jawaban tugas problem 2 litan
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TUGAS PENGOLAHAN SINYAL DIGITAL
Disusun Oleh :KELOMPOK 05TEDDY SEPTIANAH1C008030JIHAN WAHYUDIH1C008034AHADI ANGGUN RAHARJOH1C009029ELISA WIDYANINGRUMH1C009037SUWITNOH1C009039KEMENTERIAN PENDIDIKAN DAN KEBUDAYAANUNIVERSITAS JENDERAL SOEDIRMANFAKULTAS SAINS DAN TEKNIKPROGRAM STUDI TEKNIK ELEKTROPURBALINGGA2012JAWABAN SOAL 2.1
Diberikan analog sinyal
X(t) = 5 cos (2.1500t) untuk t 0Pada 8080 Hz
Ditanya : a) Spektrum dari sinyal b) Spektrum sinyal pada sampling 0-20 kHzJawab :Analog signal pada 1500 Hz
5 cos (2 x 1500t) = 5 ( +) = 2,5 + 2,5
Maka C1 = 2,5C-1 = 2,5Magnitude dari signal diperoleh :
X(f)
2,5f (kHz)
Spektrum sinyal pada sampling 0-20 kHz
n . f(s)
-1 1 7 8 9 15 16 17
JAWABAN SOAL 2.2
Diberikan sinyal analog
X(t) = 5 cos (2.2500t) + 2 cos (2.3200t) untuk t 0Ditanya : a. Spektrum dari sinyal, spektrum sinyal pada sampling 0-20 khzb. sketsa sinyal baik jika low pass filter dengan frekuensi 4 khz
Jawab:
Sinyal Analog :
X(t) = 5 cos (2 . 2500t) + 2 cos (2 . 2200t) untuk t 0Dengan Euler diperolehX(t) = + 2,5 +1 +X(t) = 2,5 + + + Maka samping spektrum pada magnitude :
Xs(f)
F kHz
-11 -10 -8 -6 -5 -3 -2 2 3 5 6 8 10 11 13 14 16 18 19b) Sinyal balik jika low pass filter f = 4 kHz
-3 -2 2 3
JAWABAN SOAL 2.3
Diberikan sinyal : x(t) = 5 cos (2. 2500t) + 2 cos (2. 4500t); t0 dengan frekuensi sample 8000 Hza. Gambarkan spectrum sinyal sample hingga 20 Khzb. Gambarkan spectrum sinyal yang tercover sebuah lowpass filter idea dengan frequency cut off 4 khz yang digunakan untuk memfilter sunyal yang ditentukan untuk meng- cover sinyal aslic. Tentukan frekuensi ganguan analog
Jawab:a). Dengan identitas Euler :
5 cos (2. 2500t) + 2 cos (2. 4500t) = + + +
Xs(f) 2,5
kHz
-8 -5,5-4,5 -3,5 - 2,5 3,5 4,5 5,5 8 10,5 12,5 16
b). Xs(f)
kHz
-4,5 -3,5 -2,5 2,5 3,5 4,5c). Frekuensi ganguan Aliansing = 3,5 KhzJAWABAN SOAL 2.4
a.formulaformula8
-8
6
6
6,5
-6,5
4
-4
0
0,5
1,5
-0,5
-1,5
-7500
7500
5500
-5500
-2,5
2,5
5
5
b.
JAWABAN SOAL 2.50,1uF
2,25 kohm
2,25 kohm
Vin
Vo
0,05 uF
JAWABAN SOAL 2.6
formula
JAWABAN SOAL 2.7
Given a DSP system in which a sampling rate 8000 Hz is used and the anti-aliasing filter is a second order butterworth lowpass filter with a cutoff frequency of 3,2 kHz. Determine :a. The precentage of aliasing level at the cutoff frequency b. The precentage of aliasing level at the frequency of 1000 Hz
fs=8000 Hzfc=3200 Hzn=2fa=fc=3200 Hz
a. aliasing noise level (%) = = formula = = 57,43%
b. aliasing noise level (%) = formula
=
= 20,55%
JAWABAN SOAL 2.8
Given a DSP system inn which a sampling rate of a 8 kHz is used
and the anti aliasing filter in a butterworth lowpass filter with a
cutoff frequency 3,2 kHz. Determine the order of the butterworth
lowpass filter for the precentage of aliasing level at the cutoff
frequency required to be less than 10 %.|H(f)| =formula
n = 1 ----> |H(f)|= formula
=
= 37,1 %
n = 2 ----> |H(f)| = formula
= 15,79 %
n = 3 ----> |H(f)| = formula
= 6,38 %
>>>>> 6,38 % < 10 % ----> n = 3
JAWABAN SOAL 2.9
Gives a DSP system with a sampling rate of 8 kHz and assuming that the hold circuit is used after DAC. Determine :a. The precentage of distortion at the frequency of 3200 Hzb. The precentage of distortion at the frequency of 1500 Hz
a. f T ----> 3200 X = 0,4
distorsi % = formulaX 100%
= formula X 100 %
= 24,32 %
b . fT = 1500 X = 0, 187
% = formula X 100 %
= 5,68 %
JAWABAN SOAL 2.10A DSP system is given with the following specifications:Design requirements:
Sampling rate 20000 Hz
Maximum allowable gain variation from 0 to 4000 Hz = 2dB
40 dB rejection at the frequency of 16000 Hz
Butterworth filter assumed
Determine the cutoff frequency and order for the anti-image filter.
Solution:f= 4000 Hz, fT= 4000 x 1/20000= 0.2
Gain = formula= formula= 0.9359 = -0.57 dB
f= 16000, fT= 16000 x 1/20000= 0.8
Gain= formula= formula= 0.2339 = -12 dB
Hence, the design requirements for the anti-image filter are:Butterworth lowpass filter
Maximum allowable gain variation from 0 to 4000 Hz = (2 0.57) = 1.43 dB
(40 12 ) = 28 dB rejection at frequency 16000 Hz.
Set up equations using log operations of the Butterworth magnitude function as:
formula
formulaFrom these two equations, we have to satisfyformulaformula
Taking the ratio of these two equations yields
formula
Then formula
Finally, the cutoff frequency can be computed as
formula
formula
choose the smaller one, that is: fc = 4679,77 Hz
JAWABAN SOAL 2.11Given the 2-bit flash ADC unit with an analog sample-and-hold voltage of 2 volts shown in Figure 2.37, determine the output bits.Solution:Seperti terlihat pada Gambar 2.37, 2-bit Flash Unit ADC terdiri dari seri tegangan referensi yang dibuat oleh resistor nilai yang sama, satu set comparator, dan logika unit. Pada soal ini tegangan referensi pada adalah 1,25 volt, 2,5 volt, 3,75 volt dan 5 volt. Analog sample dan hold voltage-nya adalah Vin = 2 volt, maka satu comparator yang mempunyai nilai lebih rendah masing-masing akan berlogika 1 pada outputnya. Melalui unit logika, hanya baris yang berlabel 01 yang bernila aktif tinggi (logic 1), dan sisanya adalah baris aktif rendah (logic 0). Oleh karena itu, rangkaian logika pengkodean output kode 2-bit biner dari 01. Dan berikut gambarnya:
JAWABAN SOAL 2.12Given the R-2R DAC unit with a 2-bit value of b1b0 = 01 shown in Figure 2.38, determine the converted voltage.
Solution:Converted voltage:
formula
formula
Jawaban Soal No 2.13
a. L = 2m = 24 = 16b. = formula= formula= 0,3125 Volt
c. ketika x = 3,2 Volt formula= 10,24
maka i = round (formula) = round (10,24) = 10
maka level quantisasinya = xq = 0 + 10 = 10 . 0,3125 = 3,125 volt
d. Quantization table for the 4-bit unipolar quantizer (stepsize = = (xmax xmin)/24, xmax = maximum voltage, and xmin = 0.
Binary CodeQuantization Level eq (V)Input Signal Subrange (V)
000100100011010001010110011110001001101010111100110111101111234567891011121314150 x < 0,50,5 x < 1,51,5 x < 2,52,5 x < 3,53,5 x < 4,54,5 x < 5,55,5 x < 6,56,5 x < 7,57,5 x < 8,58,5 x < 9,59,5 x < 10,510,5 x < 11,511,5 x < 12,512,5 x < 13,513,5 x < 14,5
e. eq = xq x = 3,125 3,2 = -0,075catatan bahwa eror kuantisasi nilainya adalah kurag dari setengah nilai resolusi|eq| = 0,075 < formula= 0,3125 volt
Jawaban Soal No 2.14
a. L = 2m = 23 = 8
b. = formula= formula= 0,625 Volt
c. ketika x = -1,2 Volt formula= -1,92
maka i = round (formula) = round (formula) = 1,6
maka level quantisasinya = xq = xmin + i = -2,5 + (1,6 . 0,625) = -1,5 volt
d. Quantization table for the 3-bit bipolar quantizer (stepsize = = (xmax xmin)/23, xmax = maximum voltage, and xmin = -xmax.
Binary CodeQuantization Level eq (V)Input Signal Subrange (V)
000001010011100101110111
-3-2-10234-2,5 x < -1,5-1,5 x < -0,5-0,5 x < 1,51,5 x < 2,52,5 x < 3,53,5 x < 4,54,5 x < 5,55,5 x < 6,5
e. eq = xq x = -1,5 (-1,2) = -0,3 Voltcatatan bahwa eror kuantisasi nilainya adalah kurag dari setengah nilai resolusi|eq| = 0,3 < formula= 0,3 < formula= 0,3125 Volt
Jawaban Soal No 2.15
a. L = 2m = 26 = 64
b. = formula= formula= 0,3125 Volt
c. SNRdB = 10,79 + 20 log10 (formula)= 10,79 + 20 log10 (xrms) 20 log10 0,3125
= 10,79 + 20 log10 (xrms) 10,102
= 0,688 + 20 log10 (xrms)