω 0 ]. Define T = 2π/ω 0 and let ) ( ) ( nT t x t y − = ∑ ∞ n −∞ = y(t) = constant. (15 %) 6. The sample rate conversion between 44.1 k design. For this problem, consider the resolution of the samples is 16 bits. (15%) i. Sketch a simple block diagram to illustrate how to achieve this conversion. You need to label all parameters in the blocks. ii. Let the number of coefficients of the filter be es c Attn N Ω ⋅ − = 1 . 0 8 , where Attn is the filter attenuation (junit: dB) and Ω c is the cutoff of the tly implemented without considering any trick, low-pass filter. Find N. iii. If the filter is to be direc
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DSP Midterm 2013-11-12
1. Solve the following difference equation for n ≥ 0 with nnx )21(][ = :
1]1[,0]0[ ],[]2[41]1[][ ===−+−− yynxnynyny . (15 %)
2. Let y[n] be the sequence obtained by a linear convolution of two causal finite-length sequences h[n] and x[n]. Given y[n] = {-1, -1, 11, -3, 30, 28, 48} and h[n] = {-1, 2, 3, 4}, determine x[n]. The first sample of each sequence starts from n = 0. (10 %)
3. An analog filter with the frequency response shown below is to be replaced by a digital filter. Find the minimum sampling rate of the digital filter. Based on the sample rate you calculate, estimate the values of attenuation (Attn), Ωp, and Ωs. (15 %)
Response of digital LPF
Ωp Ωs
96 dB
Attn
360 kHz 20 kHz
Response of analog LPF 4. A discrete causal LTI system has the following frequency response
)2cos3cos21()( 2 Ω+Ω+=Ω Ω− jeH . Sketch a block diagram (containing delays, adders, and constant multiplications)
5. whose Fourier Transform X(ω) is identically
Prove that y(t) has the form C0 + C1cos(ω0t + φ0),
where C0, C1, and φ0 ares/s and 48 ks/s is actually difficult to
timated by
to represent this system. (15 %) Let x(t) be a band-limited signal,zero outside the interval [-1.5ω0, 1.5ω0]. Define T = 2π/ω0 and let
)()( nTtxty −= ∑∞
n −∞=
y(t) = constant. (15 %)
6. The sample rate conversion between 44.1 kdesign. For this problem, consider the resolution of the samples is 16 bits. (15%) i. Sketch a simple block diagram to illustrate how to achieve this conversion.
You need to label all parameters in the blocks.
ii. Let the number of coefficients of the filter be esc
AttnNΩ⋅−
=1.0
8 ,
where Attn is the filter attenuation (junit: dB) and Ωc is the cutoff of the
tly implemented without considering any trick, low-pass filter. Find N.
iii. If the filter is to be direc
estimate the number of multiplications (per sec) required. Sketch the spectrum of the signal y[n] expressed as 0][ny =7.
. (15 %)
])[)1(][(5. nxnx n−+ . The spectrum of x[n] is shown below. Explain your work
π/2 −π/2 Ω
X(Ω)
1
