Tugas Kelompok I.docx

TUGAS KELOMPOK AMPLITUDE MODULATION (AM) Disusun Untuk Memenuhi Tugas Kelompok I Mata Kuliah Pengantar Telekomunikasi Semester 2

PEMBIMBING : Anisari Mei Prihatini, ST, MT

Penyusun:

JTD 1D Kelompok 3 No

Nama

NIM

08

Farchan Aditya I

1641160056

14

Nadhia Syafira Arifianti

1641160104

21

Taj Putra Akbar A

1641160115

22

Trushero Kharisma C

1641160058

23

Wildana Ahmad

1641160007

JARINGAN TELEKOMUNIKASI DIGITAL TEKNIK ELEKTRO POLITEKNIK NEGERI MALANG 2017

1. A carrier signal with a peak voltage of 2.0 V is amplitude modulated with a 10 kHz sine wave. The modulation voltage has an effective value of 750 mV. Compute the following: a. The percent modulation index, M b. The instantaneous voltage of the positive and negative envelope when the 10 Khz sine wave has completed 68 μs of its cycle c. Illustrate the resulting AM waveform Diketahui : 

Vc = 2,0 V



Vm = 750 mV = 0,75 V



fm = 10 kHz

Ditanya

:

a. M b. Envelope positif dan Envelope negatif c. Gambar gelombang AM Dijawab

:

a. m = =

𝑉𝑚 𝑉𝑐 0,75 2

= 0,375 b. m

= 2 fm = 210000 = 20000

Envelope positif

= Vc + Vm sin m t = 2 + 0,75 sin 20000 65 x 10-6 = 2 + 0,75 sin 1,36

Envelope Negatif

= -( Vc + Vm sin m t ) = -( 2 + 0,75 sin 1,36 ) = -2 – 0,75 sin 1,36

2. An AM broadcast station’s peak carrier voltage of 2 kV has been amplitude modulated to index of 75 % with a 2 kHz test tone. The station’s broadcast frequency is 810 kHz. Compute the following: a. The lower and upper sideband frequency, LSB and USB b. The peak modulation voltage, Vm c. The peak lower and upper sideband voltage, VLSB and VUSB d. The maximum signal amplitude, Vmax Diket : 

Vc = 2 kV



M = 75%



fc = 2 kHz



fm = 810 kHz

Ditanya : a. fLSB dan fUSB b. Vm c. VLSB dan VUSB d. Vmax Jawab : a. 𝑓𝐿𝑆𝐵 = 𝑓𝑐 − 𝑓𝑚 = 2 𝑘𝐻𝑧 − 810 𝑘𝐻𝑧 = −808 𝑘𝐻𝑧 𝑓𝐿𝑆𝐵 = 𝑓𝑐 + 𝑓𝑚 = 2 𝑘𝐻𝑧 + 810 𝑘𝐻𝑧 = 812 𝑘𝐻𝑧

b. M

= m x 100%

75%

= m x 100%

m

= 0,75

0,75 =

𝑉𝑚 𝑉𝐶

0,75 =

𝑉𝑚 2000

𝑉𝑚 = 1500 𝑉

c. 𝑉𝑈𝑆𝐵 = =

𝑚 2

. 𝑉𝐶

0,75 . 2000 2

= 0,375 𝑥 2000 = 750 𝑉 𝑚 .𝑉 2 𝐶 0,75 = . 2000 2

𝑉𝐿𝑆𝐵 =

= 0,375 𝑥 2000 = 750 𝑉 d. 𝑉𝑚𝑎𝑥 = 𝑉𝐶 + 𝑉𝐿𝑆𝐵 + 𝑉𝑈𝑆𝐵 = 2000 + 750 + 750 = 3500 𝑉

3. A spectrum analyzer with an input impedance of 50Ω is used to measure the power spectrum of an AM signal at the output of a preamplifier circuit. The AM signal has been modulated with a sine wave. The effective carrier power, Pc, is 750 mW, and each sideband, PUSB and PLSB, is 120 mW. Compute the following: a) Total effective power, PT b) The peak carrier voltage, Vc c) The modulation index, m and the percent modulation index, M d) The modulation voltage e) The lower and upper sideband voltages, VUSB and VLSB Diketahui : 

Rinput

= 50Ω



Pc

= 750mW



PUSB

= 120 mW



PLSB

= 120 mW

Ditanya : a. PT ? b. VC ?

c. m , M ? d. Vm ? e. PUSB , PLSB ? Jawab : a. 𝑃𝑇 = 𝑃𝐶 + 𝑃𝐿𝑆𝐵 + 𝑃𝑈𝑆𝐵 = 750 𝑚𝑊 + 120 𝑚𝑊 + 120 𝑚𝑊 = 990 𝑚𝑊 = 0,99 𝑚𝑊

b. 𝑃𝐶

=

0,75 = 75

𝑉𝑐 2 2𝑅

𝑉𝑐 2 2 . 50

= 𝑉𝑐 2

√75 = 𝑉𝑐 𝑉𝑐

= 8,66 𝑉

c. 𝑃𝐿𝑆𝐵 = 𝑃𝑈𝑆𝐵 =

𝑚 2 . 𝑃𝐶

2

0,12 =

𝑚 . 0,75 4

0,48 = 𝑚2 . 0,75 0,48 = 𝑚2 0,75 0,64 = 𝑚2 √0,64 = 𝑚 0,8

=𝑚

𝑀 = 𝑚 𝑥 100% = 0,8 𝑥 100% = 80 % d. 𝑚 = 0,8 =

𝑉𝑚 𝑉𝐶

𝑉𝑚 8,66

𝑉𝑚 = 6,928 𝑉

4

e. 𝑉𝑈𝑆𝐵 = =

𝑚 2

. 𝑉𝐶

0,8 . 8,66 2

= 0,4 𝑥 8,66 = 3,464 𝑉 𝑚 .𝑉 2 𝐶 0,8 = . 8,66 2

𝑉𝐿𝑆𝐵 =

= 0,4 𝑥 8,66 = 3,464 𝑉