midterm.sol

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Amplitud
Original
0.4
Frequency[MHz]
After sampling
0.4 0.6 After Reconstruction s1 s2
1
1.4
1.6
2
2.4
2.6
3 Frequency[MHz]
sin(x)/x,
x=πf/fclk s3 s4 1.6 2 s5 2.4 s6 2.6 3 Frequency[MHz]
With the sampling frequency at fs, the lowest frequency which can alias down to the bandwidth of interest is (fs-0.5MHz). Since the attenuation is at 50dB/decade then: fs-0.5MHz should be one decade beyond the filter corner frequency to attain 50dB attenuation. Fs-0.5MHz=10x7.5MHzà fs=8MHz. à fsampling =8MHz
Solution to Problem 4 (10points):: For the system shown below, what is the minimum sampling frequency fs that guarantee 50 dB attenuation of aliasing components in the discrete time signal y(kT)? (T=1/fs)
set Z = e jωT Since fclk > > f signal →Z ≈ 1 + jωT Substituting for Z V0 1 ( jω ) = CI 2π f Vi 1+ j 1+ Cs fc l k 2π f 2π x fc l k = 1 10 f fc l k
clk
1µs
fclk=1MHz
Switch closed for only brief period each clock cycle
clk
+
C
Vi
To Spectrum Analyzer
Solution: There are two functions involved: 1-Sampling- Generates aliases of the signal at nfclk+-fsignal 2- Reconstruction- Shapes the spectrum with sinx/x
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Solution to Problem 3 (10points)::
For the circuit shown below, the input is a 400KHz sinusoidal signal. Sketch the output spectrum from 0 to 3MHz showing the amplitude and frequency of each component. Note that the output is not resampled. The clock rate is 1MHz, and the switch and opamp are ideal.
Solution to Problem 1 (20points):
φ1
φ2
CI=0.9/2π PF
Vo Vi +
Cs=0.1/2π PF
a-There are two approaches to this problem: 1. Either start with deriving the recursive equations: Φ1 à Qs [(n-1)Ts]= Cs Vi [(n-1)Ts] , QI [(n-1)Ts] = QI[(n-3/2)Ts] Φ2 à , (QI+Qs) [(n-1/2) Ts] = QI [(n-3/2) Ts] + Qs [(n-1) Ts] (Cs+CI) Vo[(n-1/2)Ts =Cs Vi [(n-1)Ts +CI Vo [(n-1)Ts] Φ1 _à Qs [nTs ] = Cs Vi [nTs ] , Vo [nTs ] = Vo(n-1/2) Ts ] (Cs+CI) Vo(n)Ts =Cs Vi [(n-1)Ts] +CI Vo [(n-1)Ts] (Cs+CI) Vo= Cs ViZ-1+CI VoZ-1 Cs V0 CI (Z ) = Cs Vi 1 + Z −1 CI
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2. Or use the z block diagram derived in the lecture
z
− 12
z
−1
2
z +1
Vo
2
1 − z− 1
Vo2
V0 Vi
Cs 1 CI (Z ) = = Cs CI CI 1+ Z − 1 1+ Z − Cs CI Cs
7 6 5 4 3 2 1 0 0 1 2 3 4 5 ADC Input Voltage [LSB] 6 7 8
DNL: 0 -0.5 0 -1 +0.5 +0.5 +0.5 0 INL: 0 -0.5 -0.5 -1.5 -1 -0.5 0 0 The ADC has a missing code at position 3. Since the INL has a quadratic shape (bow shape), this ADC will produce even order harmonics of the input signal.
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UNIVERSITY OF CALIFORNIA College of Engineering Department of Electrical Engineering and Computer Sciences Midterm Exam Solution H. Khorramabadi EECS 247 FALL 2004
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2- Also, the output is a staircase signal, depending on the application, it may need a smoothing filter. Solution to Problem 2 (10points):: Find the DNL & INL for the 3bit ADC with the transfer function shown below. ADC Output Code [LSB]
b−
V0 1 ( jω ) = Vi 1 + j (1 + 9 )
1+ j
f → First o r d e r filter with f − 3d B = clk 10 |H(f)| 0dB -3dB 20dB/dec .
fclk/10
f
2
Note that if Vo2 is derived instead of Vo then Vo2=V0x Z0.5 . Then there would be one extra zero at high frequencies you can neglect the zero. c-The continuous-time opamp RC filter performing the same function would have the following configuration with the same value integrating capacitance since noise in both filters is proportional to sqrt(KT/Cintg) and thus R=1/(fclkCs)
Cintg R R +
Vin
Vo
dAdvantages of the continuous-time filter: Does not require an anti-aliasing prefilter Disadvantages for the continuous-time filter: 1-It requires an extremely large resistor and thus large area required for the implementation of two large Rs. 2-Also, since the opamp has to drive a resistive load power dissipation is going to be higher compared to the SC counterpart. 3-Main disadvantage is that the corner frequency is a function of absolute value of R & C and in the monolithic form that can vary drastically. In the case that an accurate corner frequency is required, additional tuning circuitry is needed. Advantages of the switched-capacitor filter is: 1-The frequency response is accurately set by the ratio of capacitors and a clock frequency. 2-The area required to implement the capacitors is much smaller compared to the area required for the R in the continuous-time filter. 3-Since the opamp here drives a capacitive load, the power dissipation is lower. Disadvantages for the SC filter: 1-Since it is a sampled-data circuit, it requires an anti-aliasing prefilter.